Oxford IB Study Guides: Physics for the IB Diploma: Oxford Ib Diploma Program (IB Science 2014) [2014 ed.] 0198393555, 9780198393559

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OXFORD IB STUDY GUIDES

Ti Kik

Physics  o r T h e I B d I p lo m a

2014 edition

2

3 Great Clarendon Street, Oxford, OX2 6DP, United Kingdom Oxford University Press is a department of the University of Oxford. It furthers the Universitys objective of excellence in research, scholarship, and education by publishing worldwide. Oxford is a registered trade mark of Oxford University Press in the UK and in certain other countries  Tim Kirk 2014 The moral rights of the author have been asserted First published in 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by licence or under terms agreed with the appropriate reprographics rights organization. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above. You must not circulate this work in any other form and you must impose this same condition on any acquirer British Library Cataloguing in Publication Data Data available 978-0-19-839355-9 1 3 5 7 9 10 8 6 4 2 Paper used in the production of this book is a natural, recyclable product made from wood grown in sustainable forests. The manufacturing process conforms to the environmental regulations of the country of origin. Printed in Great Britain Acknowledgements This work has been developed independently from and is not endorsed by the International Baccalaureate (IB). Cover:  James Brittain/Corbis; p191: Chase Preuninger; p211: NASA/WMAP Science Team; p117: vilax/Shutterstock; p205: NASA/WMAP Artwork by Six Red Marbles and Oxford University Press. We have tried to trace and contact all copyright holders before publication. If notied the publishers will be pleased to rectify any errors or omissions at the earliest opportunity.

Introduction and acknowledgements Many people seem to think that you have to be really clever to understand Physics and this puts some people o studying it in the rst place. So do you really need a brain the size o a planet in order to cope with IB Higher Level Physics? The answer, you will be pleased to hear, is No. In act, it is one o the worlds best kept secrets that Physics is easy! There is very little to learn by heart and even ideas that seem really difcult when you rst meet them can end up being obvious by the end o a course o study. But i this is the case why do so many people seem to think that Physics is really hard? I think the main reason is that there are no saety nets or short cuts to understanding Physics principles. You wont get ar i you just learn laws by memorising them and try to plug numbers into equations in the hope o getting the right answer. To really make progress you need to be amiliar with a concept and be completely happy that you understand it. This will mean that you are able to apply your understanding in unamiliar situations. The hardest thing, however, is oten not the learning or the understanding o new ideas but the getting rid o wrong and conused every day explanations. This book should prove useul to anyone ollowing a preuniversity Physics course but its structure sticks very

closely to the recently revised International Baccalaureate syllabus. It aims to provide an explanation (albeit very brie) o all o the core ideas that are needed throughout the whole IB Physics course. To this end each o the sections is clearly marked as either being appropriate or everybody or only being needed by those studying at Higher level. The same is true o the questions that can be ound at the end o the chapters. I would like to take the opportunity to thank the many people that have helped and encouraged me during the writing o this book. In particular I need to mention David Jones and Paul Ruth who provided many useul and detailed suggestions or improvement  unortunately there was not enough space to include everything. The biggest thanks, however, need to go to Betsan or her support, patience and encouragement throughout the whole project. Tim Kirk October 2002

Third edition Since the IB Study Guide's rst publication in 2002, there have been two signicant IB Diploma syllabus changes. The aim, to try and explain all the core ideas essential or the IB Physics course in as concise a way as possible, has remained the same. I continue to be grateul to all the teachers and students who have taken time to comment and I would welcome urther eedback. In addition to the team at OUP, I would particularly like to thank my exceptional colleagues and all the outstanding students at my current school, St. Dunstan's College, London. It goes without saying that this third edition could not have been achieved without Betsan's continued support and encouragement. This book is dedicated to the memory o my ather, Francis Kirk. Tim Kirk August 2014

I n tr o d u c tI o n an d ac kn o wle d g e m e n ts

iii

Contents (Italics denote topics which are exclusively Higher Level)

1 measurement and uncertaIntIes The realm o physics  range o magnitudes o quantities in our universe The SI system o undamental and derived units Estimation Uncertainties and error in experimental measurement Uncertainties in calculated results Uncertainties in graphs Vectors and scalars IB Questions  measurement and uncertainties

1 2 3 4 5 6 7 8

2 mechanIcs Motion Graphical representation o motion Uniormly accelerated motion Projectile motion Fluid resistance and ree-all Forces and ree-body diagrams Newtons rst law Equilibrium Newtons second law Newtons third law Mass and weight Solid riction Work Energy and power Momentum and impulse IB Questions  mechanics

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

3 thermal PhYsIcs Thermal concepts Heat and internal energy Specic heat capacity Phases (states) o matter and latent heat The gas laws 1 The gas laws 2 Molecular model o an ideal gas IB Questions  thermal physics

25 26 27 28 29 30 31 32

4 waVes Oscillations Graphs o simple harmonic motion Travelling waves Wave characteristics Electromagnetic spectrum Investigating speed o sound experimentally

iv

co n te n ts

33 34 35 36 37 38

Intensity Superposition Polarization Uses o polarization Wave behaviour  Refection Snells law and reractive index Reraction and critical angle Diraction Two-source intererence o waves Nature and production o standing (stationary) waves Boundary conditions IB Questions  waves

39 40 41 42 43 44 45 46 47 48 49 50

5 electrIcItY and magnetIsm Electric charge and Coulomb's law Electric elds Electric potential energy and electric potential dierence Electric current Electric circuits Resistors in series and parallel Potential divider circuits and sensors Resistivity Example o use o Kircho's laws Internal resistance and cells Magnetic orce and elds Magnetic orces Examples o the magnetic eld due to currents IB Questions  electricity and magnetism

51 52 53 54 55 56 57 58 59 60 61 62 63 64

6 cIrcular motIon and graVItatIon Uniorm circular motion Angular velocity and vertical circular motion Newtons law o gravitation IB Questions  circular motion and gravitation

65 66 67 68

7 atomIc, nuclear and PartIcle PhYsIcs Emission and absorption spectra Nuclear stability Fundamental orces Radioactivity 1 Radioactivity 2 Hal-lie Nuclear reactions

69 70 71 72 73 74 75

Fission and usion Structure o matter Description and classifcation o particles Quarks Feynman diagrams IB Questions  atomic, nuclear and particle physics

76 77 78 79 80 81

8 energY ProductIon Energy and power generation  Sankey diagram Primary energy sources Fossil uel power production Nuclear power  process Nuclear power  saety and risks Solar power and hydroelectric power Wind power and other technologies Thermal energy transer Radiation: Wiens law and the SteanBoltzmann law Solar power The greenhouse eect Global warming IB Questions  energy production

82 83 84 85 86 87 88 89 90 91 92 93 94

9 WavE phEnomEna Sie ri ti Eergy ges drig sie ri ti Dirti Tw-sre itereree  wes: Ygs dbe-sit exeriet mtie-sit dirti Ti re fs Resti Te Der eet xes d itis  te Der eet IB Qestis  we ee

95 96 97 98 99 100 101 102 103 104

10 IElDS pteti (gritti d eetri) Eqitetis Gritti teti eergy d teti orbit ti Eetri teti eergy d teti Eetri d Gritti ieds red IB Qestis  feds

105 106 107 108 109 110 111

11 ElEcTRomaGnETIc InDucTIon Induced electromotive force (emf) lez's w d rdy's w atertig rret (1 ) atertig rret (2) Retifti d stig irits cite

112 113 114 115 116 117

118 119 120

citr disrge citr rge IB Qestis  eetrgeti idti

12 QuanTum anD nuclEaR phYSIcS 121 122 123 124 125

pteetri eet mtter wes ati setr d ti eergy sttes Br de  te t Te Srdiger de  te t Te heiseberg ertity riie d te ss  deteriis Teig, teti brrier d trs etig teig rbbiity Te es ner eergy ees d rditie dey IB Qestis  qt d er ysis

126 127 128 129 130

13 oPtIon a  relatIVItY Reerence rames Maxwells equations Special relativity Lorentz transormations Velocity addition Invariant quantities Time dilation Length contraction and evidence to support special relativity Spacetime diagrams ( Minkowski diagrams) 1 Spacetime diagrams 2 The twin paradox 1 Twin paradox 2 Spacetime diagrams 3 mss d eergy Retiisti et d eergy Retiisti eis exes Geer retiity  te eqiee riie Gritti red sit Srtig eidee crtre  setie Bk es IB Questions  option A  relativity

131 132 133 134 135 136 137 138 139 140 140 141 142 143 144 145 146 147 148 149 150 151

14 oPtIon B  engIneerIng PhYsIcs Translational and rotational motion Translational and rotational relationships Translational and rotational equilibrium Equilibrium examples Newtons second law  moment o inertia Rotational dynamics

c o n te n ts

152 153 154 155 156 157

v

Solving rotational problems Thermodynamic systems and concepts Work done by an ideal gas The rst law o thermodynamics Second law o thermodynamics and entropy Heat engines and heat pumps lids t rest lids i motio  Berolli efect Berolli  exmples viscosity orced oscilltios d resoce (1 ) Resoce (2) IB Questions  option B  engineering physics

158 159 160 161 162 163 164 165 166 167 168 169 170

15 oPtIon c  ImagIng Image ormation Converging lenses Image ormation in convex lenses Thin lens equation Diverging lenses Converging and diverging mirrors The simple magniying glass Aberrations The compound microscope and astronomical telescope Astronomical refecting telescopes Radio telescopes Fibre optics Dispersion, attenuation and noise in optical bres Channels o communication X-rys X-ry imgig techiqes ultrsoic imgig Imgig cotied IB Questions  option C  imaging

vi

co n te n ts

171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189

16 oPtIon d  astroPhYsIcs Objects in the universe (1) Objects in the universe (2) The nature o stars Stellar parallax Luminosity Stellar spectra Nucleosynthesis The HertzsprungRussell diagram Cepheid variables Red giant stars Stellar evolution The Big Bang model Galactic motion Hubbles law and cosmic scale actor The accelerating universe ncler sio  the Jes criterio ncleosythesis of the mi seqece Types o speroe The cosmologicl priciple d mthemticl models Rottio cres d drk mtter The history o the uierse The tre o the uierse Drk eergy astrophysics reserch IB Questions  astrophysics

190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214

17 aPPendIX Graphs 215 Graphical analysis and determination o relationships 216 Grphicl lysis  logrithmic ctios 217 ANSWERS ORIGIN OF INDIVIDUAL QUESTIONS INDEX

218 218 219

1 M E AS U R E M E N T AN D U N CE R TAI N TI E S Te ream o pysics  rane o manitudes o quantities in our universe ORDERS Of MAgNITUDE  INClUDINg ThEIR RATIOS Physics seeks to explain nothing less than the Universe itself. In attempting to do this, the range of the magnitudes of various quantities will be huge. If the numbers involved are going to mean anything, it is important to get some feel for their relative sizes. To avoid getting lost among the numbers it is helpful to state them to the nearest order of magnitude or power of ten. The numbers are just rounded up or down as appropriate. Comparisons can then be easily made because working out the ratio between two powers of ten is just a matter of adding or subtracting whole numbers. The diameter of an atom, 1 0 - 1 0 m, does not sound that much larger than the diameter of a proton in its nucleus, 1 0 - 1 5 m, but the ratio between them is 1 0 5 or 1 00,000 times bigger. This is the same ratio as between the size of a railway station (order of magnitude 1 0 2 m) and the diameter of the Earth (order of magnitude 1 0 7 m) .

electrons

RANgE Of MASSES 10 52 10 48 10 44 10 40 10 36 10 32 10 28 10 24 10 20 10 16 10 12 10 8 10 4 10 0 10 -4 10 -8 10 - 12 10 - 16 10 - 20 10 - 24 10 - 28 10 - 32

Mass / kg total mass of observable Universe mass of local galaxy (Milky Way) mass of Sun mass of Earth total mass of oceans total mass of atmosphere laden oil supertanker elephant human mouse grain of sand blood corpuscle bacterium haemoglobin molecule proton electron

RANgE Of lENgThS 10 26 10 24 10 22 10 20 10 18 10 16 10 14 10 12 10 10 10 8 10 6 10 4 10 2 10 0 10 - 2 10 - 4 10 - 6 10 - 8 10 - 10 10 - 12 10 - 14 10 - 16

Size / m radius of observable Universe

radius of local galaxy (Milky Way) distance to nearest star

distance from Earth to Sun distance from Earth to Moon radius of the Earth deepest part of the ocean / highest mountain tallest building length of ngernail thickness of piece of paper human blood corpuscle wavelength of light diameter of hydrogen atom wavelength of gamma ray diameter of proton

protons

RANgE Of TIMES Carbon atom railway station Earth For example, you would probably feel very pleased with yourself if you designed a new, environmentally friendly source of energy that could produce 2.03  1 0 3 J from 0.72 kg of natural produce. But the meaning of these numbers is not clear  is this a lot or is it a little? In terms of orders of magnitudes, this new source produces 1 0 3 joules per kilogram of produce. This does not compare terribly well with the 1 0 5 joules provided by a slice of bread or the 1 0 8 joules released per kilogram of petrol. You do NOT need to memorize all of the values shown in the tables, but you should try and develop a familiarity with them.

10 20 10 18 10 16 10 14 10 12 10 10 10 8 10 6 10 4 10 2 10 0 10 - 2 10 - 4 10 - 6 10 - 8 10 - 10 10 - 12 10 - 14 10 - 16 10 - 18 10 - 20 10 - 22 10 - 24

Time / s age of the Universe

RANgE Of ENERgIES 10 44

age of the Earth

10 34

age of species  Homo sapiens

10 30

typical human lifespan 1 year 1 day

10 26 10 22 10 18 10 14

heartbeat

10 10

period of high-frequency sound

10 6 10 2

passage of light across a room

10 - 2 10 - 6

vibration of an ion in a solid period of visible light

Energy / J energy released in a supernova

energy radiated by Sun in 1 second

energy released in an earthquake energy released by annihilation of 1 kg of matter energy in a lightning discharge energy needed to charge a car battery kinetic energy of a tennis ball during game energy in the beat of a ys wing

10 - 10 10 - 14

passage of light across an atom passage of light across a nucleus

10 - 18 10 - 22

energy needed to remove electron from the surface of a metal

10 - 26

M E A S U R E M E N T A N D U N C E R TA I N T I E S

1

The SI system o undamenta and deried units fUNDAMENTAl UNITS Any measurement and every quantity can be thought o as being made up o two important parts: 1.

the number and

2.

the units.

Without both parts, the measurement does not make sense. For example a persons age might be quoted as seventeen but without the years the situation is not clear. Are they 1 7 minutes, 1 7 months or 1 7 years old? In this case you would know i you saw them, but a statement like length = 4.2 actually says nothing. Having said this, it is really surprising to see the number o candidates who orget to include the units in their answers to examination questions. In order or the units to be understood, they need to be defned. There are many possible systems o measurement that have

DERIvED UNITS Having fxed the undamental units, all other measurements can be expressed as dierent combinations o the undamental units. In other words, all the other units are derived units. For example, the undamental list o units does not contain a unit or the measurement o speed. The defnition o speed can be used to work out the derived unit.

been developed. In science we use the International System o units (SI) . In SI, the fundamental or base units are as ollows Quantity

SI unit

SI symbol

Mass

kilogram

kg

Length

metre

m

Time

second

s

Electric current

ampere

A

Amount o substance

mole

mol

Temperature

kelvin

K

(Luminous intensity

candela

cd)

You do not need to know the precise defnitions o any o these units in order to use them properly.

are so large that the SI unit (the metre) always involves large orders o magnitudes. In these cases, the use o a dierent (but non SI) unit is very common. Astronomers can use the astronomical unit (AU) , the light-year (ly) or the parsec (pc) as appropriate. Whatever the unit, the conversion to SI units is simple arithmetic. 1 AU = 1 .5  1 0 1 1 m

distance Since speed = _ time units o distance Units o speed = __ units o time metres = _ (pronounced metres per second) seconds m = _ s = m s 1 O the many ways o writing this unit, the last way (m s 1 ) is the best. Sometimes particular combinations o undamental units are so common that they are given a new derived name. For example, the unit o orce is a derived unit  it turns out to be kg m s - 2 . This unit is given a new name the newton (N) so that 1 N = 1 kg m s - 2 . The great thing about SI is that, so long as the numbers that are substituted into an equation are in SI units, then the answer will also come out in SI units. You can always play sae by converting all the numbers into proper SI units. Sometimes, however, this would be a waste o time. There are some situations where the use o SI becomes awkward. In astronomy, or example, the distances involved

1 ly = 9.5  1 0 1 5 m 1 pc = 3.1  1 0 1 6 m There are also some units (or example the hour) which are so common that they are oten used even though they do not orm part o SI. Once again, beore these numbers are substituted into equations they need to be converted. Some common unit conversions are given on page 3 o the IB data booklet. The table below lists the SI derived units that you will meet. SI derived unit

SI base unit

Alternative SI unit

newton (N)

kg m s

-

pascal (Pa)

kg m- 1 s - 2

N m- 2 -

-2

hertz (Hz)

s

joule (J)

kg m2 s - 2

Nm

watt (W)

kg m s

J s- 1

coulomb (C)

As

volt (V)

-1

2

-3

-

kg m s

-3

ohm ()

kg m s

-3

weber (Wb)

kg m2 s - 2 A- 1

Vs

tesla (T)

kg s

Wb m- 2

becquerel (Bq)

s- 1

2

2

-2

A

-1

A

-1

WA- 1

A

-2

VA- 1

-

PREfIxES To avoid the repeated use o scientifc notation, an alternative is to use one o the list o agreed prefxes given on page 2 in the IB data booklet. These can be very useul but they can also lead to errors in calculations. It is very easy to orget to include the conversion actor. 1W For example, 1 kW = 1 000 W. 1 mW = 1 0 - 3 W (in other words, ____ ) 1 000

2

M E A S U R E M E N T A N D U N C E R TA I N T I E S

Estimation ORDERS Of MAgNITUDE It is important to develop a eeling or some o the numbers that you use. When using a calculator, it is very easy to make a simple mistake (eg by entering the data incorrectly) . A good way o checking the answer is to rst make an estimate beore resorting to the calculator. The multiple-choice paper (paper 1 ) does not allow the use o calculators. Approximate values or each o the undamental SI units are given below. 1 kg

A packet o sugar, 1 litre o water. A person would be about 50 kg or more

1 m

Distance between ones hands with arms outstretched

1 s

Duration o a heart beat (when resting  it can easily double with exercise)

1 amp

Current fowing rom the mains electricity when a computer is connected. The maximum current to a domestic device would be about 1 0 A or so

1 kelvin 1 K is a very low temperature. Water reezes at 273 K and boils at 373 K. Room temperature is about 300 K 1 mol

1 2 g o carbon1 2. About the number o atoms o carbon in the lead o a pencil

The same process can happen with some o the derived units. 1 m s- 1

Walking speed. A car moving at 30 m s - 1 would be ast

1 ms

Quite a slow acceleration. The acceleration o gravity is 1 0 m s - 2

-2

1 N

A small orce  about the weight o an apple

1 V

Batteries generally range rom a ew volts up to 20 or so, the mains is several hundred volts

1 Pa

A very small pressure. Atmospheric pressure is about 1 0 5 Pa

1 J

A very small amount o energy  the work done liting an apple o the ground

POSSIblE REASONAblE ASSUMPTIONS Everyday situations are very complex. In physics we oten simpliy a problem by making simple assumptions. Even i we know these assumptions are not absolutely true they allow us to gain an understanding o what is going on. At the end o the calculation it is oten possible to go back and work out what would happen i our assumption turned out not to be true. The table below lists some common assumptions. Be careul not to assume too much! Additionally we oten have to assume that some quantity is constant even i we know that in reality it is varying slightly all the time. Assumption

Example

Object treated as point particle

Mechanics: Linear motion and translational equilibrium

Friction is negligible

Many mechanics situations  but you need to be very careul

No thermal energy (heat) loss

Almost all thermal situations

Mass o connecting string, etc. is negligible

Many mechanics situations

Resistance o ammeter is zero

Circuits

Resistance o voltmeter is innite

Circuits

Internal resistance o battery is zero

Circuits

Material obeys Ohms law

Circuits

Machine 1 00% ecient

Many situations

Gas is ideal

Thermodynamics

Collision is elastic

Only gas molecules have perectly elastic collisions

Object radiates as a perect black body

Thermal equilibrium, e.g. planets

SCIENTIfIC NOTATION

SIgNIfICANT fIgURES

Numbers that are too big or too small or decimals are oten written in scientifc notation:

Any experimental measurement should be quoted with its uncertainty. This indicates the possible range o values or the quantity being measured. At the same time, the number o signifcant fgures used will act as a guide to the amount o uncertainty. For example, a measurement o mass which is quoted as 23.456 g implies an uncertainty o  0.001 g (it has ve signicant gures) , whereas one o 23.5 g implies an uncertainty o  0.1 g (it has three signicant gures) .

a  1 0b where a is a number between 1 and 1 0 and b is an integer. e.g. 1 53.2 = 1 .532  1 0 2 ; 0.00872 = 8.72  1 0 - 3

A simple rule or calculations (multiplication or division) is to quote the answer to the same number o signicant digits as the LEAST precise value that is used. For a more complete analysis o how to deal with uncertainties in calculated results, see page 5.

M E A S U R E M E N T A N D U N C E R TA I N T I E S

3

Uncertainties and error in experimenta measurement Systematic and random errors can oten be recognized rom a graph o the results.

quantity A

ERRORS  RANDOM AND SySTEMATIC (PRECISION AND ACCURACy) An experimental error just means that there is a dierence between the recorded value and the perect or correct value. Errors can be categorized as random or systematic. Repeating readings does not reduce systematic errors.

perfect results random error systematic error

Sources o random errors include  The readability o the instrument.  The observer being less than perect.

quantity B

 The eects o a change in the surroundings. Sources o systematic errors include  An instrument with zero error. To correct or zero error the value should be subtracted rom every reading.

Perect results, random and systematic errors o two proportional quantities.

 An instrument being wrongly calibrated.  The observer being less than perect in the same way every measurement. An accurate experiment is one that has a small systematic error, whereas a precise experiment is one that has a small random error.

true value

measured true value value probability that result has a certain value

measured value

value

value

(a)

(b)

Two examples illustrating the nature o experimental results: (a) an accurate experiment o low precision (b) a less accurate but more precise experiment.

ESTIMATINg ThE UNCERTAINTy RANgE An uncertainty range applies to any experimental value. The idea is that, instead o just giving one value that implies perection, we give the likely range or the measurement. 1.

Estimating rom frst principles

All measurement involves a readability error. I we use a measuring cylinder to fnd the volume o a liquid, we might think that the best estimate is 73 cm3 , but we know that it is not exactly this value (73.000 000 000 00 cm3 ) . Uncertainty range is  5 cm3 . We say volume = 73  5 cm3 .

cm 3 100 90 80 70 60 50 40 30 20 10

Normally the uncertainty range due to readability is estimated as below. Device

Example

Uncertainty

Analogue scale

Rulers, meters with moving pointers

 (hal the smallest scale division)

gRAPhICAl REPRESENTATION Of UNCERTAINTy

Digital scale

In many situations the best method o presenting and analysing data is to use a graph. I this is the case, a neat way o representing the uncertainties is to use error bars. The graphs below explains their use.

Top-pan balances, digital meters

 (the smallest scale division)

2.

quantity C

quantity A

Since the error bar represents the uncertainty range, the bestft line o the graph should pass through ALL o the rectangles created by the error bars.

Estimating uncertainty range rom several repeated measurements

I the time taken or a trolley to go down a slope is measured fve times, the readings in seconds might be 2.01 , 1 .82, 1 .97, 2.1 6 and 1 .94. The average o these fve readings is 1 .98 s. The deviation o the largest and smallest readings can be calculated (2.1 6 - 1 .98 = 0.1 8; 1 .98 - 1 .82 = 0.1 6). The largest value is taken as the uncertainty range. In this example the time is 1 .98 s  0.1 8 s. It would also be appropriate to quote this as 2.0  0.2 s.

SIgNIfICANT fIgURES IN UNCERTAINTIES

quantity E

quantity B

mistake assumed

quantity F

4

quantity D

The best ft line is included by all the error bars in the upper two graphs. This is not true in the lower graph.

M E A S U R E M E N T A N D U N C E R TA I N T I E S

In order to be cautious when quoting uncertainties, fnal values rom calculations are oten rounded up to one signifcant fgure, e.g. a calculation that fnds the value o a orce to be 4.264 N with an uncertainty o  0.362 N is quoted as 4.3  0.4 N. This can be unnecessarily pessimistic and it is also acceptable to express uncertainties to two signifcant fgures. For example, the charge on an electron is 1 .6021 76565  1 0 - 1 9 C  0.000000035 1 0 - 1 9 C. In data booklets this is sometimes expressed as 1 .6021 76565(35)  1 0 - 1 9 C.

Uncertainties in cacuated resuts MAThEMATICAl REPRESENTATION Of UNCERTAINTIES

Then the ractional uncertainty is

For example i the mass o a block was measured as 1 0  1 g and the volume was measured as 5.0  0.2 cm3 , then the ull calculations or the density would be as ollows.

p _ p , which makes the percentage uncertainty

mass 10 Best value or density = ______ = __ = 2.0 g cm- 3 5 volume 11 The largest possible value o density = ___ = 2.292 g cm- 3 4.8 9 The smallest possible value o density = ___ = 1 .731 g cm- 3 5.2

p _ p  1 00% . In the example above, the ractional uncertainty o the density is 0.1 5 or 1 5%. Thus equivalent ways o expressing this error are density = 2.0  0.3 g cm- 3

Rounding these values gives density = 2.0  0.3 g cm- 3 We can express this uncertainty in one o three ways  using absolute, fractional or percentage uncertainties. I a quantity p is measured then the absolute uncertainty would be expressed as p.

MUlTIPlICATION, DIvISION OR POwERS Whenever two or more quantities are multiplied or divided and they each have uncertainties, the overall uncertainty is approximately equal to the addition o the percentage (ractional) uncertainties. Using the same numbers rom above, m =  1 g

(

1 g m _ _ m =  10 g

)

=  0.1 =  1 0%

V =  0.2 cm3

(

)

0.2 cm3 =  0.04 =  4% V =  _ _ 5 cm3 V The total % uncertainty in the result =  (1 0 + 4) % =  14 % 1 4% o 2.0 g cm- 3 = 0.28 g cm- 3  0.3 g cm- 3 So density = 2.0  0.3 g cm- 3 as beore.

OR density = 2.0 g cm- 3  1 5% Working out the uncertainty range is very time consuming. There are some mathematical short-cuts that can be used. These are introduced in the boxes below.

ab In symbols, i y = _ c y b _ a c [note this is ALWAYS added] _ _ Then _ y = a + b + c Power relationships are just a special case o this law. I y = an

|

|

y a (always positive) _ Then ___ y = n a For example i a cube is measured to be 4.0  0.1 cm in length along each side, then 0.1 % Uncertainty in length =  _ =  2.5 % 4.0 3 3 Volume = (length) = (4.0) = 64 cm3

% Uncertainty in [volume] = = = =

% uncertainty in [(length) 3 ] 3  (% uncertainty in [length] ) 3  ( 2.5 % )  7.5 %

Absolute uncertainty = 7.5% o 64 cm3 = 4.8 cm3  5 cm3 Thus volume o cube = 64  5 cm3

OThER MAThEMATICAl OPERATIONS

Oter unctions

I the calculation involves mathematical operations other than multiplication, division or raising to a power, then one has to fnd the highest and lowest possible values.

There are no short-cuts possible. Find the highest and lowest values.

In symbols

sin 

Addition or subtraction Whenever two or more quantities are added or subtracted and they each have uncertainties, the overall uncertainty is equal to the addition o the absolute uncertainties.

e.g. uncertainty o sin  i  = 60  5

1 0.91 0.87 0.82

I y = a  b y = a + b (note ALWAYS added) uncertainty of thickness in a pipe wall external radius of pipe = 6.1  cm  0.1  cm ( 2% )

55 60 65



 = 60  5

i

best value to sin  = 0.87 max. sin  = 0.91

internal radius of pipe = 5.3 cm  0.1  cm ( 2% ) thickness o pipe wall = 6.1 - 5.3 cm

min. sin  = 0.82 

sin  = 0.87  0.05 worst value used

= 0.8 cm uncertainty in thickness = (0.1 + 0.1 )  cm = 0.2 cm = 25%

M E A S U R E M E N T A N D U N C E R TA I N T I E S

5

Uncertainties in graphs UNCERTAINTy IN SlOPES I the gradient o the graph has been used to calculate a quantity, then the uncertainties o the points will give rise to an uncertainty in the gradient. Using the steepest and the shallowest lines possible (i.e. the lines that are still consistent with the error bars) the uncertainty range or the gradient is obtained. This process is represented below.

best t line

steepest gradient

quantity a

quantity a

ERROR bARS Plotting a graph allows one to visualize all the readings at one time. Ideally all o the points should be plotted with their error bars. In principle, the size o the error bar could well be dierent or every single point and so they should be individually worked out.

shallowest gradient

quantity b

In practice, it would oten take too much time to add all the correct error bars, so some (or all) o the ollowing short-cuts could be considered.  Rather than working out error bars or each point  use the worst value and assume that all o the other error bars are the same.  Only plot the error bar or the worst point, i.e. the point that is urthest rom the line o best ft. I the line o best ft is within the limits o this error bar, then it will probably be within the limits o all the error bars.  Only plot the error bars or the frst and the last points. These are oten the most important points when considering the uncertainty ranges calculated or the gradient or the intercept (see right) .  Only include the error bars or the axis that has the worst uncertainty.

quantity b

UNCERTAINTy IN INTERCEPTS I the intercept o the graph has been used to calculate a quantity, then the uncertainties o the points will give rise to an uncertainty in the intercept. Using the steepest and the shallowest lines possible (i.e. the lines that are still consistent with the error bars) we can obtain the uncertainty in the result. This process is represented below.

quantity a

A ull analysis in order to determine the uncertainties in the gradient o a best straight-line graph should always make use of the error bars for all of the data points.

best value for intercept

maximum value of intercept

minimum value of intercept quantity b

6

M E A S U R E M E N T A N D U N C E R TA I N T I E S

vectors and scaars DIffERENCE bETwEEN vECTORS AND SCAlARS

REPRESENTINg vECTORS

I you measure any quantity, it must have a number AND a unit. Together they express the magnitude o the quantity. Some quantities also have a direction associated with them. A quantity that has magnitude and direction is called a vector quantity whereas one that has only magnitude is called a scalar quantity. For example, all orces are vectors.

In most books a bold letter is used to represent a vector whereas a normal letter represents a scalar. For example F would be used to represent a orce in magnitude AND direction. The list below shows some other recognized methods. F,  F or F 

The table lists some common quantities. The frst two quantities in the table are linked to one another by their defnitions (see page 9). All the others are in no particular order. Vectors

Scalars

Displacement

Distance

Velocity

Speed

Acceleration

Mass

Force

Energy (all orms)

Momentum

Temperature

Electric feld strength

Potential or potential dierence

Magnetic feld strength

Density

Gravitational feld strength

Area

Although the vectors used in many o the given examples are orces, the techniques can be applied to all vectors.

Vectors are best shown in diagrams using arrows:

pull

 the relative magnitudes o the vectors involved are shown by the relative length o the arrows

friction normal reaction weight

 the direction o the vectors is shown by the direction o the arrows.

ADDITION / SUbTRACTION Of vECTORS I we have a 3 N and a 4 N orce, the overall orce (resultant orce) can be 3N anything between = 7N 4N 1 N and 7 N depending on 5N 3N the directions = involved.

4N

COMPONENTS Of vECTORS It is also possible to split one vector into two (or more) vectors. This process is called resolving and the vectors that we get are called the components o the original vector. This can be a very useul way o analysing a situation i we choose to resolve all the vectors into two directions that are at right angles to one another.

Fvertical

F

F

The way to take 3N the directions into account is to do a scale 3N diagram and use the parallelogram law o vectors. This process is the same as adding vectors in turn  the tail o one vector is drawn starting rom the head o the previous vector.

3N 4N

=

4N

1N

= b a+b

a Parallelogram o vectors

Fhorizontal Splitting a vector into components

forces

Push Surface force Weight

TRIgONOMETRy Vector problems can always be solved using scale diagrams, but this can be very time consuming. The mathematics o trigonometry oten makes it much easier to use the mathematical unctions o sine or cosine. This is particularly appropriate when resolving. The diagram below shows how to calculate the values o either o these components.

Av

A

A v = Asin 

These mutually perpendicular directions are totally independent o each other and can be analysed separately. I appropriate, both directions can then be combined at the end to work out the fnal resultant vector.

components  PV PH

SH SV

A H = Acos 

AH

See page 1 4 or an example.

W Pushing a block along a rough surace

M E A S U R E M E N T A N D U N C E R TA I N T I E S

7

Ib Questions  measurement and uncertainties An object is rolled rom rest down an inclined plane. The distance travelled by the object was measured at seven dierent times. A graph was then constructed o the distance travelled against the (time taken) 2 as shown below.

distance travelled/ (cm)

1.

3.

9 8 4.

7 6

A. 0.1 m

C. 1 .0 m

B. 0.2 m

D. 2.0 m

In order to determine the density o a certain type o wood, the ollowing measurements were made on a cube o the wood. Mass

5

= 493 g

Length o each side = 9.3 cm

4

The percentage uncertainty in the measurement o mass is 0.5% and the percentage uncertainty in the measurement o length is 1 .0% .

3 2

The best estimate or the uncertainty in the density is

1

A. 0.5%

C. 3.0%

0 0.0

B. 1 .5%

D. 3.5%

a) (i)

0.1

0.2

0.3

0.4 0.5 (time taken) 2 / s 2

What quantity is given by the gradient o such a graph?

5.

[2]

(ii) Explain why the graph suggests that the collected data is valid but includes a systematic error. [2] (iii) Do these results suggest that distance is proportional to (time taken) 2 ? Explain your answer. [2] (iv) Making allowance or the systematic error, calculate the acceleration o the object. [2] b) The ollowing graph shows that same data ater the uncertainty ranges have been calculated and drawn as error bars.

distance travelled/ (cm)

A stone is dropped down a well and hits the water 2.0 s ater it is released. Using the equation d = __12 g t2 and taking g = 9.81 m s - 2 , a calculator yields a value or the depth d o the well as 1 9.62 m. I the time is measured to 0.1 s then the best estimate o the absolute error in d is

9 8

Astronauts wish to determine the gravitational acceleration on Planet X by dropping stones rom an overhanging cli. Using a steel tape measure they measure the height o the cli as s = 7.64 m  0.01 m. They then drop three similar stones rom the cli, timing each all using a hand-held electronic stopwatch which displays readings to onehundredth o a second. The recorded times or three drops are 2.46 s, 2.31 s and 2.40 s. a) Explain why the time readings vary by more than a tenth o a second, although the stopwatch gives readings to one hundredth o a second.

[1 ]

b) Obtain the average time t to all, and write it in the orm (value  uncertainty) , to the appropriate number o signifcant digits.

[1 ]

c) The astronauts then determine the gravitational 2s acceleration a g on the planet using the ormula ag = __ . t Calculate ag rom the values o s and t, and determine the uncertainty in the calculated value. Express the result in the orm ag = (value  uncertainty) , to the appropriate number o signifcant digits. [3]

7

2

6 5 4 3

HL

2 6.

1 0 0.0

0.1

0.2

0.3

0.4 0.5 (time taken) 2 / s 2

Add two lines to show the range o the possible acceptable values or the gradient o the graph. 2.

This question is about fnding the relationship between the orces between magnets and their separations. In an experiment, two magnets were placed with their Northseeking poles acing one another. The orce o repulsion, f, and the separation o the magnets, d, were measured and the results are shown in the table below.

[2]

Separation d/m

The lengths o the sides o a rectangular plate are measured, and the diagram shows the measured values with their uncertainties.

50  0.5 mm

25  0.5 mm

Which one o the ollowing would be the best estimate o the percentage uncertainty in the calculated area o the plate? A.  0.02%

C.  3%

B.  1 %

D.  5%

8

Force o repulsion f/N

0.04

4.00

0.05

1 .98

0.07

0.74

0.09

0.32

a) Plot a graph o log (orce) against log (distance) .

[3]

b) The law relating the orce to the separation is o the orm f = kdn (i)

Use the graph to fnd the value o n.

(ii) Calculate a value or k, giving its units.

I B Q U E S T I o N S  M E A S U R E M E N T A N D U N C E R TA I N T I E S

[2] [3]

2 m e ch an i cs m Definitions These technical terms should not be conused with their everyday use. In particular one should note that  Vector quantities always have a direction associated with them.  Generally, velocity and speed are NOT the same thing. This is particularly important i the object is not going in a straight line.  The units o acceleration come rom its denition. (m s - 1 )  s = m s - 2 .  The denition o acceleration is precise. It is related to the change in velocity (not the same thing as the change in speed) . Whenever the motion o an object changes, it is called acceleration. For this reason acceleration does not necessarily mean constantly increasing speed  it is possible to accelerate while at constant speed i the direction is changed.  A deceleration means slowing down, i.e. negative acceleration i velocity is positive. Symbol Displacement Velocity

s

v or u

Defnition

Example

The distance moved in a particular direction.

The displacement rom London to Rome is 1 .43  1 0 6 m southeast.

The rate o change o displacement. change o displacement velocity = ________________ time taken

Speed

v or u

The rate o change o distance. distance gone speed = __________ time taken

Acceleration

a

The rate o change o velocity. change o velocity acceleration = _____________ time taken

Vector

The average velocity during a fight rom London to Rome is 1 60 m s - 1 southeast.

m s- 1

Vector

The average speed during a fight rom London to Rome is 1 60 m s - 1

m s- 1

Scalar

The average acceleration o a plane on the runway during take-o is 3.5 m s- 2 in a orwards direction. This means that on average, its velocity changes every second by 3.5 m s- 1

m s- 2

Vector

It should be noticed that the average value (over a period o time) is very dierent to the instantaneous value (at one particular time) . In the example below, the positions o a sprinter are shown at dierent times ater the start o a race. The average speed over the whole race is easy to work out. It is the total distance (1 00 m) divided by the total time (1 1 .3 s) giving 8.8 m s - 1 .

t = 0.0 s

t = 2.0 s

But during the race, her instantaneous speed was changing all the time. At the end o the rst 2.0 seconds, she had travelled 1 0.04 m. This means that her average speed over the rst 2.0 seconds was 5.02 m s- 1 . During these rst two seconds, her instantaneous speed was increasing  she was accelerating. I she started at rest (speed = 0.00 m s- 1 ) and her average speed (over the whole two seconds) was 5.02 m s- 1 then her instantaneous speed at 2 seconds must be more than this. In act the instantaneous speed or this sprinter was 9.23 m s- 1 , but it would not be possible to work this out rom the inormation given.

d = 28.21 m

d = 47.89 m

d = 69.12 m

t = 4.0 s

t = 6.0 s

t = 8.0 s

frames of reference I two things are moving in the same straight line but are travelling at dierent speeds, then we can work out their relative velocities by simple addition or subtraction as appropriate. For example, imagine two cars travelling along a straight road at dierent speeds. I one car (travelling at 30 m s - 1 ) overtakes the other car (travelling at 25 m s - 1 ) , then according to the driver o the slow car, the relative velocity o the ast car is +5 m s - 1 .

Vector or scalar?

m

instantaneous vs average

start d = 0.00 m d = 10.04 m

SI Unit

In technical terms what we are doing is moving rom one rame o reerence into another. The velocities o 25 m s - 1 and 30 m s - 1 were measured according

30 m s -1

nish d = 100.00 m

t = 11.3 s

to a stationary observer on the side o the road. We moved rom this rame o reerence into the drivers rame o reerence.

gap between the cars increases by 5 m s - 1

25 m s - 1 observer one car overtaking another, as seen by an observer on the side o the road.

one car overtaking another, as seen by the driver o the slow car.

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9

g    the use of graphs

2.

Graphs are very useul or representing the changes that happen when an object is in motion. There are three possible graphs that can provide useul inormation

To make things simple at the beginning, the graphs are normally introduced by considering objects that are just moving in one particular direction. I this is the case then there is not much dierence between the scalar versions (distance or speed) and the vector versions (displacement or velocity) as the directions are clear rom the situation. More complicated graphs can look at the component o a velocity in a particular direction.

 displacementtime or distancetime graphs  velocitytime or speedtime graphs  accelerationtime graphs. There are two common methods o determining particular physical quantities rom these graphs. The particular physical quantity determined depends on what is being plotted on the graph. 1.

Finding the gradient of the line.

To be a little more precise, one could fnd either the gradient o  a straight-line section o the graph (this fnds an average value) , or  the tangent to the graph at one point (this fnds an instantaneous value) .

Finding the area under the line.

I the object moves orward then backward (or up then down) , we distinguish the two velocities by choosing which direction to call positive. It does not matter which direction we choose, but it should be clearly labelled on the graph. Many examination candidates get the three types o graph muddled up. For example a speedtime graph might be interpreted as a distancetime graph or even an acceleration time graph. Always look at the axes o a graph very careully.

velocitytime graphs

accelerationtime graphs

 The gradient o a velocitytime graph is the acceleration

 The area under a displacementtime graph does not represent anything useul

 The area under a velocitytime graph is the displacement

 The gradient o an acceleration time graph is not oten useul (it is actually the rate o change o acceleration)

e

e

10 .0

= 20 m s-1

 rst 4 seconds at con stan t speed

2 0 .0

ob ject is s lo win g d own a ccelera tion 20 = 1 = - 20 m s-2

10 .0

spee d = 2 0 = 5 m s - 1 4 0 1 .0 2 .0 3 .0 4.0 5 .0 6 .0 7.0 8.0

0 1 .0 2 .0 3 .0 4.0 5 .0 6 .0 7.0 8 .0

tim e / s

h igh e st p oin t a t t = 0 .9 s

velocity / m s -1

4.0

2 .0

object re tu rn s to h a n d a t t = 1 .8 s 1 .0

level o f ha n d a s zero d ispla cem en t

2 .0 tim e / s

O bject is th ro wn vertica lly u pwa rd s.

tim e / s

a ccelera tion is co n sta n tly in crea s in g, ra te o f ch a n ge o f ve lo city is in crea s in g

in itia l u pw a rd ve lo city is + ve m a x. h eigh t = a rea u n d er gra ph 1 =  0 .9  9 .0 m = 4.0 5 m 9 .0 2 - 9 .0 a ccelera tio n = = - 10 m s - 2 0 .9

- 9 .0

1 .0

2 .0 tim e / s

in s ta n ta n eou s d own wa rd velo city = ze ro velo city is a t h igh e st p o in t ne ga tive ( t = 0 .9 s ) O bject is th ro wn vertica l ly u p wa rd s.

object a t co n s ta n t a ccelera tion o f 2 0 m s - 2 ve lo city s till ch a n gin g a ll th e tim e ra te o f a ccele ra tion is d ecrea s in g, b u t velo city co n tin u e s to in crea se

2 0 .0

10 .0

d is ta n ce tra vel led in  rs t 4 se con d s = a rea u n d er gra p h = 1  4  2 0 m = 40 m 2 O bject m ove s with co n sta n t a ccelera tion , then con sta n t velo city , then d ecelerate s.

O bje ct m ove s a t con sta n t sp eed , s top s then re tu rn s.

displacement / m

e

object's ve lo city is in crea s in g object a t con s ta n t 20 a cce lera tion = 4 sp eed ( = 2 0 m s - 1 ) = 5 m s - 2 a ccelera tion is zero

acceleration / m s -2

2 0 .0

object re tu rn s a t a fa s ter sp eed 20 spe ed = 1

0 1 .0 2 .0 3 .0 4.0 5 .0 6 .0 7.0 8.0 tim e / s 1  4  2 0 = 40 m s - 1 2 O b ject m o ve s with in crea s in g, th en co n s ta n t, th en d ecrea s in g a ccelera tion . Cha n ge in velo city =

acceleration / m s - 2

object s ta tion a ry for 3 se con d s spe ed = 0 m s - 1

 The area under an acceleration time graph is the change in velocity

velocity / m s 1

displacement / m

Displacementtime graphs  The gradient o a displacementtime graph is the velocity

+ 10 .0

1 .0  10 .0

2 .0 tim e / s

ch a n ge in velo city = a rea u n d er gra p h = - 10 .0  1 .8 m s - 1 = - 18 m s - 1 ( cha n ge from + 9 .0 to - 9 .0 m s - 1 ) O b ject is th ro wn ve rtica lly u pwa rd s.

example of equation of uniform motion A car accelerates uniormly rom rest. Ater 8 s it has travelled 1 20 m. Calculate: (i) its average acceleration speed ater 8 s 1 at2 (i) s = ut + _ 2 1 a  8 2 = 32 a  1 20 = 0  8 + _ 2 a = 3.75 m s 2

10

m ech an i cs

(ii) v2 = = =  v =

u 2 + 2 as 0 + 2  3.75  1 20 900 30 m s 1

(ii) its instantaneous

uy d  practical calculations

equations of uniform motion

In order to determine how the velocity (or the acceleration) o an object varies in real situations, it is oten necessary to record its motion. Possible laboratory methods include.

These equations can only be used when the acceleration is constant  dont orget to check i this is the case!

A strobe light gives out very brie fashes o light at xed time intervals. I a camera is pointed at an object and the only source o light is the strobe light, then the developed picture will have captured an objects motion.

t = 0.0 s t = 0.1 s t = 0.2 s t = 0.3 s

t = 0.4 s

a

acceleration (const)

t

time taken

s

distance travelled

u+v s = _ t 2

)

tk  A ticker timer can be arranged to make dots on a strip o paper at regular intervals o time (typically every tieth o a second) . I the piece o paper is attached to an object, and the object is allowed to all, the dots on the strip will have recorded the distance moved by the object in a known time.

downwards +ve

v = u + at

20 5 2 .0

3 .0 tim e / s

1 .0 2 .0 a ccelera tion / m s -2

3 .0 tim e / s

30 20 10

1 at2 s = ut + _ 2 1 at2 s = vt - _ 2 The rst equation is derived rom the denition o acceleration. In terms o these symbols, this denition would be (v - u) a= _ t This can be rearranged to give the rst equation. v = u + at

(1 )

The second equation comes rom the denition o average velocity. average velocity = _s t Since the velocity is changing uniormly we know that this average velocity must be given by

or

(u + v) _s = _ t 2

downwards +ve

v2 = u 2 + 2as

(v + u) average velocity = _ 2

t = 0.5 s

d ispla cem en t / m 45

1 .0 velocity / m s -1

The ollowing equations link these dierent quantities.

(

downwards +ve

initial velocity nal velocity

The other equations o motion can be derived by using these two equations and substituting or one o the variables (see previous page or an example o their use) .

(2)

2 .0

3 .0 tim e / s

In the absence o air resistance, all alling objects have the SAME acceleration o ree-all, INDEPENDENT o their mass. Air resistance will (eventually) aect the motion o all objects. Typically, the graphs o a alling object aected by air resistance become the shapes shown below.

d ispla cem en t / m stra igh t lin e a s velocity becom es con sta n t

20 5 1 .0 velocity / m s -1

2 .0

3 .0 tim e / s

23 20 term in a l velocity of 23 m s -1

10

This can be rearranged to give (u + v) t s=_ 2

10

1 .0

downwards +ve

s hhy

u v

Taking down as positive, the graphs o the motion o any object in ree-all are

downwards +ve

Alternatively, two light gates and a timer can be used to calculate the average velocity between the two gates. Several light gates and a computer can be joined together to make direct calculations o velocity or acceleration.

The list o variables to be considered (and their symbols) is as ollows

downwards +ve

lh  A light gate is a device that senses when an object cuts through a beam o light. The time or which the beam is broken is recorded. I the length o the object that breaks the beam is known, the average speed o the object through the gate can be calculated.

falling objects A very important example o uniormly accelerated motion is the vertical motion o an object in a uniorm gravitational feld. I we ignore the eects o air resistance, this is known as being in ree-all.

1 .0 2 .0 3 .0 tim e / s a ccelera tion = zero a ccelera tion / m s -2 a t term in a l velocity

1 .0

2 .0

3 .0 tim e / s

As the graphs show, the velocity does not keep on rising. It eventually reaches a maximum or terminal velocity. A piece o alling paper will reach its terminal velocity in a much shorter time than a alling book.

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11

p  components of projectile motion

hz 

I two children are throwing and catching a tennis ball between them, the path o the ball is always the same shape. This motion is known as projectile motion and the shape is called a parabola.

There are no orces in the horizontal direction, so there is no horizontal acceleration. This means that the horizontal velocity must be constant.

ball travels at a constant horizontal velocity v3

v2 v1 path taken by ball is a parabola

vH vH dH

vH

vH v4

vH v5 vH

dH

dH

dH

dH

v6

v 

The only orces acting during its fight are gravity and riction. In many situations, air resistance can be ignored.

There is a constant vertical orce acting down, so there is a constant vertical acceleration. The value o the vertical acceleration is 1 0 m s - 2 , which is the acceleration due to gravity.

It is moving horizontally and vertically at the same time but the horizontal and vertical components o the motion are independent o one another. Assuming the gravitional orce is constant, this is always true.

v2

vertical velocity

v3

vH

vH v4 v1

vH

vH

v5

vH

changes

vH v6

mathematics of parabolic motion

example

The graphs o the components o parabolic motion are shown below.

A projectile is launched horizontally rom the top o a cli.

     y-d 

a y / m s -2

a x / m s -2

     x-d 

0

t /s

in itia l h orizon ta l velo city uH

0 g

t /s

vy / m s -1

vx / m s -

h

ux

0

t /s

0

h eigh t o f cli

uy slope = - g t /s x

0

y/m

x/m

uH slope = u x

t /s

0

m a xim u m heigh t

v vertical motion u=0 v=? a = 1 0 m s- 2 s=h t =?

t /s

Once the components have been worked out, the actual velocities (or displacements) at any time can be worked out by vector addition. The solution o any problem involving projectile motion is as ollows:  use the angle o launch to resolve the initial velocity into components.  the time o fight will be determined by the vertical component o velocity.  the range will be determined by the horizontal component (and the time o fight) .  the velocity at any point can be ound by vector addition. Useul short-cuts in calculations include the ollowing acts:  or a given speed, the greatest range is achieved i the launch angle is 45.  i two objects are released together, one with a horizontal velocity and one rom rest, they will both hit the ground together.

12

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s = ut + so

h=0+



t2 =

vf

horizontal motion u = uH v = uH a=0 s=x t= ?

1 2 at 2 1  1 0  t2 2

2h 10

2h s 10 Since v = u + at t= 

v = 0 + 10

2h m 10

x = uH  t

2h m 10 The nal velocity vf is the vector addition o v and u H . =

 20h

m s -1

= uH 

fd  d - fluiD resistance When an object moves through a fuid (a liquid or a gas) , there will be a rictional fuid resistance that aects the objects motion. An example o this eect is the terminal velocity that is reached by a ree-alling object, e.g. a spherical mass alling through a liquid or a parachutist alling towards the Earth. See page 1 1 or how the motion graphs will be altered in these situations. Modelling the precise eect o fuid resistance on moving objects is complex but simple predictions are possible. The Engineering Physics option (see page 1 67) introduces a mathematical analysis o the rictional drag orce that acts on a perect sphere when it moves through a fuid. Key points to note are that:  Viscous drag acts to oppose motion through a fuid  The drag orce is dependent on:  Relative velocity o the object with respect to the fuid  The shape and size o the object (whether the object is aerodynamic or not)  The fuid used (and a property called its viscosity) . For example page 1 2 shows how, in the absence o fuid resistance, an object that is in projectile motion will ollow a parabolic path. When fuid resistance is taken into account, the vertical and the horizontal components o velocity will both be reduced. The eect will be a reduced range and, in the extreme, the horizontal velocity can be reduced to near zero.

parabolic path (no uid resistance) path (with uid resistance)

experiment to Determine free-fall acceleration All experiments to determine the ree-all acceleration or an object are based on the use o a constant acceleration equation with recorded measurements o displacement and time. Some experimental set-ups will be more sophisticated and use more equipment than others. This increased use o technology potentially brings greater precision but can introduce more complications. Simple equipment oten means that, with a limited time available or experimentation, it is easier or many repetitions to be attempted. I an object ree-alls a height, h, rom rest in a time, t, the acceleration, g, can be calculated using s = ut + __12 at2 which rearranges 2h to give = __ . Rather than just calculating a single value, a more reliable value comes rom taking a series o measurement o the t dierent times o all or dierent heights h = __12 gt2 . A graph o h on the y-axis against t2 on the x-axis will give a straight line graph that goes through the origin with a gradient equal to __12 g, making g twice the gradient. 2

Possible set-ups include: Set-up

Comments

Direct measurement o a alling object, e.g. ball bearing with a stop watch and a metre ruler

Very simple set-up meaning many repetitions easily achieved so random error can be eliminated. I height o all is careully controlled, great precision is possible even though equipment is standard. For a simple everyday object such as a ball bearing, the eect o air resistance will be negligible in the laboratory whereas the eect o air resistance on a Ping-Pong ball will be signicant.

Electromagnet release and electronic timing version o the above

The increased precision o the timing can improve accuracy but set-up will take longer. Introduction o technology can mean that systematic errors are harder to identiy.

Motion o alling object automatically recording on ticker-tape attached to alling object

Physical record allows detailed analysis o motion and thus allows the objects whole all to be considered (not just the overall time taken) and or the data to be graphically analysed. Addition o moving paper tape introduces riction to the motion, however.

Distance sensor and data logger

All measurements can be automated and very precise. Sotware can be programmed to perorm all the calculations and to plot appropriate graphs. Experimenter needs to understand how to operate the data logger and associated sotware.

Video analysis o alling object

Capturing a visual record o the objects all against a known scale, allows detailed measurements to be taken. Timing inormation rom the video recording needed, which oten involves ICT.

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13

f d -d d forces  what they are anD what they Do In the examples below, a orce (the kick) can cause deormation (the ball changes shape) or a change in motion (the ball gains a velocity) . There are many dierent types o orces, but in general terms one can describe any orce as the cause o a deormation or a velocity change. The SI unit or the measurement o orces is the newton (N) .

(a) deformation

(b) change in velocity

 A (resultant) orce causes a CHANGE in velocity. I the (resultant) orce is zero then the velocity is constant. Remember a change in velocity is called an acceleration, so we can say that a force causes an acceleration. A (resultant) orce is NOT needed or a constant velocity (see page 1 6) .  The act that a orce can cause deormation is also important, but the deormation o the ball was, in act, not caused by just one orce  there was another one rom the wall.  One orce can act on only one object. To be absolutely precise the description o a orce should include

kick

kick kick causes deformation of football

kick causes a change in motion of football

Eect o a orce on a ootball

 its magnitude  its direction  the object on which it acts (or the part o a large object)  the object that exerts the orce  the nature o the orce A description o the orce shown in the example would thus be a 50 N push at 20 to the horizontal acting ON the ootball FROM the boot.

Different types of forces

forces as vectors

The ollowing words all describe the orces (the pushes or pulls) that exist in nature.

Since orces are vectors, vector mathematics must be used to fnd the resultant orce rom two or more other orces. A orce can also be split into its components. See page 7 or more details.

Gravitational force Electrostatic force Magnetic force

Normal reaction Friction Tension

Compression Upthrust Lift

One way o categorizing these orces is whether they result rom the contact between two suraces or whether the orce exists even i a distance separates the objects. The origin o all these everyday orces is either gravitational or electromagnetic. The vast majority o everyday eects that we observe are due to electromagnetic orces.

(a) by vector mathematics example: block being pushed on rough surface force diagram: S, surface force P, push force resultant W S force W, P weight (b) by components example: block sliding down a smooth slope

measuring forces The simplest experimental method or measuring the size o a orce is to use the extension o a spring. When a spring is in tension it increases in length. The dierence between the natural length and stretched length is called the extension o a spring.

R, reaction  

original length extension = 5.0 cm

2N

W, weight

extension = 15.0 cm

resultant down slope = W sin  component into slope resultant into = W cos  slope = W cos  - R = zero component down slope = W sin 

Vector addition

6N

free-boDy Diagrams In a free-body diagram

extension / cm

15.0

mathematically, F x F = kx

10.0

spring constant (units N m -1 )

5.0 2.0

4.0

6.0

 one object (and ONLY one object) is chosen  all the orces on that object are shown and labelled. For example, i we considered the simple situation o a book resting on a table, we can construct ree-body diagrams or either the book or the table.

8.0 force / N situation:

Hookes law Hookes law states that up to the elastic limit, the extension, x, o a spring is proportional to the tension orce, F. The constant o proportionality k is called the spring constant. The SI units or the spring constant are N m- 1 . Thus by measuring the extension, we can calculate the orce.

14

m ech an i cs

free-body diagram free-body diagram for book: for table: P, push from RT, reaction from table book RE , reaction from Earths RE W weight of table surface

w, weight of book gravitational pull of Earth gravitational pull of Earth

n f  newtons irst law Newtons frst law o motion states that an object continues in uniorm motion in a straight line or at rest unless a resultant external orce acts. On frst reading, this can sound complicated but it does not really add anything to the description o a orce given on page 1 4. All it says is that a resultant orce causes acceleration. No resultant orce means no acceleration  i.e. uniorm motion in a straight line.

bk     

lg  hvy u

R

P, pull from person

R, reaction from ground W

W, weight of suitcase

sin ce

a ccelera tion = zero resu lta n t force = zero  R - W = zero

If the suitcase is too heavy to lift, it is not moving:  acceleration = zero  P+ R= W

c vg   gh  P

phu   

R

R

F, air friction F W F is force forwards, due to engine P is force backwards due to air resistance At all times force up (2R) = force down (W) . If F > P the car accelerates forwards. If F = P the car is at constant velocity (zero acceleration) . If F < P the car decelerates ( i.e. there is negative acceleration and the car slows down) .

parachutist free-falling downwards

p    h  mvg ud

If W > F the parachutist accelerates downwards. As the parachutist gets faster, the air friction increases until W= F The parachutist is at constant velocity (the acceleration is zero) .

lift moving upwards

W, weight

R 2

R 2

W The total force up from the oor of the lift = R. The total force down due to gravity = W. If R > W the person is accelerating upwards. If R = W the person is at constant velocity ( acceleration = zero) . If R < W the person is decelerating ( acceleration is negative) .

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e equilibrium I the resultant orce on an object is zero then it is said to be in translational equilibrium (or just in equilibrium) . Mathematically this is expressed as ollows:  F = zero

Translational equilibrium does NOT mean the same thing as being at rest. For example i the child in the previous example is allowed to swing back and orth, there are times when she is instantaneously at rest but he is never in equilibrium.

From Newtons rst law, we know that the objects in the ollowing situations must be in equilibrium. 1.

An object that is constantly at rest.

2.

An object that is moving with constant (uniorm) velocity in a straight line.

Since orces are vector quantities, a zero resultant orce means no orce IN ANY DIRECTION.

T T

For 2-dimensional problems it is sucient to show that the orces balance in any two non-parallel directions. I this is the case then the object is in equilibrium.

T

 W

tension, T

P, pull

At the end of the swing the forces are not balanced but the child is instantaneously at rest.

W W Forces are not balanced in the centre as the child is in circular motion and is accelerating (see page 65) .

weight, W

if in equilibrium: T sin  = P ( since no resultant horizontal force) T cos  = W (since no resultant vertical force)

Different types of forces Name o orce

Description

Gravitational orce

The orce between objects as a result o their masses. This is sometimes reerred to as the weight o the object but this term is, unortunately, ambiguous  see page 1 9.

Electrostatic orce

The orce between objects as a result o their electric charges.

Magnetic orce

The orce between magnets and/or electric currents.

Normal reaction

The orce between two suraces that acts at right angles to the suraces. I two suraces are smooth then this is the only orce that acts between them.

Friction

The orce that opposes the relative motion o two suraces and acts along the suraces. Air resistance or drag can be thought o as a rictional orce  technically this is known as fuid riction.

Tension

When a string (or a spring) is stretched, it has equal and opposite orces on its ends pulling outwards. The tension orce is the orce that the end o the string applies to another object.

Compression

When a rod is compressed (squashed) , it has equal and opposite orces on its ends pushing inwards. The compression orce is the orce that the ends o the rod applies to another object. This is the opposite o the tension orce.

Upthrust

This is the upward orce that acts on an object when it is submerged in a fuid. It is the buoyancy orce that causes some objects to foat in water (see page 1 64) .

Lit

This orce can be exerted on an object when a fuid fows over it in an asymmetrical way. The shape o the wing o an aircrat causes the aerodynamic lit that enables the aircrat to fy (see page 1 66) .

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n d  newtons seconD law of motion

examples of newtons seconD law

Newtons frst law states that a resultant orce causes an acceleration. His second law provides a means o calculating the value o this acceleration. The best way o stating the second law is use the concept o the momentum o an object. This concept is explained on page 23.

1.

A correct statement o Newtons second law using momentum would be the resultant orce is proportional to the rate o change o momentum. I we use SI units (and you always should) then the law is even easier to state  the resultant orce is equal to the rate o change o momentum. In symbols, this is expressed as ollows

12 N

3 kg

no friction between block and surface I a mass o 3 kg is accelerated in a straight line by a resultant orce o 1 2 N, the acceleration must be 4 m s - 2 . Since F = ma, F 12 _ -2 a= _ m = 3 =4ms . 2.

acceleration = 1.5 m s -2

Use o F = ma in a slightly more complicated situation

12 N

3 kg

friction force I a mass o 3 kg is accelerated in a straight line by a orce o 1 2 N, and the resultant acceleration is 1 .5 m s - 2 , then we can work out the riction that must have been acting. Since

p In SI units, F = _ t dp or, in ull calculus notation, F = _ dt

F = ma

p is the symbol or the momentum o a body.

resultant orce = 3  1 .5 = 4.5 N

Until you have studied what this means this will not make much sense, but this version o the law is given here or completeness. An equivalent (but more common) way o stating Newtons second law applies when we consider the action o a orce on a single mass. I the amount o mass stays constant we can state the law as ollows. The resultant orce is proportional to the acceleration. I we also use SI units then the resultant orce is equal to the product o the mass and the acceleration.

Use o F = ma in a simple situation

This resultant orce = orward orce - riction thereore, riction = orward orce - resultant orce = 1 2 - 4.5 N = 7.5 N 3.

Use o F = ma in a 2-dimensional situation

normal reaction

friction ( max. 8.0 N)

3 kg

In symbols, in SI units,

F = m a rultt for urd  wto

 urd  klogr

30 N

30

lrto urd   - 2

A mass o 3 kg eels a gravitational pull towards the Earth o 30 N. What will happen i it is placed on a 30 degree slope given that the maximum riction between the block and the slope is 8.0 N?

n orm a l rea ction friction

Note:  The F = ma version o the law only applies i we use SI units  or the equation to work the mass must be in kilograms rather than in grams.  F is the resultant orce. I there are several orces acting on an object (and this is usually true) then one needs to work out the resultant orce beore applying the law.  This is an experimental law.  There are no exceptions  Newtons laws apply throughout the Universe. (To be absolutely precise, Einsteins theory o relativity takes over at very large values o speed and mass.) The F = ma version o the law can be used whenever the situation is simple  or example, a constant orce acting on a constant mass giving a constant acceleration. I the situation is more difcult (e.g. a changing orce or a changing mass) then one needs to dp use the F = _ version. dt

3 kg

com pon en t in to slope 30

30 N

30

com pon en t d own th e slope

into slope: normal reaction = component into slope The block does not accelerate into the slope. down the slope: component down slope = 30 N  sin 30 = 15 N maximum riction orce up slope = 8 N  resultant orce down slope = 1 5 - 8 =7N F = ma F  acceleration down slope = _ m 7 = 2.3 m s - 2 =_ 3

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n ird  statement o the law

In symbols,

Newtons second law is an experimental law that allows us to calculate the effect that a force has. Newtons third law highlights the fact that forces always come in pairs. It provides a way of checking to see if we have remembered all the forces involved. It is very easy to state. When two bodies A and B interact, the force that A exerts on B is equal and opposite to the force that B exerts on A. Another way of saying the same thing is that for every action on one object there is an equal but opposite reaction on another object.

FA B = - FB A Key points to notice include  The two forces in the pair act on different objects  this means that equal and opposite forces that act on the same object are NOT Newtons third law pairs.  Not only are the forces equal and opposite, but they must be of the same type. In other words, if the force that A exerts on B is a gravitational force, then the equal and opposite force exerted by B on A is also a gravitational force.

examples o the law rc b rr-kr

push of B on A

push of A on B

A 2.0 m s - 1

B 1.5 m s - 1

a rr-kr u f r   If one roller-skater pushes another, they both feel a force. The forces must be equal and opposite, but the acceleration will be dierent (since they have dierent masses) .

The person with the smaller mass will gain the greater velocity. A

push of wall on girl

2.5 m s - 1 The force on the girl causes her to accelerate backwards.

B

a bk   b  n ird 

push of girl on wall

The mass of the wall (and Earth) is so large that the force on it does not eectively cause any acceleration.

a ccrig cr

R, reaction from table

W, weight

These two forces are not third law pairs. There must be another force (on a dierent object) that pairs with each one:

R W EARTH

F, push forward from the ground on the car In order to accelerate, there must be a forward force on the car. The engine makes the wheels turn and the wheels push on the ground. force from car on ground = - force from ground on car

If the table pushes upwards on the book with force R, then the book must push down on the table with force R.

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If the Earth pulls the book down with force W, then the book must pull the Earth up with force W.

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mass and  weight Mass and weight are two very dierent things. Unortunately their meanings have become muddled in everyday language. Mass is the amount o matter contained in an object (measured in kg) whereas the weight o an object is a orce (measured in N) .

R

2M

M

W

weight, W

new weight = 2W

Double the mass means double the weight To make things worse, the term weight can be ambiguous even to physicists. Some people choose to defne weight as the gravitational orce on an object. Other people defne it to be the reading on a supporting scale. Whichever defnition you use, you weigh less at the top o a building compared with at the bottom  the pull o gravity is slightly less!

situation:

acceleration upwards

I an object is taken to the Moon, its mass would be the same, but its weight would be less (the gravitational orces on the Moon are less than on the Earth) . On the Earth the two terms are oten muddled because they are proportional. People talk about wanting to gain or lose weight  what they are actually worried about is gaining or losing mass.

Although these two defnitions are the same i the object is in equilibrium, they are very dierent in non-equilibrium situations. For example, i both the object and the scale were put into a lit and the lit accelerated upwards then the defnitions would give dierent values.

If the lift is accelerating upwards: R> W The sae thing to do is to avoid using the term weight i at all possible! Stick to the phrase gravitational orce or orce o gravity and you cannot go wrong. Gravitational orce = m g On the surace o the Earth, g is approximately 1 0 N kg- 1 , whereas on the surace o the moon, g  1 .6 N kg- 1

Weight can be dened as either (a) the pull of gravity, W or (b) the force on a supporting scale R. OR R

W Two dierent defnitions o weight

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sd  factors affecting friction  static anD Dynamic

push = zero

friction

friction, F = zero F= 0 N

block P = 0 N stationary

P= 5 N

block stationary F= 5 N

push P = 10 N

block stationary F = 10 N ( = Fmax) block accelerates

Friction arises from the unevenness of the surfaces.

increasing push force

Friction is the orce that opposes the relative motion o two suraces. It arises because the suraces involved are not perectly smooth on the microscopic scale. I the suraces are prevented rom relative motion (they are at rest) then this is an example o static riction. I the suraces are moving, then it is called dynamic riction or kinetic riction.

P = 15 N F= 9 N The value o Fm a x depends upon  the nature o the two suraces in contact.

push causes motion to RIGHT

 the normal reaction orce between the two suraces. The maximum rictional orce and the normal reaction orce are proportional. I the two suraces are kept in contact by gravity, the value o Fm a x does NOT depend upon the area o contact

friction opposes motion, acting to LEFT A key experimental act is that the value o static riction changes depending on the applied orce. Up to a certain maximum orce, Fm a x , the resultant orce is zero. For example, i we try to get a heavy block to move, any value o pushing orce below Fm a x would ail to get the block to accelerate.

Once the object has started moving, the maximum value o riction slightly reduces. In other words, Fk < Fm a x For two suraces moving over one another, the dynamic rictional orce remains roughly constant even i the speed changes slightly.

coefficient of friction

example

Experimentally, the maximum rictional orce and the normal reaction orce are proportional. We use this to defne the coefcient o riction, .

I a block is placed on a slope, the angle o the slope can be increased until the block just begins to slide down the slope. This turns out to be an easy experimental way to measure the coefcient o static riction.

coecient of friction = 

reaction, R P pull forward

F frictional force

W gravitational attraction

Fmax = R

R, reaction component of W down slope ( W sin )

The coefcient o riction is defned rom the maximum value that riction can take Fm a x =  R where R = normal reaction orce It should be noted that  since the maximum value or dynamic riction is less than the maximum value or static riction, the values or the coefcients o riction will be dierent d <  s  the coefcient o riction is a ratio between two orces  it has no units.  i the suraces are smooth then the maximum riction is zero i.e.  = 0.  the coefcient o riction is less than 1 unless the suraces are stuck together. Ff   s R and Ff =  d R

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friction F  component of W into slope (W cos )

W  If balanced,

F = W sin  R = W cos   is increased. When block just starts moving, F = Fmax  static =

=

Fmax R W sin  W cos 

= tan 

w when is work Done?

Definition of work

Work is done when a orce moves its point o application in the direction o the orce. I the orce moves at right angles to the direction o the orce, then no work has been done.

Work is a scalar quantity. Its defnition is as ollows.

1) before

after v

at rest force

block now moving  work has been done

force distance

2) before

block now higher up  work has been done

after

F  work done = Fs cos 

s Work done = F s cos 

I the orce and the displacement are in the same direction, this can be simplifed to Work done = orce  distance From this defnition, the SI units or work done are N m. We defne a new unit called the joule: 1 J = 1 N m.

examples

force distance

(1) lifting vertically small distance

force large force

3) before force spring has been compressed  work has been done

after force

The task in the second case would be easier to perorm (it involves less orce) but overall it takes more work since work has to be done to overcome riction. In each case, the useul work is the same. I the orce doing work is not constant (or example, when a spring is compressed) , then graphical techniques can be used.

distance 4) before

(2) pushing along a rough slope is ta n ce la rge d r fo rce s m a l le

original length

after

FA book supported by shelf  no work is done

Fmax

after constant velocity v

friction-free su rface

v

friction-free su rface

object continues at constant velocity  no work is done

xmax The total work done is the area under the orcedisplacement graph.

force

5) before

x

Fmax x

In the examples above the work done has had dierent results.  In 1 ) the orce has made the object move aster.  In 2) the object has been lited higher in the gravitational feld.  In 3) the spring has been compressed.  In 4) and 5) , NO work is done. Note that even though the object is moving in the last example, there is no orce moving along its direction o action so no work is done.

F = kx

0

total work done = area under graph = 1 k x2 2

xmax Useul equations or the work done include:

extension

 work done when liting something vertically = mgh where m represents mass (in kg) g represents the Earths gravitational feld strength (1 0 N kg- 1 ) h represents the height change (in m)  work done in compressing or extending a spring = __12 k  x2

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e d  concepts of energy anD work K  0 J P = 1000 J

Energy and work are linked together. When you do work on an object, it gains energy and you lose energy. The amount o energy transerred is equal to the work done. Energy is a measure o the amount o work done. This means that the units o energy must be the same as the units o work  joules.

energy transformations  conservation of energy In any situation, we must be able to account or the changes in energy. I it is lost by one object, it must be gained by another. This is known as the principle o conservation o energy. There are several ways o stating this principle:

K = 250 J P = 750 J

 Overall the total energy o any closed system must be constant.  Energy is neither created nor destroyed, it just changes orm.

K = 250 J P = 750 J

 There is no change in the total energy in the Universe.

energy types Kinetic energy Gravitational potential Elastic potential energy Radiant energy Electrostatic potential Thermal energy Nuclear energy Solar energy Chemical energy Electrical energy Internal energy Light energy Equations or the frst three types o energy are given below. Kinetic energy = =

__1 mv

2 p2 ___ 2m

2

where m is the mass (in kg) , v is the velocity (in m s - 1 )

where p is the momentum (see page 23) (in kg m s- 1 ), and m is the mass (in kg)

Gravitational potential energy = mgh where m represents mass (in kg), g represents the Earths gravitational feld (1 0 N kg- 1 ) , h represents the height change (in m)

K = 500 J P = 500 J

K = 750 J P = 250 J

K = 1000 J P = 0 J

Elastic potential energy = the extension (in m)

__1 k x 2

2

where k is the spring constant (in N m- 1 ) ,  x is

power anD efficiency

examples

1. Power Power is defned as the RATE at which energy is transerred. This is the same as the rate at which work is done. energy transerred Power = __ time taken work done __ Power = time taken The SI unit or power is the joule per second (J s- 1 ) . Another unit or power is defned - the watt (W) . 1 W = 1 J s - 1 . I something is moving at a constant velocity v against a constant rictional orce F, the power P needed is P = F v

1.

2. Efciency Depending on the situation, we can categorize the energy transerred (work done) as useul or not. In a light bulb, the useul energy would be light energy, the wasted energy would be thermal energy (and non-visible orms o radiant energy). We defne efciency as the ratio o useul energy to the total energy transerred. Possible orms o the equation include: useul work OUT Efciency = __ total work IN useul energy OUT __ Efciency = total energy IN useul power OUT __ Efciency = total power IN Since this is a ratio it does not have any units. Oten it is expressed as a percentage.

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A grasshopper (mass 8 g) uses its hindlegs to push or 0.1 s and as a result jumps 1 .8 m high. Calculate (i) its take o speed, (ii) the power developed. (i) PE gained = mgh 1 mv2 KE at start = _ 2 1 mv2 = mgh (conservation o _ 2 energy)  =  v = 2gh 2  1 0  1 .8 = 6 m s- 1 mgh (ii) Power = _ t 0.008  1 0  1 .8 = __ 0.1  1 .4 W

2.

A 60W lightbulb has an efciency o 1 0%. How much energy is wasted every hour? Power wasted = 90% o 60W = 54W Energy wasted = 54  60  60J = 1 90 kJ

m d  Definitions  linear momentum anD impulse

conservation of momentum

Linear momentum (always given the symbol p) is defned as the product o mass and velocity.

The law o conservation o linear momentum states that the total linear momentum o a system o interacting particles remains constant provided there is no resultant external force.

Momentum = mass  velocity p=mv The SI units or momentum must be kg m s . Alternative units o N s can also be used (see below) . Since velocity is a vector, momentum must be a vector. In any situation, particularly i it happens quickly, the change o momentum p is called the impulse (p = F t) . -1

use of momentum in newtons seconD law Newtons second law states that the resultant orce is proportional to the rate o change o momentum. Mathematically we can write this as (fnal momentum - initial momentum) p F = ____ = _ time taken t Example 1 A jet o water leaves a hose and hits a wall where its velocity is brought to rest. I the hose cross-sectional area is 25 cm2 , the velocity o the water is 50 m s - 1 and the density o the water is 1 000 kg m- 3 , what is the orce acting on the wall?

ve lo city = 5 0 m s - 1 50 m d e n sity o f wa te r = 1 0 0 0 kg m - 3

cro ss-se ctio n a l a re a = 2 5 cm 2 = 0 .0 0 2 5 m 2

To see why, we start by imagining two isolated particles A and B that collide with one another.  The orce rom A onto B, FA B will cause Bs momentum to change by a certain amount.  I the time taken was t, then the momentum change (the impulse) given to B will be given by p B = FA B t  By Newtons third law, the orce rom B onto A, FB A will be equal and opposite to the orce rom A onto B, FA B = - FB A .  Since the time o contact or A and B is the same, then the momentum change or A is equal and opposite to the momentum change or B, pA = - FA B t.  This means that the total momentum (momentum o A plus the momentum o B) will remain the same. Total momentum is conserved. This argument can be extended up to any number o interacting particles so long as the system o particles is still isolated. I this is the case, the momentum is still conserved.

elastic anD inelastic collisions The law o conservation o linear momentum is not enough to always predict the outcome ater a collision (or an explosion) . This depends on the nature o the colliding bodies. For example, a moving railway truck, mA , velocity v, collides with an identical stationary truck mB . Possible outcomes are:

( a ) ela stic collision a t re st

n ew velocity = v

In one second, a jet o water 50 m long hits the wall. So volume o water hitting wall = 0.0025  50 = 0.1 25 m3 every second mass o water hitting wall = 0.1 25  1 000 = 1 25 kg every second

( b) tota lly in ela stic collision

momentum o water hitting wall = 1 25  50 = 6250 kg m s - 1 every second This water is all brought to rest,  change in momentum, p = 6250 kg m s - 1 p 6250 = 6250 N  orce = _ = _ 1 t Example 2 The graph below shows the variation with time o the orce on a ootball o mass 500 g. Calculate the fnal velocity o the ball.

force/N

The football was given an impulse of approximately 100  0.01 = 1 N s during this 0.01 s. 100 Area under graph is the total 90 impulse given to the 80 ball  5 N s p = mv 70 p v = 60 m 5Ns 50 = 0.5 kg 40 = 10 m s-1 30  nal 20 velocity v= 10 m s-1 10 0.00 0.02 0.04 0.06 0.08 0.10 time/s

mB

mA

n ew velocity = v 2 mA

( c) in ela stic collision n ew velocity = v 4 mA

mB

n ew velocity = 3 v 4 mB

In (a) , the trucks would have to have elastic bumpers. I this were the case then no mechanical energy at all would be lost in the collision. A collision in which no mechanical energy is lost is called an elastic collision. In reality, collisions between everyday objects always lose some energy  the only real example o elastic collisions is the collision between molecules. For an elastic collision, the relative velocity o approach always equals the relative velocity o separation. In (b) , the railway trucks stick together during the collision (the relative velocity o separation is zero) . This collision is what is known as a totally inelastic collision. A large amount o mechanical energy is lost (as heat and sound) , but the total momentum is still conserved. In energy terms, (c) is somewhere between (a) and (b) . Some energy is lost, but the railway trucks do not join together. This is an example o an inelastic collision. Once again the total momentum is conserved. Linear momentum is also conserved in explosions.

m ech an i cs

23

ib questons  mechancs 1.

A. 1 :  2 2.

6.

Two identical objects A and B all rom rest rom dierent heights. I B takes twice as long as A to reach the ground, what is the ratio o the heights rom which A and B ell? Neglect air resistance. B. 1 :2

C. 1 :4

D. 1 :8

A trolley is given an initial push along a horizontal foor to get it moving. The trolley then travels orward along the foor, gradually slowing. What is true o the horizontal orce(s) on the trolley while it is slowing? A. There is a orward orce and a backward orce, but the orward orce is larger. B. There is a orward orce and a backward orce, but the backward orce is larger. C. There is only a orward orce, which diminishes with time. D. There is only a backward orce.

3.

A mass is suspended by cord rom a ring which is attached by two urther cords to the ceiling and the wall as shown. The cord rom the ceiling makes an angle o less than 45 with the vertical as shown. The tensions in the three cords are labelled R, S and T in the diagram.

S

5.

The vehicles collide head-on and become entangled together. a) During the collision, how does the orce exerted by the car on the truck compare with the orce exerted by the truck on the car? Explain.

[2]

b) In what direction will the entangled vehicles move ater collision, or will they be stationary? Support your answer, reerring to a physics principle. [2] c) Determine the speed (in km h- 1 ) o the combined wreck immediately ater the collision.

[3]

d) How does the acceleration o the car compare with the acceleration o the truck during the collision? Explain.

[2]

e) Both the car and truck drivers are wearing seat belts. Which driver is likely to be more severely jolted in the collision? Explain. [2]

T

f) The total kinetic energy o the system decreases as a result o the collision. Is the principle o conservation o energy violated? Explain. [1 ]

R

7. a) A net orce o magnitude F acts on a body. Dene the impulse I o the orce.

How do the tensions R, S and T in the three cords compare in magnitude?

4.

A car and a truck are both travelling at the speed limit o 60 km h- 1 but in opposite directions as shown. The truck has twice the mass o the car.

[1 ]

b) A ball o mass 0.0750 kg is travelling horizontally with a speed o 2.20 m s - 1 . It strikes a vertical wall and rebounds horizontally.

A. R > T > S

B. S > R > T

C. R = S = T

D. R = S > T

ball mass 0.0750 kg

A 24 N orce causes a 2.0 kg mass to accelerate at 8.0 m s- 2 along a horizontal surace. The coecient o dynamic riction is: A. 0.0

B. 0.4

C. 0.6

D. 0.8

2.20 ms -1

An athlete trains by dragging a heavy load across a rough horizontal surace. Due to the collision with the wall, 20 % o the balls initial kinetic energy is dissipated. (i)

F

Show that the ball rebounds rom the wall with a speed o 1 .97 m s - 1 . [2]

(ii) Show that the impulse given to the ball by the wall is 0.31 3 N s.

25

[2]

c) The ball strikes the wall at time t = 0 and leaves the wall at time t = T. The athlete exerts a orce o magnitude F on the load at an angle o 25 to the horizontal.

The sketch graph shows how the orce F that the wall exerts on the ball is assumed to vary with time t.

a) Once the load is moving at a steady speed, the average horizontal rictional orce acting on the load is 470 N. Calculate the average value o F that will enable the load to move at constant speed.

F [2]

b) The load is moved a horizontal distance o 2.5 km in 1.2 hours. Calculate (i)

the work done on the load by the orce F.

(ii) the minimum average power required to move the load.

0

[2]

Explain, in terms o energy changes, why the minimum average power required is greater than in (b) (ii) . [2]

i B Q u esti o n s  m ech an i cs

t

The time T is measured electronically to equal 0.0894 s.

c) The athlete pulls the load uphill at the same speed as in part (a) .

24

T

[2] Use the impulse given in (b) (ii) to estimate the average value o F. [4]

33 T h e r m a l p h Ys i C s T cct TemperaTure and heaT flow

Kelvin and Celsius

Hot and cold are just labels that identiy the direction in which thermal energy (sometimes known as heat) will be naturally transerred when two objects are placed in thermal contact. This leads to the concept o the hotness o an object. The direction o the natural ow o thermal energy between two objects is determined by the hotness o each object. Thermal energy naturally ows rom hot to cold.

Most o the time, there are only two sensible temperature scales to chose between  the Kelvin scale and the Celsius scale.

Heat is not a substance that ows rom one object to another. What has happened is that thermal energy has been transerred. Thermal energy (heat) reers to the non-mechanical transer o energy between a system and its surroundings.

There is an easy relationship between a temperature T as measured on the Kelvin scale and the corresponding temperature t as measured on the Celsius scale. The approximate relationship is T (K) = t (C) + 273

700 K 630 K 600 K

COLD

mercury boils

400 K 373 K 300 K 273 K

400 C 357 C 300 C

500 K

200 C 100 C

water boils

200 K

water freezes mercury freezes carbon dioxide freezes

100 K

oxygen boils

0K HOT

Notice the size of the units is identical on each scale.

Celsius scale

This means that the size o the units used on each scale is identical, but they have dierent zero points.

Kelvin scale

The temperature o an object is a measure o how hot it is. In other words, i two objects are placed in thermal contact, then the temperature dierence between the two objects will determine the direction o the natural transer o thermal energy. Thermal energy is naturally transerred down the temperature dierence  rom high temperature to low temperature. Eventually, the two objects would be expected to reach the same temperature. When this happens, they are said to be in thermal equilibrium.

In order to use them, you do not need to understand the details o how either o these scales has been defned, but you do need to know the relation between them. Most everyday thermometers are marked with the Celsius scale and temperature is quoted in degrees Celsius (C) .

hydrogen boils

0 C -100 C -200 C -273 C

The Kelvin scale is an absolute thermodynamic temperature scale and a measurement on this scale is also called the absolute temperature. Zero Kelvin is called absolute zero (see page 29) .

direction of transfer of thermal energy

examples: Gases For a given sample o a gas, the pressure, the volume and the temperature are all related to one another.  The pressure, P, is the orce per unit area rom the gas acting at 90 on the container wall. F p= _ A The SI units o pressure are N m- 2 or Pa (Pascals) . 1 Pa = 1 N m- 2 Gas pressure can also be measured in atmospheres (1 atm  1 0 5 Pa)

In order to investigate how these quantities are interrelated, we choose:  one quantity to be the independent variable (the thing we alter and measure)  another quantity to be the dependent variable (the second thing we measure) .  The third quantity needs to be controlled (i.e. kept constant) . The specifc values that will be recorded also depend on the mass o gas being investigated and the type o gas being used so these need to be controlled as well.

 The volume, V, o the gas is measured in m3 or cm3 (1 m3 = 1 0 6 cm3 )  The temperature, t, o the gas is measured in C or K

T H E R M A L P H YS I C S

25

ht  t gy miCrosCopiC vs maCrosCopiC

KineTiC TheorY

When analysing something physical, we have a choice.

Molecules are arranged in dierent ways depending on the phase o the substance (i.e. solid, liquid or gas) .

 The macroscopic point o view considers the system as a whole and sees how it interacts with its surroundings.  The microscopic point o view looks inside the system to see how its component parts interact with each other. So ar we have looked at the temperature o a system in a macroscopic way, but all objects are made up o atoms and molecules. According to kinetic theory these particles are constantly in random motion  hence the name. See below or more details. Although atoms and molecules are dierent things (a molecule is a combination o atoms) , the dierence is not important at this stage. The particles can be thought o as little points o mass with velocities that are continually changing.

inTernal enerGY I the temperature o an object changes then it must have gained (or lost) energy. From the microscopic point o view, the molecules must have gained (or lost) this energy. The two possible orms are kinetic energy and potential energy.

speed in a random direction  molecule has KE

v

solids Macroscopically, solids have a fxed volume and a fxed shape. This is because the molecules are held in position by bonds. However the bonds are not absolutely rigid. The molecules vibrate around a mean (average) position. The higher the temperature, the greater the vibrations.

Each molecule vibrates around a mean position.

Bonds between molecules

The molecules in a solid are held close together by the intermolecular bonds.

liquids A liquid also has a fxed volume but its shape can change. The molecules are also vibrating, but they are not completely fxed in position. There are still strong orces between the molecules. This keeps the molecules close to one another, but they are ree to move around each other.

F equilibrium position

resultant force back towards equilibrium position due to neighbouring molecules  molecule has PE

 The molecules have kinetic energy because they are moving. To be absolutely precise, a molecule can have either translational kinetic energy (the whole molecule is moving in a certain direction) or rotational kinetic energy (the molecule is rotating about one or more axes) .  The molecules have potential energy because o the intermolecular orces. I we imagine pulling two molecules urther apart, this would require work against the intermolecular orces. The total energy that the molecules possess (random kinetic plus inter molecule potential) is called the internal energy o a substance. Whenever we heat a substance, we increase its internal energy.

Bonds between neighbouring molecules; these can be made and broken, allowing a molecule to move.

Each molecule is free to move throughout the liquid by moving around its neighbours.

Gases A gas will always expand to fll the container in which it is put. The molecules are not fxed in position, and any orces between the molecules are very weak. This means that the molecules are essentially independent o one another, but they do occasionally collide. More detail is given on page 31 .

Molecules in random motion; no xed bonds between molecules so they are free to move

Temperature is a measure of the average kinetic energy of the molecules in a substance. I two substances have the same temperature, then their molecules have the same average kinetic energy.

heaT and worK Many people have conused ideas about heat and work. In answers to examination questions it is very common to read, or example, that heat rises  when what is meant is that the transer o thermal energy is upwards.

same temperature

v

same average kinetic energy

V m

M molecules with large mass moving with lower average speed

molecules with small mass moving with higher average speed

 When a orce moves through a distance, we say that work is done. Work is the energy that has been transmitted rom one system to another rom the macroscopic point o view.  When work is done on a microscopic level (i.e. on individual molecules) , we say that heating has taken place. Heat is the energy that has been transmitted. It can either increase the kinetic energy o the molecules or their potential energy or, o course, both. In both cases energy is being transerred.

26

T H E R M A L P H YS I C S

scfc t ccty deiniTions and miCrosCopiC explanaTion In theory, i an object could be heated up with no energy loss, then the increase in temperature T depends on three things:

meThods o measurinG heaT CapaCiTies and speCiiC heaT CapaCiTies The are two basic ways to measure heat capacity.

 the energy given to the object Q,

1.

 the mass, m, and

The experiment would be set up as below:

Electrical method

 the substance rom which the object is made.

heater ( placed in object)

1000 J

1000 J

mass m substance X

mass m substance Y V voltmeter

dierent temperature change

ammeter A

small temperature change since more molecules

variable power supply ItV  the specifc heat capacity c = _ . m(T2 - T1 ) Sources o experimental error

large temperature change since fewer molecules

 loss o thermal energy rom the apparatus.

Two dierent blocks with the same mass and same energy input will have a dierent temperature change. We defne the thermal capacity C o an object as the energy required to raise its temperature by 1 K. Dierent objects (even dierent samples o the same substance) will have dierent values o heat capacity. Specifc heat capacity is the energy required to raise a unit mass o a substance by 1 K. Specifc here just means per unit mass.

 the container or the substance and the heater will also be warmed up.  it will take some time or the energy to be shared uniormly through the substance. 2.

Method o mixtures

The known specifc heat capacity o one substance can be used to fnd the specifc heat capacity o another substance.

before temperature TA (hot)

In symbols, Thermal capacity Specifc heat capacity

Q C = _ (J K- 1 or J C - 1 ) T Q c = _ (J kg- 1 K- 1 or J kg- 1 C - 1 ) (m T) Q = mcT

mix together

Note  A particular gas can have many dierent values o specifc heat capacity  it depends on the conditions used  see page 1 61.  These equations reer to the temperature dierence resulting rom the addition o a certain amount o energy. In other words, it generally takes the same amount o energy to raise the temperature o an object rom 25 C to 35 C as it does or the same object to go rom 402 C to 41 2 C. This is only true so long as energy is not lost rom the object.

temperature

 I an object is raised above room temperature, it starts to lose energy. The hotter it becomes, the greater the rate at which it loses energy.

temperature TB (cold)

mass m A

mass m B temperature Tmax

after

Procedure:  measure the masses o the liquids mA and m B .  measure the two starting temperatures TA and TB .

increase in temperature if no energy is lost

 mix the two liquids together.  record the maximum temperature o the mixture Tm a x . I no energy is lost rom the system then, energy lost by hot substance cooling down = energy gained by cold substance heating up

increase in temperature in a real situation

m A cA (TA - Tm a x ) = m B cB (Tm a x - TB )

time Temperature change o an object being heated at a constant rate

Again, the main source o experimental error is the loss o thermal energy rom the apparatus; particularly while the liquids are being transerred. The changes o temperature o the container also need to be taken into consideration or a more accurate result.

T H E R M A L P H YS I C S

27

p (tt)  tt  ltt t meThods of measurinG The two possible methods or measuring latent heats shown below are very similar in principle to the methods or measuring specifc heat capacities (see previous page) .

temperature C /

definiTions and miCrosCopiC view When a substance changes phase, the temperature remains constant even though thermal energy is still being transerred.

500 400

1.

molten lead

A method or measuring the specifc latent heat o vaporization o water

set-up

300

liquid and solid mix

electrical circuit heater

solid to electrical circuit

200 100

V voltmeter A ammeter

water 1 2 3 4 5 6 7 8 9 10 11 12 13 14 time / min

heater

Cooling curve or molten lead (idealized) The amount o energy associated with the phase change is called the latent heat. The technical term or the change o phase rom solid to liquid is usion and the term or the change rom liquid to gas is vaporization. The energy given to the molecules does not increase their kinetic energy so it must be increasing their potential energy. Intermolecular bonds are being broken and this takes energy. When the substance reezes bonds are created and this process releases energy. It is a very common mistake to think that the molecules must speed up during a phase change. The molecules in water vapour at 1 00 C must be moving with the same average speed as the molecules in liquid water at 1 00 C. The specifc latent heat o a substance is defned as the amount o energy per unit mass absorbed or released during a change o phase.

beaker

variable power supply

The amount o thermal energy provided to water at its boiling point is calculated using electrical energy = I t V. The mass vaporized needs to be recorded. ItV  The specifc latent heat L = _ . (m 1 - m 2 ) Sources o experimental error  Loss o thermal energy rom the apparatus.  Some water vapour will be lost beore and ater timing. 2.

A method or measuring the specifc latent heat o usion o water

Providing we know the specifc heat capacity o water, we can calculate the specifc latent heat o usion or water. In the example below, ice (at 0 C) is added to warm water and the temperature o the resulting mix is measured.

ice

water

In symbols,

temperature/C

Q Specifc latent heat L = _ (J kg- 1 ) Q = ML M In the idealized situation o no energy loss, a constant rate o energy transer into a solid substance would result in a constant rate o increase in temperature until the melting point is reached:

mix together mass: m ice temp.: 0 C

mass: m water temp.: T C

liquid solid and liquid mix

mass: m water + m ice temp.: T mix

solid

I no energy is lost rom the system then,

energy supplied/J

energy lost by water cooling down = energy gained by ice

Phase-change graph with temperature vs energy

m w a te r cw a te r (Tw a te r - Tm ix ) = m ice L u s io n + m ice cw a te r Tm ix

In the example above, the specifc heat capacity o the liquid is less than the specifc heat capacity o the solid as the gradient o the line that corresponds to the liquid phase is greater than the gradient o the line that corresponds to the solid phase. A given amount o energy will cause a greater increase in temperature or the liquid when compared with the solid.

Sources o experimental error  Loss (or gain) o thermal energy rom the apparatus.  I the ice had not started at exactly zero, then there would be an additional term in the equation in order to account or the energy needed to warm the ice up to 0 C.  Water clinging to the ice beore the transer.

28

T H E R M A L P H YS I C S

Th g  1 Gas laws

(a) constant volume graph extrapolates pressure / Pa back to -273 C

-300

-100

100 temp. / C

(2) constant pressure

0

100 temp. / C

(3) constant temperature

volume / m 3

-200

-100

(c) constant temperature

absolute temperature / K

0

absolute temperature / K

pressure / Pa

(b) constant pressure graph extrapolates back to -273 C

(1) constant volume

volume / m 3

-200

pressure / Pa

-300

The trends can be seen more clearly i this inormation is presented in a slightly dierent way.

pressure / Pa

For the experimental methods shown below, the graphs below outline what might be observed.

volume / m 3

1 -3 volume / m

Points to note:  Although pressure and volume both vary linearly with Celsius temperature, neither pressure nor volume is proportional to Celsius temperature.  A dierent sample o gas would produce a dierent straightline variation or pressure (or volume) against temperature but both graphs would extrapolate back to the same low temperature, - 273 C. This temperature is known as absolute zero.  As pressure increases, the volume decreases. In act they are inversely proportional.

experimenTal invesTiGaTions 1 . Temperature t as the independent variable; P as the dependent variable; V as the control.

temperature t measured pressure gauge to measure P surface of water air in ask

xed volume of air water (or oil) bath

 Fixed volume o gas is trapped in the ask. Pressure is measured by a pressure gauge.  Temperature o gas altered by temperature o bath  time is needed to ensure bath and gas at same temperature. 2. Temperature t as the independent variable; V as the dependent variable; P as the control.

temperature t measured capillary tube scale to measure V (length and volume) surface of water water bath bead of sulfuric acid gas (air) zero of scale volume V

From these graphs or a fxed mass o gas we can say that: p 1 . At constant V, p  T or _ = constant (the pressure law) T V = constant (Charless law) 2. At constant p, V  T or _ T 1 or p V = constant (Boyles law) 3. At constant T, p  _ V These relationships are known as the ideal gas laws. The temperature is always expressed in Kelvin (see page 25) . These laws do not always apply to experiments done with real gases. A real gas is said to deviate rom ideal behaviour under certain conditions (e.g. high pressure) .

 Volume o gas is trapped in capillary tube by bead o concentrated suluric acid.  Concentrated suluric acid is used to ensure gas remains dry.  Heating gas causes it to expand moving bead.  Pressure remains equal to atmospheric.  Temperature o gas altered by temperature o bath; time is needed to ensure bath and gas at same temperature. 3. P as the independent variable; V as the dependent variable; t as the control.

zero of scale scale to measure V (length and volume)

trapped air

pressure gauge to measure p air

oil column

pump surface of oil

oil  Volume o gas measured against calibrated scale.  Increase o pressure orces oil column to compress gas.  Temperature o gas will be altered when volume is changed; time is needed to ensure gas is always at room temperature.

T H E R M A L P H YS I C S

29

T g  2 equaTion of sTaTe

definiTions

The three ideal gas laws can be combined together to produce one mathematical relationship. pV _ = constant T

The concepts o the mole, molar mass and the Avogadro constant are all introduced so as to be able to relate the mass o a gas (an easily measurable quantity) to the number o molecules that are present in the gas.

This constant will depend on the mass and type o gas.

Ideal gas

An ideal gas is one that ollows the gas laws or all values o o P, V and T (see page 29) .

Mole

The mole is the basic SI unit or amount o substance. One mole o any substance is equal to the amount o that substance that contains the same number o particles as 0.01 2 kg o carbon1 2 ( 1 2 C) . When writing the unit it is (slightly) shortened to the mol.

Avogadro constant, NA

This is the number o atoms in 0.01 2 kg o carbon1 2 ( 1 2 C) . It is 6.02  1 0 2 3 .

Molar mass

The mass o one mole o a substance is called the molar mass. A simple rule applies. I an element has a certain mass number, A, then the molar mass will be A grams. N n=_ NA number o atoms number o moles = __ Avogadro constant

I we compare the value o this constant or dierent masses o dierent gases, it turns out to depend on the number o molecules that are in the gas  not their type. In this case we use the defnition o the mole to state that or n moles o ideal gas pV _ = a universal constant. nT The universal constant is called the molar gas constant R. The SI unit or R is J mol- 1 K- 1 R = 8.31 4 J mol- 1 K- 1 pV Summary: _ = R Or p V = n R T nT

example a) What volume will be occupied by 8 g o helium (mass number 4) at room temperature (20 C) and atmospheric pressure (1 .0  1 0 5 Pa) 8 = 2 moles n = _ 4 T = 20 + 273 = 293 K

ideal Gases and real Gases

nRT 2  8.31 4  293 = 0.049 m3 __ V = _ p = 1 .0  1 0 5 b) How many atoms are there in 8 g o helium (mass number 4)? 8 = 2 moles n = _ 4 number o atoms = 2  6.02  1 0 2 3 = 1 .2  1 0 2 4

linK beTween The maCrosCopiC and miCrosCopiC The equation o state or an ideal gas, pV = nRT, links the three macroscopic properties o a gas (p, V and T) . Kinetic theory (page 26) describes a gas as being composed o molecules in random motion and or this theory to be valid, each o these macroscopic properties must be linked to the microscopic behaviour o molecules. A detailed analysis o how a large number o randomly moving molecules interact beautiully predicts another ormula that allows the links between the macroscopic and the microscopic to be identifed. The derivation o the ormula only uses Newtons laws and a handul o assumptions. These assumptions describe rom the microscopic perspective what we mean by an ideal gas.

An ideal gas is a one that ollows the gas laws or all values o p, V and T and thus ideal gases cannot be liquefed. The microscopic description o an ideal gas is given on page 3 1 . Real gases, however, can approximate to ideal behaviour providing that the intermolecular orces are small enough to be ignored. For this to apply, the pressure/density o the gas must be low and the temperature must be moderate.

Equating the right-hand side o this ormula with the righthand side o the macroscopic equation o state or an ideal gas shows that: 2 N__ EK nRT = _ 3 N , so But n = _ NA N RT = _ 2 N__ _ EK NA 3 __



3 _ R T EK = _ 2 NA

R (the molar gas constant) and NA (Avogadro constant) are fxed numbers so this equation shows that the absolute temperature is proportional to the average KE per molecule __

The detail o this derivation is not required by the IB syllabus but the assumptions and the approach are outlined on the ollowing page. The result o this derivation is that the pressure and volume o the idealized gas are related to just two quantities: 2 N__ pV = _ EK 3  The number o molecules present, N __

 The average random kinetic energy per molecule, EK.

30

T H E R M A L P H YS I C S

T  EK R R The ratio __ is called the Boltzmanns constant kB . kB = __ N N A

__

3 k T= _ 3 _ R T EK = _ 2 B 2 NA

A

mc     g KineTiC model of an ideal Gas

before

wall

Assumptions:

 When a molecule bounces o the walls o a container its momentum changes (due to the change in direction  momentum is a vector) .

 Newtons laws apply to molecular behaviour  there are no intermolecular orces except during a collision

 There must have been a orce on the molecule rom the wall (Newton II) .

 the molecules are treated as points  the molecules are in random motion  the collisions between the molecules are elastic (no energy is lost)

A single molecule hitting the walls o the container.

after

wall

 there is no time spent in these collisions.

 There must have been an equal and opposite orce on the wall rom the molecule (Newton III) .  Each time there is a collision between a molecule and the wall, a orce is exerted on the wall.

The pressure o a gas is explained as ollows:

 The average o all the microscopic orces on the wall over a period o time means that there is eectively a constant orce on the wall rom the gas.  This orce per unit area o the wall is what we call pressure.

result

F P= _ A

overall force on wall

overall force on molecule The pressure o a gas is a result o collisions between the molecules and the walls o the container.

Since the temperature o a gas is a measure o the average kinetic energy o the molecules, as we lower the temperature o a gas the molecules will move slower. At absolute zero, we imagine the molecules to have zero kinetic energy. We cannot go any lower because we cannot reduce their kinetic energy any urther!

pressure law

Charless law

boYles law

Macroscopically, at a constant volume the pressure o a gas is proportional to its temperature in kelvin (see page 29) . Microscopically this can be analysed as ollows

Macroscopically, at a constant pressure, the volume o a gas is proportional to its temperature in kelvin (see page 29) . Microscopically this can be analysed as ollows

Macroscopically, at a constant temperature, the pressure o a gas is inversely proportional to its volume (see page 29) . Microscopically this can be seen to be correct.

 I the temperature o a gas goes up, the molecules have more average kinetic energy  they are moving aster on average.

 A higher temperature means aster moving molecules (see let) .

 The constant temperature o gas means that the molecules have a constant average speed.

 Fast moving molecules will have a greater change o momentum when they hit the walls o the container.

 Faster moving molecules hit the walls with a greater microscopic orce (see let) .

 Thus the microscopic orce rom each molecule will be greater.

 I the volume o the gas increases, then the rate at which these collisions take place on a unit area o the wall must go down.

 The molecules are moving aster so they hit the walls more oten.

 The average orce on a unit area o the wall can thus be the same.

 For both these reasons, the total orce on the wall goes up.

 Thus the pressure remains the same.

 Thus the pressure goes up.

low temperature

low temperature

high temperature

 The microscopic orce that each molecule exerts on the wall will remain constant.  Increasing the volume o the container decreases the rate with which the molecules hit the wall  average total orce decreases.  I the average total orce decreases the pressure decreases.

high pressure

low pressure

high temperature

constant volume low pressure high pressure Microscopic justifcation o the pressure law

constant pressure low volume high volume Microscopic justifcation o Charless law

constant temperature low volume high volume Microscopic justifcation o Boyles law

T H E R M A L P H YS I C S

31

ib questons  thermal physcs The following information relates to questions 1 and 2 below.

b) An electrical heater or swimming pools has the ollowing inormation written on its side:

temperature

A substance is heated at a constant rate o energy transer. A graph o its temperature against time is shown below.

50 Hz (i)

P N L

O

2.3 kW

Estimate how many days it would take this heater to heat the water in the swimming pool.

(ii) Suggest two reasons why this can only be an approximation. 6.

M

[4] [2]

a) A cylinder ftted with a piston contains 0.23 mol o helium gas.

K piston

time 1.

helium gas

Which regions o the graph correspond to the substance existing in a mixture o two phases?

The ollowing data are available or the helium with the piston in the position shown.

A. KL, MN and OP B. LM and NO

Volume

= 5.2  1 0 - 3 m3

C. All regions

Pressure

= 1 .0  1 0 5 Pa

D. No regions 2.

Temperature = 290 K

In which region o the graph is the specifc heat capacity o the substance greatest?

(i)

A. KL

(ii) State the assumption made in the calculation in (a) (i) .

B. LM 7.

C. MN D. OP 3.

B. the gas molecules repel each other more strongly C. the average velocity o gas molecules hitting the wall is greater

She records her measurements as ollows:

D. the requency o collisions with gas molecules with the walls is greater A lead bullet is fred into an iron plate, where it deorms and stops. As a result, the temperature o the lead increases by an amount T. For an identical bullet hitting the plate with twice the speed, what is the best estimate o the temperature increase? A. T

(1 )

This question is about determining the specifc latent heat o usion o ice. A student determines the specifc latent heat o usion o ice at home. She takes some ice rom the reezer, measures its mass and mixes it with a known mass o water in an insulating jug. She stirs until all the ice has melted and measures the fnal temperature o the mixture. She also measured the temperature in the reezer and the initial temperature o the water.

When the volume o a gas is isothermally compressed to a smaller volume, the pressure exerted by the gas on the container walls increases. The best microscopic explanation or this pressure increase is that at the smaller volume A. the individual gas molecules are compressed

4.

Use the data to calculate a value or the universal gas constant. (2)

Mass o ice used

mi

0.1 2 kg

Initial temperature o ice

Ti

-1 2 C

Initial mass o water

mw

0.40 kg

Initial temperature o water

Tw

22 C

Final temperature o mixture

T

1 5 C

B. 2 T The specifc heat capacities o water and ice are cw = 4.2 kJ kg- 1 C - 1 and ci = 2.1 kJ kg- 1 C - 1

C. 2 T D. 4 T 5.

a)

Set up the appropriate equation, representing energy transers during the process o coming to thermal equilibrium, that will enable her to solve or the specifc latent heat L i o ice. Insert values into the equation rom the data above, but do not solve the equation. [5]

Clearly show any estimated values. The ollowing inormation will be useul:

b)

Explain the physical meaning o each energy transfer term in your equation (but not each symbol) . [4]

Specifc heat capacity o water

41 86 J kg- 1 K- 1

c)

Density o water

1 000 kg m- 3

State an assumption you have made about the experiment, in setting up your equation in (a) .

[1 ]

d)

Why should she take the temperature o the mixture immediately ater all the ice has melted?

[1 ]

e)

Explain rom the microscopic point o view, in terms o molecular behaviour, why the temperature o the ice does not increase while it is melting. [4]

In winter, in some countries, the water in a swimming pool needs to be heated. a) Estimate the cost o heating the water in a typical swimming pool rom 5 C to a suitable temperature or swimming. You may choose to consider any reasonable size o pool.

Cost per kW h o electrical energy (i)

Estimated values

(ii) Calculations

32

$0.1 0 [4] [7]

I B Q u E S T I o n S  T H E R M A L P H YS I C S

4 w aV e s Oo DefinitiOns

simple HarmOnic mOtiOn (sHm)

Many systems involve vibrations or oscillations; an object continually moves to-and-ro about a fxed average point (the mean position) retracing the same path through space taking a fxed time between repeats. Oscillations involve the interchange o energy between kinetic and potential.

Simple harmonic motion is defned as the motion that takes place when the acceleration, a, o an object is always directed towards, and is proportional to, its displacement rom a fxed point. This acceleration is caused by a restoring orce that must always be pointed towards the mean position and also proportional to the displacement rom the mean position.

Mass moving between two horizontal springs Mass moving on a vertical spring

Simple pendulum

Buoy bouncing up and down in water An oscillating ruler as a result o one end being displaced while the other is fxed

Kinetic energy Moving mass Moving mass

Moving pendulum bob Moving buoy Moving sections o the ruler

Potential energy store Elastic potential energy in the springs Elastic potential energy in the springs and gravitational potential energy Gravitational potential energy o bob Gravitational PE o buoy and water Elastic PE o the bent ruler

a  -x or a = - (constant)  x The negative sign signifes that the acceleration is always pointing back towards the mean position.

acceleration a / m s -2

A

displacement x / m

-A

Points to note about SHM:  The time period T does not depend on the amplitude A. It is isochronous.  Not all oscillations are SHM, but there are many everyday examples o natural SHM oscillations.

Defnition Displacement, The instantaneous distance (SI x measurement: m) o the moving object rom its mean position (in a specifed direction) Amplitude, A The maximum displacement (SI measurement: m) rom the mean position Frequency, f The number o oscillations completed per unit time. The SI measurement is the number o cycles per second or Hertz (Hz). Period, T The time taken (SI measurement: s) 1 or one complete oscillation. T = __ f Phase dierence, 

F  -x or F = - (constant)  x Since F = ma

This is a measure o how in step dierent particles are. I moving together they are in phase.  is measured in either degrees () or radians (rad) . 360 or 2 rad is one complete cycle so 1 80 or  rad is completely out o phase by hal a cycle. A phase dierence o 90 or /2 rad is a quarter o a cycle.

object oscillates between extremes

example Of sHm: mass between twO springs sim ple ha rm onic m otion displacem en t velocity a ccelera tion aga inst ti m e aga inst ti m e aga inst ti m e

la rge d isplacem en t to right

m a xim um displa cem en t zero velocity m a xim um acceleration

right zero velocity mass m

la rge force to left

left

sm a ll d isplacem en t to right right small velocity to left mass m

sm a ll force to left

left

zero displa cem en t m a xim um velocity zero acceleration

right large velocity to left mass m

zero net force

left

sm a ll d isplacem en t to left right small velocity to left mass m

left

sm a ll force to right

la rge d isplacem en t to left m a xim um displa cem en t zero velocity m a xim um acceleration

right zero velocity mass m

displacement, x

la rge force to right

left

amplitude, A mean position

W AV E S

33

gh of  ho oo acceleratiOn, VelOcity anD Displacement During sHm

 acceleration leads velocity by 90  velocity leads displacement by 90  acceleration and displacement are 1 80 out of phase

displacement

 displacement lags velocity by 90

velocity T 4

3T 4

T 2

 velocity lags acceleration by 90

time

T

acceleration

energy cHanges During simple HarmOnic mOtiOn During SHM, energy is interchanged between KE and PE. Providing there are no resistive forces which dissipate this energy, the total energy must remain constant. The oscillation is said to be undamped. Energy in SHM is proportional to:  the mass m  the (amplitude) 2  the (frequency) 2

E tot p

Graph showing the variation with distance, x of the energy during SHM

k x0

- x0

x

 tot k

Graph showing the variation with time, t of the energy during SHM

p t T 4

34

W AV E S

T 2

3T 4

T

tv v intrODuctiOn  rays anD waVe frOnts

transVerse waVes

Light, sound and ripples on the surace o a pond are all examples o wave motion.

Suppose a stone is thrown into a pond. Waves spread out as shown below.

 They all transer energy rom one place to another.

situation

 They do so without a net motion o the medium through which they travel.  They all involve oscillations (vibrations) o one sort or another. The oscillations are SHM.

(1) wave front diagram

(2) ray diagram

A continuous wave involves a succession o individual oscillations. A wave pulse involves just one oscillation. Two important categories o wave are transverse and longitudinal (see below) . The table gives some examples.

direction of energy ow

The ollowing pages analyse some o the properties that are common to all waves. Example of energy transfer Water ripples (Transverse)

A foating object gains an up and down motion.

Sound waves (Longitudinal)

The sound received at an ear makes the eardrum vibrate.

Light wave (Transverse)

The back o the eye (the retina) is stimulated when light is received.

Earthquake waves (Both T and L)

Buildings collapse during an earthquake.

Waves along a stretched rope (Transverse)

A sideways pulse will travel down a rope that is held taut between two people.

Compression waves down a spring (Longitudinal)

A compression pulse will travel down a spring that is is held taut between two people.

cross-section through water wave pattern moves out from centre

wave pattern at a given instant of time

water surface moves wave pattern slightly up and down later in time centre of pond edge of pond The top o the wave is known as the crest, whereas the bottom o the wave is known as the trough. Note that there are several aspects to this wave that can be studied. These aspects are important to all waves.  The movement o the wave pattern. The wave fronts highlight the parts o the wave that are moving together.  The direction o energy transer. The rays highlight the direction o energy transer.  The oscillations o the medium. It should be noted that the rays are at right angles to the wave ronts in the above diagrams. This is always the case.

lOngituDinal waVes Sound is a longitudinal wave. This is because the oscillations are parallel to the direction o energy transer.

o

This wave is an example o a transverse wave because the oscillations are at right angles to the direction o energy transer. Transverse mechanical waves cannot be propagated through fuids (liquids or gases) .

view from above (1) wave front diagram

(2) ray diagram

A point on the wave where everything is bunched together (high pressure) is known as a compression. A point where everything is ar apart (low pressure) is known as a rarefaction.

displacement of molecules to the right

loudspeaker

distance along wave

to the left rarefaction rarefaction wave moves to right situation

loudspeaker

cross-section through wave at one instant of time direction of energy transfer motion of air molecules in same direction as energy transfer

wave pattern moves out from loudspeaker

variation of pressure

rarefaction v

compression compression

average pressure

distance along wave

Relationship between displacement and pressure graphs

W AV E S

35

wv chrcrc DefinitiOns

waVe equatiOns

There are some useul terms that need to be dened in order to analyse wave motion in more detail. The table below attempts to explain these terms and they are also shown on the graphs.

There is a very simple relationship that links wave speed, wavelength and requency. It applies to all waves.

Because the graphs seem to be identical, you need to look at the axes o the graphs careully.

The time taken or one complete oscillation is the period o the wave, T.

 The displacementtime graph on the let represents the oscillations or one point on the wave. All the other points on the wave will oscillate in a similar manner, but they will not start their oscillations at exactly the same time.

In this time, the wave pattern will have moved on by one wavelength, .

 The displacementposition graph on the right represents a snapshot o all the points along the wave at one instant o time. At a later time, the wave will have moved on but it will retain the same shape.

A time / s

Term

Symbol

Displacement

Amplitude

Period

x

A

T

Frequency

Wavelength

f



A

Defnition This measures the change that has taken place as a result o a wave passing a particular point. Zero displacement reers to the mean (or average) position. For mechanical waves the displacement is the distance (in metres) that the particle moves rom its undisturbed position. This is the maximum displacement rom the mean position. I the wave does not lose any o its energy its amplitude is constant. This is the time taken (in seconds) or one complete oscillation. It is the time taken or one complete wave to pass any given point. This is the number o oscillations that take place in one second. The unit used is the hertz (Hz). A requency o 50 Hz means that 50 cycles are completed every second. This is the shortest distance (in metres) along the wave between two points that are in phase with one another. In phase means that the two points are moving exactly in step with one another. For example, the distance rom one crest to the next crest on a water ripple or the distance rom one compression to the next one on a sound wave.

c

This is the speed (in m s - 1 ) at which the wave ronts pass a stationary observer.

Intensity

I

The intensity o a wave is the power per unit area that is received by the observer. The unit is W m- 2 . The intensity o a wave is proportional to the square o its amplitude: I  A2 .

The period and the requency o any wave are inversely related. For example, i the 1 requency o a wave is 1 00 Hz, then its period must be exactly ___ o a second. 1 00

36

W AV E S

c = f In words, velocity = requency  wavelength

example

Wave speed

In symbols, 1 T = __ f

1 =f Since __ T



position / m

-

distance  c = ________ = __ T time

A stone is thrown onto a still water surace and creates a wave. A small foating cork 1 .0 m away rom the impact point has the ollowing displacementtime graph (time is measured rom the instant the stone hits the water) :

displacement/cm

T

+

displacement / x

displacement / x

 The graphs can be used to represent longitudinal AND transverse waves because the y-axis records only the value o the displacement. It does NOT speciy the direction o this displacement. So, i this displacement were parallel to the direction o the wave energy, the wave would be a longitudinal wave. I this displacement were at right angles to the direction o the wave energy, the wave would be a transverse wave.

This means that the speed o the wave must be given by

2 1 time/s

0 1.4

1.5

1.6 1.7

1.8

-1 -2

a) the amplitude o the wave: 2 cm ......................................................... b) the speed o the wave: d = ____ 1.0 = 0.67 m s - 1 c = __ t 1.5 ......................................................... c)

the requency o the wave: 1 = ____ 1 = 3.33 Hz f = __ T 0.3 .........................................................

d) the wavelength o the wave: 0.666 = 0.2 m  = __c = ______ 3.33 f .........................................................

eo pu electrOmagnetic waVes Visible light is one part o a much larger spectrum o similar waves that are all electromagnetic.

10 21 frequency

wavelength

3  10 15 Hz

10 -7 m

radium

10 - 12 10 20 -rays

10 - 11 10 19 10 - 10

X-rays 10 18

X-ray tube

10 - 9 10 17 10 - 8 UV

the sun

10 16 10 - 7 10 15

UV

VISIBLE

10 - 6 10 14

IR

10 13

1  10 15 H z

10 12

light bulb

10 - 5 10 - 4 10 - 3

10 11

wavelength  / m

Although all electromagnetic waves are identical in their nature, they have very dierent properties. This is because o the huge range o requencies (and thus energies) involved in the electromagnetic spectrum.

10 - 13

frequency f / Hz

These oscillating felds propagate (move) as a transverse wave through space. Since no physical matter is involved in this propagation, they can travel through a vacuum. The speed o this wave can be calculated rom basic electric and magnetic constants and it is the same or all electromagnetic waves, 3.0  1 08 m s- 1 .

10 22

electric heater

10 - 2 microwaves Violet Indigo VISIBLE

Charges that are accelerating generate electromagnetic felds. I an electric charge oscillates, it will produce a varying electric and magnetic feld at right angles to one another.

possible source

 /m

f / H3

See page 1 32 (option A) or more details.

Blue Green Yellow Orange Red

10 10 10 - 1 microwave oven

10 9 1 short radio waves

10 8 10 1 10 7

standard broadcast

10 2 TV broadcast aerial

10 6 10 3 10 5

3  10 14 Hz

10 -6 m

long radio waves

10 4 10 4 10 5 10 3

radio broadcast aerial

W AV E S

37

ivgg d o od xlly 1. Direct metHODs The most direct method to measure the speed o sound is to record the time taken t or sound to cover a known distance d: speed c = __dt . In air at normal pressures and temperatures, sound travels at approximately 330 m s- 1 . Given the much larger speed o light (3  1 0 8 m s- 1 ), a possible experiment would be to use a stop watch to time the dierence between seeing an event (e.g. the ring o a starting pistol or a race or seeing two wooden planks being hit together) and hearing the same event some distance away (1 00 m or more). Echoes can be used to put the source and observer o the sound in the same place. Standing a distance d in ront o a tall wall (e.g. the side o a building that is not surrounded by other buildings) can allow the echo rom a pulse o sound (e.g. a single clap o the hands) to be heard. With practice, it is possible or an experimenter to adjust the requency o clapping to synchronize the sound o the claps with their echoes. When this is achieved, the requency o clapping f can be recorded (counting the number o claps in a given time) and the time period T between claps is just T = __1f . In this time, the sound travels to the wall and back. The speed o sound is thus c = 2df. In either o the above situations a more reliable result will be achieved i a range o distances, rather than one single value is used. A graph o distance against time will allow the speed o sound to be calculated rom the gradient o the best-t straight line (which should go through the origin) . Timing pulses o sound over smaller distances requires small time intervals to be recorded with precision. It is possible to automate the process using electronic timers and / or data loggers. This equipment would allow, or example, the speed o a sound wave along a metal rod or through water to be investigated.

2. inDirect metHODs Since c = f , the speed o sound can be calculated i we measure a sounds requency and wavelength. Frequency measurement a) A microphone and a cathode ray oscilloscope (CRO) [page 1 1 6] can display a graph o the oscillations o a sound wave. Appropriate measurements rom the graph allow the time period and hence the requency to be calculated. b) Stroboscopic techniques (e.g. fashing light o known requency) can be used to measure the requency o the vibrating object (e.g. a tuning ork) that is the source o the sound. c)

Frequency o sound can be controlled at source using a known requency source (e.g. a standard tuning ork) or a calibrated electronic requency generator.

d) Comparisons can also be made between the unknown requency and a known requency. Wavelength measurement a) The intererence o waves (see page 40) can be employed to nd the path dierence between consecutive positions o destructive intererence. The path dierence between these two situations will be .

S path A

1 2

*



D

source of frequency f path B

detector (microphone and cathode ray oscilloscope (CRO) )

b) Standing waves (see page 48) in a gas can be employed to nd the location o adjacent nodes. The positions in an enclosed tube can be revealed either:  in the period pattern made by dust in the tube  electronically using a small movable microphone. c)

A resonance tube (see page 49) allows the column length or dierent maxima to be recorded. The length distance between  adjacent maxima will be __ . 2

3. factOrs tHat affect tHe speeD Of sOunD Factors include:  Nature o material  Density  Temperature (or an ideal gas, c   T)  Humidity (or air) .

38

W AV E S

i intensity The sound intensity, I, is the amount of energy that a sound wave brings to a unit area every second. The units of sound intensity are W m- 2 . It depends on the amplitude of the sound. A more intense sound (one that is louder) must have a larger amplitude. Intensity  (amplitude) 2

displacement of a particle when a sound wave passes

This relationship between intensity and amplitude is true for all waves.

louder sound of same pitch

time amplitude, A

I  A2

inVerse square law Of raDiatiOn

The surface area A of a sphere of radius r is calculated using:

As the distance of an observer from a point source of light increases, the power received by the observer will decrease as the energy spreads out over a larger area. A doubling of distance will result in the reduction of the power received to a quarter of the original value.

area 4A

A = 4r2 If the point source radiates a total power P in all directions, then the power received per unit area (the intensity I) at a distance r away from the point source is: P I = _____ 4r2 For a given area of receiver, the intensity of the received radiation is inversely proportional to the square of the distance from the point source to the receiver. This is known as the inverse square law and applies to all waves.

area A

I  x- 2

d

d

waVefrOnts anD rays As introduced on page 35, waves can be described in terms of the motion of a wavefront and/or in terms of rays.

rays spreading out

A ray is the path taken by the wave energy as it travels out from the source. A wavefront is a surface joining neighbouring points where the oscillations are in phase with one another. In two dimensions, the wavefront is a line and in one dimension, the wavefront is a point.

wavefront

point source of wave energy

W AV E S

39

soo interference Of waVes When two waves o the same type meet, they interfere and we can work out the resulting wave using the principle o superposition. The overall disturbance at any point and at any time where the waves meet is the vector sum o the disturbances that would have been produced by each o the individual waves. This is shown below.

I the waves have the same amplitude and the same requency then the intererence at a particular point can be constructive or destructive.

graphs wave 1 displacement (at P) A

y 1 / unit

(a) wave 1

time wave 2 displacement (at P)

0

t/s

A time

(b) wave 2 y 2 / unit

time

time

resultant displacement (at P) 0

t/s

zero result

2A (c) wave 1 + wave 2 = wave 3 y / unit

time 0

t/s

constructive

time

destructive

Wave superposition

tecHnical language

examples Of interference

Constructive intererence takes place when the two waves are in step with one another  they are said to be in phase. There is a zero phase difference between them. Destructive intererence takes place when the waves are exactly out o step  they are said to be out of phase. There are several dierent ways o saying this. One could say that the phase dierence is equal to hal a cycle or 1 80 degrees or  radians.

Water waves A ripple tank can be used to view the intererence o water waves. Regions o large-amplitude waves are constructive intererence. Regions o still water are destructive intererence.

Intererence can take place i there are two possible routes or a ray to travel rom source to observer. I the path dierence between the two rays is a whole number o wavelengths, then constructive intererence will take place. path dierence = n   constructive 1 )   destructive path dierence = (n + __ 2 n = 0, 1 , 2, 3 . . . For constructive or destructive intererence to take place, the sources o the waves must be phase linked or coherent.

Sound It is possible to analyse any noise in terms o the component requencies that make it up. A computer can then generate exactly the same requencies but o dierent phase. This antisound will interere with the original sound. An observer in a particular position in space could have the overall noise level reduced i the waves superimposed destructively at that position. Light The colours seen on the surace o a soap bubble are a result o constructive and destructive intererence o two light rays. One ray is refected o the outer surace o the bubble whereas the other is refected o the inner surace.

superpOsitiOn Of waVe pulses Whenever wave pulses meet, the principle o superposition applies: At any instant in time, the net displacement that results rom dierent waves meeting at the same point in space is just the vector sum o the displacements that would have been produced by each individual wave. yo v e ra ll = y1 + y2 + y3 etc.

a) i)

b) i) A

A pulse P

A

pulse Q pulse P

pulse Q

A

ii) P + Q = 2A

P+ Q = A- A= 0

ii)

iii)

iii) A

A pulse Q

pulse P

A A pulse Q

40

W AV E S

pulse P

poo pOlarizeD ligHt

brewsters law

Light is part o the electromagnetic spectrum. It is made up o oscillating electric and magnetic elds that are at right angles to one another (or more details see page 1 32) . They are transverse waves; both elds are at right angles to the direction o propagation. The plane of vibration o electromagnetic waves is dened to be the plane that contains the electric eld and the direction o propagation.

A ray o light incident on the boundary between two media will, in general, be refected and reracted. The refected ray is always partially plane-polarized. I the refected ray and the reracted ray are at right angles to one another, then the refected ray is totally plane-polarized. The angle o incidence or this condition is known as the polarizing angle.

direction of magnetic eld oscillation

reected ray is totally plane-polarized

plane vibration of direction of electric EM wave containing electric oscillations eld oscillation

incident ray is unpolarized

transmitted ray is partially polarized represents electric eld oscillation into the paper represents electric eld oscillation in the plane of the paper

There are an innite number o ways or the elds to be oriented. Light (or any EM wave) is said to be unpolarized i the plane o vibration varies randomly whereas plane-polarized light has a xed plane o vibration. The diagrams below represent the electric elds o light when being viewed head on.

unpolarized light: over a period of time, the electric eld oscillates in random directions

A mixture o polarized light and unpolarized light is partially plane-polarized. I the plane o polarization rotates uniormly the light is said to be circularly polarized. Most light sources emit unpolarized light whereas radio waves, radar and laboratory microwaves are oten plane-polarized as a result o the processes that produce the waves. Light can be polarized as a result o refection or selective absorption. In addition, some crystals exhibit double refraction or birefringence where an unpolarized ray that enters a crystal is split into two plane-polarized beams that have mutually perpendicular planes o polarization.

medium 1 ( vacuum) medium 2 ( water) r

distance along the wave

polarized light: over a period of time, the electric eld only oscillates in one direction

i

i + r = 90 Brewsters law relates the reractive index o medium 2, n, to the incident angle i: sin  sin  n = _____i = _____i = tan i sin r cos i

A polarizer is any device that produces plane-polarized light rom an unpolarized beam. An analyser is a polarizer used to detect polarized light. Polaroid is a material which preerentially absorbs any light in one particular plane o polarization allowing transmission only in the plane at 90 to this.

a)

b)

transparent indicates the preferred directions

zero transmission

maluss law

Optically actiVe substances

When plane-polarized light is incident on an analyser, its preerred direction will allow a component o the light to be transmitted:

An optically active substance is one that rotates the plane o polarization o light that passes through it. Many solutions (e.g. sugar solutions o dierent concentrations) are optically active.

0 

analysers preferred direction



0 cos  transmitted component of electric eld after analyser  = 0 cos 

plane-polarized light seen head-on with electric eld amplitude, 0



analyser

plane of vibration rotated through angle 

 original plane of vibration optically active substance

The intensity o light is proportional to the (amplitude) 2 .

I is transmitted intensity o light in W m- 2

Transmitted intensity I  E

I0 is incident intensity o light in W m- 2

2

 I  E0 2 cos 2  as expressed by Maluss law: I = I0 cos 2 

 is the angle between the plane o vibration and the analysers preerred direction

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41

u o oo pOlarOiD sunglasses

cOncentratiOn Of sOlutiOns

Polaroid is a material containing long chain molecules. The molecules selectively absorb light that have electric elds aligned with the molecules in the same way that a grid o wires will selectively absorb microwaves.

For a given optically active solution, the angle  through which the plane o polarization is rotated is proportional to:  The length of the solution through which the planepolarized light passes.  The concentration of the solution.

grid viewed head on

electric eld

absorption

A polarimeter is a device that measures  or a given solution. It consists o two polarizers (a polarizer and an analyser) that are initially aligned. The optically active solution is introduced between the two and the analyser is rotated to nd the maximum transmitted light.

stress analysis grid viewed head on

electric eld

transmission

Glass and some plastics become bireringent (see page 41 ) when placed under stress. When polarized white light is passed through stressed plastics and then analysed, bright coloured lines are observed in the regions o maximum stress.

When worn normally by a person standing up, Polaroid dark glasses allow light with vertically oscillating electric elds to be transmitted and absorb light with horizontally oscillating electric elds.  The absorption will mean that the overall light intensity is reduced.  Light that has reected from horizontal surfaces will be horizontally plane-polarized to some extent.  Polaroid sunglasses will preferentially absorb reected light, reducing glare rom horizontal suraces.

liquiD-crystal Displays (lcDs) LCDs are used in a wide variety o dierent applications that include calculator displays and computer monitors. The liquid crystal is sandwiched between two glass electrodes and is bireringent. One possible arrangement with crossed polarizers surrounding the liquid crystal is shown below:

reector

polarizers

 With no liquid crystal between the electrodes, the second polarizer would absorb all the light that passed through the rst polarizer. The screen would appear black.  The liquid crystal has a twisted structure and, in the absence o a potential dierence, causes the plane o polarization to rotate through 90.  This means that light can pass through the second polarizer, reach the refecting surace and be transmitted back along its original direction.  With no pd between the electrodes, the LCD appears light.

liquid crystal

electrodes etched into glass light enters the LCD from the front

furtHer pOlarizatiOn examples Only transverse waves can be polarized. Page 41 has concentrated on the polarization o light but all EM waves that are transverse are able, in principle, to be polarized.  Sound waves, being longitudinal waves, cannot be polarized.  The nature of radio and TV broadcasts means that the signal is oten polarized and aerials need to be properly aligned i they are to receive the maximum possible signal strength.

42

W AV E S

 A pd across the liquid crystal causes the molecules to align with the electric eld. This means less light will be transmitted and this section o the LCD will appear darker.  The extent to which the screen appears grey or black can be controlled by the pd  Coloured lters can be used to create a colour image.  A picture can be built up from individual picture elements.

 Microwave radiation (with a typical wavelength of a few cm) can be used to demonstrate wave characteristics in the laboratory. Polarization can be demonstrated using a grid o conducting wires. I the grid wires are aligned parallel to the plane o vibration o the electric eld the microwaves will be absorbed. Rotation o the grid through 90 will allow the microwaves to be transmitted.

wv bhvou  fo relectiOn anD transmissiOn

relectiOn O twO-DimensiOnal plane waVes

In general, when any wave meets the boundary between two dierent media it is partially refected and partially transmitted.

The diagram below shows what happens when plane waves are refected at a boundary. When working with rays, by convention we always measure the angles between the rays and the normal. The normal is a construction line that is drawn at right angles to the surace.

incident ray

normal

incident ray

reected ray medium (1) medium (2)

incident normal reected angle i angle r

reected ray

transmitted ray surface

medium (2) is optically denser than medium (1)

Law of reection: i = r

law O relectiOn The location and nature o optical images can be worked out using ray diagrams and the principles o geometric optics. A ray is a line showing the direction in which light energy is propagated. The ray must always be at right angles to the waveront. The study o geometric optics ignores the wave and particle nature o light.

light leaves in all directions

diuse reection

light leaves in one direction

mirror reection

types O relectiOn When a single ray o light strikes a smooth mirror it produces a single refected ray. This type o perect refection is very dierent to the refection that takes place rom an uneven surace such as the walls o a room. In this situation, a single incident ray is generally scattered in all directions. This is an example o a diffuse refection.

rays spreading out

wavefront

point source of light We see objects by receiving light that has come rom them. Most objects do not give out light by themselves so we cannot see them in the dark. Objects become visible with a source o light (e.g. the Sun or a light bulb) because diuse refections have taken place that scatter light rom the source towards our eyes.

The surfaces of the picture scatter the light in all directions.

Light from the central bulb sets o in all directions.

When a mirror refection takes place, the direction o the refected ray can be predicted using the laws o refection. In order to speciy the ray directions involved, it is usual to measure all angles with respect to an imaginary construction line called the normal. For example, the incident angle is always taken as the angle between the incident ray and the normal. The normal to a surace is the line at right angles to the surace as shown below.

normal incident ray i

r reected ray

The laws o refection are that:  t he incident angle is equal to the refected angle

An observer sees the painting by receiving this scattered light. Our brains are able to work out the location o the object by assuming that rays travel in straight lines.

 t he incident ray, the refected ray and the normal all lie in the same plane (as shown in the diagram) . The second statement is only included in order to be precise and is oten omitted. It should be obvious that a ray arriving at a mirror (such as the one represented above) is not suddenly refected in an odd direction (e.g. out o the plane o the page) .

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43

s  d v d refractiVe inDex anD snells law

examples

Reraction takes place at the boundary between two media. In general, a wave that crosses the boundary will undergo a change o direction. The reason or this change in direction is the change in wave speed that has taken place.

1. Parallel-sided block A ray will always leave a parallel-sided block travelling in a parallel direction to the one with which it entered the block. The overall eect o the block has been to move the ray sideways. An example o this is shown below.

As with refection, the ray directions are always specied by considering the angles between the ray and the normal. I a ray travels into an optically denser medium (e.g. rom air into water) , then the ray o light is reracted towards the normal. I the ray travels into an optically less dense medium then the ray o light is reracted away from the normal.

incident ray

glass

normal

ray leaves block parallel to incident ray refracted ray

incident ray

less dense medium more dense medium

2. Ray travelling between two media I a ray goes between two dierent media, the two individual reractive indices can be used to calculate the overall reraction using the ollowing equation n sin  _____ n 1 sin 1 = n 2 sin 2 or __ n = sin  1

2

2

1

n1 reractive index o medium 1 1 angle in medium 1

ray refracted towards normal normal refracted ray more dense medium

incident ray

n2 reractive index o medium 2 2 angle in medium 2 Suppose a ray o light is shone into a sh tank that contains water. The reraction that takes place would be calculated as shown below: 1 st reraction: air water sin a = _____ (n air = 1.0) (n water = 1.3) n glass glass sin b (n glass = 1.6) 2nd reraction:

c

less dense medium a

nglass  sin b = nw ate r  sin c

b

nglass sin c _____ _____ nw ate r = sin b

ray refracted away from the normal Snells law allows us to work out the angles involved. When a ray is reracted between two dierent media, the ratio

sin(angle o incidence) ________________ sin(angle o reraction)

is a constant.

The constant is called the reractive index n between the two media. This ratio is equal to the ratio o the speeds o the waves in the two media. sin i _____ =n sin r I the reractive index or a particular substance is given as a particular number and the other medium is not mentioned then you can assume that the other medium is air (or to be absolutely correct, a vacuum) . Another way o expressing this is to say that the reractive index o air can be taken to be 1 .0. For example the reractive index or a type o glass might be given as ng la s s = 1 .34 This means that a ray entering the glass rom air with an incident angle o 40 would have a reracted angle given by

Overall the reraction is rom incident angle a to reracted angle c. sin a = _____ sin a  _____ sin b i.e. no ve rall = _____ sin c sin c sin b = nw a te r

refractiOn Of plane waVes The reason or the change in direction in reraction is the change in speed o the wave.

normal medium 1 ( e.g. air) i medium 2 (e.g. glass) wavelength smaller since speed reduced

boundary r refracted ray

Snells law (an experimental law o reraction) states that sin 40 sin r = _______ = 0.4797 1 .34  r = 28.7 ngla s s

44

n g la s s sin  a ir Va ir _______ = ____ = _____ n a ir = sin  Vg la s s g la s s

W AV E S

sin i the ratio ____ = constant, or a given requency. sin r

The ratio is equal to the ratio o the speeds in the dierent media n1 sin 2 V2  speed o wave in medium 2 ___ ______ ___ n 2 = sin  = V1  speed o wave in medium 1 1

ro d   tOtal internal reflectiOn anD critical angle

examples

In general, both refection and reraction can happen at the boundary between two media.

1.

What a sh sees under water

entire world outside water is visible in an angle of twice the critical angle

It is, under certain circumstances, possible to guarantee complete (total) refection with no transmission at all. This can happen when a ray meets the boundary and it is travelling in the denser medium.

n1< n2

partial transmission 1

grazing emergence

n1 n2 c

2

critical angle c At greater than the critical angle, the surface of the water acts like a mirror. Objects inside the water are seen by reection.

total reection

1 2 3

2.

O source

partial reection

Ray1

This ray is partially refected and partially reracted.

Ray2

This ray has a reracted angle o nearly 90. The critical ray is the name given to the ray that has a reracted angle o 90. The critical angle is the angle o incidence c or the critical ray.

Ray3

Prismatic refectors

A prism can be used in place o a mirror. I the light strikes the surace o the prism at greater than the critical angle, it must be totally internally refected. Prisms are used in many optical devices. Examples include:  periscopes  the double refection allows the user to see over a crowd.  binoculars  the double refection means that the binoculars do not have to be too long

This ray has an angle o incidence greater than the critical angle. Reraction cannot occur so the ray must be totally refected at the boundary and stay inside medium 2. The ray is said to be totally internally refected.

 SLR cameras  the view through the lens is refected up to the eyepiece.

The critical angle can be worked out as ollows. For the critical ray,

periscope

n1 sin 1 = n 2 sin 2 1 = 90 2 =  c 1  sin c = ___ n2

binoculars The prism arrangement delivers the image to the eyepiece the right way up. By sending the light along the instrument three times, it also allows the binoculars to be shorter. eyepiece lens

metHODs fOr Determining refractiVe inDex experimentally Part of ray identied in several positions objective lens glass block Part of ray in block can be inferred from measurements

1.

Locate paths taken by dierent rays either by sending a ray through a solid and measuring its position or aligning objects by eye. Uncertainties in angle measurement are dependent on protractor measurements. (See diagrams on let)

2.

Use a travelling microscope to measure real and apparent depth and apply ollowing ormula:

Part of ray identied

ray heading towards centre will not be refracted entering the block semi-circular glass block

real depth o object n = _______________________ apparent depth o object 3.

Very accurate measurements o angles o reraction can be achieved using a prism o the substance and a custom piece o equipment call a spectrometer.

centre

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45

Dfo DiractiOn

There are some important points to note rom these diagrams.

When waves pass through apertures they tend to spread out. Waves also spread around obstacles. This wave property is called diffraction.

 D  iraction becomes relatively more important when the wavelength is large in comparison to the size o the aperture (or the object) .

geometric shadow region

d> 

 T  he wavelength needs to be o the same order o magnitude as the aperture or diraction to be noticeable.

practical signiicance O DiractiOn d  geometric shadow region

d 

d d>  geometric shadow region

d

diraction more important with smaller obstacles  geometric shadow region

d

 The electron microscope  resolves items that cannot be resolved using a light microscope. The electrons have an eective wavelength that is much smaller than the wavelength o visible light (see page 1 27) .

geometric shadow region

 Radio telescopes  the size of the dish limits the maximum resolution possible. Several radio telescopes can be linked together in an array to create a virtual radio telescope with a greater diameter and with a greater ability to resolve astronomical objects. (See page 1 81 )

 d <  geometric d shadow region

Diraction eects mean that it is impossible ever to see atoms because they are smaller than the wavelength o visible light, meaning that light will diract around the atoms. It is, however, possible to image atoms using smaller wavelengths. Practical devices where diraction needs to be considered include:  CDs and DVDs  the maximum amount of information that can be stored depends on the size and the method used or recording inormation.

 d   geometric shadow region

Whenever an observer receives inormation rom a source o electromagnetic waves, diraction causes the energy to spread out. This spreading takes place as a result o any obstacle in the way and the width o the device receiving the electromagnetic radiation. Two sources o electromagnetic waves that are angularly close to one another will both spread out and interere with one another. This can aect whether or not they can be resolved (see page 1 01 ) .

geometric shadow region

examples O DiractiOn Diraction provides the reason why we can hear something even i we can not see it.

 d = width o obstacle/gap Diraction - wave energy is received in geometric shadow region.

basic ObserVatiOns Diraction is a wave eect. The objects involved (slits, apertures, etc.) have a size that is o the same order o magnitude as the wavelength.

intensity There is a central maximum intensity. Other maxima occur roughly halfway between the minima. As the angle increases, the intensity of the maxima decreases. 1st minimum Diraction of a single slit.

46

W AV E S

angle

I you look at a distant street light at night and then squint your eyes the light spreads sideways  this is as a result o diraction taking place around your eyelashes! (Needless to say, this explanation is a simplifcation.)

to-o  o v principles Of tHe twO-sOurce interference pattern Two-source interference is simply another application of the principle of superposition, for two coherent sources having roughly the same amplitude. Two sources are coherent if:  they have the same frequency

matHematics The location of the light and dark fringes can be mathematically derived in one of two ways. The derivations do not need to be recalled. Method 1 The simplest way is to consider two parallel rays setting off from the slits as shown below.

 there is a constant phase relationship between the two sources.

parallel rays

regions where waves are in phase: constructive interference destructive interference

S1 d

 S2

S2

S1

Two dippers in water moving together are coherent sources. This forms regions of water ripples and other regions with no waves. Two loudspeakers both connected to the same signal generator are coherent sources. This forms regions of loud and soft sound. A set-up for viewing two-source interference with light is shown below. It is known as Youngs double slit experiment. A monochromatic source of light is one that gives out only one frequency. Light from the twin slits (the sources) interferes and patterns of light and dark regions, called fringes, can be seen on the screen.

region in which superposition occurs

Set-up 1

separation monochromatic of slits light source

S0

S1 S2

source twin source slit slits (less than 5 mm) 0.1 m 1m

possible screen positions

Set-up 2 The use of a laser makes the set-up easier.

laser

double slit

screen

The experiment results in a regular pattern of light and dark strips across the screen as represented below.

intensity distribution intensity

view seen fringe width, d



path dierence p = dsin 

If these two rays result in a bright patch, then the two rays must arrive in phase. The two rays of light started out in phase but the light from source 2 travels an extra distance. This extra distance is called the path difference. Constructive interference can only happen if the path difference is a whole number of wavelengths. Mathematically, Path difference = n  [where n is an integer  e.g. 1 , 2, 3 etc.] From the geometry of the situation Path difference = d sin  In other words n  = d sin  Method 2 If a screen is used to make the fringes visible, then the rays from the two slits cannot be absolutely parallel, but the physical set-up means that this is effectively true. p sin  = __ s X __ tan  = P D If  is small sin   tan  p S1 X __ __ X so s = D  Xs   p = ___ D s For constructive N 2 interference: p p = n S2 Xn s D  n = ____ D nD  Xn = _____ s D fringe separation d = Xn + 1 - Xn = ___ s D ___  s= d This equation only applies when the angle is small. Example Laser light of wavelength 450 nm is shone on two slits that are 0.1 mm apart. How far apart are the fringes on a screen placed 5.0 m away? D 4.5  1 0 - 7  5 = 0.0225 m = 2.25 cm ______________ d = _____ s = 1 .0  1 0 - 4

dark bright dark bright

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47

nur d produco of d (ory) v stanDing waVes A special case o intererence occurs when two waves meet that are:  o f the same amplitude  o f the same frequency

There are some points on the rope that are always at rest. These are called the nodes. The points where the maximum movement takes place are called antinodes. The resulting standing wave is so called because the wave pattern remains xed in space  it is its amplitude that changes over time. A comparison with a normal (travelling) wave is given below.

 travelling in opposite directions. Stationary wave

Normal (travelling) wave

Amplitude

All points on the wave have dierent amplitudes. The maximum amplitude is 2A at the antinodes. It is zero at the nodes.

All points on the wave have the same amplitude.

Frequency

All points oscillate with the same requency.

All points oscillate with the same requency.

Wavelength

This is twice the distance rom one node (or antinode) to the next node (or antinode) .

This is the shortest distance (in metres) along the wave between two points that are in phase with one another.

Phase

All points between one node and the next node are moving in phase.

All points along a wavelength have dierent phases.

Energy

Energy is not transmitted by the wave, but it does have an energy associated with it.

Energy is transmitted by the wave.

In these conditions a standing wave will be ormed. The conditions needed to orm standing waves seem quite specialized, but standing waves are in act quite common. They oten occur when a wave refects back rom a boundary along the route that it came. Since the refected wave and the incident wave are o (nearly) equal amplitude, these two waves can interere and produce a standing wave. Perhaps the simplest way o picturing a standing wave would be to consider two transverse waves travelling in opposite directions along a stretched rope. The series o diagrams below shows what happens.

resultant wave wave 1 moves  wave 2 moves  displacement

a)

total distance

wave 1 wave 2

displacement

b)

total

wave 1

distance

wave 2

displacement

c) wave 2

total distance

wave 1

d) displacement

total wave 2 distance

wave 1

displacement

e) wave 2

Although the example let involved transverse waves on a rope, a standing wave can also be created using sound or light waves. All musical instruments involve the creation o a standing sound wave inside the instrument. The production o laser light involves a standing light wave. Even electrons in hydrogen atoms can be explained in terms o standing waves. A standing longitudinal wave can be particularly hard to imagine. The diagram below attempts to represent one example  a standing sound wave.

total

wave 1

antinode

zero movement

max. movement

node

distance

antinode

Production o standing waves

antinode

total displacement

antinode

etc. distance

A longitudinal standing wave

node

node

node

A standing wave  the pattern remains xed

48

node

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node

etc.

bod odo bOunDary cOnDitiOns The boundary conditions o the system speciy the conditions that must be met at the edges (the boundaries) o the system when standing waves are taking place. Any standing wave that meets these boundary conditions will be a possible resonant mode o the system.

As beore, the boundary conditions determine the standing waves that can exist in the tube. A closed end must be a displacement node. An open end must be an antinode. Possible standing waves are shown or a pipe open at both ends and a pipe closed at one end.

A

N

A

N

A ' = l

N = node A = antinode  0 = 2l, 1 st harmonic = f0

l

N = node A = a n tin o d e 1 st ha rm on ic freq u en cy = f0 0 = 2l

l

1. Transverse waves on a string I the string is xed at each end, the ends o the string cannot oscillate. Both ends o the string would refect a travelling wave and thus a standing wave is possible. The only standing waves that t these boundary conditions are ones that have nodes at each end. The diagrams below show the possible resonant modes.

f' = 2 f0

A

N

A

N

A " =

N

2l 3

f" = 3 f0

A

 ' = l, f' = 2 f0

N

A

N

A

N

A

Harmonic modes or a pipe open at both ends

N

A

N

A

N

N = node A = a n tin o d e

l

1 st ha rm on ic freq u en cy = f0  0 = 4l

2

 " = 3 l, f" = 3 f0

N

A

N

A

N

A

N

N

A ' =

l

f' = 3 f0

 "' = 2 , f"' = 4 f0

N

A

N

A

N

A

N

A

N

4l 3

N

A

N

A " =

Harmonic modes or a string The resonant mode that has the lowest requency is called the undamental or the frst harmonic. Higher resonant modes are called harmonics. Many musical instruments (e.g. piano, violin, guitar etc.) involve similar oscillations o metal strings. 2. Longitudinal sound waves in a pipe A longitudinal standing wave can be set up in the column o air enclosed in a pipe. As in the example above, this results rom the refections that take place at both ends.

example

4l 3

f" = 5 f0

N

A

N

A

N

A

Harmonic modes or a pipe closed at one end Musical instruments that involve a standing wave in a column o air include the fute, the trumpet, the recorder and organ pipes.

resOnance tube

An organ pipe (open at one end) is 1 .2 m long. Calculate its undamental requency. The speed o sound is 330 m s - 1 .  l = 1 .2 m  __ = 1 .2 m (rst harmonic) 4   = 4.8 m

Tuning fork of known frequency

A

N

x

v = f 330 f = ____  69 Hz 4.8

Resonance will occur at dierent values of x . The distance between adjacent resonance lengths = 2

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49

ib questons  waves 1.

2.

b) On the diagram above

A surfer is out beyond the breaking surf in a deep-water region where the ocean waves are sinusoidal in shape. The crests are 20 m apart and the surfer rises a vertical distance of 4.0 m from wave trough to crest, in a time of 2.0 s. What is the speed of the waves? A. 1 .0 m s - 1

B. 2.0 m s - 1

C. 5.0 m s - 1

D. 1 0.0 m s - 1

(i)

A standing wave is established in air in a pipe with one closed and one open end.

draw an arrow to indicate the direction in which the marker is moving.

[1 ]

(ii) indicate, with the letter A, the amplitude of the wave.

[1 ]

(iii) indicate, with the letter , the wavelength of the wave.

[1 ]

(iv) draw the displacement of the string a time __T4 later, where T is the period of oscillation of the wave. Indicate, with the letter N, the new position of the marker. [2]

X

The wavelength of the wave is 5.0 cm and its speed is 1 0 cm s - 1 . c) Determine

3.

The air molecules near X are

(i)

A. always at the centre of a compression.

(ii) how far the wave has moved in __T4 s.

the frequency of the wave.

[1 ] [2]

B. always at the centre of a rarefaction.

Interference of waves

C. sometimes at the centre of a compression and sometimes at the centre of a rarefaction.

d) By reference to the principle of superposition, explain what is meant by constructive interference. [4]

D. never at the centre of a compression or a rarefaction.

The diagram below (not drawn to scale) shows an arrangement for observing the interference pattern produced by the light from two narrow slits S 1 and S 2 .

This question is about sound waves. A sound wave of frequency 660 Hz passes through air. The variation of particle displacement withdistance along the wave at one instant of time is shown below.

P

0.5 displacement /mm 0

S1 0

1 .0

2 .0

monochromatic light source

distance / m

yn 



dM S2

O

x

 0.5 double slit

b) Using data from the above graph, deduce for this sound wave, (i)

4.

the wavelength.

[1 ]

(iii) the speed.

[2]

e) (i)

The diagram below represents the direction of oscillation of a disturbance that gives rise to a wave.

a transverse wave and

(iii) Deduce an expression for  in terms of D and yn .

[1 ]

A wave travels along a stretched string. The diagram below shows the variation with distance along the string of the displacement of the string at a particular instant in time. A small marker is attached to the string at the point labelled M. The undisturbed position of the string is shown as a dotted line.

direction of wave travel M

50

I B Q u E S t I o n S  W AV E S

[1 ]

For a particular arrangement, the separation of the slits is 1 .40 mm and the distance from the slits to the screen is 1 .50 m. The distance yn is the distance of the eighth bright fringe from O and the angle  = 2.70  1 0 - 3 rad.

[1 ]

(ii) a longitudinal wave.

State the condition in terms of the distance S 2 X and the wavelength of the light , for there to be a bright fringe at P. [2]

(ii) Deduce an expression for  in terms of S 2 X and d. [2]

a) By redrawing the diagram, add arrows to show the direction of wave energy transfer to illustrate the difference between (i)

screen

The distance S 1 S 2 is d, the distance between the double slit and screen is D and D  d such that the angles  and  shown on the diagram are small. M is the mid-point of S 1 S 2 and it is observed that there is a bright fringe at point P on the screen, a distance yn from point O on the screen. Light from S 2 travels a distance S 2 X further to point P than light from S 1 .

[1 ]

(ii) the amplitude.

D

single slit

a) State whether this wave is an example of a longitudinal or a transverse wave. [1 ]

f) Using your answers to (e) to determine

5.

(i) the wavelength of the light.

[2]

(ii) the separation of the fringes on the screen.

[3]

A bright source of light is viewed through two polarisers whose preferred directions are initially parallel. Calculate the angle through which one sheet should be turned to reduce the transmitted intensity to half its original value.

5 E lE Ctr i Ci ty an d m ag n E ti s m Eecc ce  C'  ConsErvation of ChargE

Coulombs law

Two types o charge exist  positive and negative. Equal amounts o positive and negative charge cancel each other. Matter that contains no charge, or matter that contains equal amounts o positive and negative charge, is said to be electrically neutral.

The diagram shows the orce between two point charges that are ar away rom the infuence o any other charges.

Charges are known to exist because o the orces that exist between all charges, called the electrostatic force: like charges repel, unlike charges attract.

F force

F

The directions o the orces are along the line joining the charges. I they are like charges, the orces are away rom each other  they repel. I they are unlike charges, the orces are towards each other  they attract.

+ + -

F

F F

F F

+

F

+

F

A very important experimental observation is that charge is always conserved. Charged objects can be created by riction. In this process electrons are physically moved rom one object to another. In order or the charge to remain on the object, it normally needs to be an insulator.

before neutral comb

neutral hair

distance r q2 charge

q1 charge

F force

Each charge must eel a orce o the same size as the orce on the other one. Experimentally, the orce is proportional to the size o both charges and inversely proportional to the square o the distance between the charges. kq1 q 2 q1 q2 F= _ = k_ r2 r2 This is known as Coulombs law and the constant k is called the Coulomb constant. In act, the law is oten quoted in a slightly dierent orm using a dierent constant or the medium called the permittivity, . value of rst charge value of second charge force between two point charges

q q F= 1 2 2 4 0 r

constants

after

attraction + + positive hair + negative + + comb -- + +

distance between the charges

permittivity of free space (a constant)

1 k= _ 4 0 I there are two or more charges near another charge, the overall orce can be worked out using vector addition.

force on q A (due to q C ) electrons have been transferred from hair to comb The total charge beore any process must be equal to the total charge aterwards. It is impossible to create a positive charge without an equal negative charge. This is the law o conservation o charge.

overall force on q A (due to q B and q C)

qB

qA force on q A (due to q B ) qC

Veo o of eeo foe

ConduCtors and insulators

Electrical conductors

Electrical insulators

A material that allows the fow o charge through it is called an electrical conductor. I charge cannot fow through a material it is called an electrical insulator. In solid conductors the fow o charge is always as a result o the fow o electrons rom atom to atom.

all metals e.g. copper aluminium brass graphite

plastics e.g. polythene nylon acetate rubber dry wood glass ceramics

ElEctri ci ty an d M ag n Eti s M

51

Eecc fe ElECtriC iElds  dEinition A charge, or combination o charges, is said to produce an electric feld around it. I we place a test charge at any point in the feld, the value o the orce that it eels at any point will depend on the value o the test charge only.

A A test charge placed at A would feel this force.

In practical situations, the test charge needs to be small so that it doesnt disturb the charge or charges that are being considered. The defnition o electric feld, E, is F E=_ q 2 = orce per unit positive point test charge. Coulombs law can be used to relate the electric feld around a point charge to the charge producing the feld. q1 E=_ 4 0 r2 When using these equations you have to be very careul:

A test charge placed at B would feel this force. q1

 not to muddle up the charge producing the feld and the charge sitting in the feld (and thus eeling a orce)  not to use the mathematical equation or the feld around a point charge or other situations (e.g. parallel plates) .

B

A test charge would eel a dierent orce at dierent points around a charge q 1 .

rEprEsEntation o ElECtriC iElds This is done using feld lines. +

At any point in a feld:



two opposite charges

 the direction o feld is represented by the direction o the feld lines closest to that point  the magnitude o the feld is represented by the number o feld lines passing near that point.

The eld here must be strong as the eld lines are close together.

The direction of the force here must be as shown.

The eld here must be weak as the eld lines are far apart.

+

+

two like charges



a negatively charged conducting sphere

F

Field around a positive point charge The resultant electric feld at any position due to a collection o point charges is shown to the right. The parallel feld lines between two plates mean that the electric feld is uniorm. Electric feld lines:

(radial eld)

two oppositely charged parallel metal plates + + + + +

+ + + + +

 begin on positive charges and end on negative charges  never cross  are close together when the feld is strong.          

parallel eld lines in the centre Patterns o electric felds

52

ElEctri ci ty an d M ag n Eti s M

Eecc e ee  eecc e feece EnErgy diErEnCE in an ElECtriC iEld

ElECtriC potEntial diErEnCE

When placed in an electric feld, a charge eels a orce. This means that i it moves around in an electric feld work will be done. As a result, the charge will either gain or lose electric potential energy. Electric potential energy is the energy that a charge has as a result o its position in an electric feld. This is the same idea as a mass in a gravitational feld. I we lit a mass up, its gravitational potential energy increases. I the mass alls, its gravitational potential energy decreases. In the example below a positive charge is moved rom position A to position B. This results in an increase in electric potential energy. Since the feld is uniorm, the orce is constant. This makes it very easy to calculate the work done.

In the example on the let, the actual energy dierence between A and B depended on the charge that was moved. I we doubled the charge we would double the energy dierence. The quantity that remains fxed between A and B is the energy dierence per unit charge. This is called the potential difference, or pd, between the points.

force needed to move charge q

+

B

+

A q

distance d

Potential dierence energy dierence between two points = per unit charge moved energy dierence work done = __ = __ charge charge W V= _ q The basic unit or potential dierence is the joule/coulomb, J C - 1 . A very important point to note is that or a given electric feld, the potential dierence between any two points is a single fxed scalar quantity. The work done between these two points does not depend on the path taken by the test charge. A technical way o saying this is the electric feld is conservative.

units position of higher electric potential energy

position of lower electric potential energy

Charge moving in an electric feld Change in electric potential energy = orce  distance = Eq d See page 52 or a defnition o electric feld, E.

The smallest amount o negative charge available is the charge on an electron; the smallest amount o positive charge is the charge on a proton. In everyday situations this unit is ar too small so we use the coulomb, C. One coulomb o negative charge is the charge carried by a total o 6.25  1 0 1 8 electrons. From its defnition, the unit o potential dierence (pd) is J C - 1 . This is given a new name, the volt, V. Thus: 1 volt = 1 J C - 1

In the example above the electric potential energy at B is greater than the electric potential energy at A. We would have to put in this amount o work to push the charge rom A to B. I we let go o the charge at B it would be pushed by the electric feld. This push would accelerate it so that the loss in electrical potential energy would be the same as the gain in kinetic energy.

Voltage and potential dierence are dierent words or the same thing. Potential dierence is probably the better name to use as it reminds you that it is measuring the dierence between two points. When working at the atomic scale, the joule is ar too big to use or a unit or energy. The everyday unit used by physicists or this situation is the electronvolt. As could be guessed rom its name, the electronvolt is simply the energy that would be gained by an electron moving through a potential dierence o 1 volt. 1 electronvolt = 1 volt  1 .6  1 0 - 1 9 C

B+

A

+

velocity v

= 1 .6  1 0 - 1 9 J The normal SI prefxes also apply so one can measure energies in kiloelectronvolts (keV) or megaelectronvolts (MeV) . The latter unit is very common in particle physics.

Exmpe A positive charge released at B will be accelerated as it travels to point A. gain in kinetic energy = loss in electric potential energy 1 __ mv2 = Eqd 2

mv2 = 2Eqd  v=

Calculate the speed o an electron accelerated in a vacuum by a pd o 1 000 V (energy = 1 KeV) . KE o electron = V  e = 1 000  1 .6  1 0 - 1 9 = 1 .6  1 0 - 1 6 J 1 mv2 = 1 .6  1 0 - 1 6 J _ 2 v = 1 .87  1 0 7 m s - 1

 2Eqd _ m

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53

Eecc ce ElECtriCal ConduCtion in a mEtal

CurrEnt

Whenever charges move we say that a current is owing. A current is the name or moving charges and the path that they ollow is called the circuit. Without a complete circuit, a current cannot be maintained or any length o time.

Current is defned as the rate o fow o electrical charge. It is always given the symbol, I. Mathematically the defnition or current is expressed as ollows: charge owed Current = __ time taken Q dQ _ I= or (in calculus notation) I = _ dt t 1 coulomb 1 ampere = 1 second

Current ows THROUGH an object when there is a potential dierence ACROSS the object. A battery (or power supply) is the device that creates the potential dierence. By convention, currents are always represented as the ow o positive charge. Thus conventional current, as it is known, ows rom positive to negative. Although currents can ow in solids, liquids and gases, in most everyday electrical circuits the currents ow through wires. In this case the things that actually move are the negative electrons  the conduction electrons. The direction in which they move is opposite to the direction o the representation o conventional current. As they move the interactions between the conduction electrons and the lattice ions means that work needs to be done. Thereore, when a current ows, the metal heats up. The speed o the electrons due to the current is called their drit velocity.

conventional current, I

1 A = 1 C s- 1 I a current ows in just one direction it is known as a direct current. A current that constantly changes direction (frst one way then the other) is known as an alternating current or ac. In SI units, the ampere is the base unit and the coulomb is a derived unit

metal wire positive ions held in place

conduction electrons drift velocity Electrical conduction in a metal It is possible to estimate the drit velocity o electrons using the generalized drit speed equation. All currents are comprised o the movement o charge-carriers and these could be positive or negative; not all currents involve just the movement o electrons. Suppose that the number density o the charge-carriers (the number per unit volume that are available to move) is n, the charge on each carrier is q and their average speed is v. In a time t, the average distance moved by a charge-carrier = v  t so volume o charge moved past a point = A  vt so number o charge-carriers moved past a point = n Avt so charge moved past a point, Q = nAvt  q Q current I = _ t I = nAvq It is interesting to compare:  A typical drit speed o an electron: 1 0 - 4 m s - 1 (5A current in metal conductor o cross section 1 mm2 )  The speeds o the electrons due to their random motion: 1 0 6 m s - 1  The speed o an electrical signal down a conductor: approx. 3  1 0 8 m s - 1

54

__

ElEctri ci ty an d M ag n Eti s M

1 C=1 As

Eecc cc ohms law  ohmiC and non-ohmiC bEhaviour

rEsistanCE

The graphs below show how the current varies with potential dierence or some typical devices.

Resistance is the mathematical ratio between potential dierence and current. I something has a high resistance, it means that you would need a large potential dierence across it in order to get a current to ow.

(c) diode

potential dierence

current

current

(b) lament lamp

current

(a) metal at constant temperature

potential dierence

potential dierence

potential dierence Resistance = __ current V In symbols, R = _ I We defne a new unit, the ohm, , to be equal to one volt per amp. 1 ohm = 1 V A- 1

I current and potential dierence are proportional (like the metal at constant temperature) the device is said to be ohmic. Devices where current and potential dierence are not proportional (like the flament lamp or the diode) are said to be non-ohmic.

Ohms law states that the current owing through a piece o metal is proportional to the potential dierence across it providing the temperature remains constant. In symbols, V  I [i temperature is constant] A device with constant resistance (in other words an ohmic device) is called a resistor.

powEr dissipation

resistor. All this energy is going into heating up the resistor. In symbols:

energy dierence Since potential dierence = __ charge owed charge owed __ And current = time taken This means that potential dierence  current

P = V I Sometimes it is more useul to use this equation in a slightly dierent orm, e.g. P = V  I but V = I  R so

(energy dierence) (charge owed) energy dierence = __  __ = __ time (charge owed) (time taken)

P = (I  R)  I P = I2 R Similarly

This energy dierence per time is the power dissipated by the

V2 P= _ R

CirCuits  KirChoffs CirCuit laws

ExamplE

An electric circuit can contain many dierent devices or components. The mathematical relationship V = IR can be applied to any component or groups o components in a circuit.

A 1 .2 kW electric kettle is plugged into the 250 V mains supply. Calculate

When analysing a circuit it is important to look at the circuit as a whole. The power supply is the device that is providing the energy, but it is the whole circuit that determines what current ows through the circuit. Two undamental conservation laws apply when analysing circuits: the conservation o electric charge and the conservation o energy. These laws are collectively known as Kirchos circuit laws and can be stated mathematically as: First law:

 I = 0 (junction)

Second law:  V = 0 (loop) The frst law states that the algebraic sum o the currents at any junction in the circuit is zero. The current owing into a junction must be equal to the current owing out o a junction. In the example (right) the unknown current x = 5.5 + 2.7  3.4 = 4.8 A

3.4A 5.5A 2.7A

(i) the current drawn (ii) its resistance 1 200 (i) I = _ = 4.8 A 250

x The second law states that around any loop, the total energy per unit charge must sum to zero. Any source o potential dierence within the loop must be completely dissipated across the components in the loop (potential drop across the component) . Care needs to be taken to get the sign o any pd correct.  I the chosen loop direction is rom the negative side o a battery to its positive side, this is an increase in potential and the value is positive when calculating the sum.  I the direction around the loop is in the same direction as the current owing through the component, this is a potential drop and the value is negative when calculating the sum.

250 (ii) R = _ = 52  4.8

The example below shows one loop in a larger circuit. Anti-clockwise consideration o the loop means that: 1 2.0 - 5.3 - x + 2.7 - 3.2 = 0. The potential dierence across the bulb, x = 6.2 V pd = +12v pd = -3.2v

+ -

+

+ pd = -5.3v +- -

pd = +2.7v

pd = -x An example o the use o Kirchos circuit laws is shown on page 59.

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55

re  ee d e rEsistors in sEriEs

ElECtriCal mEtErs

A series circuit has components connected one ater another in a continuous chain. The current must be the same everywhere in the circuit since charge is conserved. The total potential dierence is shared among the components.

A current-measuring meter is called an ammeter. It should be connected in series at the point where the current needs to be measured. A perect ammeter would have zero resistance.

power supply (24 V) + -

I (2 A)

electrical energy is converted into:

R1 (3 )

R2 (4 )

resistor thermal energy

bulb light energy and thermal energy

potential dierence:

6V

R3 (5 ) M motor mechanical energy and thermal energy

8V

I (2 A)

10 V (6 + 8 + 10 = 24 V) pd of power supply

Total resistance = 3 + 4  + 5  = 1 2 

rEsistors in parallEl A parallel circuit branches and allows the charges more than one possible route around the circuit.

Vtotal

I total

Itotal V I1

I2 + I3

R1

I1 I2 + I3

V I2 I3

R2 V M

R3

I2 I3

Example o a parallel circuit Since the power supply fxes the potential dierence, each component has the same potential dierence across it. The total current is just the addition o the currents in each branch. 1 =_ 1 +_ 1 +_ 1 - 1 _ Ito ta l = I1 + I2 + I3 Rto ta l 5 3 4 20 + 1 5 + 1 2 - 1 __ V V V _ _ _ =  = + + 60 R1 R2 R3 47  - 1 =_ 1 = _ 1 +_ 1 +_ 1 _ 60 Rto ta l R1 R2 R3 60   Rto ta l = _ 47 = 1 .28 

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A meter that measures potential dierence is called a voltmeter. It should be placed in parallel with the component or components being considered. A perect voltmeter has infnite resistance.

Example o a series circuit We can work out what share they take by looking at each component in turn, e.g. The potential dierence across the resistor = I  R1 The potential dierence across the bulb = I  R2 Rto ta l = R1 + R2 + R3 This always applies to a series circuit. Note that V = IR correctly calculates the potential dierence across each individual component as well as calculating it across the total.

pe e cc  e potEntial dividEr CirCuit

ExamplE

The example on the right is an example o a circuit involving a potential divider. It is so called because the two resistors divide up the potential dierence o the battery. You can calculate the share taken by one resistor rom the ratio o the resistances but this approach does not work unless the voltmeters resistance is also considered. An ammeters internal resistance also needs to be considered. One o the most common mistakes when solving problems involving electrical circuits is to assume the current or potential dierence remains constant ater a change to the circuit. Ater a change, the only way to ensure your calculations are correct is to start again.

In the circuit below the voltmeter has a resistance o 20 k. Calculate:

A variable potential divider (a potentiometer) is oten the best way to produce a variable power supply. When designing the potential divider, the smallest resistor that is going to be connected needs to be taken into account: the potentiometers resistance should be signifcantly smaller.

A potentiometer has 3 terminals  the 2 ends and the central connection

(a) the pd across the 20 k resistor with the switch open (b) the reading on the voltmeter with the switch closed.

6.0 V 10 k

20 k

V 20 k 20 (a) pd = _________  6.0 = 4.0 V (20 + 10) (b) resistance o 20 k resistor and voltmeter combination, R, given by: 1 =_ 1 k - 1 1 +_ _ R 20 20  R = 1 0 k 10  pd = _  6.0 = 3.0 V (1 0 + 1 0)

sEnsors A light-dependent resistor (LDR) , is a device whose resistance depends on the amount o light shining on its surace. An increase in light causes a decrease in resistance.

output voltage

In order to measure the VI characteristics o an unknown resistor R, the two circuits (A and B) below are constructed. Both will both provide a range o readings or the potential dierence, V, across and current, I, through R. Providing that R >> the resistance o the potentiometer, this circuit (circuit B) is preerred because the range o readings is greater.

When light shines on the LDR LDR, there will be a decrease in pd across the LDR. pd Vtotal

 Circuit B allows the potential dierence across R (and hence the current through R) to be reduced down to zero. Circuit A will not go below the minimum value achieved when the variable resistor is at its maximum value.  Circuit B allows the potential dierence across R (and hence the current through R) to be increased up to the maximum value Vsu p p ly that can be supplied by the power supply in regular intervals. The range o values obtainable by Circuit A depends on a maximum o resistance o the variable resistor.

Circuit A  variable resistor variable resistor A

Vsupply

R

When light shines on the 10 k LDR, there will be an increase in pd across the xed resistor. A thermistor is a resistor whose value o resistance depends on its temperature. Most are semi-conducting devices that have a negative temperature coefcient (NTC) . This means that an increase in temperature causes a decrease in resistance. Both o these devices can be used in potential divider circuits to create sensor circuits. The output potential dierence o a sensor circuit depends on an external actor.

V 10 k

Circuit B  potentiometer

pd Vtotal A

Vsupply

When the temperature of the thermistor increases, there will be an increase in pd across the xed resistor.

R

V

When the temperature of the thermistor NTC increases, there will be a thermistor decrease in pd across the thermistor.

potentiometer

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57

re rEsistivity The resistivity, , o a material is defned in terms o its resistance, R, its length l and its cross-sectional area A. l R= _ A The units o resistivity must be ohm metres ( m) . Note that this is the ohm multiplied by the metre, not ohms per metre.

Exmpe The resistivity o copper is 3.3  1 0 - 7  m; the resistance o a 1 00 m length o wire o cross-sectional area 1 .0 mm2 is: 1 00 = 0.3  R = 3.3  1 0 - 7  _ 1 0- 4

invEstigating rEsistanCE The resistivity equation predicts that the resistance R o a substance will be: a) Proportional to the length l o the substance b) Inversely proportional to the cross-sectional area A o the substance. These relationships can be predicted by considering resistors in series and in parallel: a) Increasing l is like putting another resistor in series. Doubling l is the same as putting an identical resistor in series. R in series with R has an overall resistance o 2R. Doubling l means doubling R. So R  l. A graph o R vs I will be a straight line going through the origin. b) Increasing A is like putting another resistor in parallel. Doubling A is the same as putting an identical resistor in parallel. R in R 1 1 parallel with R has an overall resistance o __ . Doubling A means halving R. So R  __ . A graph o R vs __ will be a straight line A A 2 going through the origin. To practically investigate these relationships, we have: Independent variable: Control variables:

Either l or A A or l (depending on above choice) ; Temperature; Substance.

Data collection:

For each value o independent variable:  a range o values or V and I should be recorded  R can be calculated rom the gradient o a V vs I graph.

Data analysis

Values o R and the independent variable analysed graphically.

Possible sources o error/uncertainty include:  Temperature variation o the substance (particularly i currents are high) . Circuits should not be let connected. d  The cross-sectional area o the wire is calculated by measuring the wires diameter, d, and using A = r2 = ___ . Several sets o 4 measurements should be taken along the length o the wire and the readings in a set should be mutually perpendicular. 2

 The small value o the wires diameter will mean that the uncertainties generated using a ruler will be large. This will be improved using a vernier calliper or a micrometer.

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Ee  e  Kcf'  KirCho CirCuit laws ExamplE Great care needs to be taken when applying Kirchos laws to ensure that every term in the equation is correctly identifed as positive or negative. The concept o em ( see page 60) as sources o electrical energy can be used along with V = IR to provide an alternative statement o the second law which may help avoid conusion: Round any closed circuit, the sum o the ems is equal to the sum o the products o current and resistance. (em) = (IR) Process to follow  Draw a ull circuit diagram.  It helps to set up the equations in symbols beore substituting numbers and units.  It helps to be as precise as possible. Potential dierence V is a dierence between two points in the circuit so speciy which two points are being considered (use labels) .  Give the unknown currents symbols and mark their directions on the diagram. I you make a mistake and choose the wrong direction or a current, the solution to the equations will be negative.  Use Kirchos frst law to identiy appropriate relationships between currents.  Identiy a loop to apply Kirchos second law. Go all around the loop in one direction (clockwise or anticlockwise) adding the ems and I  R in senses shown below:

With chosen direction around loop in the direction shown,  and IR are both positive in the Kircho equation:

emf  I

I

chosen direction around loop R I I

 (emf) =  (IR) (If chosen direction opposite to that shown, values are negative)

 The total number o dierent equations generated by Kirchos laws needs to be the same as the number o unknowns or the problem to be able to be solved.  Use simultaneous equations to substitute and solve or the unknown values.  A new loop can be identifed to check that calculated values are correct.

Exmpe 6v

Sub (1 ) into (4)

20

A

B

i1

30i2 + 1 0(i1 + i2 ) = 5

i1 10

C

D i3

i3

(3)  4

1 0i1 + 40i2 = 5

(5)

1 20i1 + 40i2 = 24

(6)

1 1 0i1 = 1 9

(6) - (5) 

i2

i1 = 0.1 727 A = 172.7 mA

i2

(3)  1 0i2 = 6 - 30i1

E 5v

= 0.81 82

F

30



Kircho 1 st law junction C(or D) i1 + i2 = i3

= 81.8 mA (1 )

Kircho 2nd law and ACDB 1 0i3 + 20i1 = 6

i2 = 0.081 82 A i3 = 1 72.7 + 81 .8 mA = 254.5 mA

(2)

Sub (1 ) into (2) 1 0 (i1 + i2 ) + 20i1 = 6 

30i1 + 1 0i2 = 6

(3)

Kircho 2nd law and CEFD - 30i2 - 1 0i3 = -5

(4)

ElEctri ci ty an d M ag n Eti s M

59

ie ece  ce ElECtromotivE forCE and intErnal rEsistanCE

perfect battery  (e m f) = 6 V

When a 6V battery is connected in a circuit some energy will be used up inside the battery itsel. In other words, the battery has some internal resistance. The TOTAL energy dierence per unit charge around the circuit is still 6 volts, but some o this energy is used up inside the battery. The energy dierence per unit charge rom one terminal o the battery to the other is less than the total made available by the chemical reaction in the battery.

internal resistance r

terminals of battery R external resistance

For historical reasons, the TOTAL energy dierence per unit charge around a circuit is called the electromotive force (emf) . However, remember that it is not a orce (measured in newtons) but an energy dierence per charge (measured in volts) .

e m  = I  Rto ta l = I(r + R) = Ir + IR

In practical terms, em is exactly the same as potential dierence i no current fows.

IR = em - Ir terminal p d, V lost volts

 = I (R + r)

V =  - Ir

An electric battery is a device consisting o one or more cells joined together. In a cell, a chemical reaction takes place, which converts stored chemical energy into electrical energy. There are two dierent types o cell: primary and secondary. A primary cell cannot be recharged. During the lietime o the cell, the chemicals in the cell get used in a non-reversible reaction. Once a primary cell is no longer able to provide electrical energy, it is thrown away. Common examples include zinccarbon batteries and alkaline batteries. A secondary cell is designed to be recharged. The chemical reaction that produces the electrical energy is reversible. A reverse electrical current charges the cell allowing it to be reused many times. Common examples include a leadacid car battery, nickelcadmium and lithium-ion batteries. The charge capacity o a cell is how much charge can fow beore the cells stops working. Typical batteries have charge capacities that are measured in Amp-hours (A h). 1 A h is the charge that fows when a current o 1 A fows or one hour i.e. 1 A h = 3600 C.

disChargE CharaCtEristiCs

terminal voltage (V)

When current (and thus electrical energy) is drawn rom a cell, the terminal potential dierence varies with time. A perect cell would maintain its terminal pd throughout its lietime; real cells, however, do not. The terminal potential dierence o a typical cell:  loses its initial value quickly,  has a stable and reasonably constant value or most o its lietime. This is ollowed by a rapid decrease to zero (cell discharges) . The graph below shows the discharge characteristics or one particular type o leadacid car battery.

13 12 11 10 9 8 7 0

discharge characteristics ambient temperature: 25 C

10.8 10.5

14.3A9.5A 5.6A 33A 55A

9.6

165A 110A 2 3

5 10 20 30 60 2 3 discharge time

To experimentally determine the internal resistance r o a cell (and its em ) , the circuit below can be used:

battery terminal

terminal pd, V V

battery terminal

internal emf,  resistance, r

external resistance, R I

Procedure:  Vary external resistance R to get a number (ideally 1 0 or more) o matching readings o V and I over as wide a range as possible.  Repeat readings.  Do not leave current running or too long (especially at high values o I) .  Take care that nothing overheats. Data analysis:  The relevant equation, V = - Ir was introduced above.  A plot o V on the y-axis and I on the x-axis gives a straight line graph with  gradient = - r  y-intercept = 

rECharging sECondary CElls In order to recharge a secondary cell, it is connected to an external DC power source. The negative terminal o the secondary cell is connected to the negative terminal o the power source and the positive terminal o the power source with the positive terminal o the secondary cell. In order or a charging current, I, to fow, the voltage output o the power source must be slightly higher than that o the battery. A large dierence between the power source and the cell's terminal potential dierence means that the charging process will take less time but risks damaging the cell.

secondary cell being charged I I

7.6

(min)

60

3.0A

dEtErmining intErnal rEsistanCE ExpErimEntally

current, I

CElls and battEriEs

5 10 20 (h)

ElEctri ci ty an d M ag n Eti s M

- + power source (slightly higher pd)

mec orce  fe magnEtiC iEld linEs There are many similarities between the magnetic orce and the electrostatic orce. In act, both orces have been shown to be two aspects o one orce  the electromagnetic interaction (see page 78) . It is, however, much easier to consider them as completely separate orces to avoid conusion. Page 52 introduced the idea o electric felds. A similar concept is used or magnetic felds. A table o the comparisons between these two felds is shown below. Electric eld Symbol

Magnetic eld

E

B

Caused by 

Charges

Magnets (or electric currents)

Aects 

Charges

Magnets (or electric currents)

Two types o 

Charge: positive and negative

Pole: North and South

Simple orce rule:

Like charges repel, unlike charges attract

Like poles repel, unlike poles attract

In order to help visualize a magnetic feld we, once again, use the concept o feld lines. This time the feld lines are lines o magnetic feld  also called fux lines. I a test magnetic North pole is placed in a magnetic feld, it will eel a orce.

geographic North Pole Earth A magnet free to move in all S directions would line up pointing along the eld lines. A compass is normally only free to N move horizontally, so it ends up pointing along the horizontal component of the eld. geographic The magnetic North pole of the compass points towards the geographic South Pole North Pole  hence its name. An electric current can also cause a magnetic feld. The mathematical value o the magnetic felds produced in this way is given on page 63. The feld patterns due to dierent currents can be seen in the diagrams below.

 The direction o the orce is shown by the direction o the feld lines.  The strength o the orce is shown by how close the lines are to one another.

A test South pole here would feel a force in the opposite direction.

Force here strong since eld lines are close together. N

Overall force is in direction shown because a North pole would feel a repulsion and an attraction as shown.

S

Force here weak since eld lines are far apart. rotate

thumb (current direction)

I current

I

curl of ngers gives direction of eld lines

The feld lines are circular around the current. The direction o the feld lines can be remembered with the righthand grip rule. I the thumb o the right hand is arranged to point along the direction o a current, the way the fngers o the right hand naturally curl will give the direction o the feld lines. Field pattern o a straight wire carrying current

cross-section current into page

N

N S

S

A small magnet placed in the feld would rotate until lined up with the feld lines. This is how a compass works. Small pieces o iron (iron flings) will also line up with the feld lines  they willbe induced to become little magnets.

current out of page Field pattern o a at circular coil A long current-carrying coil is called a solenoid.

Field pattern o an isolated bar magnet Despite all the similarities between electric felds and magnetic felds, it should be remembered that they are very dierent. For example:  A magnet does not eel a orce when placed in an electric feld.

eld pattern of solenoid is the same as a bar magnet cross-section

 A positive charge does not eel a orce when placed stationary in a magnetic feld.  Isolated charges exist whereas isolated poles do not.  The Earth itsel has a magnetic feld. It turns out to be similar to that o a bar magnet with a magnetic South pole near the geographic North Pole as shown below.

N

S

poles of solenoid can be predicted using right-hand grip rule Field pattern or a solenoid

ElEctri ci ty an d M ag n Eti s M

61

mec ces magnEtiC forCE on a CurrEnt

force on current

When a current-carrying wire is placed in a magnetic feld the magnetic interaction between the two results in a orce. This is known as the motor effect. The direction o this orce is at right angles to the plane that contains the feld and the current as shown below.

I

N

S

zero force

N

I

I thumb (force) F

force (F) eld ( B)

rst nger (eld) B second nger (current) I

current (I)

F I



N

S

force at right angles to plane of current and eld lines

Flemings let-hand rule Experiments show that the orce is proportional to:  the magnitude o the magnetic feld, B  the magnitude o the current, I  the length o the current, L, that is in the magnetic feld

F I

N

S

S

force maximum when current and eld are at right angles

 the sine o the angle, , between the feld and current. The magnetic feld strength, B is defned as ollows: F = BIL sin 

or

F B=_ IL sin  A new unit, the tesla, is introduced. 1 T is defned to be equal to 1 N A- 1 m- 1 . Another possible unit or magnetic feld strength is Wb m- 2 . Another possible term is magnetic ux density.

magnEtiC forCE on a moving ChargE A single charge moving through a magnetic feld also eels a orce in exactly the same way that a current eels a orce. In this case the orce on a moving charge is proportional to:

Since the orce on a moving charge is always at right angles to the velocity o the charge the resultant motion can be circular. An example o this would be when an electron enters a region where the magnetic feld is at right angles to its velocity as shown below.

 the magnitude o the magnetic feld, B

S

 the magnitude o the charge, q  the velocity o the charge, v  the sine o the angle, , between the velocity o the charge and the feld. We can use these relationships to give an alternative defnition o the magnetic feld strength, B. This defnition is exactly equivalent to the previous defnition. F = Bqv sin 

or

F B=_ qv sin 

electron

F F

r

F

N

F

An electron moving at right angles to a magnetic feld

62

ElEctri ci ty an d M ag n Eti s M

Exe  e ec fe ue  cue The formulae used on this page do not need to be remembered.

straight wirE The feld pattern around a long straight wire shows that as one moves away rom the wire, the strength o the feld gets weaker. Experimentally the feld is proportional to:  the value o the current, I  the inverse o the distance away rom the wire, r. I the distance away is doubled, the magnetic feld will halve.

two parallEl wirEs  dEinition o thE ampErE Two parallel current-carrying wires provide a good example o the concepts o magnetic feld and magnetic orce. Because there is a current owing down the wire, each wire is producing a magnetic feld. The other wire is in this feld so it eels a orce. The orces on the wires are an example o a Newtons third law pair o orces. r length l1

 The feld also depends on the medium around the wire. These actors are summarized in the equation:

length l2 I1

I B= _ 2r

I2 B1

B1

B1 F

r B1 B1 = eld produced by I1  I1 = 2r

I r

force felt by I2 = B1  I2  l2  force per unit length of I2

I

= Magnetic feld o a straight current

= B1 I2

The constant  is called the permeability and changes i the medium around the wire changes. Most o the time we consider the feld around a wire when there is nothing there  so we use the value or the permeability o a vacuum, 0 . There is almost no dierence between the permeability o air and the permeability o a vacuum. There are many possible units or this constant, but it is common to use N A- 2 or T m A- 1 . Permeability and permittivity are related constants. In other words, i you know one constant you can calculate the other. In the SI system o units, the permeability o a vacuum is defned to have a value o exactly 4   1 0 - 7 N A- 2 . See the defnition o the ampere (right) or more detail.

length l1

 I1 I2 2r

length l2

I1

I2 B2

force felt by I1 B 2 = B2  I1  l1

F

 force per unit length of I1 B2 I1 l1 l1

B2

r B2 B2 = eld produced by I2  I2 = 2r

= B2 I1

The magnetic feld o a solenoid is very similar to the magnetic feld o a bar magnet. As shown by the parallel feld lines, the magnetic feld inside the solenoid is constant. It might seem surprising that the feld does not vary at all inside the solenoid, but this can be experimentally verifed near the centre o a long solenoid. It does tend to decrease near the ends o the solenoid as shown in the graph below.

axis

I

=

r

=

magnEtiC iEld in a solEnoid

B1 I2 l2 l2

I

magnetic eld along axis constant eld B in centre

=

 I1 I2 2r

I I Magnitude o orce per unit length on either wire = ____ 2r 1 2

This equation is experimentally used to defne the ampere. The coulomb is then defned to be one ampere second. I we imagine two infnitely long wires carrying a current o one amp separated by a distance o one metre, the equation would predict the orce per unit length to be 2  1 0- 7 N. Although it is not possible to have infnitely long wires, an experimental setup can be arranged with very long wires indeed. This allows the orces to be measured and ammeters to be properly calibrated.

The mathematical equation or this constant feld at the centre o a long solenoid is n I B= _ l

( )

Thus the feld only depends on:  the current, I

distance (n = number of turns, l = length)

n  the number o turns per unit length, _ l  the nature o the solenoid core,  It is independent o the cross-sectional area o the solenoid.

Variation o magnetic feld in a solenoid

ElEctri ci ty an d M ag n Eti s M

63

ib Questons  electrcty and magnetsm 1.

2.

Which one o the feld patterns below could be produced by two point charges? A.

C.

B.

D.

Two long, vertical wires X and Y carry currents in the same direction and pass through a horizontal sheet o card.

X

12 V battery

a) On the circuit above, add circuit symbols showing the correct positions o an ideal ammeter and an ideal voltmeter that would allow the V-I characteristics o this lamp to be measured. [2] The voltmeter and the ammeter are connected correctly in the circuit above.

Y

b) Explain why the potential dierence across the lamp (i)

cannot be increased to 1 2 V.

[2]

(ii) cannot be reduced to zero.

Iron flings are scattered on the card. Which one o the ollowing diagrams best shows the pattern ormed by the iron flings? (The dots show where the wires X and Y enter the card.) A.

[2]

An alternative circuit or measuring the V-I characteristic uses a potential divider. c) (i)

C.

Draw a circuit that uses a potential divider to enable the V-I characteristics o the flament to be ound.

(ii) Explain why this circuit enables the potential dierence across the lamp to be reduced to zero volts.

B.

3.

[3]

[3]

[2]

The graph below shows the V-I characteristic or two 1 2 V flament lamps A and B.

D.

potential dierence / V 12

lamp A

lamp B

This question is about the electric feld due to a charged sphere and the motion o electrons in that feld. The diagram below shows an isolated metal sphere in a vacuum that carries a negative electric charge o 9.0 nC.

-

0 0

a) On the diagram draw arrows to represent the electric feld pattern due to the charged sphere. [3] b) The electric feld strength at the surace o the sphere and at points outside the sphere can be determined by assuming that the sphere acts as though a point charge o magnitude 9.0 nC is situated at its centre. The radius o the sphere is 4.5  1 0 - 2 m. Deduce that the magnitude o the feld strength at the surace o the sphere is 4.0  1 0 4 V m- 1 . [1 ]

0.5

1.0 current / A

d) State and explain which lamp has the greater power dissipation or a potential dierence o 1 2 V.

[3]

The two lamps are now connected in series with a 1 2 V battery as shown below.

12 V battery

An electron is initially at rest on the surace o the sphere. c) (i) Describe the path ollowed by the electron as it leaves the surace o the sphere. [1 ] (ii) Calculate the initial acceleration o the electron.

[3]

(iii) State and explain whether the acceleration o the electron remains constant, increases or decreases as it moves away rom the sphere. [2] (iv) At a certain point P, the speed o the electron is 6.0  1 0 6 m s - 1 . Determine the potential dierence between the point P and the surace o the sphere. [2] 4.

In order to measure the voltage-current (V-I) characteristics o a lamp, a student sets up the ollowing electrical circuit.

64

lamp A e) (i)

lamp B

State how the current in lamp A compares with that in lamp B.

[1 ]

(ii) Use the V-I characteristics o the lamps to deduce the total current rom the battery.

[4]

(iii) Compare the power dissipated by the two lamps.

[2]

i B Q u Esti o n s  ElEctri ci ty an d M ag n Eti s M

6 c i r c U l a r M o t i o n a n d g r av i t at i o n Um u m Mechanics of circUlar Motion

MatheMatics of circUlar Motion

The phrase uniorm circular motion is used to describe an object that is going around a circle at constant speed. Most o the time this also means that the circle is horizontal. An example o uniorm circular motion would be the motion o a small mass on the end o a string as shown below.

The diagram below allows us to work out the direction o the centripetal acceleration  which must also be the direction o the centripetal orce. This direction is constantly changing.

situation diagram

vector diagram change in velocity directed in towards centre of circle

vB B vA

vB

vA

A

The object is shown moving between two points A and B on a horizontal circle. Its velocity has changed rom vA to vB . The magnitude o velocity is always the same, but the direction has changed. Since velocities are vector quantities we need to use vector mathematics to work out the average change in velocity. This vector diagram is also shown above.

mass moves at constant speed Example o uniorm circular motion

In this example, the direction o the average change in velocity is towards the centre o the circle. This is always the case and thus true or the instantaneous acceleration. For a mass m moving at a speed v in uniorm circular motion o radius r, v2 [in towards the centre o the circle] Centripetal acceleration ace n trip e ta l = __ r A orce must have caused this acceleration. The value o the orce is worked out using Newtons second law:

It is important to remember that even though the speed o the object is constant, its direction is changing all the time.

v m s -1

vA + change = vB

v m s -1

v m s- 1

Centripetal orce (CPF) Fce n trip e ta l = m ace n trip e ta l m v2 [in towards the centre o the circle] = ____ r For example, i a car o mass 1 500 kg is travelling at a constant speed o 20 m s - 1 around a circular track o radius 50 m, the resultant orce that must be acting on it works out to be

s -1

vm v m s -1 speed is constant but the direction is constantly changing Circular motion  the direction o motion is changing all the time

1 500(20 ) 2 F = __________ = 1 2 000 N 50 It is really important to understand that centripetal orce is NOT a new orce that starts acting on something when it goes in a circle. It is a way o working out what the total orce must have been. This total orce must result rom all the other orces on the object. See the examples below or more details.

This constantly changing direction means that the velocity o the object is constantly changing. The word acceleration is used whenever an objects velocity changes. This means that an object in uniorm circular motion MUST be accelerating even i the speed is constant.

One fnal point to note is that the centripetal orce does NOT do any work. (Work done = orce  distance in the direction of the force.)

The acceleration o a particle travelling in circular motion is called the centripetal acceleration. The orce needed to cause the centripetal acceleration is called the centripetal force.

exaMples Earth's gravitational attraction on Moon

Moon

Earth R

R cos 

 F

T cos  T

friction forces between tyres and road

T sin 

R sin  

mg W A conical pendulum  centripetal force At a particular speed, the horizontal component provided by horizontal component of the normal reaction can provide all the of tension. centripetal force (without needing friction) .

C i r C u l a r m o t i o n a n d g r a v i tat i o n

65

au    u m radians

angUlar velocity, , and tiMe period, T

Angles measure the raction o a complete circle that has been achieved. They can, o course, be measured in degrees (symbol: ) but in studying circular motion, the radian (symbol: rad) is a more useul measure.

radius

An object travelling in circular motion must be constantly changing direction. As a result its velocity is constantly changing even i its speed is constant (uniorm circular motion) . We defne the average angular velocity, symbol  (omega) as: angle turned  a v e ra ge = ____________ = ____ time taken t The units o angular velocity are radians per second (rad s 1 ) .

r



s distance along circular arc

The instantaneous angular velocity is the rate o change o angle: d  = rate o change o angle = ___ dt 1 . Link between  and v

The raction o the circle that has been achieved is the ratio o arc length s to the circumerence: s raction o circle = ____ 2r In degrees, the whole circle is divided up into 360 which defnes the angle  as: s (in degrees) = ____  360 2r In radians, the whole circle is divided up into 2 radians which defnes the angle  as: Angle/ Angle/radian s (in radians) = ____  2 = __rs 2r 0 0.00 For small angles (less than 5 0.09 about 0.1 rad or 5) , the arc and  45 0.74 = __ the two radii orm a shape that 4 60 1 .05 approximates to a triangle. Since  radians are just a ratio, the 90 1 .57 = __ 2 ollowing relationship applies i 1 80 3.1 4 =  working in radians: 3 270 4.71 = ___ 2 sin  tan   360 6.28 = 2

y

In a time t, the object rotates an angle   = __rs  s = r r s v = ___ = = r t t v = r

_

2.

v



x

Link between  and time period T

The time period T is the time taken to complete one ull circle. In this time, the total angle turned is 2 radians, so: 2 or T = ___ 2  = ___  T 3. Circular motion equations Substitution o the above equations into the ormulae or centripetal orce and centripetal acceleration (page 65) provide versions that are sometime more useul: v 2 = r 2 = _____ 4 2 r centripetal acceleration, a = ___ r T2 m v2 4 2 mr 2 _______ centripetal orce, F = ____ r = mr = T2

circUlar Motion in a vertical plane

1.

Uniorm circular motion o a mass on the end o a string in a horizontal plane requires a constant centripetal orce to act and the magnitude o the tension in the string will not change. Circular motion in the vertical plane is more complicated as the weight o the object always acts in the same vertical direction. The object will speed up and slow down during its motion due to the component o its weight that acts along the tangent to the circle. The maximum speed will be when the object is at the bottom and the minimum speed will occur at the top. The tension in the string will also change during one revolution.

The tension in the string, T, and the weight, mg, are in the same direction and add together to provide the CPF: mvto p 2 Tto p + mg = ______ r To remain in the vertical circle, the object must be moving with a certain minimum speed. At this minimum top speed, vto p m in , the tension is zero and the centripetal orce is provided by the objects weight:

In a vertical circle, the tension o the string will always act at 90 to the objects velocity so this orce does no work in speeding it up or slowing it down. The conservation o energy means that: 1 mv2 = constant mgy + __ 2

a) SITUATION DIAGRAM

H

r small mass m

s t ri

ng

O path

y

At the top o the circle:,

m( vto p m in ) 2 mg = __________ r vto p m in = rg  2.

At the bottom o the circle:,

The tension in the string, T, and the weight, mg, are in opposite directions and the resultant orce provides the CPF: mvb o tto m 2 Tb o tto m -mg = ________ r In order to complete the vertical circle, the KE at the bottom o the circle must be large enough or the object to arrive at the top o the circle with sufcient speed( vto p m in = rg  ) to complete the circle. Energetically the object gains PE (= mg  2r) so it must lose the same amount o KE: 1 m( v 1 m( v 1 mrg __ ) 2 - mg2r = __ ) 2 = __ b o tto m m in to p m in 2 2 2

L b) FREE-BODY DIAGRAM

 ( vb o tto m m in ) 2 -4gr = rg  vb o tto m m in = 5rg 

F mg

66

instantaneous acceleration

The mathematics in the above example (a mass on the end o a string) can also apply or any vehicle that is looping the loop. In place o T, the tension in the rope, there is N, the normal reaction rom the surace.

C i r C u l a r m o t i o n a n d g r a v i tat i o n

n    newtons law of Universal gravitation

Universal gravitational constant G = 6.67  1 0 - 1 1 N m2 kg- 2

I you trip over, you will all down towards the ground.

The ollowing points should be noticed:

Newtons theory o universal gravitation explains what is going on. It is called universal gravitation because at the core o this theory is the statement that every mass in the Universe attracts all the other masses in the Universe. The value o the attraction between two point masses is given by an equation.

 There is a orce acting on each o the masses. These orces are EQUAL and OPPOSITE (even i the masses are not equal) .

m1

 The law only deals with point masses.

 The orces are always attractive.

m2 r F

m1m2

F=

r2

The interaction between two spherical masses turns out to be the same as i the masses were concentrated at the centres o the spheres.

Gm 1 m 2 r2

gravitational field strength The table below should be compared with the one on page 61 . Gravitational feld strength

In the example on the let the numerical value or the gravitational feld can be calculated using Newtons law: GMm F = _____ r2

g

Caused by

Masses

Aects

Masses

One type o

Mass

Simple orce rule:

All masses attract

orce Acceleration = _____ mass (rom F = ma) For the Earth

The gravitational feld is thereore defned as the orce F per unit mass. g = __ m m = small point test mass

F

M = 6.0  1 0 2 4 kg r = 6.4  1 0 6 m

test mass m 2

value of g =

F m2

mass M1 producing gravitational eld g The SI units or g are N kg- 1 . These are the same as m s - 2 . Field strength is a vector quantity and can be represented by the use o feld lines.

gravitational eld lines

6.67  1 0 - 1 1  6.0  1 0 2 4 = 9.8 m s - 2 g = ________________________ (6.4  1 0 6 ) 2

exaMple In order to calculate the overall gravitational feld strength at any point we must use vector addition. The overall gravitational feld strength at any point between the Earth and the Moon must be a result o both pulls. There will be a single point somewhere between the Earth and the Moon where the total gravitational feld due to these two masses is zero. Up to this point the overall pull is back to the Earth, ater this point the overall pull is towards the Moon.

Earth sphere

GM g = ____ r2

The gravitational feld strength at the surace o a planet must be the same as the acceleration due to gravity on the surace. orce Field strength is defned to be _____ mass

Symbol

F

 Gravitation orces act between ALL objects in the Universe. The orces only become signifcant i one (or both) o the objects involved are massive, but they are there nonetheless.

up to X overall pull is back to Earth

beyond X overall pull is towards Moon Moon

point mass r1

Field strength around masses (sphere and point)

gravitational eld

X r2

distance between Earth and Moon = (r1 + r2 ) I resultant gravitational feld at X = zero,

EARTH Gravitational feld near surace o the Earth

GME a rth GM M oon _______ = _______ r1 2 r2 2

C i r C u l a r m o t i o n a n d g r a v i tat i o n

67

iB Questons  crcular moton and gravtaton 1.

A ball is tied to a string and rotated at a uniorm speed in a vertical plane. The diagram shows the ball at its lowest position. Which arrow shows the direction o the net orce acting on the ball? [1 ]

a) (i) On the diagram above, draw and label arrows to represent the orces on the ball in the position shown.

A

b) Determine the speed o rotation o the ball. 5.

A particle o mass m is moving with constant speed v in uniorm circular motion. What is the total work done by the centripetal orce during one revolution? mv2 A. Zero B. ____ 2 C. mv2 D. 2mv2

Which diagram correctly shows the direction o the velocity v and acceleration a o the particle P in the position shown? B.

a

v

6.

D.

v

a

[1 ]

a

P

P

C.

4.

[1 ]

A particle P is moving anti-clockwise with constant speed in a horizontal circle.

A.

P

(i) Derive an expression or the gravitational feld strength at the surace o a planet in terms o its mass M, its radius R and the gravitational constant G.

[2]

(ii) Use your expression in (b) (i) above to estimate the mass o Jupiter.

[2]

Gravitational felds and potential a) Derive an expression or the gravitational feld strength as a unction o distance away rom a point mass M.

[3]

b) The radius o the Earth is 6400 km and the gravitational feld strength at its surace is 9.8 N kg- 1 . Calculate a value or the mass o the Earth.

[2]

c) On the diagram below draw lines to represent the gravitational feld outside the Earth.

[2]

d) A satellite that orbits the Earth is in the gravitational feld o the Earth. Discuss why an astronaut inside the satellite eels weightless.

[3]

v

a and v

P

This question is about circular motion. A ball o mass 0.25 kg is attached to a string and is made to rotate with constant speed v along a horizontal circle o radius r = 0.33 m. The string is attached to the ceiling and makes an angle o 30 with the vertical.

30 vertical

r = 0.33 m

68

[2]

b) The gravitational feld strength at the surace o Jupiter is 25 N kg- 1 and the radius o Jupiter is 7.1  1 0 7 m.

D

3.

[3]

This question is about gravitational felds. a) Defne gravitational feld strength.

B C

2.

[2]

(ii) State and explain whether the ball is in equilibrium. [2]

i B Q u e s t i o n s  C i r C u l a r m o t i o n a n d g r a v i tat i o n

7 ato m i c , n u clE ar an d par ti clE ph ys i cs E   e Emission spEctra and absorption spEctra

Explanation of atomic spEctra

When an element is given enough energy it emits light. This light can be analysed by splitting it into its various colours (or requencies) using a prism or a diraction grating. I all possible requencies o light were present, this would be called a continuous spectrum. The light an element emits, its emission spectrum, is not continuous, but contains only a ew characteristic colours. The requencies emitted are particular to the element in question. For example, the yellow-orange light rom a street lamp is oten a sign that the element sodium is present in the lamp. Exactly the same particular requencies are absent i a continuous spectrum o light is shone through an element when it is in gaseous orm. This is called an absorption spectrum.

In an atom, electrons are bound to the nucleus. See page 77, the atomic model. This means that they cannot escape without the input o energy. I enough energy is put in, an electron can leave the atom. I this happens, the atom is now positive overall and is said to be ionized. Electrons can only occupy given energy levels  the energy o the electron is said to be quantized. These energy levels are fxed or particular elements and correspond to allowed obitals. The reason why only these energies are allowed orms a signifcant part o quantum theory (see HL topic 1 2).

spectra: emission set-up

prism (or diraction grating)

slit

sample of gas

spectral lines

light emitted from gas spectra: absorption set-up

prism ( or diraction grating)

slit

light source

spectral lines

all frequencies sample of gas (continuous spectrum) 330

mercury 313 helium

415 420

334

365366

398405

389 403 361 371 382 396 412

319

384 397 410 380 389

hydrogen

546 579

439 447 471 492502

434

energy in joules E = hf Plancks constant 6.63  1 0 - 3 4 J s

requency o light in Hz

Speed o light in m s - 1 Since c = f hc  = ___ E Wavelength in m

Thus the requency o the light, emitted or absorbed, is fxed by the energy dierence between the levels. Since the energy levels are unique to a given element, this means that the emission (and the absorption) spectrum will also be unique.

569 590 615

436

The energy o a photon is given by

energy

sodium

When an electron moves between energy levels it must emit or absorb energy. The energy emitted or absorbed corresponds to the dierence between the two allowed energy levels. This energy is emitted or absorbed as packets o light called photons. A higher energy photon corresponds to a higher requency (shorter wavelength) o light.

588

668

486

allowed energy levels

656

wavelength,  / nm 300

310

320

330

340

350 360 370 380 390 400

450

500

550

600 650 700 yellow

invisible (UV)

approximate colour

violet indigo

blue

green

red orange invisible (IR)

Emission spectra

m e rcu ry

33 0

334

3 13

h e liu m

415 4 2 0

3 65 3 6 6

39 8 405

389 403 3 61 3 71 3 8 2 3 9 6 412

3 19

384 397 380 389 410

h y d ro ge n

photon of pa rticular frequency absorbed

5 69 5 9 0 615

43 6

energy

s o d iu m

electron promoted from low energy level to higher energy level

5 4 6 57 9

43 9 4 47

471 49 2 5 0 2

43 4

588

486

668

allowed energy levels

65 6

wa ve le n gth ,  / n m 300

3 10

320

330

340

35 0 3 60 370 3 8 0 39 0 40 0

45 0

50 0

550

60 0

65 0

70 0

y e l lo w a p p ro xim a te c o lo u r

i n vi s i b le ( U V)

vio le t in d igo

blue

gre e n

re d o ra n ge i n vi s i b le ( IR )

Absorption spectra

electron falls from high energy level to lower energy level

photon of pa rticular frequency emitted

Ato m i c , n u c l e A r A n d pA r t i c l e p h ys i c s

69

ne  isotopEs

nuclEar stability

When a chemical reaction takes place, it involves the outer electrons o the atoms concerned. Dierent elements have dierent chemical properties because the arrangement o outer electrons varies rom element to element. The chemical properties o a particular element are fxed by the amount o positive charge that exists in the nucleus  in other words, the number o protons. In general, dierent nuclear structures will imply dierent chemical properties. A nuclide is the name given to a particular species o atom (one whose nucleus contains a specifed number o protons and a specifed number o neutrons) . Some nuclides are the same element  they have the same chemical properties and contain the same number o protons. These nuclides are called isotopes  they contain the same number o protons but dierent numbers o neutrons.

Many atomic nuclei are unstable. The process by which they decay is called radioactive decay (see page 72) . It involves emission o alpha () , beta () or gamma () radiation. The stability o a particular nuclide depends greatly on the numbers o neutrons present. The graph below shows the stable nuclides that exist.  For small nuclei, the number o neutrons tends to equal the number o protons.  For large nuclei there are more neutrons than protons.  Nuclides above the band o stability have too many neutrons and will tend to decay with either alpha or beta decay (see page 72) .

neutron number, N

 Nuclides below the band o stability have too ew neutrons and will tend to emit positrons (see page 73) .

notation A

mass number  equal to number of nucleons chemical symbol

Z

160 150 140 130 120 110

atomic number  equal to number of protons in the nucleus

100

Nuclide notation

90 80

ExamplEs Notation

70 Description

Comment

1

12 6

C

carbon-1 2

isotope o 2

2

13 6

C

carbon-1 3

isotope o 1

3

238 92

U

uranium-238

4

1 98 78

Pt

platinum-1 98

same mass number as 5

5

1 98 80

Hg

mercury-1 98

same mass number as 4

60 50 40 30 20 10

Each element has a unique chemical symbol and its own atomic number. No.1 and No.2 are examples o two isotopes, whereas No.4 and No.5 are not. In general, when physicists use this notation they are concerned with the nucleus rather than the whole atom. Chemists use the same notation but tend to include the overall charge on the atom. Thus 1 26 C can represent the carbon nucleus to a physicist or the carbon atom to a chemist depending on the context. I the charge is present the situation becomes unambiguous. 31 57 C l must reer to a chlorine ion  an atom that has gained one extra electron.

70

0

10 20 30 40 50 60 70 80 90 100 atomic number, Z

Key N number o neutrons Z number o protons 

naturally occurring stable nuclide



naturally occurring -emitting nuclide



artifcially produced -emitting nuclide



naturally occurring - -emitting nuclide



artifcially produced + -emitting nuclide



artifcially produced - -emitting nuclide



artifcially produced electron-capturing nuclide



artifcial nuclide decaying by spontaneous fssion

At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s

fe e strong nuclEar forcE

WEak nuclEar forcE

The protons in a nucleus are all positive. Since like charges repel, they must be repelling one another all the time. This means there must be another orce keeping the nucleus together. Without it the nucleus would fy apart. We know a ew things about this orce.

The strong nuclear orce (see box let) explains why nuclei do not fy apart and thus why they are stable. Most nuclei, however, are unstable. Mechanisms to explain alpha and gamma emission (see page 72) can be identied but another nuclear orce must be involved i we wish to be able to explain all aspects o the nucleus including beta emission. We know a ew things about this orce:

 It must be strong. I the proton repulsions are calculated it is clear that the gravitational attraction between the nucleons is ar too small to be able to keep the nucleus together.  It must be very short-ranged as we do not observe this orce anywhere other than inside the nucleus.

 It must be weak. Many nuclei are stable and beta emission does not always occur.  It must be very short-ranged as we do not observe this orce anywhere other than inside the nucleus.

 It is likely to involve the neutrons as well. Small nuclei tend to have equal numbers o protons and neutrons. Large nuclei need proportionately more neutrons in order to keep the nucleus together.

 Unlike the strong nuclear orce, it involves the lighter particles (e.g. electrons, positrons and neutrinos) as well as the heavier ones (e.g. protons and neutrons) .

The name given to this orce is the strong nuclear force.

The name given to this orce is the weak nuclear force.

othEr fundamEntal forcEs/intEractions

Electromagnetic

The standard model o particle physics is based around the orces that we observe on a daily basis along with the two new orces that have been identied as being involved in nuclear stability (above) . As a result in the standard model, there are only our undament orces (or interactions) that are known to exist. These are Gravity, Electromagnetic, Strong and Weak. More detail about all these orces is discussed on page 78. Outline inormation about two everyday interactions is listed below:

 This single orce includes all the orces that we normally categorize as either electrostatic or magnetic.

Gravity  Gravity is the orce o attraction between all objects that have mass.  Gravity is always attractive  masses are pulled together.  The range o the gravity orce is innite.  Despite the above, the gravity orce is relatively quite weak. At least one o the masses involved needs to be large or the eects to be noticeable. For example, the gravitational orce o attraction between you and this book is negligible, but the orce between this book and the Earth is easily demonstrable (drop it) .  Newtons law o gravitation describes the mathematics governing this orce.

 Electromagnetic orces involve charged matter.  Electromagnetic orces can be attractive or repulsive.  The range o the electromagnetic orce is innite.  The electromagnetic orce is relatively strong  tiny imbalances o charges on an atomic level give rise to signicant orces on the laboratory scale.  At the end o the 1 9th century, Maxwell showed that the electrostatic orce and the magnetic orce were just two dierent aspects o the more undamental electromagnetic orce.  The mathematics o the electromagnetic orce is described by Maxwells equations.  Friction (and many other everyday orces) is simply the result o the orce between atoms and this is governed by the electromagnetic interaction. The electromagnetic orce and the weak nuclear orce are now considered to be aspects o the single electroweak orce.

particlEs that ExpEriEncE and mEdiatE thE fundamEntal forcEs. See page 78 onwards or more details about the standard model or the undamental structure o matter. The ollowing table summarizes which particles experience these orces and how they are mediated.

Particles experience Particles mediate

Gravitational

Weak

Electromagnetic

Strong

All

Quark, Gluon

Charged

Quark, Gluon

Graviton

W+ , W- , Z 0



Gluon

Ato m i c , n u c l e A r A n d pA r t i c l e p h ys i c s

71

rv 1 ioniZing propErtiEs

EffEcts of radiation

Many atomic nuclei are unstable. The process by which they decay is called radioactive decay. Every decay involves the emission o one o three dierent possible radiations rom the nucleus: alpha () , beta () or gamma () .

At the molecular level, an ionization could cause damage directly to a biologically important molecule such as DNA or RNA. This could cause it to cease unctioning. Alternatively, an ionization in the surrounding medium is enough to interere with the complex chemical reactions (called metabolic pathways) taking place.

 

Molecular damage can result in a disruption to the unctions that are taking place within the cells that make up the organism. As well as potentially causing the cell to die, this could just prevent cells rom dividing and multiplying. On top o this, it could be the cause o the transormation o the cell into a malignant orm. As all body tissues are built up o cells, damage to these can result in damage to the body systems that have been aected. The non-unctioning o these systems can result in death or the animal. I malignant cells continue to grow then this is called cancer.



radiation safEty Alpha, beta and gamma all come rom the nucleus All three radiations are ionizing. This means that as they go through a substance, collisions occur which cause electrons to be removed rom atoms. Atoms that have lost or gained electrons are called ions. This ionizing property allows the radiations to be detected. It also explains their dangerous nature. When ionizations occur in biologically important molecules, such as DNA, unction can be aected.

There is no such thing as a sae dose o ionizing radiation. Any hospital procedures that result in a patient receiving an extra dose (or example having an X-ray scan) should be justiable in terms o the inormation received or the benet it gives. There are three main ways o protecting onesel rom too large a dose. These can be summarized as ollows:  Run away! The simplest method o reducing the dose received is to increase the distance between you and the source. Only electromagnetic radiation can travel large distances and this ollows an inverse square relationship with distance.  Dont waste time! I you have to receive a dose, then it is important to keep the time o this exposure to a minimum.  I you cant run away, hide behind something! Shielding can always be used to reduce the dose received. Lead-lined aprons can also be used to limit the exposure or both patient and operator.

propErtiEs of alpha, bEta and gamma radiations Property

Alpha, 

Beta, 

Gamma, 

Eect on photographic lm

Yes

Yes

Yes

Approximate number o ion pairs produced in air

1 0 4 per mm travelled

1 0 2 per mm travelled

1 per mm travelled

Typical material needed to absorb it

1 0 - 2 mm aluminium; piece o paper

A ew mm aluminium

1 0 cm lead

Penetration ability

Low

Medium

High

Typical path length in air

A ew cm

Less than one m

Eectively innite

Defection by E and B elds

Behaves like a positive charge

Behaves like a negative charge

Not defected

Speed

About 1 0 7 m s - 1

About 1 0 8 m s - 1 , very variable

3  1 08 m s- 1

naturE of alpha, bEta and gamma dEcay When a nucleus decays the mass numbers and the atomic numbers must balance on each side o the nuclear equation.  Alpha particles are helium nuclei, 42   or 42 He 2 + . In alpha decay, a chunk o the nucleus is emitted. The portion that remains will be a dierent nuclide. A Z

X 

e.g.

72

( A - 4) (Z - 2)

2 41 95

4 2

Y+ 

Am  

237 93

4 2

Np + 

The atomic numbers and the mass numbers balance on each side o the equation. (95 = 93 + 2 and 241 = 237 + 4)  Beta particles are electrons, -01  or -01 e - , emitted rom the nucleus. The explanation is that the electron is ormed when a neutron decays. At the same time, another particle is emitted called an antineutrino. 0 1

n  11 p +

0 -1

_

+

Since an antineutrino has no charge and virtually no mass it does not aect the equation.

At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s

A Z

X  ( Z + A1 ) Y +

e.g.

90 38

Sr 

90 39

0 -1

_

+

Y+

0 -1

_

+

 Gamma rays are unlike the other two radiations in that they are part o the electromagnetic spectrum. Ater their emission, the nucleus has less energy but its mass number and its atomic number have not changed. It is said to have changed rom an excited state to a lower energy state. A Z

X*  AZ X + 00 

Excited state

Lower energy state

rv 2 antimattEr The nuclear model given on page 77 is somewhat simplifed. One important thing that is not mentioned there is the existence o antimatter. Every orm o matter has its equivalent orm o antimatter. I matter and antimatter came together they would annihilate each other. Not surprisingly, antimatter is rare but it does exist. For example, another orm o radioactive decay that can take place is beta plus or positron decay. In this decay a proton decays into a neutron,

and the antimatter version o an electron, a positron, is emitted.

background radiation

Some cosmic gamma rays will be responsible, but there will also be ,  and  radiation received as a result o radioactive decays that are taking place in the surrounding materials. The pie chart below identifes typical sources o background radiation, but the actual value varies rom country to country and rom place to place.

Radioactive decay is a natural phenomenon and is going on around you all the time. The activity o any given source is measured in terms o the number o individual nuclear decays that take place in a unit o time. This inormation is quoted in becquerels (Bq) with 1 Bq = 1 nuclear decay per second. Experimentally this would be measured using a Geiger counter, which detects and counts the number o ionizations taking place inside the GM tube. A working Geiger counter will always detect some radioactive ionizations taking place even when there is no identifed radioactive source: there is a background count as a result o the background radiation. A reading o 30 counts per minute, which corresponds to the detector registering 30 ionizing events, would not be unusual. To analyse the activity o a given radioactive source, it is necessary to correct or the background radiation taking place. It would be necessary to record the background count without the radioactive source present and this value can then be subtracted rom all readings with the source present.

random dEcay Radioactive decay is a random process and is not aected by external conditions. For example, increasing the temperature o a sample o radioactive material does not aect the rate o decay. This means that is there no way o knowing whether or not a particular nucleus is going to decay within a certain period o time. All we know is the chances o a decay happening in that time. Although the process is random, the large numbers o atoms involved allows us to make some accurate predictions. I we

1 1

p  10 n +

19 10

Ne 

19 9

0 +1

+ + 

F+

0 +1

+ + 

The positron,  , emission is accompanied by a neutrino. +

The antineutrino is the antimatter orm o the neutrino. For more details see page 78.

foo d ra d on

m ed icin e n u clea r in d u stry bu ild ings/soil cosm ic

m ed icin e  14% n u clea r in d u stry  1 % bu ild in gs/soil  18 % cosm ic  14% ra d on  42 % foo d / d rin king water  11 %

n a tu ra l ra d ia tion 85 %

start with a given number o atoms then we can expect a certain number to decay in the next minute. I there were more atoms in the sample, we would expect the number decaying to be larger. On average the rate o decay o a sample is proportional to the number o atoms in the sample. This proportionality means that radioactive decay is an exponential process. The number o atoms o a certain element, N, decreases exponentially over time. Mathematically this is expressed as: dN ___  -N dt

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73

h-e

amount o su stance

ExamplE In simple situations, working out how much radioactive material remains is a matter o applying the hal-lie property several times. A common mistake is to think that i the hallie o a radioactive material is 3 days then it will all decay in six days. In reality, ater six days (two hal-lives) a hal o a hal will remain, i.e. a quarter.

number of nuclides available to decay

half-lifE There is a temptation to think that every quantity that decreases with time is an exponential decrease, but exponential curves have a particular mathematical property. In the graph shown below, the time taken or hal the number o nuclides to decay is always the same, whatever starting value we choose. This allows us to express the chances o decay happening in a property called the half-life, T__1 . The 2 hal-lie o a nuclide is the time taken or hal the number o nuclides present in a sample to decay. An equivalent statement is that the hal-lie is the time taken or the rate o decay (or activity) o a particular sample o nuclides to halve. A substance with a large hal-lie takes a long time to decay. A substance with a short hal-lie will decay quickly. Hal-lives can vary rom ractions o a second to millions o years.

T1

2

(x)

decay of radioactive parent nuclei

increase of stable daughter nuclei

after 2 half-lives 3 of the nuclei are  daughter nuclei 4 after 2 half-lives 1 of the original parent nuclei will remain 4

1 2

number = x

N0

1

The time taken to halve from any point is always T1 . 2

number = x 2

N1/2 ( 2x ) N 1/4

3

6

9

12 time / days

The decay o parent into daughter e.g. The hal-lie o

14 6

C is 5570 years.

Approximately how long is needed beore less than 1 % o a sample o 1 64 C remains?

N 1/8 N1/16 T1

2

T1

T1

2

2

T1

2

time

half-life Hal-lie o an exponential decay

Time

Fraction left

T__1

50%

2

2 T__1

25%

2

3 T__1

1 2.5%

2

4T__1

invEstigating half-lifE ExpErimEntally When measuring the activity o a source, the background rate should be subtracted.  I the hal-lie is short, then readings can be taken o activity against time.  A simple graph o activity against time would produce the normal exponential shape. Several values o hal-lie could be read rom the graph and then averaged. This method is simple and quick but not the most accurate.  A graph o ln (activity) against time could be produced. This should give a straight line and the decay constant can be calculated rom the gradient. See page 21 7.  I the hal-lie is long, then the activity will eectively be constant over a period o time. In this case one needs to fnd a way to calculate the number o nuclei present, N, and then use

5 T__1

~ 3.1 %

2

6 T__1

~ 1 .6%

2

7 T__1

~ 0.8%

2

6 hal lives

= 33420 years

7 hal lives

= 38990 years

 approximately 37000 years needed

simulation The result o the throw o a die is a random process and can be used to simulate radioactive decay. The dice represent nuclei available to decay. Each throw represents a unit o time. Every six represents a nucleus decaying meaning this die is no longer available.

dN ___ = - N. dt

74

~ 6.3%

2

At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s

ne e artificial transmutations

units

There is nothing that we can do to change the likelihood o a certain radioactive decay happening, but under certain conditions we can make nuclear reactions happen. This can be done by bombarding a nucleus with a nucleon, an alpha particle or another small nucleus. Such reactions are called artifcial transmutations. In general, the target nucleus frst captures the incoming object and then an emission takes place. The frst ever artifcial transmutation was carried out by Rutherord in 1 91 9. Nitrogen was bombarded by alpha particles and the presence o oxygen was detected spectroscopically.

Using Einsteins equation, 1 kg o mass is equivalent to 9  1 0 1 6 J o energy. This is a huge amount o energy. At the atomic scale other units o energy tend to be more useul. The electronvolt (see page 53) , or more usually, the megaelectronvolt are oten used.

4 2

He 2 + +

14 7

17 8

N

O + 11 p

The mass numbers (4 + 1 4 = 1 7 + 1 ) and the atomic numbers (2 + 7 = 8 + 1 ) on both sides o the equation must balance.

1 eV = 1 .6  1 0

-19

1 MeV = 1 .6  1 0

J

-13

J

1 u o mass converts into 931 .5 MeV Since mass and energy are equivalent it is sometimes useul to work in units that avoid having to do repeated multiplications by the (speed o light) 2 . A new possible unit or mass is thus MeV c- 2 . It works like this: I 1 MeV c- 2 worth o mass is converted you get 1 MeV worth o energy.

unifiEd mass units The individual masses involved in nuclear reactions are tiny. In order to compare atomic masses physicists oten use unifed mass units, u. These are defned in terms o the most common isotope o carbon, carbon-1 2. There are 1 2 nucleons in the carbon-1 2 atom (6 protons and 6 neutons) and one unifed mass unit is defned as exactly one twelth the mass o a carbon-1 2 atom. Essentially, the mass o a proton and the mass o a neutron are both 1 u as shown in the table below. 1 mass o a (carbon-1 2) atom = 1 .66  1 0 - 2 7 kg 1 u = ___ 12 mass* o 1 proton = 1 .007 276 u mass* o 1 neutron = 1 .008 665 u

WorkEd ExamplEs Question: How much energy would be released i 1 4 g o carbon-1 4 decayed as shown in the equation below? 14 6

C

14 7

N+

0 -1

_

+

Answer: Inormation given atomic mass o carbon-1 4 = 1 4.003242 u; atomic mass o nitrogen-1 4 = 1 4.003074 u; mass o electron = 0.000549 u mass o let-hand side = nuclear mass o

mass* o 1 electron = 0.000 549 u

= 1 3.999948 u nuclear mass o

mass dEfEct and binding EnErgy

14 7

N = 1 4.003074 - 7(0.000549) u = 1 3.999231 u

The table above shows the masses o neutrons and protons. It should be obvious that i we add together the masses o 6 protons, 6 neutrons and 6 electrons we will get a number bigger than 1 2 u, the mass o a carbon-1 2 atom. What has gone wrong? The answer becomes clear when we investigate what keeps the nucleus bound together. The dierence between the mass o a nucleus and the masses o its component nucleons is called the mass deect. I one imagined assembling a nucleus, the protons and neutrons would initially need to be brought together. Doing this takes work because the protons repel one another. Creating the bonds between the protons and neutrons releases a greater amount o energy than the work done in bringing them together. This energy released must come rom somewhere. The answer lies in Einsteins amous massenergy equivalence relationship.

E = mc2 mass in kg

C

= 1 4.003242 - 6(0.000549) u

* = Technically these are all rest masses  see option A

energy in joules

14 6

mass o right-hand side = 1 3.999231 + 0.000549 u = 1 3.999780 u mass dierence = LHS - RHS = 0.0001 68 u energy released per decay = 0.0001 68  931 .5 MeV = 0.1 56492 MeV 1 4g o C-1 4 is 1 mol  Total number o decays = NA = 6.022  1 0 2 3  Total energy release = 6.022  1 0 2 3  0.1 56492 MeV = 9.424  1 0 2 2 MeV = 9.424  1 0 2 2  1 .6  1 0 - 1 3 J = 1 .51  1 0 1 0 J  1 5 GJ

speed o light in m s - 1

In Einsteins equation, mass is another orm o energy and it is possible to convert mass directly into energy and vice versa. The binding energy is the amount o energy that is released when a nucleus is assembled rom its component nucleons. It comes rom a decrease in mass. The binding energy would also be the energy that needs to be added in order to separate a nucleus into its individual nucleons. The mass deect is thus a measure o the binding energy.

NB Many examination calculations avoid the need to consider the masses o the electrons by providing you with the nuclear mass as opposed to the atomic mass.

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75

f a  fission Fission is the name given to the nuclear reaction whereby large nuclei are induced to break up into smaller nuclei and release energy in the process. It is the reaction that is used in nuclear reactors and atomic bombs. A typical single reaction might involve bombarding a uranium nucleus with a neutron. This can cause the uranium nucleus to break up into two smaller nuclei. A typical reaction might be: 1 0

n+

235 92

U

1 41 56

Ba +

92 36

Since the one original neutron causing the reaction has resulted in the production o three neutrons, there is the possibility o a chain reaction occurring. It is technically quite difcult to get the neutrons to lose enough energy to go on and initiate urther reactions, but it is achievable.

Kr + 3 10 n + energy

Ba-141

n U-235 Kr-92 A fssion reaction

A chain reaction

fusion

binding EnErgy pEr nuclEon

Fusion is the name given to the nuclear reaction whereby small nuclei are induced to join together into larger nuclei and release energy in the process. It is the reaction that uels all stars including the Sun. A typical reaction that is taking place in the Sun is the usion o two dierent isotopes o hydrogen to produce helium.

Whenever a nuclear reaction (fssion or usion) releases energy, the products o the reaction are in a lower energy state than the reactants. Mass loss must be the source o this energy. In order to compare the energy states o dierent nuclei, physicists calculate the binding energy per nucleon. This is the total binding energy or the nucleus divided by the total number o nucleons. One o the nuclei with the largest binding energy per nucleon is iron-56, 52 66 Fe.

H + 31 H  42 He + 10 n + energy

hydrogen-2 helium-4 fusion

hydrogen-3

neutron

One o the usion reactions happening in the Sun

binding energy per nucleon / MeV

2 1

A reaction is energetically easible i the products o the reaction have a greater binding energy per nucleon when compared with the reactants.

iron-56

10 8 6 4 2

fusion energetically possible 20

40

60

80

Graph o binding energy per nucleon

76

ssion energetically possible

At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s

100 120 140 160 180 200

nucleon number

A Head f e se introduction

atomic modEl

All matter that surrounds us, living or otherwise, is made up o dierent combinations o atoms. There are only a hundred, or so, dierent types o atoms present in nature. Atoms o a single type orm an element. Each o these elements has a name and a chemical symbol; e.g. hydrogen, the simplest o all the elements, has the chemical symbol H. Oxygen has the chemical symbol O. The combination o two hydrogen atoms with one oxygen atom is called a water molecule - H2 O. The ull list o elements is shown in a periodic table. Atoms consist o a combination o three things: protons, neutrons and electrons.

The basic atomic model, known as the nuclear model, was developed during the last century and describes a very small central nucleus surrounded by electrons arranged in dierent energy levels. The nucleus itsel contains protons and neutrons (collectively called nucleons) . All o the positive charge and almost all the mass o the atom is in the nucleus. The electrons provide only a tiny bit o the mass but all o the negative charge. Overall an atom is neutral. The vast majority o the volume is nothing at all  a vacuum. The nuclear model o the atom seems so strange that there must be good evidence to support it. Protons Relative mass Charge

1 +1

nucleus

Neutrons

Electrons

1

Negligible

Neutral

-1

Electron clouds. The positions of the 6 electrons are not exactly known but they are most likely to be found in these orbitals. The dierent orbitals correspond to dierent energy levels.

10 -10 m cell

10 -14 m

DNA protons neutron nucleus Atomic model of carbon

atom

This simple model has limitations. Accelerated charges are known to radiate energy so orbital electrons should constantly lose energy (the changing direction means the electrons are accelerating) .

In the basic atomic model, we are made up of protons, neutrons, and electrons  nothing more.

EvidEncE One o the most convincing pieces o evidence or the nuclear model o the atom comes rom the RutherordGeigerMarsden experiment. Positive alpha particles were red at a thin gold lea. The relative size and velocity o the alpha particles meant that most o them were expected to travel straight through the gold lea. The idea behind this experiment was to see i there was any detectable structure within the gold atoms. The amazing discovery was that some o the alpha particles were defected through huge angles. The mathematics o the experiment showed that numbers being defected at any given angle agreed with an inverse square law o repulsion rom the nucleus. Evidence or electron energy levels comes rom emission and absorption spectra. The existence o isotopes provides evidence or neutrons.

positive nucleus

vacuum gold foil target about 10 -8 m thick

screens source of -particles

beam of -particles 

most pass straight through

detector about 1 in 8000 is repelled back

some are deviated through a large angle 

RutherordGeigerMarsden experiment

1 in 8000 particles rebound from the foil.

gold atom

stream of -particles positive -particle deected by nucleus N B no t to sca le. On ly a m in u te percentage of -p a rticles a re scattered or rebou nd . Atomic explanation o RutherordGeigerMarsden experiment

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77

de  f  e classiication o particlEs

consErvation laWs

Particle accelerator experiments identiy many, many new particles. Two original classes o particles were identifed  the leptons (= light) and the hadrons (= heavy) . Protons and neutrons are hadrons whereas electrons are leptons. The hadrons were subdivided into mesons and baryons. Protons and neutrons are baryons. Another class o particles is involved in the mediation o the interactions between the particles. These were called gauge bosons or exchange bosons.

Not all reactions between particles are possible. The study o the reactions that did take place gave rise to some experimental conservation laws that applied to particle physics. Some o these laws were simply confrmation o conservation laws that were already known to physicists  charge, momentum (linear and angular) and mass-energy. On top o these undamental laws there appeared to be other rules that were never broken e.g. the law o conservation o baryon number. I all baryons were assigned a baryon number o 1 (and all antibaryons were assigned a baryon number o -1 ) then the total number o baryons beore and ater a collision was always the same. A similar law o conservation o lepton number applies.

Particles are called elementary i they have no internal structure, that is, they are not made out o smaller constituents. The classes o elementary particles are quarks, leptons and the exchange particles. Another particle, the Higgs boson, is also an elementry particle. Combinations o elementary particles are called composite particles. All hadrons are composed o combinations o quarks. Inside all baryons there are three quarks (or three antiquarks) ; inside all mesons there is one quark and one antiquark.

Other reactions suggested new and dierent particle properties that were oten, but not always, conserved in reactions. Strangeness and charm are examples o two such properties. Strangeness is conserved in all electromagnetic and strong interactions, but not always in weak interactions. All particles, whether they are elementary or composite, can be specifed in terms o their mass and the various quantum numbers that are related to the conservation laws that have been discovered. The quantum numbers that are used to identiy particles include:  electric charge, strangeness, charm, lepton number, baryon number and colour (this property is not the same as an objects actual colour  see page 79). Every particle has its own antiparticle. An antiparticle has the same mass as its particle but all its quantum numbers (including charge, etc.) are opposite. There are some particles (e.g. the photon) that are their own antiparticle.

thE standard modEl  lEptons There are six dierent leptons and six dierent antileptons. The six leptons are considered to be in three dierent generations or amilies in exactly the same way that there are considered to be three dierent generations o quarks (see page 79) .

Similar principles are used to assign lepton numbers o +1 or -1 to the muon and the tau amily members. Lepton family number is also conserved in all reactions. For example, whenever a muon is created, an antimuon or an antimuon neutrino must also be created so that the total number o leptons in the muon amily is always conserved.

ExchangE particlEs There are only our undamental interactions that exist: Gravity, Electromagnetic, Strong and Weak. All our interactions can be thought o as being mediated by an exchange o particles. Each interaction has its own exchange particle or particles. The bigger the mass o the exchange boson, the smaller the range o the orce concerned.

Generation 1

2

3

0

e ( electronneutrino) M = 0 or almost 0

 ( muonneutrino) M = 0 or almost 0

 ( tau- neutrino) M = 0 or almost 0

-1

e ( electron) M = 0.51 1 MeV c- 2

 ( muon) M = 1 05 MeV c- 2

 ( tau) M = 1 784 MeV MeV c- 2

Lepton

The electron and the electron neutrino have a lepton (electron amily) number o +1 . The antielectron and the antielectron neutrino have a lepton (electron amily) number o -1 .

E lectric charge

The greater the mass o the exchange particle, the smaller the time or which it can exist. The range o the weak interaction is small as the masses o its exchange particles (W+ , W- and Z0 ) are large. In particle physics, all real particles can be thought o as being surrounded by a cloud o virtual particles that appear and disappear out o the surrounding vacuum. The lietime o these particles is inversely proportional to their mass. The interaction between two particles takes place when one or more o the virtual particles in one cloud is absorbed by the other particle. Interaction

Relative Range Exchange Particles strength (m) particle experience

Strong

1

Electromagnetic 1 0 - 2

~1 0 - 1 5

infnite photon

The exchange results in repulsion between the two particles From the point o view o quantum mechanics, the energy needed to create these virtual particles, E is available so long as the energy o the particle does not exist or a longer time t than is proscribed by the uncertainty principle (see page 1 26) .

78

8 dierent gluons

W ,W ,Z +

-

Weak

1 0- 1 3

~1 0 - 1 8

Gravity

1 0- 3 9

infnite graviton

Quarks, gluons Charged

0

Quarks, lepton All

Leptons and bosons are unaected by the strong orce.

At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s

Q standard modEl  Quarks

Isolated quarks cannot exist. They can exist only in twos or threes. Mesons are made rom two quarks (a quark and an antiquark) whereas baryons are made up o a combination o three quarks (either all quarks or all antiquarks) .

The standard model o particle physics is the theory that says that all matter is considered to be composed o combinations o six types o quark and six types o lepton. This is the currently accepted theory. Each o these particles is considered to be undamental, which means they do not have any deeper structure. Gravity is not explained by the standard model. All hadrons are made up rom dierent combinations o undamental particles called quarks. There are six dierent types o quark and six types o antiquark. This very neatly matches the six leptons that are also known to exist. Quarks are aected by the strong orce (see below) , whereas leptons are not. The weak interaction can change one type o quark into another.

Quarks

E lectric Generation charge 1 u 2 + __ e ( up) 3 M = 5 MeV c- 2

2 c ( charm) M = 1 500 MeV c- 2

Name o particle

Quark structure

Baryons

proton (p) neutron (n) lambda  _ antiproton ( p )

u u u __ u

Mesons

- (pi-minus) + (pi-plus) K 0 (Kz e ro )

u d __ u __d d s

u d d __ u

d d s_ d

__

The orce between quarks is still the strong interaction but the ull description o this interaction is termed QCD theory  quantum chromodynamics. The quantum dierence between the quarks is a property called colour. All quarks can be red (r) , _ _ green (g) or blue (b) . Antiquarks can be antired ( r) , antigreen ( g) __ or antiblue ( b ) . The two up quarks in a proton are not identical because they have dierent colours.

3 t ( top) M = 1 74 MeV c- 2

d s b ( down) ( strange) ( bottom) M = 1 0 MeV c- 2 M = 200 MeV c- 2 M = 4700 MeV c- 2 1 All quarks have a baryon number o + __, 3 1 All antiquarks have a baryon number o -  __ 3 All quarks have a strangeness number o 0 except the s quark that has a strangeness number o - 1 .

1 -  __ e 3

Only white (colour neutral) combinations are possible. _ _ __ Baryons must contain r, g and b quarks (or r, g, b ) whereas _ mesons contain a colour and the anticolour (e.g. r and r or b __ and b , etc.) The orce between quarks is sometimes called the colour orce. Eight dierent types o gluon mediate it. The details o QCD do not need to be recalled.

The c quark is the only quark with a charm number = +1 , all other quarks have charm number o 0.

Quantum chromodynamics (Qcd)

increases. Isolated quarks and gluons cannot be observed. I sufcient energy is supplied to a hadron in order to attempt to isolate a quark, then more hadrons or mesons will be produced rather than isolated quarks. This is known as quark confnement.

The interaction between objects with colour is called the colour interaction and is explained by a theory called quantum chromodynamics. The orce-carrying particle is called the gluon. There are eight dierent types o gluon each with zero mass. Each gluon carries a combination o colour and anticolour and their emission and absorption by dierent quarks causes the colour orce.

The six colour-changing gluons are: Gr _b , Gr _g , Gb _g , Gb _r , Gg _b , Gg _r . For example when a blue up quark emits the gluon Gb _r it loses its blue colour and becomes a red up quark (the gluon contains antired, so red colour must be let behind) . A red down quark absorbing this gluon will become a blue down quark.

As the gluons themselves are coloured, there will be a colour interaction between gluons themselves as well as between quarks. The overall eect is that they bind quarks together. The orce between quarks increases as the separation between quarks

u

r

+

b G rb

g G gb

g G0

b G bg

u p

d r

b

g

g

There are two additional colour-neutral gluons: G0 and G 0 , making a total o eight gluons.

u

r b

b G rb

g

r

G gb

g G rg

b d

g

b

r r

G gr

g

G rb r

b

strong intEraction

higgs boson

The colour interaction and the strong interaction are essentially the same thing. Properly, the colour interaction is the undamental orce that binds quarks together into baryons and mesons. It is mediated by gluons. The residual strong interaction is the orce that binds colour-neutral particles (such as the proton and neutron) together in a nucleus. The overall eect o the interactions between all the quarks in the nucleons is a short-range interaction between colour-neutral nucleons.

In addition to the three generations o leptons and quarks in the standard model there are the our classes o gauge boson and an additional highly massive boson, the Higgs boson. This was proposed in 1 964 to explain the process by which particles can acquire mass. In 201 3 scientists working with the Large Hadron Collider announced the experimental detection o a particle that that matched the standard models predictions or the Higgs boson.

The particles mediating the strong interaction can be considered to involve the exchange o composite particles ( mesons: + , - or ) whereas the undamental colour interaction is always seen as the exchange o gluons.

Ato m i c , n u c l e A r A n d pA r t i c l e p h ys i c s

79

fe  rulEs for draWing fEynman diagrams

Some simple rules help in the construction o correct diagrams:

Feynman diagrams can be used to represent possible particle interactions. The diagrams are used to calculate the overall probability o an interaction taking place. In quantum mechanics, in order to nd out the overall probability o an interaction, it is necessary to add together all the possible ways in which an interaction can take place. Used properly they are a mathematical tool or calculations but, at this level, they can be seen as a simple pictorial model o possible interactions.

 Each junction in the diagram (vertex) has an arrow going in and one going out. These will represent a leptonlepton transition or a quarkquark transition.  Quarks or leptons are solid straight lines.  Exchange particles are either wavy or broken (photons, W or Z) or curly (gluons) .  Time fows rom let to right. Arrows rom let to right represent particles travelling orward in time. Arrows rom right to let represent antiparticles travelling orward in time.  The labels or the dierent particles are shown at the end o the line.  The junctions will be linked by a line representing the exchange particle involved.

In the Feynman diagrams below the x-axis represents time going rom let to right and the y-axis represents space (some books reverse these two axes) . To view them in the alternative way, turn the page anti-clockwise by 90.

ExamplEs e-

An electron emits a photon.

e-

e-

An electron absorbs a photon.

e-



 A positron emits a photon.

e+

e+

e+

A positron absorbs a photon.

e+   

e-

A photon produces an electron and a positron (an electron- positron pair) .

e+ u d before

W-

after

e-

Beta decay. A down quark changes into an up quark with the emission o a Wparticle. This decays into an e electron and an antineutrino. The top vertex involves quarks, the bottom vertex einvolves leptons.

An electron and positron annihilate to produce two photons.



An electron and a positron meet and annihilate (disappear) , producing a  photon.

ee+ d

W+

+

+

u

Pion decay. The quark and antiquark annihilate to produce a W+ particle. This decays into an antimuon and a muon neutrino.

u

u g

n

p

Simple diagrams can also be drawn with exchanges between e - hadrons.

W-

p

An up quark (in a proton) emits a gluon which in turn transorms +  into a down/antidown quark pair. This reaction could take place as a result o a protonproton collision: p + p  p + n + + . A  mediates the strong nuclear orce between a proton and a neutron in a nucleus.

p 

Beta decay (hadron version)

e

usEs of fEynman diagrams Once a possible interaction has been identied with a Feynman diagram, it is possible to use it to calculate the probabilities or certain undamental processes to take place. Each line and vertex corresponds to a mathematical term. By adding together all the terms, the probability o the interaction can be calculated using the diagram. More complicated diagrams with the same overall outcome need to be considered in order to calculate the overall

80

d d



e+



n

n

probability o a chosen outcome. The more diagrams that are included in the calculation, the more accurate the answer. In a Feynman diagram, lines entering or leaving the diagram represent real particles and must obey mass, energy and momentum relationships. Lines in intermediate stages in the diagram represent virtual particles and do not have to obey energy conservation providing they exist or a short enough time or the uncertainty relationship to apply. Such virtual particles cannot be detected.

At o m i c , n u c l e A r A n d pA r t i c l e p h ys i c s

ib Questons  atomc, nuclear and partcle physcs 1.

A sample o radioactive material contains the element Ra 226. The hal-lie o Ra 226 can be dened as the time it takes or

(ii) Explain why it is necessary to give the deuterons a certain minimum kinetic energy beore they can react with the magnesium nuclei.

A. the mass o the sample to all to hal its original value. B. hal the number o atoms o Ra 226 in the sample to decay. C. hal the number o atoms in the sample to decay. D. the volume o the sample to all to hal its original value. 2.

Oxygen-1 5 decays to nitrogen-1 5 with a hal-lie o approximately 2 minutes. A pure sample o oxygen-1 5, with a mass o 1 00 g, is placed in an airtight container. Ater 4 minutes, the masses o oxygen and nitrogen in the container will be Mass of oxygen

3.

4.

Mass of nitrogen

A. 0 g

1 00 g

B. 25 g

25 g

C. 50 g

50 g

D. 25 g

75 g

B. , , 

C. , 

D. , , 

a) Carbon-1 4 decays by beta-minus emission to nitrogen-1 4. Write the equation or this decay. [2]

(i) Explain why the beta activity rom the bowl diminishes with time, even though the probability o decay o any individual carbon-1 4 nucleus is constant. [3] [3]

1 0. This question is about a nuclear ssion reactor or providing electrical power.

In the Rutherord scattering experiment, a stream o  particles is red at a thin gold oil. Most o the  particles

In a nuclear reactor, power is to be generated by the ssion o uranium-235. The absorption o a neutron by 2 3 5 U results in the splitting o the nucleus into two smaller nuclei plus a number o neutrons and the release o energy. The splitting can occur in many ways; or example

B. rebound. C. are scattered uniormly. D. go through the oil.

n+

A piece o radioactive material now has about 1 /1 6 o its previous activity. I the hal-lie is 4 hours the dierence in time between measurements is approximately

235 92

U

90 38

Sr +

1 43 54

Xe + neutrons + energy

a) The nuclear fssion reaction (i) How many neutrons are produced in this reaction?

A. 8 hours.

[1 ]

(ii) Explain why the release o several neutrons in each reaction is crucial or the operation o a ssion reactor. [2]

B. 1 6 hours. C. 32 hours. D. 60 hours. 6.

The carbon in trees is mostly carbon-1 2, which is stable, but there is also a small proportion o carbon-1 4, which is radioactive. When a tree is cut down, the carbon-1 4 present in the wood at that time decays with a hal-lie o 5,800 years.

(ii) Calculate the approximate age o the wooden bowl.

A. are scattered randomly.

5.

[2]

Radioactive carbon dating

b) For an old wooden bowl rom an archaeological site, the average count-rate o beta particles detected per kg o carbon is 1 3 counts per minute. The corresponding count rate rom newly cut wood is 52 counts per minute.

A radioactive nuclide Z X undergoes a sequence o radioactive decays to orm a new nuclide Z + 2 Y. The sequence o emitted radiations could be A. , 

9.

(iii) The sum o the rest masses o the uranium plus neutron beore the reaction is 0.22 u greater than the sum o the rest masses o the ssion products. What becomes o this missing mass? [1 ]

a) Use the standard model to describe, in terms o undamental particles, the internal structure o: (i) A proton

(iv) Show that the energy released in the above ssion reaction is about 200 MeV.

(ii) An electron (iii) Baryons

[2]

b) A nuclear fssion power station

(iv) Mesons

(i)

b) Draw Feynman diagram or  + decay. 7.

A proton undergoes a strong interaction with a  - particle (quark content:  ud) to produce a neutron and another particle. Use conservation laws to deduce the structure o the particle produced in this reaction.

8.

a) Two properties o the isotope o uranium, (i) it decays radioactively (to

234 90

238 92

U are:

Th)

(ii) it reacts chemically (e.g. with fuorine to orm UF 6 ) . What eatures o the structure o uranium atoms are responsible or these two widely dierent properties?

Suppose a nuclear ssion power station generates electrical power at 550 MW. Estimate the minimum number o ssion reactions occurring each second in the reactor, stating any assumption you have made about eciency. [4]

1 1 . Which o the ollowing is a correct list o particles upon which the strong nuclear orce may act? A. protons and neutrons

B. protons and electrons

C. neutrons and electrons

D. protons, neutrons and electrons

[2]

2 1

b) A beam o deuterons (deuterium nuclei, H) are accelerated through a potential dierence and are then incident on a magnesium target ( 21 62 Mg) . A nuclear reaction occurs resulting in the production o a sodium nucleus and an alpha particle. (i) Write a balanced nuclear equation or this reaction.

[2]

i B Q u e s t i o n s  Ato m i c , n u c l e A r A n d pA r t i c l e p h ys i c s

81

8 EN ERGY PRODUCTI ON Energy nd poer genertion  snkey digrm ENERGY CONvERsIONs

ElECTRICal POwER PRODUCTION

The production o electrical power around the world is achieved using a variety o dierent systems, oten starting with the release o thermal energy rom a uel. In principle, thermal energy can be completely converted to work in a single process, but the continuous conversion o this energy into work implies the use o machines that are continuously repeating their actions in a xed cycle. Any cyclical process must involve the transer o some energy rom the system to the surroundings that is no longer available to perorm useul work. This unavailable energy is known as degraded energy, in accordance with the principle o the second law o thermodynamics (see page 1 62) .

In all electrical power stations the process is essentially the same. A uel is used to release thermal energy. This thermal energy is used to boil water to make steam. The steam is used to turn turbines and the motion o the turbines is used to generate electrical energy. Transormers alter the potential dierence (see page 1 1 4) .

useful electrical output energy in from fuel

Energy conversions are represented using Sankey diagrams. An arrow (drawn rom let to right) represents the energy changes taking place. The width o the arrow represents the power or energy involved at a given stage. Created or degraded energy is shown with an arrow up or down.

heating and sound in transformers

friction and heating losses cooling tower losses (condenser)

Note that Sankey diagrams are to scale. The width o the useul electrical output in the diagram on the right is 2.0 mm compared with 1 2.0 mm or the width o the total energy rom the uel. This represents an overall eciency o 1 6.7%.

Sankey diagram representing the energy fow in a typical power station

steam

fuel (coal) turbine to transformer

generator boiler

water condenser

Electrical energy generation

POwER Power is dened as the rate at which energy is converted. The units o power are J S - 1 or W. energy Power = _ time

82

EN ERG Y PRO D U CTI O N

Priry energy ource RENEwablE / NON-RENEwablE ENERGY sOURCEs The law o conservation o energy states that energy is neither created nor destroyed, it just changes orm. As ar as human societies are concerned, i we wish to use devices that require the input o energy, we need to identiy sources o energy. Renewable sources o energy are those that cannot be used up, whereas non-renewable sources o energy can be used up and eventually run out. Renewable sources

Non-renewable sources

hydroelectric

coal

photovoltaic cells

oil

active solar heaters

natural gas

wind

nuclear

 It is possible or a uel to be managed in a renewable or a non-renewable way. For example, i trees are cut down as a source o wood to burn then this is clearly non-renewable. It is, however, possible to replant trees at the same rate as they are cut down. I this is properly managed, it could be a renewable source o energy. O course these possible sources must have got their energy rom somewhere in the rst place. Most o the energy used by humans can be traced back to energy radiated rom the Sun, but not quite all o it. Possible sources are:  the Suns radiated energy  gravitational energy o the Sun and the Moon

biouels

 nuclear energy stored within atoms

Sometimes the sources are hard to classiy so care needs to be taken when deciding whether a source is renewable or not. One point that sometimes worries students is that the Sun will eventually run out as a source o energy or the Earth, so no source is perectly renewable! This is true, but all o these sources are considered rom the point o view o lie on Earth. When the Sun runs out, then so will lie on Earth. Other things to keep in mind include:  Nuclear sources (both ssion and usion) consume a material as their source so they must be non-renewable.

sPECIfIC ENERGY aND ENERGY DENsITY

On the other hand, the supply available can make the source eectively renewable (usion) .

 the Earths internal heat energy. Although you might think that there are other sources o energy, the above list is complete. Many everyday sources o energy (such as coal or oil) can be shown to have derived their energy rom the Suns radiated energy. On the industrial scale, electrical energy needs to be generated rom another source. When you plug anything electrical into the mains electricity you have to pay the electricity-generating company or the energy you use. In order to provide you with this energy, the company must be using one (or more) o the original list o sources.

COmPaRIsON Of ENERGY sOURCEs Fuel

Renewable?

Two quantities are useul to consider when making comparisons between dierent energy sources  the specifc energy and the energy density. Specic energy provides a useul comparison between uels and is dened as the energy liberated per unit mass o uel consumed. Specic energy is measured in J kg- 1 specic energy energy released rom uel = ___ mass o uel consumed Fuel choice can be particularly infuenced by specic energy when the uel needs to be transported: the greater the mass o uel that needs to be transported, the greater the cost. Energy density is dened as the energy liberated per unit volume o uel consumed. The unit is J m- 3 energy density

CO 2 emission

Specifc energy(MJ kg - 1 ) (values vary depending on type)

Energy density (MJ m - 3 )

Coal

No

Yes

2233

23,000

Oil

No

Yes

42

36,500

Gas

No

Yes

54

37

Nuclear (uranium)

No

No

8.3  1 0 7

1 .5  1 0 1 2

Waste

No

Yes

10

variable n/a

Solar

Yes

No

n/a

Wind

Yes

No

n/a

n/a

Hydro  water stored in dams

Yes

No

n/a

n/a

Tidal

Yes

No

n/a

n/a

Pumped storage

n/a

No

n/a

n/a

Wave

Yes

No

n/a

n/a

Geothermal

Yes

No

n/a

n/a

Biouels e.g. ethanol

Some types

Yes

30

21 ,000

energy release rom uel = ___ volume o uel consumed

EN ERGY PRO D U CTI O N

83

foi ue poer production ORIGIN Of fOssIl fUEl Coal, oil and natural gas are known as fossil fuels. These uels have been produced over a timescale that involves tens or hundreds o millions o years rom accumulations o dead matter. This matter has been converted into ossil uels by exposure to the very high temperatures and pressure that exist beneath the Earths surace. Coal is ormed rom the dead plant matter that used to grow in swamps. Layer upon layer o decaying matter decomposed.

As it was buried by more plant matter and other substances, the material became more compressed. Over the geological timescale this turned into coal. Oil is ormed in a similar manner rom the remains o microscopic marine lie. The compression took place under the sea. Natural gas, as well as occurring in underground pockets, can be obtained as a by-product during the production o oil. It is also possible to manuacture gas rom coal.

ENERGY TRaNsfORmaTIONs Fossil uel power stations release energy in uel by burning it. The thermal energy is then used to convert water into steam that once again can be used to turn turbines. Since all ossil uels were originally living matter, the original source o this energy was the Sun. For example, millions o years ago energy radiated rom the Sun was converted (by photosynthesis) into living plant matter. Some o this matter has eventually been converted into coal.

solar energy

photosynthesis

chemical energy in plants

chemical energy in fossil fuels

compression

Energy storage in ossil uels

ExamPlE

EffICIENCY Of fOssIl fUEl POwER sTaTIONs

Use the data on this page and the previous page to calculate the typical rate (in tonnes per hour) at which coal must be supplied to a 500 MW coal fred power station.

The efciency o dierent power stations depends on the design. At the time o publishing, the ollowing fgures apply.

Answer Electrical power supply

= 500 MW = 5  1 0 J s

Power released rom uel

= 5  1 0 8 / efciency = 5  1 0 8 / 0.35 = 1 .43  1 0 9 J s - 1

8

Rate o consumption o coal = 1 .43  1 0 / 3.3  1 0 kg s 9

Fossil uel

Typical efciency

Current maximum efciency

Coal

35%

42%

-1

7

-1

= 43.3 kg s - 1 = 43.3  60  60 kg hr- 1

Natural gas

45%

52%

Oil

38%

45%

Note that thermodynamic considerations limit the maximum achievable efciency (see page 1 63) .

= 1 .56  1 0 5 kg hr- 1  1 60 tonnes hr- 1

aDvaNTaGEs aND DIsaDvaNTaGEs Advantages

Disadvantages

 Very high specifc energy and energy density  a great deal o energy is released rom a small mass o ossil uel.

 Combustion products can produce pollution, notably acid rain.

 Fossil uels are relatively easy to transport.

 Combustion products contain greenhouse gases.

 Still cheap when compared to other sources o energy.

 Extraction o ossil uels can damage the environment.

 Power stations can be built anywhere with good transport links and water availability.

 Non-renewable.

 Can be used directly in the home to provide heating.

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 Coal-fred power stations need large amounts o uel.

Nucer power  proce PRINCIPlEs Of ENERGY PRODUCTION Many nuclear power stations use uranium-235 as the uel. This uel is not burned  the release o energy is achieved using a fssion reaction. An overview o this process is described on page 76. In each individual reaction, an incoming neutron causes a uranium nucleus to split apart. The ragments are moving ast. In other words the temperature is very high. Among the ragments are more neutrons. I these neutrons go on to initiate urther reactions then a chain reaction is created.

The design o a nuclear reactor needs to ensure that, on average, only one neutron rom each reaction goes on to initiate a urther reaction. I more reactions took place then the number o reactions would increase all the time and the chain reaction would run out o control. I ewer reactions took place, then the number o reactions would be decreasing and the fssion process would soon stop. The chance that a given neutron goes on to cause a fssion reaction depends on several actors. Two important ones are:  the number o potential nuclei in the way  the speed (or the energy) o the neutrons. As a general trend, as the size o a block o uel increases so do the chances o a neutron causing a urther reaction (beore it is lost rom the surace o the block). As the uel is assembled together a stage is reached when a chain reaction can occur. This happens when a so-called critical mass o uel has been assembled. The exact value o the critical mass depends on the exact nature o the uel being used and the shape o the assembly. There are particular neutron energies that make them more likely to cause nuclear fssion. In general, the neutrons created by the fssion process are moving too ast to make reactions likely. Beore they can cause urther reactions the neutrons have to be slowed down.

mODERaTOR, CONTROl RODs aND hEaT ExChaNGER Three important components in the design o all nuclear reactors are the moderator, the control rods and the heat exchanger.

core

coolant

steam turbine kinetic energy

nuclear energy

electrical energy

thermal energy

distributors to electricity consumers

thermal energy losses

 Collisions between the neutrons and the nuclei o the moderator slow them down and allow urther reactions to take place.  The control rods are movable rods that readily absorb neutrons. They can be introduced or removed rom the reaction chamber in order to control the chain reaction.  The heat exchanger allows the nuclear reactions to occur in a place that is sealed o rom the rest o the environment. The reactions increase the temperature in the core. This thermal energy is transerred to heat water and the steam that is produced turns the turbines.

generators

concrete shields

control rods (moveable) pressurizer

moderator

A general design or one type o nuclear reactor (PWR or pressurized water reactor) is shown here. It uses water as the moderator and as a coolant.

HOT WATER

aDvaNTaGEs aND DIsaDvaNTaGEs

steam to turbines

secondary coolant circuit

Advantages  Extremely high specifc energy  a great deal o energy is released rom a very small mass o uranium.  Reserves o uranium large compared to oil.

heat exchange steel pressure vessel

pump

fuel rods

Disadvantages  Process produces radioactive nuclear waste that is currently just stored.

pump

primary coolant circuit

 Larger possible risk i anything should go wrong.  Non-renewable (but should last a long time) .

Pressurized water nuclear reactor (PWR)

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85

Nucer poer  ety nd ri ENRIChmENT aND REPROCEssING

NUClEaR wEaPONs

Naturally occurring uranium contains less than 1 % o uranium-235. Enrichment is the process by which this percentage composition is increased to make nuclear fssion more likely.

A nuclear power station involves controlled nuclear fssion whereas an uncontrolled nuclear fssion produces the huge amount o energy released in nuclear weapons. Weapons have been designed using both uranium and plutonium as the uel. Issues associated with nuclear weapons include:

In addition to uranium-235, plutonium-239 is also capable o sustaining fssion reactions. This nuclide is ormed as a by-product o a conventional nuclear reactor. A uranium-238 nucleus can capture ast-moving neutrons to orm uranium-239. This undergoes -decay to neptunium-239 which undergoes urther -decay to plutonium-239: 238 92 239 92

U + 10 n 

U

239 93

239 93

Np 

239 92

Np +

239 94

U 0 -1

Pu +

_

+

0 -1

_

+

Reprocessing involves treating used uel waste rom nuclear reactors to recover uranium and plutonium and to deal with other waste products. A ast breeder reactor is one design that utilizes plutonium-239.

hEalTh, safETY aND RIsk Issues associated with the use o nuclear power stations or generation o electrical energy include:  I the control rods were all removed, the reaction would rapidly increase its rate o production. Completely uncontrolled nuclear fssion would cause an explosion and thermal meltdown o the core. The radioactive material in the reactor could be distributed around the surrounding area causing many atalities. Some argue that the terrible scale o such a disaster means that the use o nuclear energy is a risk not worth taking. Nuclear power stations could be targets or terrorist attacks.  The reaction produces radioactive nuclear waste. While much o this waste is o a low level risk and will radioactively decay within decades, a signifcant amount o material is produced which will remain dangerously radioactive or millions o years. The current solution is to bury this waste in geologically secure sites.  The uranium uel is mined rom underground and any mining operation involves signifcant risk. The ore is also radioactive so extra precautions are necessary to protect the workers involved in uranium mines.  The transportation o the uranium rom the mine to a power station and o the waste rom the nuclear power station to the reprocessing plant needs to be secure and sae.  By-products o the civilian use o nuclear power can be used to produce nuclear weapons.

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 Moral issues associated with any weapon o aggression that is associated with warare. Nuclear weapons have such destructive capability that since the Second World War the threat o their deployment has been used as a deterrent to prevent non-nuclear aggressive acts against the possessors o nuclear capability.  The unimaginable consequences o a nuclear war have orced many countries to agree to non-prolieration treaties, which attempt to limit nuclear power technologies to a small number o nations.  A by-product o the peaceul use o uranium or energy production is the creation o plutonium-239 which could be used or the production o nuclear weapons. Is it right or the small number o countries that already have nuclear capability to prevent other countries rom acquiring that knowledge?

fUsION REaCTORs Fusion reactors oer the theoretical potential o signifcant power generation without many o the problems associated with current nuclear fssion reactors. The uel used, hydrogen, is in plentiul supply and the reaction (i it could be sustained) would not produce signifcant amounts o radioactive waste. The reaction is the same as takes place in the Sun (as outlined on page 76) and requires creating temperatures high enough to ionize atomic hydrogen into a plasma state (this is the ourth state o matter, in which electrons and protons are not bound in atoms but move independently) . Currently the principal design challenges are associated with maintaining and confning the plasma at sufciently high temperature and density or usion to take place.

sor poer nd ydroeectric poer sOlaR POwER (TwO TYPEs)

solar radiation

There are two ways o harnessing the radiated energy that arrives at the Earths surace rom the Sun. A photovoltaic cell (otherwise known as a solar cell or photocell) converts a portion o the radiated energy directly into a potential dierence (voltage) . It uses a piece o semiconductor to do this. Unortunately, a typical photovoltaic cell produces a very small voltage and it is not able to provide much current. They are used to run electrical devices that do not require a great deal o energy. Using them in series would generate higher voltages and several in parallel can provide a higher current.

glass/plastic cover warmer water out

cold water in

solar radiation slices of semiconductor

copper pipe (black)

solar energy metal layer

output voltage

reective insulator behind pipe

active solar heater

thermal energy in water

aDvaNTaGEs aND DIsaDvaNTaGEs Advantages  Very clean production  no harmul chemical by-products.

solar energy

photocell

 Renewable source o energy.

electrical energy

 Source o energy is ree.

Disadvantages  Can only be utilized during the day.

An active solar heater (otherwise known as a solar panel) is designed to capture as much thermal energy as possible. The hot water that it typically produces can be used domestically and would save on the use o electrical energy.

 A very large area would be needed or a signicant amount o energy.

 Source o energy is unreliable  could be a cloudy day.

hYDROElECTRIC POwER

aDvaNTaGEs aND DIsaDvaNTaGEs

The source o energy in a hydroelectric power station is the gravitational potential energy o water. I water is allowed to move downhill, the fowing water can be used to generate electrical energy.

Advantages

The water can gain its gravitational potential energy in several ways.

 Renewable source o energy.

 As part o the water cycle, water can all as rain. It can be stored in large reservoirs as high up as is easible.

 Source o energy is ree.

 Tidal power schemes trap water at high tides and release it during a low tide.  Water can be pumped rom a low reservoir to a high reservoir. Although the energy used to do this pumping must be more than the energy regained when the water fows back down hill, this pumped storage system provides one o the ew large-scale methods o storing energy.

 Very clean production  no harmul chemical by-products.

Disadvantages  Can only be utilized in particular areas.  Construction o dams will involve land being submerged under water.

energy lost due to friction throughout gravitational PE of water

KE of water

+

KE of turbines

electrical energy

EN ERGY PRO D U CTI O N

87

wind poer nd oter tecnoogie ENERGY TRaNsfORmaTIONs

maThEmaTICs density of air 

There is a great deal o kinetic energy involved in the winds that blow around the Earth. The original source o this energy is, o course, the Sun. Dierent parts o the atmosphere are heated to dierent temperatures. The temperature dierences cause pressure dierences, due to hot air rising or cold air sinking, and thus air fows as a result.

r wind speed 

blades turn wind

The area swept out by the blades o the turbine = A = r 2 In one second the volume o air that passes the turbine = v A

heating Earth solar energy

KE of wind

energy lost due to friction electric energy throughout

KE of turbine

So mass o air that passes the turbine in one second = v A 1 mv2 Kinetic energy m available per second = _ 2 1 _ = (vA) v2 2 1 _ = Av3 2 1 _ 3 In other words, power available = Av 2 In practice, the kinetic energy o the incoming wind is easy to calculate, but it cannot all be harnessed as the air must continue to move  in other words the wind turbine cannot be one hundred per cent ecient. A doubling o the wind speed would mean that the available power would increase by a actor o eight.

aDvaNTaGEs aND DIsaDvaNTaGEs Disadvantages Advantages  Very clean production  no harmul chemical by-products.  Renewable source o energy.

 Some consider large wind generators to spoil the countryside.

 Source o energy is unreliable  could be a day without wind.

 Can be noisy.

 A very large area would need be covered or a signicant amount o energy.

 Best positions or wind generators are oten ar rom centres o population.

 Source o energy is ree.

sECONDaRY ENERGY sOURCEs By ar the most common primary energy sources in use worldwide are the three main ossil uels: oil, coal and natural gas. With the inclusion o uranium, at the time o writing this guide, this accounts or 90% o the worlds energy consumption. Other primary uels include the renewables: solar, wind, tidal, biomass and geothermal. With global energy demand expected to rise in the uture, the hope is that developments with renewable energy can help to reduce the dependence on ossil uels. Primary energy sources are not convenient or individual users and typically a conversion process takes place that results in a

NEw aND DEvElOPING TEChNOlOGIEs It is impossible to predict technological developments that are going to take place over the coming years. Current models, however, predict a continuing dependence on the use o ossil uels or many years to come. The hope is that we will be able to decrease this dependency over time. It is important to be

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EN ERG Y PRO D U CTI O N

secondary energy source that can be widely used in society. The most common secondary sources are electrical energy (a very versatile secondary source) or rened uels (e.g. petrol) . The storage o electrical energy is a challenge, with everyday devices (e.g. batteries or capacitors) having a very limited capability when compared with typical everyday demands. Power companies need to vary the generation o electrical energy to match consumer demand. Currently pumped storage hydroelectric systems are the only viable large-scale method o storing spare electrical energy capacity or uture use. The eciency o a typical system is approximately 75% meaning that one quarter o the energy supplied is wasted.

aware o the development o new technologies particularly those associated with:  renewable energy sources  improving the eciency o our energy conversion process.

Ter energy trner PROCEssEs Of ThERmal ENERGY TRaNsfER

CONvECTION

There are several processes by which the transer o thermal energy rom a hot object to a cold object can be achieved. Three very important processes are called conduction, convection and radiation. Any given practical situation probably involves more than one o these processes happening at the same time. There is a ourth process called evaporation. This involves the aster moving molecules leaving the surace o a liquid that is below its boiling point. Evaporation causes cooling.

In convection, thermal energy moves between two points because o a bulk movement o matter. This can only take place in a fuid (a liquid or a gas). When part o the fuid is heated it tends to expand and thus its density is reduced. The colder fuid sinks and the hotter fuid rises up. Central heating causes a room to warm up because a convection current is set up as shown below.

Cool air is denser and sinks downwards.

Hot air is less dense and is forced upwards.

CONDUCTION In thermal conduction, thermal energy is transerred along a substance without any bulk (overall) movement o the substance. For example, one end o a metal spoon soon eels hot i the other end is placed in a hot cup o tea. Conduction is the process by which kinetic energy is passed rom molecule to molecule.

macroscopic view

Convection in a room

HOT

COLD

thermal energy

thermal energy

RESERVOIR

RESERVOIR

Thermal energy ows along the material as a result of the temperature dierence across its ends. microscopic view

HOT COLD The faster-moving molecules at the hot end pass on their kinetic energy to the slower-moving molecules as a result of intermolecular collisions. Points to note:  Poor conductors are called thermal insulators.  Metals tend to be very good thermal conductors. This is because a dierent mechanism (involving the electrons) allows quick transer o thermal energy.  All gases (and most liquids) tend to be poor conductors. Examples:  Most clothes keep us warm by trapping layers o air  a poor conductor.  I one walks around a house in bare eet, the foors that are better conductors (e.g. tiles) will eel colder than the foors that are good insulators (e.g. carpets) even i they are at the same temperature. (For the same reason, on a cold day a piece o metal eels colder than a piece o wood.)  When used or cooking ood, saucepans conduct thermal energy rom the source o heat to the ood.

ExamPlE cork  a poor conductor hot liquid surfaces silvered so as to reduce radiation air gap (poor conductor) A thermos fask prevents heat loss

The ow of air around a room Air is warmed is called a convection current. by the heater.

outer plastic cover partial vacuum between glass walls to prevent convection and conduction insulating space

Points to note:  Convection cannot take place in a solid. Examples:  The pilots o gliders (and many birds) use naturally occurring convection currents in order to stay above the ground.  Sea breezes (winds) are oten due to convection. During the day the land is hotter than the sea. This means hot air will rise rom above the land and there will be a breeze onto the shore. During the night, the situation is reversed.  Lighting a re in a chimney will mean that a breeze fows in the room towards the re.

RaDIaTION Matter is not involved in the transer o thermal energy by radiation. All objects (that have a temperature above zero kelvin) radiate electromagnetic waves. I you hold your hand up to a re to eel the heat, your hands are receiving the radiation. For most everyday objects this radiation is in the inra-red part o the electromagnetic spectrum. For more details o the electromagnetic spectrum, see page 37.

HOT OBJECT Electromagnetic radiation is given o from all surfaces.

Points to note:  An object at room temperature absorbs and radiates energy. I it is at constant temperature (and not changing state) then the rates are the same.  A surace that is a good radiator is also a good absorber.  Suraces that are light in colour and smooth (shiny) are poor radiators (and poor absorbers) .  Suraces that are dark and rough are good radiators (and good absorbers) .  I the temperature o an object is increased then the requency o the radiation increases. The total rate at which energy is radiated will also increase.  Radiation can travel through a vacuum (space) . Examples:  The Sun warms the Earths surace by radiation.  Clothes in summer tend to be white  so as not to absorb the radiation rom the Sun.

EN ERGY PRO D U CTI O N

89

Rdition: wien  nd the stenbotnn  blaCk-bODY RaDIaTION: sTEfaN-bOlTzmaNN law

wIENs law

In general, the radiation given out rom a hot object depends on many things. It is possible to come up with a theoretical model or the perect emitter o radiation. The perect emitter will also be a perect absorber o radiation  a black object absorbs all o the light energy alling on it. For this reason the radiation rom a theoretical perect emitter is known as black-body radiation.

Wiens displacement law relates the wavelength at which the intensity o the radiation is a maximum  m a x to the temperature o the black body T. This states that

intensity / arbitrary units

5

2.90  1 0 _ -3

 m a x (metres) =

intensity

Black-body radiation does not depend on the nature o the emitting surace, but it does depend upon its temperature. At any given temperature there will be a range o dierent wavelengths (and hence requencies) o radiation that are emitted. Some wavelengths will be more intense than others. This variation is shown in the graph below.

 m a x T = constant The value o the constant can be ound by experiment. It is 2.9  1 0 3 m K. It should be noted that in order to use this constant, the wavelength should be substituted into the equation in metres and the temperature in kelvin. T(kelvin)

The peak wavelength rom the Sun is approximately 500 nm.  m a x = 500 nm = 5  1 0- 7 m

6000 K

2.9  1 0 _ K -3

4 3

so T =

visible region

5  1 0- 7

= 5800 K

5000 K

2

wavelength / nm 1

 max = 5 0 0 n m

4000 K 3000 K

0 1200

800

violet blue green yellow orange red

400

1600 wavelength / nm

To be absolutely precise, it is not correct to label the y-axis on the above graph as the intensity, but this is oten done. It is actually something that could be called the intensity unction. This is dened so that the area under the graph (between two wavelengths) gives the intensity emitted in that wavelength range. The total area under the graph is thus a measure o the total power radiated. The power radiated by a Black-body (See page 1 95) is given by: Surface area in m 2 absolute temperature in kelvins

We can analyse light rom a star and calculate a value or its surace temperature. This will be much less than the temperature in the core. Hot stars will give out all requencies o visible light and so will tend to appear white in colour. Cooler stars might well only give out the higher wavelengths (lower requencies) o visible light  they will appear red. Radiation emitted rom planets will peak in the inra-red.

INTENsITY, I The intensity o radiation is the power per unit area that is received by the object. The unit is W m- 2 . Power I = _. A

P = AT4 Total power radiated in W

Stefan-Boltzmann constant

Although stars and planets are not perect emitters, their radiation spectrum is approximately the same as black-body radiation.

EqUIlIbRIUm aND EmIssIvITY I the temperature o a planet is constant, then the power being absorbed by the planet must equal the rate at which energy is being radiated into space. The planet is in thermal equilibrium. I it absorbs more energy than it radiates, then the temperature must go up and i the rate o loss o energy is greater than its rate o absorption then its temperature must go down. In order to estimate the power absorbed or emitted, the ollowing concepts are useul.

Emissivity The Earth and its atmosphere are not a perect black body. Emissivity, e, is dened as the ratio o power radiated per unit area by an object to the power radiated per unit area by a black body at the same temperature. It is a ratio and so has no units.

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EN ERG Y PRO D U CTI O N

power radiated by object per unit area e = ____________________________________________ power radiated per unit area by black body at same temperature

thus p = e A T 4

albEDO Some o the radiation received by a planet is refected straight back into space. The raction that is refected back is called the albedo, . The Earths albedo varies daily and is dependent on season (cloud ormations) and latitude. Oceans have a low value but snow has a high value. The global annual mean albedo is 0.3 (30% ) on Earth. total scattered power albedo = __ total incident power

sor power sOlaR CONsTaNT The amount o power that arrives rom the Sun is measured by the solar constant. It is properly defned as the amount o solar energy that alls per second on an area o 1 m2 above the Earths atmosphere that is at right angles to the Suns rays. Its average value is about 1 400 W m- 2 . This is not the same as the power that arrives on 1 m2 o the Earths surace. Scattering and absorption in the atmosphere means that oten less than hal o this arrives at the Earths surace. The amount that arrives depends greatly on the weather conditions.

1% absorbed in stratosphere

stratosphere 15 km

incoming solar radiation 100% NB These gures are only guidelines because gures vary with cloud cover, water vapour, etc.

tropopause

troposphere

clouds reect 23% 24% absorbed in troposphere clouds absorb 3%

4% reected from the Earths surface 24% direct

surface of the Earth

21% diuse

45% reaches Earths surface Fate o incoming radiation Dierent parts o the Earths surace (regions at dierent latitudes) will receive dierent amounts o solar radiation. The amount received will also vary with the seasons since this will aect how spread out the rays have become.

atmosphere is a near-uniform thickness all around the Earth

re

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he

rfa

Tro p

' s su

o sp

M

North Pole

Ea rth

a tm

ed ge o f

23.5

ic of

Eq u a Tro

pi c

of C

MN > PQ RS > TU R N

S 60

C anc

er

30

to r

a p ri c

incoming solar radiation travelling in near parallel lines

0 TP U Q

o rn

South Pole

30 Radiation has to travel through a greater depth of atmosphere ( RS as compared with TU) in high latitudes. When it reaches the surface the radiation is also spread out over a greater area (MN as compared with PQ) than in lower latitudes.

60

90

The eect o latitude on incoming solar radiation Tropic of Cancer

SUN

Summer in northern hemisphere

Tropic of Capricorn

Summer in southern hemisphere

The Earths orbit and the seasons

EN ERGY PRO D U CTI O N

91

Te greenoue efect PhYsICal PROCEssEs

A

Short wavelength radiation is received rom the Sun and causes the surace o the Earth to warm up. The Earth will emit inra-red radiation (longer wavelengths than the radiation coming rom the Sun) because the Earth is cooler than the Sun. Some o this inra-red radiation is absorbed by gases in the atmosphere and re-radiated in all directions. This is known as the greenhouse eect and the gases in the atmosphere that absorb inra-red radiation are called greenhouse gases. The net eect is that the upper atmosphere and the surace o the Earth are warmed. The name is potentially conusing, as real greenhouses are warm as a result o a dierent mechanism. The temperature o the Earths surace will be constant i the rate at which it radiates energy equals the rate at which it absorbs energy. The greenhouse eect is a natural process and without it the temperature o the Earth would be much lower; the average temperature o the Moon is more than 30 C colder than the Earth.

SUN

S P H E R E Some solar radiation is Some of the infra-red reected by the atmosphere radiation passes through and Earths surface the atmosphere and is lost in space Outgoing solar radiation: Solar radiation passes 103 W m - 2 through the clear atmosphere. Net outgoing infra-red Incoming solar radiation: radiation: 240 W m - 2 -2 343 W m

Net incoming solar radiation: 240 W m - 2 Solar energy absorbed by atmosphere: 72 W m - 2

M

O

G R E E N H O U S E G A S E S Some of the infra-red radiation is absorbed and re-emitted by the greenhouse gas molecules. The direct eect is the warming of the Earths surface and the troposphere. Surface gains more heat and infra-red radiation is emitted again

Solar energy is absorbed by the Earths surface and warms it... 168 W m - 2

...and is converted into heat causing the emission of longwave (infra-red) rediation back to the atmosphere E

A

R

T

H

Sources: Okanagan University College in Canada; Department of Geography, University of Oxford; United States Environmental Protection Agency (EPA) , Washington; Climate change 1995, The science of climate change, contribution of working group 1 to the second assessment report of the Intergovernmental Panel on Climate Change, UNEP and WMO, Cambridge Press, 1996

GREENhOUsE GasEs

  Chlorofuorocarbons (CFCs) . Used as rerigerants, propellants and cleaning solvents. They also have the eect o depleting the ozone layer.

The main greenhouse gases are naturally occurring but the balance in the atmosphere can be altered as a result o their release due to industry and technology. They are:  Methane, CH4 . This is the principal component o natural gas and the product o decay, decomposition or ermentation. Livestock and plants produce signifcant amounts o methane.

  Carbon dioxide, CO 2 . Combustion releases carbon dioxide into the atmosphere which can signifcantly increase the greenhouse eect. Overall, plants (providing they are growing) remove carbon dioxide rom the atmosphere during photosynthesis. This is known as carbon xation.   Nitrous oxide, N2 O. Livestock and industries (e.g. the production o Nylon) are major sources o nitrous oxide. Its eect is signifcant as it can remain in the upper atmosphere or long periods. In addition the ollowing gases also contribute to the greenhouse eect:

Absorptivity

 Water, H2 O. The small amounts o water vapour in the upper atmosphere (as opposed to clouds which are condensed water vapour) have a signifcant eect. The average water vapour levels in the atmosphere do not appear to alter greatly as a result o industry, but local levels can vary. 1

  Ozone, O 3 . The ozone layer is an important region o the atmosphere that absorbs high energy UV photons which would otherwise be harmul to living organisms. Ozone also adds to the greenhouse eect.

T

Each o these gases absorbs inra-red radiation as a result o resonance (see page 168). The natural requency o oscillation o the bonds within the molecules o the gas is in the inra-red region. I the driving requency (rom the radiation emitted rom the Earth) is equal to the natural requency o the molecule, resonance will occur. The amplitude o the molecules vibrations increases and the temperature will increase. The absorption will take place at specifc requencies depending on the molecular energy levels.

Absorption spectra for major natural greenhouse gases in the Earths atmosphere Methane CH 4

0 1

Nitrous oxide N 2O

0 1

Oxygen, O 2 & Ozone, O 3

0 1

Carbon dioxide CO 2

0 1

Water vapour H 2O

0 1 0 0.1

Total atmosphere

0.2 0.30.4 0.60.8 1 1.5 2

3 4 5 6 8 10 20 30 Wavelength (m)

[After J.N. Howard, 1959: Proc. I. R.E . 47, 1459: and R.M. Goody and G.D. Robinson, 1951: Quart. J. Roy Meteorol. Soc. 77, 153]

92

EN ERG Y PRO D U CTI O N

Go ring POssIblE CaUsEs Of GlObal waRmING Records show that the mean temperature o the Earth has been increasing in recent years.

annual mean 5-year mean

In 201 3, the IPCC (Intergovernmental Panel on Climate Change) report stated that It is extremely likely that human infuence has been the dominant cause o the observed warming since the mid20th century.

0.2 0 -0.2 -0.4 1880

 Cyclical changes in the Earths orbit and volcanic activity. The rst suggestion could be caused by natural eects or could be caused by human activities (e.g. the increased burning o ossil uels) . An enhanced greenhouse effect is an increase in the greenhouse eect caused by human activities.

0.6 0.4

 Changes in the intensity o the radiation emitted by the Sun linked to, or example, increased solar fare activity.

1900

1920

1940

1960

1980

2000

Although it is still being debated, the generally accepted view is that that the increased combustion o ossil uels has released extra carbon dioxide into the atmosphere, which has enhanced the greenhouse eect.

All atmospheric models are highly complicated. Some possible suggestions or this increase include.  Changes in the composition o greenhouse gases in the atmosphere.

EvIDENCE fOR GlObal waRmING One piece o evidence that links global warming to increased levels o greenhouse gases comes rom ice core data. The ice core has been drilled in the Russian Antarctic base at Vostok. Each years new snow all adds another layer to the ice.

Antarctic Ice Core Temperature Variation CO 2 Concentration

380 340 300 260 220 180 40 0,0 00 350,0 00

3 00 ,00 0 250,0 00 200 ,0 00 150 ,0 00

ppmv = parts per million by volume

mEChaNIsms Predicting the uture eects o global warming involves a great deal o uncertainty, as the interactions between dierent systems in the Earth and its atmosphere are extremely complex. There are many mechanisms that may increase the rate o global warming.  Global warming reduces ice/snow cover, which in turn reduces the albedo. This will result in an increase in the overall rate o heat absorption.  Temperature increase reduces the solubility o CO 2 in the sea and thus increases atmospheric concentrations.

4 2 0 -2 -4 -6 -8 -10 10 0,00 0 50,00 0 0 Years before present

C

CO 2 / ppmv

Isotopic analysis allows the temperature to be estimated and air bubbles trapped in the ice cores can be used to measure the atmospheric concentrations o greenhouse gases. The record provides data rom over 400,000 years ago to the present. The variations o temperature and carbon dioxide are very closely correlated.

 Regions with rozen subsoil exist (called tundra) that support simple vegetation. An increase in temperature may cause a signicant release o trapped CO 2 .  Not only does deorestation result in the release o urther CO 2 into the atmosphere, the reduction in number o trees reduces carbon xation. The rst our mechanisms are examples o processes whereby a small initial temperature increase has gone on to cause a urther increase in temperature. This process is known as positive feedback. Some people have suggested that the current temperature increases may be corrected by a process which involves negative eedback, and temperatures may all in the uture.

 Continued global warming will increase both evaporation and the atmospheres ability to hold water vapour. Water vapour is a greenhouse gas.

EN ERGY PRO D U CTI O N

93

Ib questions  energy production 1.

A wind generator converts wind energy into electric energy. The source o this wind energy can be traced back to solar energy arriving at the Earths surace. a) Outline the energy transormations involved as solar energy converts into wind energy.

[2]

b) List one advantage and one disadvantage o the use o wind generators.

[2]

Calculate a) the energy per second carried away by the water in the cooling tower;

5.

The expression or the maximum theoretical power, P, available rom a wind generator is 1 Av3 P= _ 2 where

 is the density o air and v is the wind speed.

[2]

d) In practice, under these conditions, the generator only provides 3 MW o electrical power. (i)

Calculate the eciency o this generator.

(ii) Give two reasons explaining why the actual power output is less than the maximum theoretical power output. 2.

[2]

6.

4.

[2] [2]

d) the mass o coal burnt each second.

[1 ]

This question is about tidal power systems. a) Describe the principle o operation o such a system.

[2]

b) Outline one advantage and one disadvantage o using such a system.

[2]

Height between high tide and low tide

4m

Trapped water would cover an area o

1 .0  1 0 6 m2

Density o water

1 .0  1 0 3 kg m3

Number o tides per day

2

[2]

[4]

Solar power and climate models. a) Distinguish, in terms o the energy changes involved, between a solar heating panel and a photovoltaic cell.

[2]

b) State an appropriate domestic use or a

This question is about energy sources.

(i)

a) Give one example o a renewable energy source and one example o a non-renewable energy source and explain why they are classied as such.

(ii) photovoltaic cell. [4]

b) A wind arm produces 35,000 MWh o energy in a year. I there are ten wind turbines on the arm show that the average power output o one turbine is about 400 kW.

[3]

c) State two disadvantages o using wind power to generate electrical power. [2] 3.

b) the energy per second produced by burning the coal; c) the overall eciency o the power station;

c) A small tidal power system is proposed. Use the data in the table below to calculate the total energy available and hence estimate the useul output power o this system.

A is the area swept out by the blades,

c) Calculate the maximum theoretical power, P, or a wind generator whose blades are 30 m long when a 20 m s - 1 wind blows. The density o air is 1 .3 kg m- 3 .

[2]

solar heating panel.

[1 ] [1 ]

c) The radiant power o the Sun is 3.90  1 0 2 6 W. The average radius o the Earths orbit about the Sun is 1 .50  1 0 1 1 m. The albedo o the atmosphere is 0.300 and it may be assumed that no energy is absorbed by the atmosphere. Show that the intensity incident on a solar heating panel at the Earths surace when the Sun is directly overhead [3] is 966 W m2 .

Wind power can be used to generate electrical energy.

d) Show, using your answer to (c) , that the average intensity incident on the Earths surace is 242 Wm2 . [3]

Construct an energy fow diagram which shows the energy transormations, starting with solar energy and ending with electrical energy, generated by windmills. Your diagram should indicate where energy is degraded. [7]

e) Assuming that the Earths surace behaves as a black-body and that no energy is absorbed by the atmosphere, use your answer to (d) to show that the average temperature o the Earths surace is predicted to be 256 K. [2]

This question is about a coal-red power station which is water cooled.

) Outline, with reerence to the greenhouse eect, why the average surace temperature o the Earth is higher than 256 K. [4]

This question is about energy transormations.

Data: Electrical power output rom the station Temperature at which water enters cooling tower

= 200 MW = 288 K

Temperature at which water leaves cooling tower

= 348 K

Rate o water fow through tower

= 4000 kg s - 1

Energy content o coal

= 2.8  1 0 7 J kg- 1

Specic heat o water

= 4200 J kg1 K- 1

94

I B Q U EsTI O N s  EN ERG Y PRO D U CTI O N

9 W av e p h e n o m e n a hL Sil i i SImpLe harmonIc motIon (Shm) equatIon

tWo exampLeS of Shm

SHM occurs when the orces on an object are such that the resultant acceleration, a, is directed towards, and is proportional to, its displacement, x, rom a fxed point. a  - x or a = - (constant)  x The mathematics o SHM is simplifed i the constant o proportionality between a and x is identifed as the square o another constant  which is called the angular requency. Thus the general orm or the equation that defnes SHM is: a = - 2 x The solutions or this equation ollow below. The angular requency  has the units o rad s - 1 and is related to the time period, T, o the oscillation by the ollowing equation. 2 =_ T

Two common situations that approximate to SHM are: 1.

Provided that:  the mass o the spring is negligible compared to the mass o the load  riction (air riction) is negligible  the spring obeys Hookes law with spring constant, k at all times (i.e. elastic limit is not exceeded)  the gravitational feld strength g is constant  the fxed end o the spring cannot move. Then it can be shown that: k 2 = _ m __ m Or T = 2 _ k The simple pendulum o length l and mass m



IdentIfIcatIon of Shm In order to analyse a situation to decide i SHM is taking place, the ollowing procedure should be ollowed.

2.

Provided that:

 Identiy all the orces acting on an object when it is displaced an arbitrary distance x rom its rest position using a ree-body diagram.

 the mass o the string is negligible compared with the mass o the load  riction (air riction) is negligible

 Calculate the resultant orce using Newtons second law. I this orce is proportional to the displacement and always points back towards the mean position (i.e. F  - x) then the motion o the object must be SHM.

 the maximum angle o swing is small ( 5 or 0.1 rad)  the gravitational feld strength g is constant  the length o the pendulum is constant.

 Once SHM has been identifed, the equation o motion must be in the ollowing orm: restoring orce per unit displacement, k a = - ____  x oscillating mass, m

(

Mass, m, on a vertical spring

Then it can be shown that: g  2 = _ l __ l Or T = 2 _ g Note that the mass o the pendulum bob, m, is not in this equation and thus does not aect the time period o the pendulum, T.

)



( )

k  This identifes the angular requency  as 2 = _ m or  k _ = m . Identifcation o  allows quantitative equations to be applied.

( )

exampLe acceLeratIon, veLocIty and dISpLacement durIng Shm

A 600 g mass is attached to a light spring with spring constant 30 N m1 .

The variation with time o the acceleration, a, velocity, v, and displacement, x, o an object doing SHM depends on the angular requency .

(b) Calculate the requency o its oscillation.

The precise ormat o the relationships depends on where the object is when the clock is started (time t = zero) . The let hand set o equations correspond to an oscillation when the object is in the mean position when t = 0. The right hand set o equations correspond to an oscillation when the object is at maximum displacement when t = 0. x = x0 sin t x = x0 cos t v = x0 cos t v = -x0 sin t a = - 2 x0 sin t a = - 2 x0 cos t The frst two equations can be rearranged to produce the ollowing relationship:

(a) Show that the mass does SHM. (a) Weight o mass = mg = 6.0 N Additional displacement x down means that resultant orce on mass = k x upwards. Since F  - x, the mass will oscillate with SHM. _____ ____ m = 2 _ 0.6 = 0.889s (b) Since SHM, T = 2 _ 30 k

( )

( )

1 = _ 1 f= _ = 1 .1 Hz T 0.889

displacement

x0 x0

_______

velocity

v =   (x0 - x2 )

x0 is the amplitude o the oscillation measured in m t is the time taken measured in s  is the angular requency measured in rad s 1  t is an angle that increases with time measured in radians. A ull oscillation is completed when ( t) = 2 rad. The angular requency is related to the time period T by the ollowing equation. ___

2 m _ T= _  = 2 k



 2 x0     

T 4

T 2

3T 4

T

time acceleration

acceleration leads velocity by 90 velocity leads displacement by 90 acceleration and displacement are 1 80 out o phase displacement lags velocity by 90 velocity lags acceleration by 90

W av e p h e n o m e n a

95

enrgy cangs during simpl armonic motion

hL

During SHM, energy is interchanged between KE and PE. Providing there are no resistive forces which dissipate this energy, the total energy must remain constant. The oscillation is said to be undamped. The kinetic energy can be calculated from

The total energy is 1 m  2 (x - x2 ) + _ 1 m  2 x2 = _ 1 m  2 x E = Ek + Ep = _ 0 0 2 2 2 Energy in SHM is proportional to:  the mass m

1 m  2 (x - x2 ) 1 mv2 = _ Ek = _ 0 2 2

 the (amplitude) 2  the (frequency) 2

The potential energy can be calculated from 1 m  2 x2 Ep = _ 2

E total p Graph showing the variation with distance, x of the energy,  during SHM

k x0

-x0

x

 total k

Graph showing the variation with time, t of the energy,  during SHM

p t T 4

96

W av e p h e n o m e n a

T 2

3T 4

T 2

hL

difi

BaSIc oBServatIonS

The intensity plot or a single slit is:

Diraction is a wave eect. The objects involved (slits, apertures, etc.) have a size that is o the same order o magnitude as the wavelength o visible light.

nu  bsl

gmil sw

intensity There is a central maximum intensity. Other maxima occur roughly halfway 10 between the minima.

difi 

As the angle increases, the intensity of the maxima decreases.

() straight edge

1.1 0.4 b = slit width

(b) single long slit b ~ 3

angle 1st minimum 2 3  = b =  = b b

The angle of the rst minimum is given by sin  = b . For small angles, this can be simplied to  = b .

() circular aperture () single long slit b ~ 5

expLanatIon The shape o the relative intensity versus angle plot can be derived by applying an idea called Huygens principle. We can treat the slit as a series o secondary wave sources. In the orward direction ( = zero) these are all in phase so they add up to give a maximum intensity. At any other angle, there is a path dierence between the rays that depends on the angle. The overall result is the addition o all the sources. The condition or the frst minimum is that the angle must make all o the sources across the slit cancel out. The condition or the frst maximum out rom the centre is 3 . At this when the path dierence across the whole slit is ___ 2 angle the slit can be analysed as being three equivalent sections  each having a path dierence o __ across its length. Together, 2 two o these sections will destructively interere leaving the resulting amplitude to be __13 o the maximum. Since intensity  (amplitude) 2 , the frst maximum intensity out rom the centre will be __19 o the central maximum intensity. By a similar argument, the second maximum intensity out rom the centre will

1 o the central have __15 o the maximum amplitude and thus be __ 25 maximum intensity.

For 1 st m in im u m : b sin  =    sin  =

b

 b

Sin ce a n gle is sm a ll, sin   





=

 b

for 1 st m in im u m

path dierence across slit = b sin 

SIngLe-SLIt dIractIon WIth WhIte LIght When a single slit is illuminated with white light, each component colour has a specifc wavelength and so the associated maxima and minima or each wavelength will be located at a dierent angle. For a given slit width, colours with longer wavelengths (red, orange, etc.) will diract more than colours with short wavelengths (blue, violet, etc.) . The maxima or the resulting diraction pattern will show all the colours o the rainbow with blue and violet nearer to the central position and red appearing at greater angles.

incident white light

Red rst order Violet zero order Violet rst order Red

W av e p h e n o m e n a

97

hL

tw-s i  ws: ys blsli i

douBLe-SLIt Interference The double-slit intererence pattern shown on page 47 was derived assuming that each slit was behaving like a perect point source. This can only take place i the slits are infnitely small. In practice they have a fnite width. The diraction pattern o each slit needs to be taken into account when working out the overall double slit intererence pattern as shown below. Decreasing the slit width will mean that the observed pattern becomes more and more idealized. Unortunately, it will also mean that the total intensity o light will be decreased. The intererence pattern will become harder to observe.

(a) Youngs fringes for innitely narrow slits relative intensity

(c) Youngs fringes for slits of nite width intensity

angle 

bright fringes

(b) diraction pattern for a nite-width slit intensity

angle  s = D still applies but dierent fringes d will have dierent intensities with it being possible for some fringes to be missing.

angle 

InveStIgatIng youngS douBLe-SLIt experImentaLLy Possible set-ups or the double-slit experiment are shown on page 47.

Set-up 1

region in which superposition occurs

separation monochromatic of slits light source

S0

In the simplifed version (set-up 2) o the experiment, ringes can still be bright enough to be viewed several metres away rom the slits and thus they can be projected onto an opaque screen (it is dangerous to look into a laser beam) . Their separation can be then be directly measured with a ruler.

S1 S2

source twin source slit slits (less than 5 mm) 0.1 m 1m

possible screen positions

In the original set-up (set-up 1 ) light rom the monochromatic source is diracted at S 0 so as to ensure that S 1 and S 2 are receiving coherent light. Diraction takes place providing S 1 and S 2 are narrow enough. The slit separations need to be approximately 1 mm (or less) thus the slit widths are o the order o 0.1 mm (or less) . This would provide ringes that were separated by approximately 0.5mm on a screen (semitransparent or translucent) situated 1 m away. The laboratory will need to be darkened to allow the ringes to be visible and they can be viewed using a microscope.

98

W av e p h e n o m e n a

The most accurate measurements or slit separation and ringe width are achieved using a travelling microscope. This is a microscope that is mounted on a rame so that it can be moved perpendicular to the direction in which it is pointing. The microscope is moved across ten or more ringes and the distance moved by the microscope can be read o rom the scale. The precision o this measurement is oten improved by utilizing a vernier scale.

Set-up 2 laser

double slit

screen

mlipl-sli ifi

hL

the dIractIon gratIng

(a) 2 slits

The diraction that takes place at an individual slit aects the overall appearance o the ringes in Youngs doubleslit experiment (see page 98 or more details) . This section considers the eect on the fnal intererence pattern o adding urther slits. A series o parallel slits (at a regular separation) is called a diffraction grating. Additional slits at the same separation will not aect the condition or constructive intererence. In other words, the angle at which the light rom slits adds constructively will be unaected by the number o slits. The situation is shown below.



path dierence between slits = d sin 

(b) 4 slits

(c) 50 slits

Grating patterns

uSeS d

 For constructive interference: path dierence = n between slits [, 2, 3]

n = d sin  This ormula also applies to the Youngs double-slit arrangement. The dierence between the patterns is most noticeable at the angles where perect constructive intererence does not take place. I there are only two slits, the maxima will have a signifcant angular width. Two sources that are just out o phase interere to give a resultant that is nearly the same amplitude as two sources that are exactly in phase.

resultant interference pattern source A time source B The addition o more slits will mean that each new slit is just out o phase with its neighbour. The overall intererence pattern will be totally destructive.

overall interference pattern is totally destructive

One o the main uses o a diraction grating is the accurate experimental measurement o the dierent wavelengths o light contained in a given spectrum. I white light is incident on a diraction grating, the angle at which constructive intererence takes place depends on wavelength. Dierent wavelengths can thus be observed at dierent angles. The accurate measurement o the angle provides the experimenter with an accurate measurement o the exact wavelength (and thus requency) o the colour o light that is being considered. The apparatus that is used to achieve this accurate measurement is called a spectrometer.

third (and part of the fourth) order spectrum not shown

R 2nd order V R 1st order V

white light

white central maximum V

diraction grating

R V

R

time The addition o urther slits at the same slit separation has the ollowing eects:  the principal maxima maintain the same separation  the principal maxima become much sharper  the overall amount o light being let through is increased, so the pattern increases in intensity.

W av e p h e n o m e n a

99

ti lll fls

hL

phaSe changeS

condItIonS or Intererence patternS

There are many situations when intererence can take place that also involve the refection o light. When analysing in detail the conditions or constructive or destructive intererence, one needs to take any phase changes into consideration. A phase change is the inversion o the wave that can take place at a refection interace, but it does not always happen. It depends on the two media involved.

A parallel-sided lm can produce intererence as a result o the refections that are taking place at both suraces o the lm.

E D

A

C

The technical term or the inversion o a wave is that it has undergone a phase change o .  When light is refected back rom an optically denser medium there is a phase change o .

thickness d

transmitted wave (no phase change)

n1 < n2

lm (refractive index = n)



 When light is refected back rom an optically less dense medium there is no phase change.

air



B

air

 = ze ro wh e n vie we d a lo n g th e n o rm a l

n1 n2

F reected wave (no phase change)

incident wave n1 < n2 incident wave

reected wave ( phase change)

1. along path AE in air 2. along ABCD in the lm of thickness d These rays then interfere and we need to calculate the optical path difference. The path AE in air is equivalent to CD in the lm So path difference = (AB + BC) in the lm.

n1

 In addition, the phase change at A is equivalent to path 2 difference.

n2 transmitted wave (no phase change)

exampLe The equations in the box on the right work out the angles or which constructive and destructive intererence take place or a given wavelength. I the source o light is an extended source, the eye receives rays leaving the lm over a range o values or . I white light is used then the situation becomes more complex. Provided the thickness o the lm is small, then one or two colours may reinorce along a direction in which others cancel. The appearance o the lm will be bright colours, such as can be seen when looking at  an oil lm on the surace o water or

rays from an extended source

W av e p h e n o m e n a

So total path difference = (AB + BC) in lm +  2 = n(AB + BC) +  2 By geometry: (AB + BC) = FC = 2 d cos   path difference = 2dn cos  +  2 if 2dn cos  = m : destructive or when  = 0, 2dn = m: destructive if 2dn cos  = m +   : constructive 2 or when  = 0, 2dn = m: constructive m = 0,1 ,2,3,4

appLIcatIonS

 soap bubbles.

100

From point A, there are two possible paths:

eye focused at innity

Applications o parallel thin lms include:  The design o non-refecting radar coatings or military aircrat. I the thickness o the extra coating is designed so that radar signals destructively interere when they refect rom both suraces, then no signal will be refected and an aircrat could go undetected.  Measurements o thickness o oil slicks caused by spillage. Measurements o the wavelengths o electromagnetic signals that give constructive and destructive intererence (at known angles) allow the thickness o the oil to be calculated.  Design o non-refecting suraces or lenses (blooming), solar panels and solar cells. A strong refection at any o these suraces would reduce the amount o energy being useully transmitted. A thin surace lm can be added so that destructive intererence takes place or a typical wavelength and thus maximum transmittance takes place at this wavelength.

hL

rsli

These examples look at the situation o a line source o light and the diraction that takes place at a slit. A more common situation would be a point source o light, and the diraction that takes place at a circular aperture. The situation is exactly the same, but diraction takes place all the way around the aperture. As seen on page 97, the diraction pattern o the point source is thus concentric circles around the central position. The geometry o the situation results in a slightly dierent value or the frst minimum o the diraction pattern.

relative intensity

dIffractIon and reSoLutIon (a) resolved

I two sources o light are very close in angle to one another, then they are seen as one single source o light. I the eye can tell the two sources apart, then the sources are said to be resolved. The diraction pattern that takes place at apertures aects the eyes ability to resolve sources. The examples to the right show how the appearance o two line sources will depend on the diraction that takes place at a slit. The resulting appearance is the addition o the two overlapping diraction patterns. The graph o the resultant relative intensity o light at dierent angles is also shown.

angle  appearance two sources clearly separate resultant intensity diraction pattern of source B

(b) just resolved diraction pattern of source A

angle  slightly dimmer appearance

For a slit, the frst minimum was at the angle  =_ b

two maxima visible (c) not resolved

resultant intensity

For a circular aperture, the frst minimum is at the angle 1 .22  =_ b

diraction pattern of source B

diraction pattern of source A

I two sources are just resolved, then the frst minimum o one diraction pattern is located on top o the maximum o the other diraction pattern. This is known as the Rayleigh criterion.

angle  appearance appears as one source

exampLe Late one night, a student was observing a car approaching rom a long distance away. She noticed that when she frst observed the headlights o the car, they appeared to be one point o light. Later, when the car was closer, she became able to see two separate points o light. I the wavelength o the light can be taken as 500 nm and the diameter o her pupil is approximately 4 mm, calculate

how ar away the car was when she could frst distinguish two points o light. Take the distance between the headlights to be 1 .8 m.

Since  small 1 .8 =_ x [x is distance to car]

When just resolved

1 .8  x = __ 1 .525  1 0 - 4

1 .22   =_ b

= 1 1 .803  1 2 km

1 .22  5  1 0 - 7 = __ 0.004 = 1 .525  1 0 - 4

reSoLvance of dIffractIon gratIngS

Example:

As a result o Rayleighs criterion, there is a limit placed on a gratings ability to resolve dierent wavelengths. The resolvance, R, o a diraction grating is defned as the ratio between a wavelength being investigated, , and the smallest possible resolvable wavelength dierence, .

In the sodium emission spectrum there are two wavelengths that are close to one another (the Na D-lines) . These are 589.00 nm and 589.59 nm. In order or these to be resolved by a diraction grating, the resolvance must be

 R= _  For any given grating, R is dependent on the diraction order, m, being observed (frst order: m = 1 ; second order: m = 2, etc.) and the total number o slits, N, on the grating that are being illuminated.

 = _ 589.00 = 1 000 R= _ 0.59  In the frst order spectrum, at least 1 000 slits must be illuminated whereas in the second order spectrum, the requirement drops to only 500 slits.

  = mN R= _ 

W av e p h e n o m e n a

101

hL

t dl f

doppLer eect The Doppler eect is the name given to the change o requency o a wave as a result o the movement o the source or the movement o the observer. When a source o sound is moving:  Sound waves are emitted at a particular requency rom the source.  The speed o the sound wave in air does not change, but the motion o the source means that the wave ronts are all bunched up ahead o the source.  This means that the stationary observer receives sound waves o reduced wavelength.  Reduced wavelength corresponds to an increased requency o sound. The overall eect is that the observer will hear sound at a higher requency than it was emitted by the source. This applies when the source is moving towards the observer. A similar

analysis quickly shows that i the source is moving away rom the observer, sound o a lower requency will be received. A change o requency can also be detected i the source is stationary, but the observer is moving.  When a police car or ambulance passes you on the road, you can hear the pitch o the sound change rom high to low requency. It is high when it is approaching and low when it is going away.  Radar detectors can be used to measure the speed o a moving object. They do this by measuring the change in the requency o the refected wave.  For the Doppler eect to be noticeable with light waves, the source (or the observer) needs to be moving at high speed. I a source o light o a particular requency is moving away rom an observer, the observer will receive light o a lower requency. Since the red part o the spectrum has lower requency than all the other colours, this is called a red shift.  I the source o light is moving towards the observer, there will be a blue shift.

movIng Source

mathematIcS o the doppLer eect

Source moves rom A to D with velocity, u s , speed o waves is v.

Mathematical equations that apply to sound are stated on this page. Unortunately the same analysis does not apply to light  the velocities can not be worked out relative to the medium. It is, however, possible to derive an equation or light that turns out to be in exactly the same orm as the equation or sound as long as two conditions are met:

o

moving source

Q

A BC D

stationary observer receives sound at lower frequency

f' = f

v v  us

P

u s t

 the relative velocity o source and detector is used in the equations.

stationary observer receives sound at higher frequency

Received frequency at P Received frequency at Q

f' = f

v v - us

f' = f

v v + us

 this relative velocity is a lot less than the speed o light. Providing v of formation surroundings

h    

work done W

HOT reservoir Thot

thermal energy Q hot

ENGINE

COLD reservoir

thermal energy Q cold

In order to do this, some thermal energy must have been taken rom a hot reservoir (during the isovolumetric increase in pressure and the isobaric expansion) . A dierent amount o thermal energy must have been ejected to a cold reservoir (during the isovolumetric decrease in pressure and the isobaric compression) .

pressure p

heat engines A central concept in the study o thermodynamics is the heat engine. A heat engine is any device that uses a source o thermal energy in order to do work. It converts heat into work. The internal combustion engine in a car and the turbines that are used to generate electrical energy in a power station are both examples o heat engines. A block diagram representing a generalized heat engine is shown below.

A

isobaric expansion

B

isovolumetric increase in pressure

total work done by the gas isovolumetric decrease in pressure

Tcold C isobaric compression D volume V The thermal efciency o a heat engine is defned as

Heat engine In this context, the word reservoir is used to imply a constant temperature source (or sink) o thermal energy. Thermal energy can be taken rom the hot reservoir without causing the temperature o the hot reservoir to change. Similarly thermal energy can be given to the cold reservoir without increasing its temperature. An ideal gas can be used as a heat engine. The pV diagram right represents a simple example. The our-stage cycle returns the gas to its starting conditions, but the gas has done work. The area enclosed by the cycle represents the amount o work done.

work done  = ____ (thermal energy taken rom hot reservoir) This is equivalent to rate o doing work  = ____ (thermal power taken rom hot reservoir) useul work done  = __ energy input The cycle o changes that results in a heat engine with the maximum possible efciency is called the Carnot cycle.

Carnot CyCles and Carnot theorem The Carnot cycle represents the cycle o processes or a theoretical heat engine with the maximum possible efciency. Such an idealized engine is called a Carnot engine.

input work W

HOT reservoir

COLD reservoir

pressure p

heat pumps A heat pump is a heat engine being run in reverse. A heat pump causes thermal energy to be moved rom a cold reservoir to a hot reservoir. In order or this to be achieved, mechanical work must be done.

A Q hot thermal energy taken in B

HEAT PUMP

Thot thermal energy Q hot

Tcold D thermal energy given out Q cold

thermal energy Q cold

area = work done by gas during Carnot cycle C volume V

Carnot cycle

Once again an ideal gas can be used as a heat pump. The thermodynamic processes can be exactly the same ones as were used in the heat engine, but the processes are all opposite. This time an anticlockwise circuit will represent the cycle o processes.

It consists o an ideal gas undergoing the ollowing processes.

pressure p

Heat pump

isobaric compression A D isovolumetric decrease in pressure

 Isothermal expansion (A  B)  Adiabatic expansion (B  C)  Isothermal compression (C  D)

total work done on the gas isovolumetric increase in pressure

 Adiabatic compression (D  A) The temperatures o the hot and cold reservoirs fx the maximum possible efciency that can be achieved. The efciency o a Carnot engine can be shown to be Tco ld C a rn o t = 1 - _ (where T is in kelvin) Th o t An engine operates at 300 C and ejects heat to the surroundings at 20 C. The maximum possible theoretical efciency is

B isobaric expansion C volume V

293 = 0.49 = 49% C a rn o t = 1 - _ 573

o p t i o n B  E n g i n E E r i n g p h ys i c s

163

HL

f  

definitions of density and pressure average density

m =_ V

normal orce

pressure

F p = _ A

The symbol representing density is the Greek letter rho, . The average density o a substance is dened by the ollowing equation:

area

mass  Pressure is a scalar quantity  the orce has a direction but the pressure does not. Pressure acts equally in all directions. volume

 Density is a scalar quantity.

 The SI unit o pressure is N m- 2 or pascals (Pa). 1 Pa = 1 N m- 2

 The SI units o density are kg m- 3 .

 Atmospheric pressure  1 0 5 Pa

 Densities can also be quoted in g cm- 3 (see conversion actor below)

 Absolute pressure is the actual pressure at a point in a fuid. Pressure gauges oten record the difference between absolute pressure and atmospheric pressure. Thus i a dierence pressure gauge gives a reading o 2  1 0 5 Pa or a gas, the absolute pressure o the gas is 3  1 0 5 Pa.

 The density o water is 1 g cm- 3 = 1 ,000 kg m- 3 Pressure at any point in a fuid (a gas or a liquid) is dened in terms o the orce, F, that acts normally (at 90) to a small area, A, that contains the point.

variation of fluid pressure

BuoyanCy and arChimedes prinCiple

The pressure in a fuid increases with depth. I two points are separated by a vertical distance, d, in a fuid o constant density, f, then the pressure dierence, p, between these two points is:

Archimedes principle states that when a body is immersed in a fuid, it experiences a buoyancy upthrust equal in magnitude to the weight o the fuid displaced. B = fVf g

density o fuid

gravitational eld strength

p = fgd pressure dierence due to depth

22N

depth

17N

The total pressure at a given depth in a liquid is the addition o the pressure acting at the surace (atmospheric pressure) and the additional pressure due to the depth: Atmospheric pressure

P = P0 + fgd

gravitational eld strength

W

W

(a)

Note that:  Pressure can be expressed in terms o the equivalent depth (or head) in a known liquid. Atmospheric pressure is approximately the same as exerted by a 760 mm high column o mercury (Hg) or a 1 0 m column o water.

volume of uid displaced (w = 5N)

AB

B2

atmospheric pressure the water column exerts a pressure at B equal to the excess pressure of the gas supply: P = hg

W

volume of uid displaced (w = 10N)

A consequence o this principle is that a foating object displaces its own weight o fuid.

 As pressure is dependent on depth, the pressures at two points that are at the same horizontal level in the same liquid must be the same provided they are connected by that liquid and the liquid is static.

h

density of uid

B1

density o fuid depth

Total pressure

excess gas pressure P

12N

weight of uid displaced = total weight of duck

pasCals prinCiple Pascals principle states that the pressure applied to an enclosed liquid is transmitted to every part o the liquid, whatever the shape it takes. This principle is central to the design o many hydraulic systems and is dierent to how solids respond to orces.

 The pressure is independent o the cross-sectional area  this means that liquids will always nd their own level.

When a solid object (e.g. an incompressible stick) is pushed at one end and its other end is held in place, then the same orce will be exerted on the restraining object.

hydrostatiC equiliBrium

Incompressible solids transmit orces whereas incompressible liquids transmit pressures.

A fuid is in hydrostatic equilibrium when it is at rest. This happens when all the orces on a given volume o fuid are balanced. Typically external orces (e.g. gravity) are balanced by a pressure gradient across the volume o fuid (pressure increases with depth  see above) . downward force due to pressure from uid above volume of uid weight of uid W contained in volume

164

upward force due to pressure from uid below

o p t i o n B  E n g i n E E r i n g p h ys i c s

A2 A1 load platform

load = F 

applied force F (eort)

piston of area A 1

piston of area A 2

hydraulic liquid

HL

    B fc

the ideal luid In most real situations, fuid fow is extremely complicated. The ollowing properties dene an ideal fuid that can be used to create a simple model. This simple model can be later rened to be more realistic.

 Is non-viscous  as a result o fuid fow, no energy gets converted into thermal energy. See page 1 67 or the denition o the viscosity o a real fuid.

An ideal fuid:

 Involves a steady fow (as opposed to a turbulent, or chaotic, fow) o fuid. Under these conditions the fow is laminar (see box below). See page 1 67 or an analysis o turbulent fow.

 Is incompressible  thus its density will be constant.

 Does not have angular momentum  it does not rotate.

laminar low, streamlines and the Continuity equation When the fow o a liquid is steady or laminar, dierent parts o the fuid can have dierent instantaneous velocities. The fow is said to be laminar i every particle that passes through a given point has the same velocity whenever the observation is made. The opposite o laminar fow, turbulent fow, takes place when the particles that pass through a given point have a wide variation o velocities depending on the instant when the observation is made (see page 1 67) . A streamline is the path taken by a particle in the fuid and laminar fow means that all particles that pass through a given point in the fuid must ollow the same streamline. The direction o the tangent to a streamline gives the direction o the instantaneous velocity that the particles o the fuid have at that point. No fuid ever crosses a streamline. Thus a collection o streamlines can together dene a tube o fow. This is tubular region o fuid where fuid only enters and leaves the tube through its ends and never through its sides.

speed 2 speed 1 area A 1 density 1

boundary (streamlines)

area A 2 density 2

In a time t, the mass, m 1 , entering the cross-section A 1 is m1 = 1 A 1 v1 t Similarly the mass, m 2 , leaving the cross-section A 2 is m2 = 2 A 2 v2 t Conservation o mass applies to this tube o fow, so 1 A 1 v1 = 2 A 2 v2 This is an ideal fuid and thus incompressible meaning 1 = 2 , so A1 v1 = A 2 v2 or Av = constant This is the continuity equation.

the Bernoulli eeCt

the Bernoulli equation

When a fuid fows into a narrow section o a pipe:

The Bernoulli equation results rom a consideration o the work done and the conservation o energy when an ideal fuid changes:

 The fuid must end up moving at a higher speed (continuity equation).   This means the fuid must have been accelerated orwards.

 its speed (as a result o a change in cross-sectional area)  its vertical height as a result o work done by the fuid pressure. The equation identies a quantity that is always constant along any given streamline: density o fuid

higher pressure lower speed

lower pressure higher speed

higher pressure lower speed

 This means there must be a pressure dierence orwards with a lower pressure in the narrow section and a higher pressure in the wider section. Thus an increase in fuid speed must be associated with a decrease in fuid pressure. This is the Bernoulli eect  the greater the speed, the lower the pressure and vice versa.

speed o fuid particles

vertical height fuid pressure

1 v2 + gz + p = constant 2 density gravitational

_

o fuid

eld strength

Note that: 1 2  The rst term (__ v ), can be thought o as the dynamic pressure. 2

 The last two terms (gz + p) , can be thought o as the static pressure.  Each term in the equation has several possible units: N m- 2 , Pa, J m- 3 .  The last o the above units leads to a new interpretation or the Bernoulli equation: KE gravitational PE per unit + per unit + pressure = constant volume volume

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B  xm

HL

appliCations of the Bernoulli equation a) Flow out o a container

d) Pitot tube to determine the speed o a plane A

liquid density 

A pitot tube is attached acing orward on a plane. It has two separate tubes:

streamline

direction of airow

h

small static pressure openings

B

arbitrary zero

impact opening

To calculate the speed o fuid fowing out o a container, we can apply Bernoullis equation to the streamline shown above.

total pressure tube

At A, p = atmospheric and v = zero At B, p = atmospheric and v = ? 1 v2 + gz + p = constant _ 2 1 v2 + 0 + p  0 + hg + p = _ 2 ___

v = 2gh

b) Venturi tubes A Venturi meter allows the rate o fow o a fuid to be calculated rom a measurement o pressure dierence between two dierent cross-sectional areas o a pipe.

to metal end area A constriction of area a A 

 The ront hole (impact opening) is placed in the airstream and measures the total pressure (sometimes called the stagnation pressure) , PT.  The side hole(s) measures the static pressure, Ps .  The dierence between PT and Ps , is the dynamic pressure. The Bernoulli equation can be used to calculate airspeed: 1 v2 PT - Ps = _ 2 ________

v=

2(P - P ) _  T

s

e) Aerooil (aka airoil)

dynamic lift F

B

pressure P1

ow of (e.g.) water, density 1

h

1

manometer liquid (e.g. mercury) , density 2

aerofoil 2

 The pressure dierence between A and B can be calculated by taking readings o h and 2 rom the attached manometer:

air ow pressure P2

PA - PB = h2 g  This value and measurements o A, a and 1 allows the fuid speed at A to be calculated by using Bernoullis equation and the equation o continuity v=

c)

[ ( )

2h2 g  ________ A 2 - 1 1 __ a

static pressure tube

]

Note that:  Streamlines closer together above the aerooil imply a decrease in cross-sectional area o equivalent tubes o fow above the aerooil.

 The rate o fow o fuid through the pipe is equal to A  v

 Decrease in cross-sectional area o tube o fow implies increased velocity o fow above the aerooil (equation o continuity) . v1 > v2

Fragrance spray

 Since v1 > v2 , P1 < P2

b. Constriction in tube causes low pressure region as air travels faster in this section below-pressure zone squeezebulb c. Liquid is drawn up tube by pressure dierence and forms little droplets a. Squeezing as it enters the air jet bulb d. Fine spray of fragrance forces air is emitted from nozzle through tube

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 Bernoulli equation can be used to calculate the pressure dierent (height dierence not relevant) which can support the weight o the aeroplane.  When angle o attack is too great, the fow over the upper surace can become turbulent. This reduces the pressure dierence and leads to the plane stalling.

HL

vc

definition of visCosity

A) Tangential stress

An ideal fuid does not resist the relative motion between dierent layers o fuid. As a result there is no conversion o work into thermal energy during laminar fow and no external orces are needed to maintain a steady rate o fow. Ideal fuids are nonviscous whereas real fuids are viscous. In a viscous fuid, a steady external orce is needed to maintain a steady rate o fow (no acceleration) . Viscosity is an internal riction between dierent layers o a fuid which are moving with dierent velocities. The denition o the viscosity o a fuid, , (Greek letter Nu) is in terms o two new quantities, the tangential stress, , and v the velocity gradient, ___ (see RH side). y The coecient o viscosity  is dened as:

relative velocity v

area of contact A

retarding force -F accelerating force F The tangential stress is dened as: F =_ A  Units o tangential stress are N m- 2 or Pa B) Velocity gradient

y velocity

tangential stress FA  = __ = _ velocity gradient vy

(v + v)

v y

 The units o  are N s m- 2 or kg m- 1 s - 1 or Pa s

v

 Typical values at room temperature:  Water: 1 .0  1 0 - 3 Pa s  Thick syrup: 1 .0  1 0 2 Pa s  Viscosity is very sensitive to changes o temperature. For a class o fuid, called Newtonian fuids, experimental measurements show that tangential stress is proportional to velocity gradient (e.g. many pure liquids). For these fuids the coecient o viscosity is constant provided external conditions remain constant.

stokes law Stokes law predicts the viscous drag orce FD that acts on a perect sphere when it moves through a fuid:

sphere has uniform velocity v

-F equal opposing viscous drag

Drag orce acting on sphere in N

v velocity gradient = _ y  Units o velocity gradient are s- 1

 The fuid is innite in volume. Real spheres alling through columns o fuid can be aected by the proximity o the walls o the container.  The size o the particles o the fuid is very much smaller than the size o the sphere.

uid at this point moves with body (boundary layer)

F r driving force innite expanse of uid 

The velocity gradient is dened as:

The orces on a sphere alling through a fuid at terminal velocity are as shown below:

uid upthrust sphere velocity v

sphere density  uid density 

W

velocity o sphere in m s - 1

At terminal velocity vt ,

Note Stokes law assumes that:

W = U + FD

 The speed o the sphere is small so that:

FD = U - W

 the fow o fuid past the sphere is streamlined  there is no slipping between the fuid and the sphere

4 r3 ( - ) g 6rvt = _ 3 2 ( - ) g 2r  vt = __ 9

turBulent flow  the reynolds numBer Streamline fow only occurs at low fuid fow rates. At high fow rates the fow becomes turbulent:

laminar

viscous drag

r pull of Earth

viscosity o fuid in Pa s

FD = 6 rv radius o sphere in m

U FD

Reynolds number

turbulent

It is extremely dicult to predict the exact conditions when fuid fow becomes turbulent. When considering fuid fow down a pipe, a useul number to consider is the Reynolds number, R, which is dened as:

speed o bulk fow

vr R= _ 

radius o pipe density o fuid viscosity o fuid

Note that:  The Reynolds number does not have any units  it is just a ratio.  Experimentally, fuid fow is oten laminar when R < 1 000 and turbulent when R > 2000 but precise predictions are dicult.

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167

fc c  c (1)

damping

displacement, x

Damping involves a rictional orce that is always in the opposite direction to the direction o motion o an oscillating particle. As the particle oscillates, it does work against this resistive (or dissipative) orce and so the particle loses energy. As the total energy o the particle is proportional to the (amplitude) 2 o the SHM, the amplitude decreases exponentially with time.

exponential envelope  

2 

4 

time, t

Heavy damping or overdamping involves large resistive orces (e.g. the SHM taking place in a viscous liquid) and can completely prevent the oscillations rom taking place. The time taken or the particle to return to zero displacement can again be long. Critical damping involves an intermediate value or resistive orce such that the time taken or the particle to return to zero displacement is a minimum. Eectively there is no overshoot. Examples o critically damped systems include electric meters with moving pointers and door closing mechanisms.

displacement

HL

overdamped critical damping

time 0.2

The above example shows the eect o light damping (the system is said to be underdamped) where the resistive orce is small so a small raction o the total energy is removed each cycle. The time period o the oscillations is not aected and the oscillations continue or a signifcant number o cycles. The time taken or the oscillations to die out can be long.

I a system is temporarily displaced rom its equilibrium position, the system will oscillate as a result. This oscillation will be at the natural frequency of vibration o the system. For example, i you tap the rim o a wine glass with a knie, it will oscillate and you can hear a note or a short while. Complex systems tend to have many possible modes o vibration each with its own natural requency. It is also possible to orce a system to oscillate at any requency that we choose by subjecting it to a changing orce that varies with the chosen requency. This periodic driving orce must be provided rom outside the system. When this driving frequency is frst applied, a combination o natural and orced oscillations take place which produces complex transient oscillations. Once the amplitude o the transient oscillations die down, a steady condition is achieved in which:

0.6

0.8

1.0

1.2

1.4

1.6

overshoot underdamped

 The amplitude o the orced oscillations depends on:  the comparative values o the natural requency and the driving requency  the amount o damping present in the system.

amplitude of oscillation

natural frequenCy and resonanCe

0.4

light damping

increased damping heavy damping

 The system oscillates at the driving requency.  The amplitude o the orced oscillations is fxed. Each cycle energy is dissipated as a result o damping and the driving orce does work on the system. The overall result is that the energy o the system remains constant.

q faCtor and damping The degree o damping is measured by a quantity called the quality actor or Q actor. It is a ratio (no units) and the defnition is: energy stored Q = 2 __ energy lost per cycle Since the energy stored is proportional to the square o amplitude o the oscillation, measurements o decreasing amplitude with time can be used to calculate the Q actor. The Q actor is approximately equal to the number o oscillations that are completed beore damping stops the oscillation.

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o p t i o n B  E n g i n E E r i n g p h ys i c s

driving frequency, fdriving natural frequency, fnatural Resonance occurs when a system is subject to an oscillating orce at exactly the same requency as the natural requency o oscillation o the system.

Typical orders o magnitude or dierent Q-actors: Car suspension: Simple pendulum: Guitar string: Excited atom:

1 1 03 1 03 1 07

When a system is in resonance and its amplitude is constant, the energy provided by the driving requency during one cycle is all used to overcome the resistive orces that cause damping. In this situation, the Q actor can be calculated as: energy stored Q = 2  resonant requency  __ power loss

HL

rc (2)

phase of forCed osCillations Ater transient oscillations have died down, the requency o the orced oscillations equals the driving requency. The phase relationship between these two oscillations is complex and depends on how close the driven system is to resonance:

phase lag /rad

driven vibration period behind

1 2



 2

heavy damping light damping in phase

driven vibration period behind

1 4

0 natural frequency

f/Hz forcing frequency

examples of resonanCe Comment Vibrations in machinery

When in operation, the moving parts o machinery provide regular driving orces on the other sections o the machinery. I the driving requency is equal to the natural requency, the amplitude o a particular vibration may get dangerously high. e.g. at a particular engine speed a trucks rear view mirror can be seen to vibrate.

Quartz oscillators

A quartz crystal eels a orce i placed in an electric feld. When the feld is removed, the crystal will oscillate. Appropriate electronics are added to generate an oscillating voltage rom the mechanical movements o the crystal and this is used to drive the crystal at its own natural requency. These devices provide accurate clocks or microprocessor systems.

Microwave generator

Microwave ovens produce electromagnetic waves at a known requency. The changing electric feld is a driving orce that causes all charges to oscillate. The driving requency o the microwaves provides energy, which means that water molecules in particular are provided with kinetic energy  i.e. the temperature is increased.

Radio receivers

Electrical circuits can be designed (using capacitors, resistors and inductors) that have their own natural requency o electrical oscillations. The ree charges (electrons) in an aerial will eel a driving orce as a result o the requency o the radio waves that it receives. Adjusting the components o the connected circuit allows its natural requency to be adjusted to equal the driving requency provided by a particular radio station. When the driving requency equals the circuits natural requency, the electrical oscillations will increase in amplitude and the chosen radio stations signal will dominate the other stations.

Musical instruments

Many musical instruments produce their sounds by arranging or a column o air or a string to be driven at its natural requency which causes the amplitude o the oscillations to increase.

Greenhouse eect

The natural requency o oscillation o the molecules o greenhouse gases is in the inra-red region. Radiation emitted rom the Earth can be readily absorbed by the greenhouse gases in the atmosphere. See page 92 or more details.

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169

iB questons  opton B  engneerng physcs 1.

A sphere o mass m and radius r rolls, without slipping, rom rest down an inclined plane. When it reaches the base o the plane, it has allen a vertical distance h. Show that the speed o the sphere, v, when it arrives at the base o the incline is given by:

4.

1 An adiabatic compression to 1 /20th o its original volume.

_____

v=



1 0gh _ 7

In a diesel engine, air is initially at a pressure o 1  1 0 5 Pa and a temperature o 27 C. The air undergoes the cycle o changes listed below. At the end o the cycle, the air is back at its starting conditions.

2 A brie isobaric expansion to 1 /1 0th o its original volume.

[4]

3 An adiabatic expansion back to its original volume. 2.

A fywheel o moment o inertia 0.75 kg m2 is accelerated uniormly rom rest to an angular speed o 8.2 rad s  1 in 6.5 s. a) Calculate the resultant torque acting on the fywheel during this time.

4 A cooling down at constant volume. a) Sketch, with labels, the cycle o changes that the gas undergoes. Accurate values are not required.

[2]

b) I the pressure ater the adiabatic compression has risen to 6.6  1 0 6 Pa, calculate the temperature o the gas. [2]

b) Calculate the rotational kinetic energy o the fywheel when it rotates at 8.2 rad s  1 [2]

c) In which o the our processes:

c) The radius o the fywheel is 1 5 cm. A breaking orce applied on the circumerence and brings it to rest rom an angular speed o 8.2 rad s  1 in exactly 2 revolutions. Calculate the value o the breaking orce. [2]

(i)

A xed mass o a gas undergoes various changes o temperature, pressure and volume such that it is taken round the pV cycle shown in the diagram below.

pressure/10 5 Pa

3.

[3]

is work done on the gas?

[1 ]

(ii) is work done by the gas?

[1 ]

(iii) does ignition o the air-uel mixture take place?

[1 ]

d) Explain how the 2nd law o thermodynamics applies to this cycle o changes.

[2]

HL 2.0

5.

X

With the aid o diagrams, explain a) What is meant by laminar fow b) The Bernoulli eect c) Pascals principle

1.0

Z

Y

1.0 2.0 3.0 4.0 5.0

d) An ideal fuid

[8]

6.

Oil, o viscosity 0.35 Pa s and density 0.95 g cm- 3 , fows through a pipe o radius 20 cm at a velocity o 2.2 m s - 1 . Deduce whether the fow is laminar or turbulent.

[4]

7.

A pendulum clock maintains a constant amplitude by means o an electric power supply. The ollowing inormation is available or the pendulum:

volume/10 3 m 3

The ollowing sequence o processes takes place during the cycle. X  Y the gas expands at constant temperature and the gas absorbs energy rom a reservoir and does 450 J o work. Y  Z the gas is compressed and 800 J o thermal energy is transerred rom the gas to a reservoir.

Maximum kinetic energy:

5  1 0- 2 J

Frequency o oscillation:

2 Hz

Q actor: 30

Z  X the gas returns to its initial stage by absorbing energy rom a reservoir.

Calculate: a) The driving requency o the power supply

[3]

a) Is there a change in internal energy o the gas during the processes X  Y? Explain.

b) The power needed to drive the clock.

[3]

[2]

b) Is the energy absorbed by the gas during the process X  Y less than, equal to or more than 450 J? Explain. [2] c) Use the graph to determine the work done on the gas during the process Y  Z.

[3]

d) What is the change in internal energy o the gas during the process Y  Z?

[2]

e) How much thermal energy is absorbed by the gas during the process Z  X? Explain your answer.

[2]

) What quantity is represented by the area enclosed by the graph? Estimate its value.

[2]

g) The overall eciency o a heat engine is dened as net work done by the gas during a cycle Eciency = ____ total energy absorbed during a cycle I this pV cycle represents the cycle or a particular heat engine determine the eciency o the heat engine. [2]

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15 o p t I o n C  I m a g I n g I fri Ray dIagRams

In order to fnd the location and nature o this image a ray diagram is needed.

I an object is placed in ront o a plane mirror, an image will be ormed.

image

object

upright same size as object laterally inverted

x

x

The process is as ollows:

The image ormed by reection in a plane mirror is always:

 Light sets o in all directions rom every part o the object. (This is a result o diuse reections rom a source o light.)

 the same distance behind the mirror as the object is in front

 Each ray o light that arrives at the mirror is reected according to the law o reection.

 upright (as opposed to being inverted)  the same size as the object (as opposed to being magnifed or diminished)

 These rays can be received by an observer.  The location o the image seen by the observer arises because the rays are assumed to have travelled in straight lines.

 laterally inverted (i.e. let and right are interchanged)  virtual (see below) .

Real and vIRtual Images object real image

(a) real image

(b) virtual image converging rays

O point object

 Upside down  Diminished  Real.

optical system

The opposite o a virtual image is a real image. In this case, the rays o light do actually pass through a single point. Real images cannot be ormed by plane mirrors, but they can be ormed by concave mirrors or by lenses. For example, i you look into the concave surace o a spoon, you will see an image o yoursel. This particular image is

concave mirror

I virtual point image

I real point image

O point object

optical system

The image ormed by reection in a plane mirror is described as a virtual image. This term is used to describe images created when rays o light seem to come rom a single point but in act they do not pass through that point. In the example above, the rays o light seem to be coming rom behind the mirror. They do not, o course, actually pass behind the mirror at all.

diverging rays

stICk In wateR The image ormed as a result o the reraction o light leaving water is so commonly seen that most people orget that the objects are made to seem strange. A straight stick will appear bent i it is placed in water. The brain assumes that the rays that arrive at ones eyes must have been travelling in a straight line.

air water

A straight stick appears bent when placed in water

The image o the end o the pen is:  Nearer to the surace than the pen actually is.  Virtual.

o pti o n C  i m ag i n g

171

Cri  ConveRgIng lenses

air

air

A converging lens brings parallel rays into one ocus point.

normal

converging lens

normal

parallel rays refraction at 1st surface

glass refraction at 2nd surface

The rays o light are all brought together in one point because o the particular shape o the lens. Any one lens can be thought o as a collection o dierent-shaped glass blocks. It can be shown that any thin lens that has suraces ormed rom sections o spheres will converge light into one ocus point.

focal point

The reason that this happens is the reraction that takes place at both suraces o the lens.

poweR of a lens

A converging lens will always be thicker at the centre when compared with the edges.

wave model of Image foRmatIon

The power o a lens measures the extent to which light is bent by the lens. A higher power lens bends the light more and thus has a smaller ocal length. The defnition o the power o a lens, P, is the reciprocal o the ocal length, f:

region in which waves are made to travel more slowly O

I

f is the ocal length measured in m

object (source of wave energy)

real image (point to which wave energy is concentrated)

P is the power o the lens measured in m- 1 or dioptres (dpt)

Formation o a real image by reraction (ignoring diraction)

1 P= _ f

A lens o power = +5 dioptre is converging and has a ocal length o 20 cm. When two thin lenses are placed close together their powers approximately add.

defInItIons When analysing lenses and the images that they orm, some technical terms need to be defned.  The curvature o each surace o a lens makes it part o a sphere. The centre o curvature or the lens surace is the centre o this sphere.

 The ocal point (principal ocus) o a lens is the point on the principal axis to which rays that were parallel to the principal axis are brought to ocus ater passing through the lens. A lens will thus have a ocal point on each side.  The ocal length is the distance between the centre o the lens and the ocal point.  The linear magnifcation, m, is the ratio between the size (height) o the image and the size (height) o the object. It has no units. h image size linear magnifcation, m = _ = _i object size ho

 The principal axis is the line going directly through the middle o the lens. Technically it joins the centres o curvature o the two suraces.

lens

centre of curvature

focal point

c

f

f

focal length

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o pti o n C  i m ag i n g

c principal axis (PA)

Ig fri i cvx  ImpoRtant Rays In order to determine the nature and position o the image created o a given object, we need to construct a scaled ray diagram o the set-up. In order to do this, we concentrate on the paths taken by three particular rays. As soon as the paths taken by two o these rays have been constructed, the paths o all the other rays can be inerred. These important rays are described below.

lens

distant object O

f

Any ray that was travelling parallel to the principal axis will be reracted towards the ocal point on the other side o the lens.

I

f

f

O f

I real image inverted same size

object between 2f and f

f

PA

PA real inverted magnied I

object at f f

O 2.

PA

f

O f

f

PA

object at 2f

Converging lens 1.

real image inverted diminished

f

Any ray that travelled through the ocal point will be reracted parallel to the principal axis.

PA virtual image upright image at innity

object closer than f f f 3.

I PA

f

O

PA

f virtual image upright magnied

Any ray that goes through the centre o the lens will be undeviated. Converging lens images

f f

PA

possIble sItuatIons A ray diagram can be constructed as ollows:  An upright arrow on the principal axis represents the object.  The paths o two important rays rom the top o the object are constructed.  This locates the position o the top o the image.  The bottom o the image must be on the principal axis directly above (or below) the top o the image. A ull description o the image created would include the ollowing inormation:  i it is real or virtual  i it is upright or inverted  i it is magnifed or diminished  its exact position. It should be noted that the important rays are just used to locate the image. The real image also consists o all the other rays rom the object. In particular, the image will still be ormed even i some o the rays are blocked o. An observer receiving parallel rays sees an image located in the ar distance (at infnity) .

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thi  i lens equatIon There is a mathematical method o locating the image ormed by a lens. An analysis o the angles involved shows that the ollowing equation can be applied to thin spherical lenses: 1 = _ 1 + _ 1 _ f v u

lIneaR magnIfICatIon In all cases, linear magnifcation, h v m = _i = -  _ u h o

height o image m = __ height o object h = _i ho

object f

f

v = - _ u

image f

object distance u Suppose

For real images, m is negative and image is inverted

f

For virtual images m is positive and image is upright

image distance v

u = 25 cm f = 1 0 cm

1 = _ 5 - _ 3 1 - _ 1 = _ 1 - _ 1 =_ 2 = _ This would mean that _ v u 10 25 50 50 50 f 50 In other word, v = _ = 1 6.7 cm i.e. image is real 3 - 1 6.7 In this case m = _ = - 1 .67 and inverted. 10

Real Is posItIve Care needs to be taken with virtual images. The equation does work but or this to be the case, the ollowing convention has to be ollowed:  Distances are taken to be positive i actually traversed by the light ray (i.e. distances to real object and image) .  Distances are taken to be negative i apparently traversed by the light ray (distances to virtual objects and images) .  Thus a virtual image is represented by a negative value or v  in other words, it will be on the same side o the lens as the object.

image object f

object distance u negative image distance v

Suppose

u = 1 0 cm f = 25 cm

1 = _ 5 = - _ 3 1 - _ 1 = _ 1 - _ 1 = _ 2 -_ This would mean that _ v u 25 10 50 50 50 f 50 In other word, v = - _ = - 1 6.7 cm i.e. image is virtual 3 1 6.7 = +1 .67 and upright In this case m = + _ 10

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o pti o n C  i m ag i n g

f

diri  dIveRgIng lenses A diverging lens spreads parallel rays apart. These rays appear to all come rom one ocus point on the other side o the lens.

concave lens

When constructing ray diagrams or diverging lenses, the important rays whose paths are known (and rom which all other ray paths can be inerred) are: 1 . Any ray that was travelling parallel to the principal axis will be reracted away rom a ocal point on the incident side o the lens.

focal point f

focal length

f

PA

2. Any ray that is heading towards the ocal point on the other side o the lens, will be reracted so as to be parallel to the principal axis.

The reason that this happens is the reraction that takes place at both suraces. A diverging lens will always be thinner at the centre when compared with the edges.

f

f

PA

defInItIons and ImpoRtant Rays Diverging lenses have the same analogous defnitions as converging lenses or all o the ollowing terms:

3. Any ray that goes through the centre o the lens will be undeviated.

Centre o curvature, principal axis, ocal point, ocal length, linear magnifcation.

f

Note that:

f

 The ocal point is the point on the principal axis from which rays that were parallel to the principal axis appear to come ater passing through the lens.

PA

 As the ocal point is behind the diverging lens, the focal length of a diverging lens is negative.

Images CReated by a dIveRgIng lens Whatever the position o the object, a diverging lens will always create an upright, diminished and virtual image located between the ocal point and the lens on the same side o the lens as the object.

If you look at an object through a concave lens, it will look smaller and closer.

If you move the object further out, the image will not move as much.

object f f

u image

f

v

object inside focal length

object

image

object outside focal length

The thin lens equation will still work providing one remembers the negative ocal length o a diverging lens. For example, i an object is placed at a distance 2l away rom a diverging lens o ocal length l, the image can be calculated as ollows: Given: u = 2l,  = - l, v = ? 1 + _ 1 = _ 1 _ u v  1 = _ -3 1 - _ 1 = _ 1 - _ 1 = _ _ v u  -l 2l 2l 2l _  v = - 3 1 This is a virtual diminished and upright image with m = + _ 3

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Cvri  ivri irrr geometRy of mIRRoRs and lenses

Image foRmatIon In mIRRoRs

The geometry o the paths o rays ater reection by a spherical concave or convex mirror is exactly analogous to the paths o rays through converging or diverging lenses. The only dierence is that mirrors reect all rays backwards whereas rays pass through lenses and continue orwards.

(1) Concave object at innity F

2f

(a) Convex lens

real PA inverted diminished

I object between innity and 2f

PA

O 2f

f (b) Concave mirror

real PA inverted diminished

F

I

object at 2f

O 2f

PA

F

real PA inverted same size

F

real PA inverted magnifed

I object between 2f and f f

O

(c) Concave lens

2f I object at f

PA

O 2f f

PA

F

virtual upright image at infnity

object between f and mirror

(d) Convex mirror

2f F

O

I

PA

virtual upright magnifed

PA f

(2) Convex object at innity

This analogous behaviour means that all the defnitions and equations or lenses can be used (with suitable attention to detail with the sign conventions) with mirrors.

I F

An additional important ray or mirrors is the ray that travels through (or towards) the centre o curvature o the mirror (located at twice the ocal length) . This ray will be reected back along the same path.

object near lens

O

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o pti o n C  i m ag i n g

2f

virtual upright PA diminished

I

F

virtual PA upright 2f diminished

th i iyi  neaR and faR poInt

angulaR sIze

The human eye can ocus objects at dierent distances rom the eye. Two terms are useul to describe the possible range o distances  the near point and the ar point distance.

I we bring an object closer to us (and our eyes are still able to ocus on it) then we see it in more detail. This is because, as the object approaches, it occupies a bigger visual angle. The technical term or this is that the object subtends a larger angle.

 The distance to the near point is the distance between the eye and the nearest object that can be brought into clear ocus (without strain or help rom, or example, lenses) . It is also known as the least distance o distinct vision. By convention it is taken to be 25 cm or normal vision.  The distance to the ar point is the distance between the eye and the urthest object that can be brought into ocus. This is taken to be infnity or normal vision.

objects are the same size angle subtended by close object

distant object

angulaR magnIfICatIon

close object

angle subtended by distant object 1.

The angular magnifcation, M, o an optical instrument is defned as the ratio between the angle that an object subtends normally and the angle that its image subtends as a result o the optical instrument. The normal situation depends on the context. It should be noted that the angular magnifcation is not the same as the linear magnifcation.

Image ormed at infnity In this arrangement, the object is placed at the ocal point. The resulting image will be ormed at infnity and can be seen by the relaxed eye.

i = hf

top rays from object at specied distance

o

h

i

bottom

f i

top

eye focused on innity

f rays from nal image formed by optical instrument

i

In this case the angular magnifcation would be h __

 f D Min f n ity = _i = _ = _ h __ f o D

bottom

This is the smallest value that the angular magnifcation can be.

 Angular magnifcation, M = _i o

2.

The largest visual angle that an object can occupy is when it is placed at the near point. This is oten taken as the normal situation.

o = h D

Image ormed at near point In this arrangement, the object is placed nearer to the lens. The resulting virtual image is located at the near point. This arrangement has the largest possible angular magnifcation.

M=

i h i /D h i D = = = o h/D h a

1 1 1 + = u v f 1 1 1  - = a D f D D = + 1  a f

h o hi D A simple lens can increase the angle subtended. It is usual to consider two possible situations.

h f

 i a



f

i

D D +1 So the magnitude o Mn e a r p o in t = _ f

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177

abrrin spheRICal A lens is said to have an aberration if, for some reason, a point object does not produce a perfect point image. In reality, lenses that are spherical do not produce perfect images. Spherical aberration is the term used to describe the fact that rays striking the outer regions of a spherical lens will be brought to a slightly different focus point from those striking the inner regions of the same lens. This is not to be confused with barrel distortion.

ray striking outer regions

ray striking inner regions

In general, a point object will focus into a small circle of light, rather than a perfect point. There are several possible ways of reducing this effect:  the shape of the lens could be altered in such a way as to correct for the effect. The lens would, of course, no longer be spherical. A particular shape only works for objects at a particular distance away.

ray striking outer regions

range of focal points outer sections of lens not used

 the effect can be reduced for a given lens by decreasing the aperture. The technical term for this is stopping down the aperture. The disadvantage is that the total amount of light is reduced and the effects of diffraction (see page 46) would be made worse.

aperture

The effect for mirrors can be eliminated for all point objects on the axis by using a parabolic (as opposed to a spherical) mirror. For mirrors, the effect can again be reduced by using a smaller aperture. Spherical aberration

ChRomatIC Chromatic aberration is the term used to describe the fact that rays of different colours will be brought to a slightly different focus point by the same lens. The refractive index of the material used to make the lens is different for different frequencies of light.

white light

A point object will produce a blurred image of different colours. The effect can be eliminated for two given colours (and reduced for all) by using two different materials to make up a compound lens. This compound lens is called an achromatic doublet. The two types of glass produce equal but opposite dispersion.

white light

converging lens of crown glass (low dispersion)

Achromatic doublet

o pti o n C  i m ag i n g

violet

R

violet

R red

Mirrors do not suffer from chromatic aberration.

178

V

red

V violet focus

Canada balsam cement diverging lens of int glass ( high dispersion)

red focus

th c icrc  ric c Compound mICRosCope A compound microscope consists o two lenses  the objective lens and the eyepiece lens. The frst lens (the objective lens) orms a real magnifed image o the object being viewed. This real image can then be considered as the object or the second lens (the eyepiece lens) which acts as a magniying lens. The rays rom this real image travel into the eyepiece lens and they orm a virtual magnifed image. In normal adjustment, this virtual image is arranged to be located at the near point so that maximum angular magnifcation is obtained.

objective lens fo

top half of object B h O

real image formed by objective lens h 1 i

fo

virtual image of top half of object

construction line

eyepiece lens fe

construction line

i fe eye focused on near point to see virtual image (in practice it would be much nearer to the eyepiece lens than implied here)

h2

M D h2

__ h  h h D M = _i = _ = _2 = _2  _1 = linear magnifcation produced by eyepiece  linear magnifcation produced by objective h __ h h1 h o D

astRonomICal telesCope An astronomical telescope also consists o two lenses. In this case, the objective lens orms a real but diminished image o the distant object being viewed. Once again, this real image can then be considered as the object or the eyepiece lens acting as a magniying lens. The rays rom this real image travel into the eyepiece lens and they orm a virtual magnifed image. In normal adjustment, this virtual image is arranged to be located at infnity.

objective lens

eyepiece lens fo

parallel rays all from top of distant object o

o

fe

real image formed in mutual focal plane of lenses fe fo h 1 i

construction line i

eye focused on innity

virtual image at innity h1 __

 f f M = _i = _ = _o h f o __ e e

1

fo

The length o the telescope  fo + fe .

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aric rfci c CompaRIson o ReleCtIng and ReRaCtIng telesCopes A reracting telescope uses an objective (converging) lens to orm a real diminished image o a distant object. This image is then viewed by the eyepiece lens (converging) which, acting as a simple magniying glass, produces a virtual but magnifed fnal image.

newtonIan mountIng A small at mirror is placed on the principal axis o the mirror to reect the image ormed to the side:

concave mirror

small at mirror Fo

In an analogous way, a reecting telescope uses a concave mirror set up so as to orm a real, diminished image o a distant object. This image, however, would be difcult to view as it would be produced in ront o the concave mirror. Thus mirrors are used to produce a viewable image that can, like the reracting telescope, be viewed by the eyepiece lens (converging) . Once again the eyepiece acts as a simple magniying glass and produces the virtual, but magnifed, fnal image. Two common mountings or reecting telescopes are the Newtonian mounting and the Cassegrain mounting.

F'o eyepiece lens

All telescopes are made to have large apertures in order to:

CassegRaIn mountIng

a) reduce diraction eects, and

A small convex mirror is mounted on the principal axis o the mirror. The mirror has a central hole to allow the image to be viewed.

b) collect enough light to make bright images o low power sources. Large telescopes are reecting because:  Mirrors do not suer rom chromatic aberration

The convex mirror will add to the angular magnifcation achieved.

concave m irror

 It is difcult to get a uniorm reractive index throughout a large volume o glass  Mounting a large lens is harder to achieve than mounting a large mirror.

Fext

 Only one surace needs to be the right shape. Reecting telescopes can easily suer damage to the mirror surace.

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o pti o n C  i m ag i n g

eyepiece lens

small convex m irror

Fo

Ri c sIngle dIsh RadIo telesCopes

RadIo InteRfeRometRy telesCopes

A single dish radio telescope operates in a very similar way to a reecting telescope. Rather than reecting visible light to orm an image, the much longer wavelengths o radio waves are reected by the curved receiving dish. The antenna that is the receiver o the radio waves can be tuned to pick up specifc wavelengths under observation and are used to study naturally occurring radio emission rom stars, galaxies, quasars and other astronomical objects between wavelengths o about 1 0 m and 1 mm.

The angular resolution o a radio telescope can be improved using a principle called intererometry. This process analyses signals received at two (or more) radio telescopes that are some distance apart but pointing in the same direction. This eectively creates a virtual radio telescope that is much larger than any o the individual telescopes.

Radio telescope incoming radio waves

Radio waves reect o the dish and focus at the tip. Receivers detect and amplify radio signals. Diraction eects can signifcantly limit the accuracy with which a radio telescope can locate individual sources o radio signals. Increasing the diameter o a radio telescope improves the telescopes ability to resolve dierent sources and ensure that more power can be received (see resolution on page 1 01 ) .

The technique is complex as it involves collecting signals rom two or more radio telescopes (an array telescope) in one central location. The arrival o each signal at an individual antenna needs to be careully calibrated against a single shared reerence signal so that dierent signals can be combined as though they arrived at one single antenna. When the signals rom the dierent telescopes are added together, they interere. The result is to create a combined telescope that is equivalent in resolution (though not in sensitivity) to a single radio telescope whose diameter is approximately equal to the maximum separations o the antennae. The principle can be extended, in a process called Very Long Baseline Interferometry, to allow recordings o radio signals (originally made hundreds o km apart) to be synchronized to within a millionth o a second thus allowing scientists rom dierent countries to collaborate to create a virtual radio telescope o huge size and high resolving power.

CompaRatIve peRfoRmanCe of eaRth-bound and satellIte-boRne telesCopes The ollowing points about Earth-based (EB) and satellite-borne (SB) telescopes can be made:  SB observations are ree rom intererence and/or absorptions due to the Earths atmosphere that hinder EB observations, giving better resolution or SB telescopes.  Modern computer techniques can eectively correct or many atmospheric eects making new ground-based telescopes similar in resolution to some SB telescopes.  Many signifcant wavelengths o EM radiation (UV, IR and long wavelength radio) are absorbed by the Earths atmosphere so SB telescopes are the only possibility in their wavelengths.  SB observations do not suer rom light pollution / radio intererence as a result o nearby human activity.  SB acilities are not subject to continual wear and tear as a result o the Earths atmosphere (storms etc.) .  The possibility o damage rom space debris exists or SB telescopes.  There is a great deal o added cost in getting the telescope into orbit and controlling it remotely, meaning that SB telescopes are signifcantly more expensive to build and this places a limit on their size and weight.  There is an added difcultly o eecting repairs / alterations to a SB telescope once operational.  SB telescopes need to withstand wider temperature variations than EB telescopes.  EB optical telescopes can only operate at night whereas SB telescopes can operate at all times.

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181

fir ic optIC fIbRe Optic fbres use the principle o total internal reection (see page 45) to guide light along a certain path. The idea is to make a ray o light travel along a transparent fbre by bouncing between the walls o the fbre. So long as the incident angle o the ray on the wall o the fbre is always greater than the critical angle, the ray will always remain within the fbre even i the fbre is bent (see right) .

 In the medical world. Bundles o optic fbres can be used to carry images back rom inside the body. This instrument is called an endoscope.  This type o optic fbre is known as a step-index optic fbre. Cladding o a material with a lower reractive index surrounds the fbre. This cladding protects and strengthens the fbre.

As shown on page 45, the relation between critical angle, c, and reractive index n is given by 1 n=_ sin c Two important uses o optic fbres are:  In the communication industry. Digital data can be encoded into pulses o light that can then travel along the fbres. This is used or telephone communication, cable TV etc.

types of optIC fIbRes The simplest fbre optic is a step-index fbre. Technically this is known as a multimode stepindex fbre. Multimode reers to the act that light can take dierent paths down the fbre which can result in some distortion o signals (see waveguide dispersion, page 1 83) . The (multimode) graded-index fbre is an improvement. This uses a graded reractive index profle in the fbre meaning that rays travel at dierent speeds depending on their distance rom the centre. This has the eect o reducing the spreading out o the pulse. Most fbres used in data communications have a graded index. The optimum solution is to have a very narrow core  a singlemode step-index fbre.

index index output prole pulse pulse n2 n1 multimode step-index n2 n1

multimode graded-index n2 n1 singlemode step-index

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diri, i  i i ic fbr mateRIal dIspeRsIon

waveguIde dIspeRsIon

The reractive index o any substance depends on the requency o electromagnetic radiation considered. This is the reason that white light is dispersed into dierent colours when it passes through a triangular prism.

I the optical fbre has a signifcant diameter, another process called waveguide dispersion that can cause the stretching o a pulse is multipath or modal dispersion. The path length along the centre o a fbre is shorter than a path that involves multiple reections. This means that rays rom a particular pulse will not all arrive at the same time because o the dierent distances they have travelled.

As light travels along an optical fbre, dierent requencies will travel at slightly dierent speeds. This means that i the source o light involves a range o requencies, then a pulse that starts out as a square wave will tend to spread out as it travels along the fbre. This process is known as material dispersion.

B C A cladding

after transmission

attenuatIon As light travels along an optic fbre, some o the energy can be scattered or absorbed by the glass. The intensity o the light energy that arrives at the end o the fbre is less than the intensity that entered the fbre. The signal is said to be attenuated. The amount o attenuation is measured on a logarithmic scale in decibels (dB) . The attenuation is given by I attenuation (dB) = 1 0log _ Io

I is the intensity o the output power measured in W Io is the intensity o the original input power measured in W A negative attenuation means that the signal has been reduced in power. A positive attenuation would imply that the signal has been amplifed. See page 1 88 or another example o the use o the decibel scale. It is common to quote the attenuation per unit length as measured in dB km- 1 . For example, 5 km o fbre optic cable causes an input power o 1 00 mW to decrease to 1 mW. The attenuation per unit length is calculated as ollows: attenuation = 1 0 log (1 0 /1 0 ) = 1 0 log (1 0 ) = -20 dB -3

-1

-2

core

The problems caused by modal dispersion have led to the development o monomode (or singlemode) step-index fbres. These optical fbres have very narrow cores (o the same order o magnitude as the wavelength o the light being used (approximately 5 m) so that there is only one eective transmission path  directly along the fbre.

The attenuation o a 1 0 km length o this fbre optic cable would thereore be - 40 dB. The overall attenuation resulting rom a series o actors is the algebraic sum o the individual attenuations. The attenuation in an optic fbre is a result o several processes: those caused by impurities in the glass, the general scattering that takes place in the glass and the extent to which the glass absorbs the light. These last two actors are aected by the wavelength o light used. A typical the overall attenuation is shown below:

6 5 4 3 2 1

attenuation per unit length / dB km -1

before transmission

paths

0.6

0.8

1.0

1.2

1.4

1.6 1.8 wavelength / m

attenuation per unit length = -20 dB/5 km = -4 dB km- 1

CapaCIty

noIse, amplIIeRs and ReshapeRs

Attenuation causes an upper limit to the amount o digital inormation that can be sent along a particular type o optical fbre. This is oten stated in terms o its capacity.

Noise is inevitable in any electronic circuit. Any dispersions or scatterings that take place within an optical fbre will also add to the noise.

capacity o an optical fbre = bit rate  distance A fbre with a capacity o 80 Mbit s km can transmit 80 Mbit s - 1 along a 1 km length o fbre but only 20 Mbit s - 1 along a 4 km length. -1

An amplifer increases the signal strength and thus will tend to correct the eect o attenuation  these are also sometimes called regenerators. An amplifer will also increase any noise that has been added to the electrical signal. A reshaper can reduce the eects o noise on a digital signal by returning the signal to a series o 1 s and 0s with sharp transitions between the allowed levels.

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Channels of communication The table below shows some common communication links.

Wire pairs (twisted pair)

copper wire insulation Coaxial cables

copper wire insulation copper mesh outside insulation

Options for communication

Uses

Advantages and disadvantages

Two wires can connect the sender and receiver o inormation. For example a simple link between a microphone, an amplifer and a loudspeaker.

Very simple communication systems e.g. intercom

Very simple and cheap.

This arrangement o two wires reduces electrical intererence. A central wire is surrounded by the second wire in the orm o an outer cylindrical copper tube or mesh. An insulator separates the two wires.

Coaxial cables are used to transer signals rom TV aerials to TV receivers. Historically they are standard or underground telephone links.

Susceptible to noise and intererence. Unable to transer inormation at the highest rates.

Simple and straightorward. Less susceptible to noise compared to simple wire pair but noise still a problem.

Wire links can carry requencies up to about 1 GHz but the higher requencies will be attenuated more or a given length o wire. A typical 1 00 MHz signal sent down low-loss cable would need repeaters at intervals o approximately 0.5 km. The upper limit or a single coaxial cable is approximately 1 40 Mbit s - 1 .

Optical fbres

Laser light can be used to send signals down optical fbres with approximately the same requency limit as cables 1  GHz. The attenuation in an optical fbre is less than in a coaxial cable. The distance between repeaters can easily be tens (or even hundreds) o kilometres.

Long-distance telecommunication and high volume transer o digital data including video data.

Compared to coaxial cables with equivalent capacity, optical fbres:  have a higher transmission capacity  are much smaller in size and weight  cost less  allow or a wider possible spacing o regenerators  oer immunity to electromagnetic intererence  suer rom negligible cross talk (signals in one channel aecting another channel)  are very suitable or digital data transmission  provide good security  are quiet  they do not hum even when carrying large volumes o data. There are some disadvantages:  the repair o fbres is not a simple task  regenerators are more complex and thus potentially less reliable.

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x-r

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IntensIty, qualIty and attenuatIon

basIC x-Ray dIsplay teChnIques

The eects o X-rays on matter depend on two things, the intensity and the quality o the X-rays.

The basic principle o X-ray imaging is that some body parts (or example bones) will attenuate the X-ray beam much more than other body parts (or example skin and tissue) . Photographic flm darkens when a beam o X-rays is shone on them so bones show up as white areas on an X-ray picture.

 The intensity, I, is the amount o energy per unit area that is carried by the X-rays.  The quality o the X-ray beam is the name given to the spread o wavelengths that are present in the beam. Low-energy photons will be absorbed by all tissues and potentially cause harm without contributing to orming the image. It is desirable to remove these rom the beam. I the energy o the beam is absorbed, then it is said to be attenuated. I there is nothing in the way o an X-ray beam, it will still be attenuated as the beam spreads out. Two processes o attenuation by matter, simple scattering and the photoelectric eect are the dominant ones or low-energy X-rays.

scattering

photoelectric eect X-ray photon

low  X-ray photon

electron electron

low  X-ray photon

X-ray beam

X-ray photograph The sharpness o an X-ray image is a measure o how easy it is to see edges o dierent organs or dierent types o tissue. X-ray beams will be scattered in the patient being scanned and the result will be to blur the fnal image and to reduce the contrast and sharpness. To help reduce this eect, a metal flter grid is added below the patient:

X-ray beam

light photon

Simple scattering aects X-ray photons that have energies between zero and 30 keV.  In the photoelectric eect, the incoming X-ray has enough energy to cause one o the inner electrons to be ejected rom the atom. It will result in one o the outer electrons alling down into this energy level. As it does so, it releases some light energy. This process aects X-ray photons that have energies between zero and 1 00 keV.

X transmission

Both attenuation processes result in a near exponential transmission o radiation as shown in the diagram below. For a given energy o X-rays and given material there will be a certain thickness that reduces the intensity o the X-ray by 50% . This is known as the hal-value thickness.

100%

50%

I = e -x I0

half-value thickness

thickness of absorber, x

The attenuation coefcient  is a constant that mathematically allows us to calculate the intensity o the X-rays given any thickness o material. The equation is as ollows: I = I0 e -  x The relationship between the attenuation coefcient and the hal-value thickness is  x_1 = ln 2 2

x_1

The hal-value thickness o the material (in m)

2

ln 2 The natural log o 2. This is the number 0.6931 

The attenuation coefcient (in m- 1 )

 depends on the wavelength o the X-rays  short wavelengths are highly penetrating and these X-rays are hard. Sot x-rays are easily attenuated and have long wavelengths.

patient

metal grid X-ray lm Alternatively computer sotware can be used to detect and enhance edges. Since X-rays cause ionizations, they are dangerous. This means that the intensity used needs to be kept to an absolute minimum. This can be done by introducing something to intensiy (to enhance) the image. There are two simple techniques o enhancement:  When X-rays strike an intensiying screen the energy is reradiated as visible light. The photographic flm can absorb this extra light. The overall eect is to darken the image in the areas where X-rays are still present (see page 1 87) .  In an image-intensifer tube, the X-rays strike a uorescent screen and produce light. This light causes electrons to be emitted rom a photocathode. These electrons are then accelerated towards an anode where they strike another uorescent screen and give o light to produce an image.

mass attenuatIon CoeffICIent An alternative way o writing the equation or the attenuation coefcient is shown below: - _ x I = I0 e (  ) 



Where  is the density o the substance. In this ormat, _  is  known as the mass attenuation coefcient _ , and x is  known as the area density or mass thickness. 

2 -1 Units o mass attenuation coefcient, _  = m kg -2 Units o area density, x = kg m

I = I0 e

()

 - _  x

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x-ray imaging techniques

1 ) Intensiying screens

3) Tomography

The arrangement o the intensiying screens described on page 1 85 are shown below.

X-rays emerging from patient cassette front (plastic) front intensifying screen: phosphor double-sided lm rear intensifying screen: phosphor felt padding cassette

Tomography is a technique that makes the X-ray photograph ocus on a certain region or slice through the patient. All other regions are blurred out o ocus. This is achieved by moving the source o X-rays and the flm together.

motion about 12 mm

X-ray tube

pivot point plane of cut A B

With a simple X-ray photograph it is hard to identiy problems within sot tissue, or example in the gut. There are two general techniques aimed at improving this situation. 2) Barium meals In a barium meal, a dense substance is swallowed and its progress along the gut can be monitored. The contrast between the gut and surrounding tissue is increased. Typically the patient is asked to swallow a harmless solution o barium sulate. The result is an increase in the sharpness o the image.

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X-ray table

lm A B B

A B motion

An extension o basic tomography is the computed tomography scan or CT scan. In this set-up a tube sends out a pulse o X-rays and a set o sensitive detectors collects inormation about the levels o X-radiation reaching each detector. The X-ray source and the detectors are then rotated around a patient and the process is repeated. A computer can analyse the inormation recorded and is able to reconstruct a 3-dimensional map o the inside o the body in terms o X-ray attenuation.

HL

uric igig

ultRasound

a- and b-sCans

The limit o human hearing is about 20 kHz. Any sound that is o higher requency than this is known as ultrasound. Typically ultrasound used in medical imaging is just within the MHz range. The velocity o sound through sot tissue is approximately 1 500 m s - 1 meaning that typical wavelengths used are o the order o a ew millimetres.

There are two ways o presenting the inormation gathered rom an ultrasound probe, the A-scan or the B-scan. The A-scan (amplitude-modulated scan) presents the inormation as a graph o signal strength versus time. The B-scan (brightness-modulated scan) uses the signal strength to aect the brightness o a dot o light on a screen.

pulse vertebra

echo A-scan display

reections from boundaries

abdominal wall

organ

to scan display

strength of signal

Unlike X-rays, ultrasound is not ionizing so it can be used very saely or imaging inside the body  with pregnant women or example. The basic principle is to use a probe that is capable o emitting and receiving pulses o ultrasound. The ultrasound is reected at any boundary between dierent types o tissue. The time taken or these reections allows us to work out where the boundaries must be located.

vertebra organ

time B-scan display

probe

path of ultrasound

aCoustIC ImpedanCe The acoustic impedance o a substance is the product o the density, , and the speed o sound, c. Z = c

A-scans are useul where the arrangement o the internal organs is well known and a precise measurement o distance is required. I several B-scans are taken o the same section o the body at one time, all the lines can be assembled into an image which represent a section through the body. This process can be achieved using a large number o transducers.

(i) probe

placenta

(ii)

(iii)

unit o Z = kg m- 2 s - 1 Very strong reections take place when the boundary is between two substances that have very dierent acoustic impedances. This can cause some difculties.  In order or the ultrasound to enter the body in the frst place, there needs to be no air gap between the probe and the patients skin. An air gap would cause almost all o the ultrasound to be reected straight back. The transmission o ultrasound is achieved by putting a gel or oil (o similar density to the density o tissue) between the probe and the skin.  Very dense objects (such as bones) can cause a strong reection and multiple images can be created. These need to be recognized and eliminated.

original reection

2nd reection reected by bone back to probe

path of ultrasound organ

beam reected from bone 2nd reection back ward b n

foetal skull

limbs (i)

( ii) (iii)

Building a picture rom a series o B-scan lines

ChoICe of fRequenCy The choice o requency o ultrasound to use can be seen as the choice between resolution and attenuation.  Here, the resolution means the size o the smallest object that can be imaged. Since ultrasound is a wave motion, diraction eects will be taking place. In order to image a small object, we must use a small wavelength. I this was the only actor to be considered, the requency chosen would be as large as possible.

pIezoeleCtRIC CRystals

 Unortunately attenuation increases as the requency o ultrasound increases. I very high requency ultrasound is used, it will all be absorbed and none will be reected back. I this was the only actor to be considered, the requency chosen would be as small as possible.

These quartz crystals change shape when an electric current ows and can be used with an alternating pd to generate ultrasound. They also generate pds when receiving sound pressure waves so one crystal is used or generation and detection.

On balance the requency chosen has to be somewhere between the two extremes. It turns out that the best choice o requency is oten such that the part o the body being imaged is about 200 wavelengths o ultrasound away rom the probe.

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Igig ci

RelatIve IntensIty levels of ultRasound The relative intensity levels o ultrasound between two points are compared using the decibel scale (dB) . As its name suggests, the decibel unit is simply one tenth o a base unit that is called the bel (B) . The decibel scale is logarithmic.

Mathematically, Relative intensity level in bels, intensity level o ultrasound at measurement point L I = log _____ intensity level o ultrasound at reerence point I I0

or Relative intensity level in bels = log _1 Since 1 bel = 1 0 dB, I I0

Relative intensity level in decibels, LI = 1 0 log _1

nmR Nuclear Magnetic Resonance (NMR) is a very complicated process but one that is extremely useul. It can provide detailed images o sections through the body without any invasive or dangerous techniques. It is o particular use in detecting tumours in the brain. It involves the use o a non-uniorm magnetic feld in conjuction with a large uniorm feld. In outline, the process is as ollows:

 I a pulse o radio waves is applied at the Larmor requency, the nuclei can absorb this energy in a process called resonance. The protons make a spin transition.  Ater the pulse, the nuclei return to their lower energy state by emitting radio waves.  The time over which radio waves are emitted is called the relaxation time.

 The nuclei o atoms have a property called spin.

 The radio waves emitted and their relaxation times can be processed to produce the NMR scan image.

 The spin o these nuclei means that they can act like tiny magnets.

 The signal analysis is targeted at the hydrogen nuclei (protons) present.

 These nuclei will tend to line up in a strong magnetic feld.

 The number o H nuclei varies with the chemical composition so dierent tissues extract dierent amounts o energy rom the applied signal.

RF generator eld gradient coils

 Thus RF signal orces protons to make a spin transition and

S

 The gradient feld allows determination o the point rom which the photons are emitted.

RF coil receiver body

 The proton spin relaxation time depends on the type o tissue at the point where the radiation is emitted.

N

permanent magnet

relaxation time oscilloscope

 They do not, however, perectly line up  they oscillate in a particular way that is called precession. This happens at a very high requency  the same as the requency o radio waves.  The particular requency o precession depends on the magnetic feld and the particular nucleus involved. It is called the Larmor frequency.

CompaRIson between ultRasound and nmR The ollowing points can be noted:

 Ultrasound waves do not enter the body easily and multiple reections can reduce the clarity o the image.

 NMR imaging is very expensive when compared with ultrasound equipment and is very bulky  patient needs to be brought to the NMR machine and process is time consuming.

 Both wave energies carry energy but the energy associated with the ultrasound is greater that the energy associated with the radio requencies used in NMR.

 Ultrasound measurements are easy to perorm (equipment can be brought to patient at the point o care) and can be repeated as required but quality o image can rely on skill o operator.

 At the radio requencies used in NMR there is no danger o resonance but some ultrasound energy can cause heating.

 NMR produces a 3-dimensional scan, ultrasound typically produces a 2-dimensional scan.  Detail produced by NMR is greater than by ultrasound.  NMR particularly useul or very delicate areas o body e.g. brain.  NMR patients have to remain very still, ultrasound images can be more dynamic.

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 Ultrasound can cause cavitation  the production o small gas bubbles which will absorb energy and can damage surrounding tissue. The requencies and intensities used or diagnostics avoid this possibility as much as possible.  The strong magnetic felds used in NMR present problems or patients with surgical implants and / or pacemakers.

Ib questions  option C  imaging 1.

For each o the ollowing situations, locate and describe the fnal image ormed. Solutions should ound using scale diagrams and mathematically.

The purpose o this particular scan is to fnd the depth d o the organ labelled O below the skin and also to fnd its length, I.

a) An object is placed 7 cm in ront o a concave mirror o ocal length 1 4 cm. [4]

b) (i) Suggest why a layer o gel is applied between the ultrasound transmitter/receiver and the skin.

b) A diverging lens o ocal length 1 2.0 cm is placed at the ocal point o a converging lens o ocal length 8.0 cm. An object is placed 1 6.0 cm in ront o the converging lens. [4]

On the graph below the pulse strength o the reected pulses is plotted against the time lapsed between the pulse being transmitted and the time that the pulse is received, t.

2.

pulse strength / relative units

c) An object is placed 1 8.0 cm in ront o a convex lens o ocal length 6.0 cm. A second convex lens o ocal length 3.0 cm is an additional 1 8 cm behind the frst lens. [4] A student is given two converging lenses, A and B, and a tube in order to make a telescope. a) Describe a simple method by which she can determine the ocal length o each lens. [2] b) She fnds the ocal lengths to be as ollows: 1 0 cm

Focal length o lens B

50 cm

(iii) The mean speed in tissue and muscle o the ultrasound used in this scan is 1 .5  1 0 3 ms - 1 . Using data rom the above graph, estimate the depth d o the organ beneath the skin and the length l o the organ O.

labels or each lens;

(iii) the position o the eye when the telescope is in use. [4] c) On your diagram, mark the location o the intermediate image ormed in the tube. [1 ] d) Is the image seen through the telescope upright or upside-down?

[1 ]

e) Approximately how long must the telescope tube be?

[1 ]

Explain what is meant by a) Material dispersion

e) A Cassegrain mounting

b) Waveguide dispersion

) Total Internal reection

c) Spherical aberrations

g) Step-index fbres

d) Chromatic aberrations 4.

gain = 20 dB

[1 ]

d) State one advantage and one disadvantage o using ultrasound as opposed to using X-rays in medical diagnosis.

[2]

a) State and explain which imaging technique is normally used

output

[2]

20 15 10

gain = 30 dB

5 0

a) the overall gain o the system b) the output power.

[2]

The graph below shows the variation o the intensity I o a parallel beam o X-rays ater it has been transmitted through a thickness x o lead.

Calculate

0

2

4

6

8

[2] b) ( i)

HL 5.

to detect a broken bone

(ii) to examine the growth o a etus.

[2 each]

optical bre

6.

[4]

c) The above scan is known as an A-scan. State one way in which a B-scan diers rom an A-scan.

(i)

A 1 5 km length o optical fbre has an attenuation o 4 dB km- 1 . A 5 mW signal is sent along the wire using two amplifers as represented by the diagram below.

input power = 5 mW

C

(ii) Indicate on the diagram the origin o the reected pulses A, B and C and D. [2]

(ii) the ocal points or each lens;

3.

D

B

0 25 50 75 100 125 150 175 200 225 250 275 300 t / s

Draw a diagram to show how the lenses should be arranged in the tube in order to make a telescope. Your diagram should include: (i)

A

intensity

Focal length o lens A

[2]

10 12 x / mm

D efne half-value thickness, x _1 .

[2 ]

2

This question is about ultrasound scanning. a) State a typical value or the requency o ultrasound used in medical scanning. The diagram below shows an ultrasound transmitter and receiver placed in contact with the skin.

d ultrasound transmitter and receiver layer of skin and fat

[1 ]

( ii) Use the graph to estimate x _1 or this beam in 2 lead.

[2 ]

( iii) D etermine the thickness o lead required to reduce the intensity transmitted to 2 0% o its initial value.

[2 ]

( iv) A second metal has a hal- value thickness x _1 2 or this radiation o 8 mm. C alculate what thickness o this metal is required to reduce the intensity o the transmitted beam by 80% . [3 ]

O

l

i B Q u esti o n s  o pti o n C  i m ag i n g

189

16 o p t i o n D  a S t r o p h yS i c S ojs   vs (1) Solar SyStem We live on the Earth. This is one o eight planets that orbit the Sun  collectively this system is known as the Solar System. Each planet is kept in its elliptical orbit by the gravitational attraction between the Sun and the planet. Other smaller masses such as dwarf planets like Pluto or planetoids also exist.

diameter / km distance to Sun /  10 8 m

Mercury

Venus

Earth

Mars

Jupiter

Saturn

Uranus

Neptune

4,880

1 2,1 04

1 2,756

6,787

1 42,800

1 20,000

51 ,800

49,500

58

1 07.5

1 49.6

228

778

1 ,427

2,870

4,497

Relative positions of the planets Jupiter Venus Mars Sun

Earth Mercury

Uranus Saturn

Neptune Some o these planets (including the Earth) have other small objects orbiting around them called moons. Our Moon is 3.8  1 0 8 m away and its diameter is about 1 /4 o the Earths.

Mercury

Earth

An asteroid is a small rocky body that drits around the Solar System. There are many orbiting the Sun between Mars and Jupiter  the asteroid belt. An asteroid on a collision course with another planet is known as a meteoroid.

Mars

Small meteors can be vaporized due to the riction with the atmosphere (shooting stars) whereas larger ones can land on Earth. The bits that arrive are called meteorites.

Venus

Sun Jupiter

Comets are mixtures o rock and ice (a dirty snowball) in very elliptical orbits around the Sun. Their tails always point away rom the Sun.

Saturn

View from one place on earth Uranus Neptune Relative sizes of the planets

nebulae In many constellations there are diuse but relatively large structures which are called nebulae. These are interstellar clouds o dust, hydrogen, helium and other ionized gases. An example is M42 otherwise known as the Orion Nebula.

I we look up at the night sky we see the stars  many o these stars are, in act, other galaxies but they are very ar away. The stars in our own galaxy appear as a band across the sky  the Milky Way. Patterns o stars have been identifed and 88 dierent regions o the sky have been labelled as the dierent constellations. Stars in a constellation are not necessarily close to one another. Over the period o a night, the constellations seem to rotate around one star. This apparent rotation is a result o the rotation o the Earth about its own axis. On top o this nightly rotation, there is a slow change in the stars and constellations that are visible rom one night to the next. This variation over the period o one year is due to the rotation o the Earth about the Sun. Planetary systems have been discovered around many stars.

View from place to place on earth I you move rom place to place around the Earth, the section o the night sky that is visible over a year changes with latitude. The total pattern o the constellations is always the same, but you will see dierent sections o the pattern.

190

O p t i O n D  A s t r O p h ys i c s

objcs   vs (2) During one Day The most important observation is that the pattern o the stars remains the same rom one night to the next. Patterns o stars have been identifed and 88 dierent regions o the sky have been labelled as the dierent constellations. A particular pattern is not always in the same place, however. The constellations appear to move over the period o one night. They appear to rotate around one direction. In the Northern Hemisphere everything seems to rotate about the pole star. It is common to reer measurements to the fxed stars the patterns o the constellations. The fxed background o stars always appears to rotate around the pole star. During the night, some stars rise above the horizon and some stars set beneath it.

maximum height at midday. At this time in the Northern Hemisphere the S un is in a southerly direction.

Sun

looking south

east

west

During the year Every night, the constellations have the same relative positions to each other, but the location o the pole star (and thus the portion o the night sky that is visible above the horizon) changes slightly rom night to night. Over the period o a year this slow change returns back to exactly the same position.

pole star

west

looking north

east

The same movement is continued during the day. The S un rises in the east and sets in the west, reaching its

The Sun continues to rise in the east and set in the west, but as the year goes rom winter into summer, the arc gets bigger and the Sun climbs higher in the sky.

unitS When comparing distances on the astronomical scale, it can be quite unhelpul to remain in SI units. Possible other units include the astronomical unit (AU) , the parsec (pc) or the light year (ly) . See page 1 93 or the defnition o the frst two o these. The light year is the distance travelled by light in one year (9.5  1 0 1 5 m) . The next nearest star to our Sun is about 4 light years away. Our galaxy is about 1 00,000 light years across. The nearest galaxy is about a million light years away and the observable Universe is 1 3.7 billion light years in any given direction.

the milky way galaxy When observing the night sky a aint band o light can be seen crossing the constellations. This path (or way) across the night sky became known as the Milky Way. What you are actually seeing is some o the millions o stars that make up our own galaxy but they are too ar away to be seen as individual stars. The reason that they appear to be in a band is that our galaxy has a spiral shape.

The centre o our galaxy lies in the direction o the constellation Sagittarius. The galaxy is rotating  all the stars are orbiting the centre o the galaxy as

a result o their mutual gravitational attraction. The period o orbit is about 250 million years.

plan view

side view 100 000 light years 25 000 light years central bulge galactic nucleus

Sun

Sun

direction of rotation

disc globular clusters The Milky Way galaxy

the uniVerSe Stars are grouped together in stellar clusters. These can be open containing 1 0 3 stars e.g. located in the disc o our galaxy or globular containing 1 0 5 stars. Our Sun is just one o the billions o stars in our galaxy (the Milky Way galaxy) . The galaxy rotates with a period o about 2.5  1 0 8 years. Beyond our galaxy, there are billions o other galaxies. Some o them are grouped together into clusters or super clusters o galaxies, but the vast majority o space (like the gaps between the planets or between stars) appears to be empty  essentially a vacuum. Everything together is known as the Universe.

1 .5  1 0 2 6 m (= 1 5 billion light years)

the visible Universe

5  1 02 2 m (= 5 million light years)

local group o galaxies

1 02 1 m (= 1 00,000 light years)

our galaxy

1 01 3 m (= 0.001 light years)

our Solar System

O p t i O n D  A s t r O p h ys i c s

191

th   ss energy flow for StarS

equilibrium

The stars are emitting a great deal o energy. The source or all this energy is the usion o hydrogen into helium. See page 1 96. Sometimes this is reerred to as hydrogen burning but it this is not a precise term. The reaction is a nuclear reaction, not a chemical one (such as combustion) . Overall the reaction is

The Sun has been radiating energy or the past 4 billion years. It might be imagined that the powerul reactions in the core should have orced away the outer layers o the Sun a long time ago. Like other stars, the Sun is stable because there is a hydrostatic equilibrium between this outward pressure and the inward gravitational orce (see page 1 64) .

4 11 p  42 He + 2 01 e + + 2 The mass o the products is less than the mass o the reactants. Using E = m c2 we can work out that the Sun is losing mass at a rate o 4  1 0 9 kg s - 1 . This takes place in the core o a star. Eventually all this energy is radiated rom the surace  approximately 1 0 2 6 J every second. The structure inside a star does not need to be known in detail.

outward radiation pressure

convective zone inward pull of gravity

surface

radiative zone core (nuclear reactions)

binary StarS Our Sun is a single star. Many stars actually turn out to be two (or more) stars in orbit around each other. (To be precise they orbit around their common centre o mass.) These are called binary stars.

A stable star is in equilibrium

B observer night 12

centre of mass

binary stars  two stars in orbit around their common centre of mass

2.

A spectroscopic binary star is identifed rom the analysis o the spectrum o light rom the star. Over time the wavelengths show a periodic shit or splitting in requency. An example o this is shown (below) .

wavelength

Both stars are moving at 90 to observer Star A is moving towards observer B whereas star B is moving away light from B will be red shifted

A observer 3.

An eclipsing binary star is identifed rom the analysis o the brightness o the light rom the star. Over time the brightness shows a periodic variation. An example o this is shown below.

brightness

A visual binary is one that can be distinguished as two separate stars using a telescope.

A

light from A will be blue shifted observer B night 24

There are dierent categories o binary star  visual, spectroscopic and eclipsing. 1.

A

night 0

night 0 time (nights)

night 12 AB

ABAB

Each wavelength splits into two A B separate wavelengths.

star B When one star blocks the light coming from the other star, the overall brightness is reduced

night 24 The explanation or the shit in requencies involves the Doppler eect. As a result o its orbit, the stars are sometimes moving towards the Earth and sometimes they are moving away. When a star is moving towards the Earth, its spectrum will be blue shited. When it is moving away, it will be red shited.

192

The explanation or the dip in brightness is that as a result o its orbit, one star gets in ront o the other. I the stars are o equal brightness, then this would cause the total brightness to drop to 50% .

O p t i O n D  A s t r O p h ys i c s

star A

observer

S  principleS of meaSurement

mathematicS  unitS

As you move rom one position to another objects change their relative positions. As ar as you are concerned, near objects appear to move when compared with ar objects. Objects that are very ar away do not appear to move at all. You can demonstrate this eect by closing one eye and moving your head rom side to side. An object that is near to you (or example the tip o your fnger) will appear to move when compared with objects that are ar away (or example a distant building) .

The situation that gives rise to a change in apparent position or close stars is shown below.

This apparent movement is known as parallax and the eect can used to measure the distance to some o the stars in our galaxy. All stars appear to move over the period o a night, but some stars appear to move in relation to other stars over the period o a year.

distant stars close star



stellar distance d

parallax angle  If ca refu lly obser ved , over th e perio d of a y ea r som e sta rs ca n appea r to m ove be tween two extrem es.

Earth (July)

Sun

 Earth (January)

1 AU

orbit of Earth

The reason or this apparent movement is that the Earth has moved over the period o a year. This change in observing position has meant that a close star will have an apparent movement when compared with a more distant set o stars. The closer a star is to the Earth, the greater will be the parallax shit.

The parallax angle, , can be measured by observing the changes in a stars position over the period o a year. From trigonometry, i we know the distance rom the Earth to the Sun, we can work out the distance rom the Earth to the star, since

Since all stars are very distant, this eect is a very small one and the parallax angle will be very small. It is usual to quote parallax angles not in degrees, but in seconds. An angle o 1 second o arc (' ' ) is equal to one sixtieth o 1 minute o arc (' ) and 1 minute o arc is equal to one sixtieth o a degree.

Since  is a very small angle, tan   sin    (in radians)

In terms o angles, 3600' ' = 1  360 = 1 ull circle.

example The star alpha Eridani (Achemar) is 1 .32  1 0 1 8 m away. Calculate its parallax angle. d = 1 .32  1 0 1 8 m 1 .32  1 0 1 8 pc = ___________ 3.08  1 0 1 6 = 42.9 pc 1 parallax angle = _____ 42.9 = 0.02' '

(distance rom Earth to Sun) tan  = __________________________ (distance rom Sun to Star)

1 This means that   __________________________ (distance rom Earth to star) In other words, parallax angle and distance away are inversely proportional. I we use the right units we can end up with a very simple relationship. The units are defned as ollows. The distance rom the Sun to the Earth is defned to be one astronomical unit (AU). It is 1 .5  1 0 1 1 m. Calculations show that a star with a parallax angle o exactly one second o arc must be 3.08  1 0 1 6 m away (3.26 light years) . This distance is defned to be one parsec (pc). The name parsec represents parallax angle o one second. I distance = 1 pc,  = 1 second I distance = 2 pc,  = 0.5 second etc. 1 Or, distance in pc = _________________________ (parallax angle in seconds) 1 d = __ p The parallax method can be used to measure stellar distances that are less than about 100 parsecs. The parallax angle or stars that are at greater distances becomes too small to measure accurately. It is common, however, to continue to use the unit. The standard SI prefxes can also be used even though it is not strictly an SI unit. 1 000 parsecs = 1 kpc 1 0 6 parsecs = 1 Mpc etc.

O p t i O n D  A s t r O p h ys i c s

193

ls luminoSity anD apparent brightneSS The total power radiated by a star is called its luminosity (L) . The SI units are watts. This is very different to the power received by an observer on the Earth. The power received per unit area is called the apparent brightness of the star. The SI units are W m- 2 . If two stars were at the same distance away from the Earth then the one with the greater luminosity would be brighter. Stars are, however, at different distances from the Earth. The brightness is inversely proportional to the (distance) 2 .

area 4A

It is thus possible for two very different stars to have the same apparent brightness. It all depends on how far away the stars are.

close star ( small luminosity) distant star (high luminosity) Two stars can have the same apparent brightness even if they have different luminosities

area A

alternatiVe unitS x

x

As d ista n ce in crea ses, th e brigh tn ess d ecrea ses sin ce the ligh t is spread over a bigger a rea . d ista n ce x

The magnitude scale can also be used to compare the luminosity of different stars, provided the distance to the star is taken into account. Astronomers quote values of absolute magnitude in order to compare luminosities on a familiar scale.

brigh tn ess b

2x 3x 4x 5x a n d so on

b 4 b 9 b 16 b 25

The SI units for luminosity and brightness have already been introduced. In practice astronomers often compare the brightness of stars using the apparent magnitude scale. A magnitude 1 star is brighter than a magnitude 3 star. This measure of brightness is sometimes shown on star maps.

inverse sq u a re

L apparent brightness b = _____ 4r2

example on luminoSity

black-boDy raDiation

The star Betelgeuse has a parallax angle of 7.7  1 0 - 3 arc seconds and an apparent brightness of 2.0  1 0- 7 W m- 2 . Calculate its luminosity.

Stars can be analysed as perfect emitters, or black bodies. The luminosity of a star is related to its brightness, surface area and temperature according to the StefanBoltzmann law. Wiens law can be used to relate the wavelength at which the intensity is a maximum to its temperature. See page 90 for more details.

Distance to Betelgeuse d 1 = __ p 1 __________ = pc 7.7  1 0 - 3 = 1 29.9 pc = 1 29.9  3.08  1 0 1 6 m = 4.0  1 0 1 8 m L = b  4d2 = 4.0  1 0 3 1 W

194

O p t i O n D  A s t r O p h ys i c s

Example: e.g. our suns temperature is 5,800k So the wavelength at which the intensity of its radiation is at a 2.9  1 0 - 3 = 500 nm maximum is  m a x = __________ 5800

S s abSorption lineS The radiation from stars is not a perfect continuous spectrum  there are particular wavelengths that are missing.

bands of wavelengths emitted by the Sun

missing wavelength

wavelength red

violet

The missing wavelengths correspond to the absorption spectra of a number of elements. Although it seems sensible to assume that the elements concerned are in the Earths atmosphere, this assumption is incorrect. The wavelengths would still be absent if light from the star was analysed in space.

A star that is moving relative to the Earth will show a Doppler shift in its absorption spectrum. Light from stars that are receding will be red shifted whereas light from approaching stars will be blue shifted.

The absorption is taking place in the outer layers of the star. This means that we have a way of telling what elements exist in the star  at least in its outer layers.

claSSification of StarS

Stefanboltzmann law

Different stars give out different spectra of light. This allows us to classify stars by their spectral class. Stars that emit the same type of spectrum are allocated to the same spectral class. Historically these were just given a different letter, but we now know that these different letters also correspond to different surface temperatures.

The StefanBoltzmann law links the total power radiated by a black body (per unit area) to the temperature of the black body. The important relationship is that

The seven main spectral classes (in order of decreasing surface temperature) are O, B, A, F, G, K and M. The main spectral classes can be subdivided. Class

Effective surface temperature/K

Colour

Total power radiated  T4 In symbols we have, Total power radiated =  A T 4 Where  is a constant called the StefanBoltzmann constant.  = 5.67  1 0 - 8 W m- 2 K- 4 A is the surface area of the emitter (in m2 ) T is the absolute temperature of the emitter (in kelvin) e.g. The radius of the Sun = 6.96  1 0 8 m.

O

30,00050,000

blue

B

1 0,00030,000

blue-white

A

7,5001 0,000

white

Surface area

= 4 r 2 = 6.09  1 0 1 0 m2 = 5800 K

F

6,0007,500

yellow-white

If temperature

G

5,2006,000

yellow

then total power radiated =  A T 4

K

3,7005,200

orange

M

2,4003,700

red

Spectral classes do not need to be mentioned but are used in many text books.

Summary If we know the distance to a star we can analyse the light from the star and work out:  the chemical composition (by analysing the absorption spectrum)

= 5.67  1 0 - 8  6.09  1 0 1 8  (5800 4 ) = 3.9  1 0 2 6 W The radius of the star r is linked to its surface area, A, using the equation A = 4 r 2 .

 the luminosity (using measurements of the brightness and the distance away)  the surface area of the star (using the luminosity, the surface temperature and the StefanBoltzmann law) .

 the surface temperature (using a measurement of  m a x and Wiens law  see page 90)

O p t i O n D  A s t r O p h ys i c s

195

nsss Stellar typeS anD black holeS The source o energy or our Sun is the usion o hydrogen into helium. This is also true or many other stars. There are however, other types o object that are known to exist in the Universe. Type of object

Description

Red giant stars

As the name suggests, these stars are large in size and red in colour. Since they are red, they are comparatively cool. They turn out to be one o the later possible stages or a star. The source o energy is the usion o some elements other than hydrogen. Red supergiants are even larger.

White dwarf stars

As the name suggests, these stars are small in size and white in colour. Since they are white, they are comparatively hot. They turn out to be one o the fnal stages or some smaller mass stars. Fusion is no longer taking place, and a white dwar is just a hot remnant that is cooling down. Eventually it will cease to give out light when it becomes sufciently cold. It is then known as a brown dwarf.

Cepheid variables

These are stars that are a little unstable. They are observed to have a regular variation in brightness and hence luminosity. This is thought to be due to an oscillation in the size o the star. They are quite rare but are very useul as there is a link between the period o brightness variation and their average luminosity. This means that astronomers can use them to help calculate the distance to some galaxies.

Neutron stars

Neutron stars are the post-supernova remnants o some larger mass stars. The gravitational pressure has orced a total collapse and the mass o a neutron star is not composed o atoms  it is essentially composed o neutrons. The density o a neutron star is enormous. Rotating neutron stars have been identifed as pulsars.

Black holes

Black holes are the post-supernova remnant o larger mass stars. There is no known mechanism to stop the gravitational collapse. The result is an object whose escape velocity is greater than the speed o light. See page 1 50.

main Sequence StarS The general name or the creation o nuclei o dierent elements as a result o fssion reactions is nucleosynthesis. Details o how this overall reaction takes place in the Sun do not need to be recalled by SL candidates, but HL candidates do need this inormation. One process is known as the proton proton cycle or pp cycle. 1 1H

step 1

+ 11 H  21 H + 01 e + + 00  e+

p

In order or any o these reactions to take place, two positively charged particles (hydrogen or helium nuclei) need to come close enough or interactions to take place. Obviously they will repel one another. This means that they must be at a high temperature. I a large cloud o hydrogen is hot enough, then these nuclear reactions can take place spontaneously. The power radiated by the star is balanced by the power released in these reactions  the temperature is eectively constant.

The star remains a stable size because the outward pressure o the radiation is balanced by the inward gravitational pull. But how did the cloud o gas get to be at a high temperature in the frst place? As the cloud comes together, the loss o gravitational potential energy must mean an increase in kinetic energy and hence temperature. In simple terms the gas molecules speed up as they all in towards the centre to orm a proto-star. Once ignition has taken place, the star can remain stable or billions o years. See page 205 or more details.

n p

p

 step 2 n p

2 1H

+ 11 H  32 He + 00 

Fg n p p

p

step 3 p pn p pn

Fg

3 2 He



Fg collapse of cloud under gravity gives molecular KE

+ 32 He  42 He + 2 11 p n p pn

p p

the proton-proton cycle (p-p cycle)

196

Fg

O p t i O n D  A s t r O p h ys i c s

cloud of gas

With sucient KE, nuclear reactions can take place.

t hzsprss d hr Diagram The point of classifying the various types of stars is to see if any patterns exist. A useful way of making this comparison is the HertzsprungRussell diagram. Each dot on the diagram represents a different star. The following axes are used to position the dot.  The vertical axis is the luminosity of the star as compared with the luminosity of the Sun. It should be noted that the scale is logarithmic.  The horizontal axis a scale of decreasing temperature. Once again, the scale is not a linear one. (It is also the spectral class of the star OBAFGKM) The result of such a plot is shown below.

10 6

10 2 our Sun 10 0

luminosity/L.

10 4

0 00

0 3

4

50

0 6

7

00

0 50

00 10

25

50

00

00

0

0

0

10 -2 10 -4 surface temperature/K

A large number of stars fall on a line that (roughly) goes from top left to bottom right. This line is known as the main sequence and stars that are on it are known as main sequence stars. Our Sun is a main sequence star. These stars are normal stable stars  the only difference between them is their mass. They are fusing hydrogen to helium. The stars that are not on the main sequence can also be broadly put into categories. In addition to the broad regions, lines of constant radius can be added to show the size of stars in comparison to our Suns radius. These are lines going from top left to bottom right.

10 2 10 s

ol a r

1 so

ola r

10 - 2

10 3

red su pergia n ts

r ra d

ii

ra d i i

l a r ra

0 .1 s

sola

sol a

r ra d

ii

red gia n ts

instability strip

d ius ra d i u

sol a

s

r ra d

m a in seq u en ce

Su n

ius

0 3

00

0 6

00

0 00 10

50

00

0

wh ite d wa rfs

eective temperature/K

maSS-luminoSity relation for main Sequence StarS For stars on the main sequence, there is a correlation between the star' s mass, M, and its luminosity, L. Stars that are brighter on the main sequence (i.e. higher up) are more massive and the relationship is: L  M3 . 5

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197

cd vbs principleS

mathematicS

Very small parallax angles can be measured using satellite observations (e.g. Gaia mission) but even these measurement are limited to stars that are about 1 00 kpc away. The essential difculty is that when we observe the light rom a very distant star, we do not know the dierence between a bright source that is ar away and a dimmer source that is closer. This is the principal problem in the experimental determination o astronomical distances to other galaxies.

The process o estimating the distance to a galaxy (in which the individual stars can be imaged) might be as ollows:

A Cepheid variable star is quite a rare type o star. Its outer layers undergo a periodic compression and contraction and this produces a periodic variation in its luminosity.

A Cepheid variable star undergoes periodic compressions and contractions. increased luminosity

lower luminosity

 Use the luminosityperiod relationship or Cepheids to estimate the average luminosity.

apparent brightness

 Use the average luminosity, the average brightness and the inverse square law to estimate the distance to the star.

O

1

2

3

4 5

6

7

8

10 6 10 5 10 4 10 3 10 2 10 1 1

2

5

10 20 50 100 period / days

General luminosityperiod graph These stars are useul to astronomers because the period o this variation in luminosity turns out to be related to the average absolute magnitude o the Cepheid. Thus the luminosity o a Cepheid can be calculated by observing the variations in brightness.

example A Cepheid variable star has a period o 1 0.0 days and apparent peak brightness o 6.34  1 0 - 1 1 W m- 2 The luminosity o the Sun is 3.8  1 0 2 6 W. Calculate the distance to the Cepheid variable in pc. Using the luminosityperiod graph (above)  peak luminosity = 1 0 3 . 7  L s u n = 501 2  3.8  1 0 2 6 = 1 .90  1 0 3 0 W L=b 4r2 ____ L   r = ____ 4b _________________ 1 .90  1 0 3 0 = ____________________ 4    6.34  1 0 - 1 1

 

= 4.88  1 0 1 9 m 4.88  1 0 1 9 pc = ___________ 3.08  1 0 1 6 = 1 590 pc

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O p t i O n D  A s t r O p h ys i c s

9 10 11 time/days

Variation o apparent magnitude or a particular Cepheid variable

peak luminosity/L SUN

When we observe another galaxy, all o the stars in that galaxy are approximately the same distance away rom the Earth. What we really need is a light source o known luminosity in the galaxy. I we had this then we could make comparisons with the other stars and judge their luminosities. In other words we need a standard candle  that is a star o known luminosity. Cepheid variable stars provide such a standard candle.

 Locate a Cepheid variable in the galaxy.  Measure the variation in brightness over a given period o time.

rd g ss after the main Sequence The massluminosity relation (page 1 97) can be used to compare the amount o time dierent mass stars take beore the hydrogen uel is used. Consider a star that is 1 0 times more massive than our Sun. This means that the luminosity o the larger star will be (1 0) 3 . 5 = 3,1 62 times more luminous that our Sun. Since the source o this luminosity is the mass o hydrogen in the star, then the larger star eectively has 1 0 times more uel but is using the uel at more than 3000 times the rate. The more massive star will fnish 1 its uel in ___ o the time. A star that has more mass exists or a 300 shorter amount o time.

I it has sufcient mass, a red giant can continue to use higher and higher elements and the process o nucleosynthesis can continue.

newly formed red giant star

helium-burning shell

A star cannot continue in its main sequence state orever. It is using hydrogen into helium and at some point hydrogen in the core will become rare. The usion reactions will happen less oten. This means that the star is no longer in equilibrium and the gravitational orce will, once again, cause the core to collapse. This collapse increases the temperature o the core still urther and helium usion is now possible. The net result is or the star to increase massively in size  this expansion means that the outer layers are cooler. It becomes a red giant star.

red giant star

dormant hydrogenburning shell

400 million km

carbonoxygen core core of star nucleosynthesis

old, high-mass red giant star 700 million km

hydrogen-burning shell helium-burning shell carbon-burning shell neon-burning shell oxygen-burning shell silicon-burning shell iron core

star runs out of hydrogen  collapses further

helium fusion possible due to increased temperature  expansion

This process o usion as a source o energy must come to an end with the nucleosynthesis o iron. The iron nucleus has one o the greatest binding energies per nucleon o all nuclei. In other words the usion o iron to orm a higher mass nucleus would need to take in energy rather than release energy. The star cannot continue to shine. What happens next is outlined on the ollowing page.

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199

S v Page 199 showed that the red giant phase or a star must eventually come to an end. There are essentially two possible routes with dierent fnal states. The route that is ollowed depends on the initial mass o the star and thus the mass o the remnant that the red giant star leaves behind: with no urther nuclear reactions taking place gravitational orces continue the collapse o the remnant. An important critical mass is called the Chandrasekhar limit and it is equal to approximately 1 .4 times the mass o our Sun. Below this limit a process called electron degeneracy pressure prevents the urther collapse o the remnant.

h  r Diagram interpretation All o the possible evolutionary paths or stars that have been described here can be represented on a H  R diagram. A common mistake in examinations is or candidates to imply that a star somehow moves along the line that represents the main sequence. It does not. Once ormed it stays at a stable luminosity and spectral class  i.e. it is represented by one fxed point in the H  R diagram.

evolution of a low-mass star luminosity

poSSible fateS for a Star (after reD giant phaSeS)

I a star has a mass less than 4 Solar masses, its remnant will be less than 1 .4 Solar masses and so it is below the Chandrasekhar limit. In this case the red giant orms a planetary nebula and becomes a white dwarf which ultimately becomes invisible. The name planetary nebula is another term that could cause conusion. The ejected material would not be planets in the same sense as the planets in our Solar System.

ejection of planetary nebula

to white dwarf

ma

in s

equ

red giant phase en c

e

surface temperature

planetary nebula

luminosity

evolution of a high-mass star

low-mass star (e.g. type G)

to black hole / neutron star

white dwarf

red giant

red giant phase

I a star is greater than 4 Solar masses, its remnant will have a mass greater than 1 .4 Solar masses. It is above the Chandrasekhar limit and electron degeneracy pressure is not sufcient to prevent collapse. In this case the red supergiant experiences a supernova. It then becomes a neutron star or collapses to a black hole. The fnal state again depends on mass.

ma i

n se

que

n ce

surface temperature

pulSarS anD quaSarS larger-mass star (e.g. type A, B, O)

very large-mass supernova red supergiant black hole large-mass supernova neutron star

A neutron star is stable due to neutron degeneracy pressure. It should be emphasized that white dwars and neutron stars do not have a source o energy to uel their radiation. They must be losing temperature all the time. The act that these stars can still exist or many millions o years shows that the temperatures and masses involved are enormous. The largest mass a neutron star can have is called the Oppenheimer Volkoff limit and is 23 Solar masses. Remnants above this limit will orm black holes.

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Pulsars are cosmic sources o very weak radio wave energy that pulsate at a very rapid and precise requency. These have now been theoretically linked to rotating neutron stars. A rotating neutron star would be expected to emit an intense beam o radio waves in a specifc direction. As a result o the stars rotation, this beam moves around and causes the pulsation that we receive on Earth. Quasi-stellar objects or quasars appear to be point-like sources o light and radio waves that are very ar away. Their red shits are very large indeed, which places them at the limits o our observations o the Universe. I they are indeed at this distance they must be emitting a great deal o power or their size (approximately 1 0 4 0 W! ) . The process by which this energy is released is not well understood, but some theoretical models have been developed that rely on the existence o super-massive black holes. The energy radiated is as a result o whole stars alling into the black hole.

t b b dl expanSion of the uniVerSe I a galaxy is moving away rom the Earth, the light rom it will be red shited. The surprising act is that light rom almost all galaxies shows red shits  almost all o them are moving away rom us. The Universe is expanding.

At frst sight, this expansion seems to suggest that we are in the middle o the Universe, but this is a mistake. We only seem to be in the middle because it was we who worked out the velocities o the other galaxies. I we imagine being in a dierent galaxy, we would get exactly the same picture o the Universe.

vA

the uniVerSe in the paSt  the big bang

vD A B

D us (at rest) C vC

vB

As far as we are concerned, most galaxies are moving away from us. vA

This point, the creation o the Universe, is known as the Big Bang. It pictures all the matter in the Universe being crushed together (very high density) and being very hot indeed. Since the Big Bang, the Universe has been expanding  which means that, on average, the temperature and density o the Universe have been decreasing. The rate o expansion would be expected to decrease as a result o the gravitational attraction between all the masses in the Universe.

vD vus D

A

vC B (at rest) C Any galaxy would see all the other galaxies moving away from it.

A good way to picture this expansion is to think o the Universe as a sheet o rubber stretching o into the distance. The galaxies are placed on this huge sheet. I the tension in the rubber is increased, everything on the sheet moves away rom everything else.

As the section of rubber sheet expands, everything moves away from everything else.

A urther piece o evidence or the Big Bang model came with the discovery o the Cosmic microwave background (CMB) radiation by Penzias and Wilson. They discovered that microwave radiation was coming towards us rom all directions in space. The strange thing was that the radiation was the same in all directions (isotopic) and did not seem to be linked to a source. Further analysis showed that this radiation was a very good match to theoretical black-body radiation produced by an extremely cold object  a temperature o just 2.73 K. This is in perect agreement with the predictions o Big Bang. There are two ways o understanding this. 1.

All objects give out electromagnetic radiation. The requencies can be predicted using the theoretical model o black-body radiation. The

background radiation is the radiation rom the Universe itsel which has now cooled down to an average temperature o 2.73 K.

Intensity per unit wavelength range

coSmic microwaVe backgrounD raDiation

I the Universe is currently expanding, at some time in the past all the galaxies would have been closer together. I we examine the current expansion in detail we fnd that all the matter in the observable universe would have been together at the SAME point approximately 1 5 billion years ago.

Note that this model does not attempt to explain how the Universe was created, or by Whom. All it does is analyse what happened ater this creation took place. The best way to imagine the expansion is to think o the expansion o space itsel rather than the galaxies expanding into a void. The Big Bang was the creation o space and time. Einsteins theory o relativity links the measurements o space and time so properly we need to imagine the Big Bang as the creation o space and time. It does not make sense to ask about what happened beore the Big Bang, because the notion o beore and ater (i.e. time itsel) was created in the Big Bang.

2.

Some time ater the Big Bang, radiation became able to travel through the Universe (see page 21 0 or details). It has been travelling towards us all this time. During this time the Universe has expanded  this means that the wavelength o this radiation will have increased (space has stretched) . See page 21 0 or anisotropies in the CMB.

peak wavelength ~1.1 mm (microwaves) individual data points for background microwave radiation theoretical line for 2.73 K black-body radiation wavelength

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201

g  DiStributionS of galaxieS Galaxies are not distributed randomly throughout space. They tend to be ound clustered together. For example, in the region o the Milky Way there are twenty or so galaxies in less than 2.5 million light years. The Virgo galactic cluster (50 million light years away rom us) has over 1 ,000 galaxies in a region 7 million light years across. On an even larger scale, the galactic clusters are grouped into huge superclusters o galaxies. In general, these superclusters oten involve galaxies arranged together in joined flaments (or bands) that are arranged as though randomly throughout empty space.

motion of galaxieS As has been seen on page 201 it is a surprising observational act that the vast majority o galaxies are moving away rom us. The general trend is that the more distant galaxies are moving away at a greater speed as the Universe expands. This does not, however, mean that we are at the centre o the Universe  this would be observed wherever we are located in the Universe. As explained on page 201 , a good way to imagine this expansion is to think o space itsel expanding. It is the expansion o space (as opposed to the motion o the galaxies through space) that results in the galaxies relative velocities. In this model, the red shit o light can be thought o as the expansion o the wavelength due to the stretching o space.

time and expansion of Universe

light wave as emitted from a distant galaxy

light wave when it arrives at Earth

mathematicS

Examle

I a star or a galaxy moves away rom us, then the wavelength o the light will be altered as predicted by the Doppler eect (see page 1 02) . I a galaxy is going away rom the Earth, the speed o the galaxy with respect to an observer on the Earth can be calculated rom the red shit o the light rom the galaxy. As long as the velocity is small when compared with the velocity o light, a simplifed red shit equation can be used.

A characteristic absorption line oten seen in stars is due to ionized helium. It occurs at 468.6 nm. I the spectrum o a star has this line at a measured wavelength o 499.3 nm, what is the recession speed o the star?

v  Z = ___  __  c 0

Where

0

= 6.55  1 0 2 

v = 6.55  1 0 - 2  3  1 0 8 m s - 1 = 1 .97  1 0 7 m s - 1

 = change in wavelength o observed light (positive i wavelength is increased)  0 = wavelength o light emitted v = relative velocity o source o light c = speed o light Z = red shit.

202

(499.3  468.6)  Z = ___ = _______________  468.6

O p t i O n D  A s t r O p h ys i c s

hs w d s s  experimental obSerVationS

hiStory of the uniVerSe

Although the uncertainties are large, the general trend or galaxies is that the recessional velocity is proportional to the distance away rom Earth. This is Hubbles law.

I a galaxy is at a distance x, then Hubbles law predicts its velocity to be H0 x. I it has been travelling at this constant speed since the beginning o the Universe, then the time that has elapsed can be calculated rom

recessional velocity / km s -1

10 000

distance Time = ________ speed x = ____ H0 x

8 000 6 000

1 = ___ H0

4 000

This is an upper limit or the age o the Universe. The gravitational attraction between galaxies predicts that the speed o recession decreases all the time.

2 000 0 0

20 40 60 80 100 120 140 distance / Mpc

size of R observable universe

Mathematically this is expressed as v d or v = H0 d

T 1 H0

where H0 is a constant known as the Hubble constant. The uncertainties in the data mean that the value o H0 is not known to any degree o precision. The SI units o the Hubble constant are s - 1 , but the unit o km s - 1 Mpc- 1 is oten used.

1 H0

time

now

the coSmic Scale factor (R)  Page 202 shows how the Doppler red shit equation, = ___  _vc , can be used to calculate the recessional velocity, v, o certain  0

galaxies. This equation can only be used when v  c or in other words, the recessional velocity, v, has to be small in comparison to the speed o light, c. There are however plenty o objects in the night sky (e.g. quasars) or which the observed red shit, z, is greater than 1 .0. This implies that their speed o recession is greater than the speed o light. In these situations it is helpul to consider a quantity called the cosmic scale actor (R) . As introduced on page 201 , the expansion o the Universe is best pictured as the expansion o space itsel. The expansion o the Universe means that a measurement undertaken at some time in the distant past, or example the wavelength o light emitted by an object 1 0 million years ago, will be stretched and will be recorded as a larger value when measured now. All measurement will be stretched over time and this can be considered as a rescaling o the Universe (the Universe getting bigger) . The cosmic scale actor, R, is a way o quantiying the expansion that has taken place. In the above example, i the wavelength was emitted 1 0 million years ago with wavelength 0 when the scale actor was R0 , the wavelength measured today would have increased by  to a larger value  ( =  0 + ) . This is because the cosmic scale actor has increased by R (to the larger value R  . The ratio o the measured wavelengths, __ , is equal to the R = R0 + R) . All measurements will have increased by the ratio, __ R  R __ ratio o the cosmic scale actors, R , so the red shit ratio, z is given by: 0

0

0

- 0  = _____  - 1 = ___ R -1 z = ____ = ___ R0 0 0 0 R - 1 or z = ___ R0 R So a measured red shit o 4 means that __ = 5. I we consider R to be the present size o the observable Universe, then the light R must have been emitted when the Universe was one fth o its current size. 0

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203

t  vs SupernoVae anD the accelerating uniVerSe Supernovae are catastrophic explosions that can occur in the development o some stars (see page 200) . Supernovae are rare events (the last one to occur in our galaxy took place in 1 604) but the large number o stars in the Universe means that many have been observed. An observer on the Earth sees a rapid increase in brightness (hence the word nova = new star) which then diminishes over a period o some weeks or months. Huge amounts o radiated energy are emitted in a short period o time and, at its peak, the apparent brightness o a single supernova oten exceeds many local stars or individual galaxies. Supernovae have been categorized into two dierent main types (see page 207 or more details) according to a spectral analysis o the light that they emit. The light rom a type II supernova indicates the presence o hydrogen (rom the absorption spectra) whereas there is no hydrogen in a type I supernova. There are urther subdivisions o these types (Ia, Ib, etc.) based on dierent aspects o the light spectrum. Type Ia supernovae are explosions involving white dwar stars. When these events take place, the amount o energy released can be predicted accurately and these supernovae can be used as standard candles. By comparing the known luminosity o a type Ia supernova and its apparent brightness as observed in a given galaxy, a distance measurement to that galaxy can be calcuated. This technique can be used with galaxies up to approximately 1 ,000 Mpc away. The expanding Universe (which is consistent with the Big Bang model) means that that the cosmic scale actor, R, is increasing. As a result o gravitational attraction, we might expect the rate at which R increases to be slowing down. Analysis o a large number o type Ia supernovae has, however, provided strong evidence that not only is the cosmic scale actor, R, increasing but the rate at which it increases is getting larger as time passes. In other words the expansion o the Universe is accelerating. The evidence rom type Ia supernovae identies this eect rom a time when the universe was approximately __23 o its current size. Note that this acceleration is dierent to the very rapid period o expansion o the early Universe which is called infation. The mechanisms that cause an accelerating Universe are not ully understood but must involve an outward accelerating orce to counteract the inward gravitational pull. There must also be a source o energy which has been given the name dark energy (see page 21 2) .

dark energy accelerated expansion

dark ages

development of galaxies

ination

1 st stars

13.7 billion years Source: NASA/WMAP Science Team

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HL

n s   Js 

the JeanS criterion

nuclear fuSion

As seen on page 1 96, stars orm out o interstellar clouds o hydrogen, helium and other materials. Such clouds can exist in stable equilibrium or many years until an external event (e.g. a collision with another cloud or the infuence o another incident such as a supernova) starts the collapse. At any given point in time, the total energy associated with the gas cloud can be thought o as a combination o:

A star on the main sequence is using hydrogen nuclei to produce helium nuclei. One process by which this is achieved is the protonproton chain as outlined on page 1 96. This is the predominant method or nuclear usion to take place in small mass stars (up to just above the mass o our Sun) . An alternative process, called the CNO (carbonnitrogenoxygen) process takes place at higher temperatures in larger mass stars. In this reaction, carbon, nitrogen and oxygen are used as catalysts to aid the usion o protons into helium nuclei. One possible cycle is shown below:

 The negative gravitational potential energy, EP, which the cloud possesses as a result o its mass and how it is distributed in space. Important actors are thus the mass and the density o the cloud.

4

 The positive random kinetic energy, EK , that the particles in the cloud possess. An important actor is thus the temperature o the cloud. The cloud will remain gravitationally bound together i EP + EK < zero. Using this inormation allows us to predict that the collapse o an interstellar cloud may begin i its mass is greater than a certain critical mass, MJ . This is the Jeans criterion. For a given cloud o gas, MJ is dependent on the clouds density and temperature and the cloud is more likely to collapse i it has:

1

START

He

1

H

H

 12 15 N

C

13 N

 15

O

13 C

 large mass

14

N

 small size  low temperature.



In symbols, the Jeans criterion is that collapse can start i M > MJ

1

H

proton neutron positron



1

H

 gamma ray  neutrino

time Spent on the main Sequence For so long as a star remains on the main sequence, hydrogen burning is the source o energy that allows the star to remain in hydrostatic equilibrium (see page 1 92) and have a constant luminosity L. A star that exists on the main sequence or a time TM S must in total radiate an energy E given by: E = L  TM S This energy release comes rom the nuclear synthesis that has taken place over its lietime. A certain raction f o the mass o the star M has been converted into energy according to Einsteins amous relationship: E = f  Mc2 

L  TM S = f  Mc2

f  Mc2 TM S = _______ L But the massluminosity relationship applies, L  M3 . 5 M  TM S  ____ M3 . 5 

TM S  M- 2 . 5

Thus the higher the mass of a star, the shorter the lifetime that it spends on the main sequence

(

- 2 .5

) (

Time on main sequence for star A Mass of star A ____________________________ = ____________ Time on main sequence for star B Mass of star B

Mass of star B = ____________ Mass of star A

)

2 .5

For example our Sun is expected to have a main sequence lietime o approximate 1 0 1 0 years. How long would a star with 1 00 times its mass be expected to last? 2 .5 1 = 1 0 5 years Time on MS or 1 00 solar mass star = 1 0 1 0  ____ 1 00

( )

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nsss f   s

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nucleoSyntheSiS o the main Sequence For so long as a star remains on the main sequence, hydrogen burning is the source o energy that allows the star to continue emitting energy whilst remaining in a stable state. More and more helium exists in the core. A nuclear synthesis involving helium (helium burning) does release energy (since the binding energy per nucleon o the products is greater than that o the reactants) but can only take place at high temperatures. For high mass stars, the helium burning process can begin gradually and spread throughout the core whereas in small mass stars this process starts suddenly. Whatever the mass o the star, a new equilibrium state is created: the red giant or red supergiant phase (see page 200) . A common process by which helium is converted is a series o nuclear reactions called the triple alpha process in which carbon is produced. 1.

Two helium nuclei use into a beryllium nucleus (and a gamma ray) , releasing energy. 4 2

2.

4 2

8 4

He 

Be + 

The beryllium nucleus uses with another helium nucleus to produce a carbon nucleus (and a gamma ray) , releasing energy. 8 4

3.

He +

Be + 42 He 

12 6

C+ 

Some o the carbon produced in the triple alpha process can go on to use with another helium nucleus to produce oxygen. Again this process releases energy: 12 6

C + 42 He 

16 8

In high and very high mass stars, gravitational contraction means that the temperature o the core can continue to rise and more massive nuclei can continue to be produced. These reactions all involve the release o energy. Typical reactions include: Production o neon:

12 6

Production o magnesium:

12 6

Production o oxygen:

12 6

C+

12 6

C

20 10

C+

12 6

C

24 12

C+

12 6

4 2

He

Mg + 

16 8

C

Ne +

O + 2 42 He

In addition i the temperatures are high enough, neon and oxygen burning can occur: 20 10 20 10

Production o sulur:

Ne +   Ne +

16 8

O+

4 2 16 8

16 8

He  O

O+ 24 12 32 16

4 2

He

Mg + 

S+ 

Many reactions are possible and other heavy nuclei such as silicon and phosphorus are also produced. Some o these alternative nuclear reactions also produce neutrons, which can easily be captured by other nuclei to orm new isotopes. This process o neutron capture is explored urther below. In very high mass stars, silicon burning can also take place which results in the ormation o iron, 52 66 Fe. As explained on page 1 99, iron has one o the highest binding energies per nucleon and represents the largest nucleus that can be created in a usion process that releases energy. Heavier nuclei can be acquired, but the reactions require an energy input.

O+ 

nuclear SyntheSiS o heaVy elementS  neutron capture Many o the reactions that take place in the core o stars also involve the release o neutrons. Since neutrons are without any charge, it is easy or them to interact with other nuclei that are present in the star. When a nucleus captures a neutron, the resulting nucleus is said to be neutron rich. Given enough time, most o these neutron-rich nuclei would undergo beta decay. In this process, the neutron changes into a proton, emitting an electron and an antineutrino: 1 0

n

1 1

A Z

X+

1 0

p+

n

0 -1

_

+ v

A + 1 Z

X

A + 1 Z + 1

Y+

0 -1

_

+ v

This is known as slow neutron capture or the s-process. The overall result o the s-process is a new element. Typically the

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s-process takes place during the helium burning stage o a red giant star. Typically this means that elements that are heavier than helium but lighter than iron are able to be created. The alternative process, rapid neutron capture or the r-process, takes place when the neutrons are present in such vast numbers that there is not sufcient time or the neutron-rich nuclei to undergo beta decay beore several more neutrons are captured. The result is or very heavy nuclei to be created. Typically the r-process takes place during the catastrophic explosion that is a supernova. Elements that are heavier than iron, such as uranium and thorium, can only be created in this way at very high temperatures and densities.

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tys f sv

SupernoVae Supernovae are among the most gigantic explosions in the Universe (see page 200) . The two categories o supernova are based on their light curves  a plot o how their brightness varies with time and a spectral analysis o the light that they emit. Type I supernovae quickly reach a maximum brightness (and an equivalent luminosity o 1 0 1 0 Suns) which then gradually decreases over time. Type II supernovae oten have lower peak luminosities (equivalent to, say, 1 0 9 Suns) .

Type II

Luminosity

Luminosity

Type I

100 200 300 0 days after maximum brightness

0

100 200 300 days after maximum brightness

400

Supernovae types are distinguish by analysis o their light spectra. All type I supernovae do not include the hydrogen spectrum in the elements identifed and the dierent subdivisions (Ia, Ib and Ic) are based on a more detailed spectral analysis:  Type Ia shows the presence o singly ionized silicon.  Type Ib shows the presence o non-ionized helium.  Type Ic does not show the presence o helium. All type II supernovae show the presence o hydrogen. The dierent subdivisions (IIP, IIL, IIn and IIb) again depend on the presence, or not, o dierent elements. The reasons or these dierences are the dierent mechanisms that are taking place: Supernova Type Ia

Supernova Type II

Spectra

Does not show hydrogen but does show singly ionized silicon.

Shows hydrogen.

Cause

White dwar exploding.

Large mass red giant star collapsing.

Context

Binary star system with white dwar and red giant orbiting each other.

Large star (greater than 8 Solar masses) at the end o its lietime, using lighter elements up to the production o iron.

Process

The gravity feld o the white dwar star attracts material rom the red giant star, thus increasing the mass o the white dwar.

When the star runs out o uel, the iron centre core cannot release any urther energy by nuclear usion. The star collapses under its own gravity orming a neutron star.

Explosion

The extra mass gained by the white dwar takes the total mass o the star beyond the Chandrasekhar limit (1 .4 Solar masses) or a white dwar. Electron degeneracy pressure is no longer sufcient to halt the gravitational collapse. Nuclear usion o heavier elements (up to iron) starts and the resulting sudden release o energy causes the star to explode with the matter being distributed throughout space.

Electron degeneracy pressure is not sufcient to halt the gravitational collapse o the core, but neutron degeneracy pressure is and the core becomes a stable and rigid neutron star. The rest o the inalling material bounces o the core creating a shock wave moving outwards. This causes all o the outer layers to be ejected.

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t s  d  ds

the coSmological principle The cosmological principle is a pair of assumptions about the structure of the Universe upon which current models are based. The two assumptions are that the Universe, providing one only considers the large scale structures in the Universe, is isotropic and homogeneous. An isotropic universe is one that looks the same in every direction  no particular direction is different to any other. From the perspective of an observer on Earth, this appears to be a true statement about the large scale structure of the universe, but the assumption does not only apply to observers on the Earth. In an isotropic universe all observers, wherever they are in the universe, are expected to see the same basic random distribution of galaxies and galaxy clusters as we do on Earth and this is true in whatever direction they observe. A homogeneous universe is one where the local distribution of galaxies and galaxy clusters that exists in one region of the universe turns out to be the same distribution in all regions of the universe. Provided one is considering a reasonably large

rotation curVeS  mathematical moDelS The stars in a galaxy rotate around their common centre of mass. Different models can be used to predict how the speed varies with distance from the galactic centre. 1.

section of space (e.g. a sphere of radius equal to several hundreds of Mpc) , then the number of galaxies in that volume of space will be effectively the same wherever we choose to look in the universe. Recent discoveries of apparently very large scale structures in the Universe cause some astrophysicists to question the validity of the cosmological principle. Einstein used the cosmological principle to develop a model of the Universe in which the Universe was static. He did this by proposing that the gravitational attraction between galaxies would be balanced by a yet-to-be-discovered cosmological repulsion. Subsequent analysis of the equations of general relativity showed that, if the cosmological principle is correct, the Universe must be non-static. Hubbles observational discovery of the expansion of the Universe and the existence of CMB has meant that many physicists now agree that the Universe is non-static based around the Big Bang model of an expanding universe. The cosmological principle is also linked to three possible models for the future of the Universe (see page 21 1 ) .

The star at a given distance r from the centre will orbit in circular motion because its centripetal force is provided by the gravitational attraction: GMm mv2 _ = _ 2 r r

Near the galactic centre

GM  v2 = _ r

A simple model to explain the different speeds of rotation of stars near the galactic centre assumes that density of the galaxy near its centre, , is constant. A star of mass m feels a resultant force of gravitational attraction in towards the centre. The value of this resultant force is the same as if the total mass M of all the stars that are closer to the galactic centre were concentrated in the centre. An important point to note is that the net effect of all the stars that are orbiting at radius that is greater than r sums to zero.

speed 

density of stars in galactic centre = 

The total mass of stars that orbit closer than of this star, M, is given by 4 3 r   M = volume  density = _ 3

4 3 G_ r 

4G 3 v2 = _ = _ r2 r _____

 v=

3

_

4G r 3

i.e. v  r 2.

Far away from the galactic centre

Far away from the galactic centre, observations of the number of visible stars show that the effective density of the galaxy has reduced so much that individual stars at these distances can be considered to be freely orbiting the central mass and to be unaffected by their neighbouring stars. In this situation,

mass of star, m radius r

GM v2 = _ r where M is the mass of the galaxy __

i.e. v 

 stars outside r have overall no net eect M r

m

stars inside r have eect of total mass M at centre

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_1r

Comparisons with observations of real galaxies show good agreement with mathematical model (1 ) but no agreement with mathematical model (2) . The proposed solution is discussed on page 209.

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r vs d d 

 an initial linear increase in orbital velocity with distance within the galactic centre  a fat or slightly increasing curve showing a roughly constant speed o rotation away rom the galactic centre.

Orbital speed (km/s)

rotation curVeS Galaxies rotate around their centre o mass and the speeds o this rotation can be calculated or individual stars rom an analysis o the stars spectra. A rotation curve or a galaxy show how this orbital speed varies with distance rom the galactic centre. Most galaxies show:

350 300 250 200 150 100 50

NGC 4378 NGC 3145 NGC 1620 NGC 7664

5 10 15 20 25 distance from centre of galaxy (kpc)

eViDence for Dark matter As shown above, observed rotation curves or real galaxies agree with theoretical models within the galactic centre (v  r) but the orbital velocity o stars is not observed to decrease with distance away rom the centre as would be expected. Instead, the orbital velocity is roughly constant whatever the radius. I the orbital velocity v o a star is constant at dierent values o radius, then GM since v2 = _ r M _ = constant or M  r r

Thus the total mass that is keeping the star orbiting in its galaxy must be increasing with distance rom the galactic centre. This is certainly not true o the visible mass (the stars emitting light)

machoS, wimpS anD other theorieS Astrophysicists are attempting to come up with theories to explain why there is so much dark matter and what it consists o. There are a number o possible theories:  The matter could be ound in Massive Astronomical Compact Halo Objects or MACHOs or short. There is some evidence that lots o ordinary matter does exist in these groupings. These can be thought o as low-mass ailed stars or high-mass planets. They could even be black holes. These would produce little or no light. Evidence suggests that these could only account or a small proportion.

that we can see so the suggestion is that there must be dark matter. In this situation it would have to be concentrated outside the galactic centre orming a halo around the galaxy. Further evidence suggests that only a very small amount o this matter could be imagined to be made up o the protons and neutrons that constitute ordinary, or baryonic, matter. Dark matter:  gravitationally attracts ordinary matter  does not emit radiation and cannot be inerred rom its interactions  is unknown in structure  makes up the majority o the Universe with less than 5% o the Universe made up o ordinary baryonic matter.

 There could be new particles that we do not know about. These are the Weakly Interacting Massive Particles. Many experimenters around the world are searching or these so-called WIMPs.  Perhaps our current theories o gravity are not completely correct. Some theories try to explain the missing matter as simply a ailure o our current theories to take everything into account.

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t s   uvs

fluctuationS in cmb The cosmic microwave background radiation (CMB) is essentially isotropic (the same in all directions) . This implies that the matter in the early Universe was uniormly distributed throughout space with no random temperature variations at all. I this was precisely the case then the development o the Universe would be expected to be absolutely identical everywhere and matter would be uniormly distributed throughout the Universe  it would be without any structure. We know, however, that matter is not uniormly distributed as it is concentrated into stars and galaxies. Further analysis o the CMB reveals tiny fuctuations (anisotropies) in the temperature distribution o the early Universe in dierent directions. These temperature variations are typically a ew K compared with the background eective temperature o 2.73 K. The diagram right is an enhanced projection which highlights the minor observed variations in the CMB (with the eects o our own galaxy removed) . Just like a map includes all the countries o the world, this

projection shows the variation in received CMB rom the whole Universe.

Variation in CMB as observed by the Wilkinson Microwave Anisotropy Probe (WMAP) The minute dierences in temperature imply minor dierences in densities, which allow structures to be developed as the Universe expands.

the hiStory of the uniVerSe We can work backwards and imagine the process that took place soon ater the Big Bang.  Very soon ater the Big Bang, the Universe must have been very hot.  As the Universe expanded it cooled. It had to cool to a certain temperature beore atoms and molecules could be ormed.  The Universe underwent a short period o huge expansion (Infation) that would have taken place rom about 1 0 - 3 6 s ater the Big Bang to 1 0 - 3 2 s. Time 1 0- 4 5 s  1 0 - 3 6 s 1 0- 3 6 s  1 0 - 3 2 s

What is happening Unication o orces Infation

1 0- 3 2 s  1 0 - 5 s

Quarklepton era

1 0- 5 s  1 0 - 2 s

Hadron era

1 0- 2 s  1 0 3 s

Nucleosynthesis

1 03 s  3  1 0 5 years 3  1 0 5 years  1 0 9 years

Plasma era (radiation era)

1 0 9 years  now

Formation o stars, galaxies and galactic clusters

Formation o atoms

Comments This is the starting point. A rapid period o expansion  the so-called infationary epoch. The reasons or this rapid expansion are not ully understood. Matter and antimatter (quarks and leptons) are interacting all the time. There is slightly more matter than antimatter. At the beginning o this short period it is just cool enough or hadrons (e.g. protons and neutrons) to be stable. During this period some o the protons and neutrons have combined to orm helium nuclei. The matter that now exists is the small amount that is let over when matter and antimatter have interacted. The ormation o light nuclei has now nished and the Universe is in the orm o a plasma with electrons, protons, neutrons, helium nuclei and photons all interacting. At the beginning o this period, the Universe has become cool enough or the rst atoms to exist. Under these conditions, the photons that exist stop having to interact with the matter. It is these photons that are now being received as part o the background microwave radiation. The Universe is essentially 75% hydrogen and 25% helium. Some o the matter can be brought together by gravitational interactions. I this matter is dense enough and hot enough, nuclear reactions can take place and stars are ormed.

coSmic Scale factor anD temperature The expansion o the Universe means that the wavelength o any radiation that has been emitted in the past will be stretched over time (see page 202) . Thus the radiation that was emitted approximately 1 2 billion years ago (shortly ater the Big Bang) at very short wavelengths is now being received as much longer microwaves  the CMB radiation. The spectrum o CMB radiation received corresponds to blackbody radiation at a temperature o 2.73 K. The calculation uses Wiens law to link the peak wavelength,  m a x , o the radiation to the temperature, T, o the black body in kelvins: 2.9  1 0 - 3 m a x = __________ T 1 m a x  _ T

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When the radiation was emitted the temperature o the universe was much hotter, the cosmic scale actor, R, was much smaller and  m a x was also proportionally much smaller. Since the stretching o the Universe is the cause o the change in wavelength, then the ratio o cosmic scale actors at two dierent times must be the same as the ratio o peak wavelengths so m a x  R 1 1 _  R or T  _ T

R

t    uvs

future of the uniVerSe (without Dark energy)

cosmic scale factor

I the Universe is expanding at the moment, what is it going to do in the uture? As a result o the Big Bang, other galaxies are moving away rom us. I there were no orces between the galaxies, then this expansion could be thought o as being constant.

R current rate of expansion

1

now

time

The expansion o the Universe cannot, however, have been uniorm. The orce o gravity acts between all masses. This means that i two masses are moving apart rom one another there is a orce o attraction pulling them back together. This orce must have slowed the expansion down in the past. What it is going to do in the uture depends on the current rate o expansion and the density o matter in the Universe.

cosmic scale factor

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R

open

at

closed now

time

An open Universe is one that continues to expand orever. The orce o gravity slows the rate o recession o the galaxies down a little bit but it is not strong enough to bring the expansion to a halt. This would happen i the density in the Universe were low. A closed Universe is one that is brought to a stop and then collapses back on itsel. The orce o gravity is enough to bring the expansion to an end. This would happen i the density in the Universe were high. A fat Universe is the mathematical possibility between open and closed. The orce o gravity keeps on slowing the expansion down but it takes an innite time to get to rest. This would only happen i the Universe were exactly the right density. One electron-positron pair more, and the gravitational orce would be a little bit bigger. Just enough to start the contraction and make the Universe closed.

critical DenSity, c The theoretical value o density that would create a fat Universe is called the critical density, c. Its value is not certain because the current rate o expansion is not easy to measure. Its order o magnitude is 1 0 - 2 6 kg m- 3 or a ew proton masses every cubic metre. I this sounds very small remember that enormous amounts o space exist that contain little or no mass at all. The density o the Universe is not an easy quantity to measure. It is reasonably easy to estimate the mass in a galaxy by estimating the number o stars and their average mass but the majority o the mass in the Universe is dark matter. The value o c can be estimated using Newtonian gravitation. We consider a galaxy at a distance r away rom an observer with a recessional velocity o v with respect to the observer.

radius r total mass in sphere, M

recessional velocity = 

The total energy ET o the galaxy is the addition o its kinetic energy EK and gravitational potential energy, EP given by: ET = EK + EP 1 EK = _ mv2 but Hubbles law gives v = H0 r 2

1  EK = _ m(H0 r) 2 2

average density of universe inside sphere =  radius r observer

recessional velocity = 

The net eect o all the masses in the Universe outside the sphere on the galaxy is zero (see page 208 or an analogous situation) . The galaxy is thus gravitationally attracted in by a total mass M which acts as though it was located at the observer as shown (above) .

GMm 4 3 EP = - _ but M = volume  density = _ r  r 3

G4r3 m 4Gr2 m EP = - _ = - _ 3r 3

I ET is positive, the galaxy will escape the inward attraction  the universe is open. I ET is negative, the galaxy will eventually all back in  the universe is closed. I ET is exactly zero, the galaxy will take an innite time to be brought to rest  the universe is fat. This will occur when the density o the universe  is equal to the critical density c. 4Gr2  cm 1  _ m(H0 r) 2 = _ 2

3

2

8Gr  cm  mH0 2 r2 = _ 3

2

3H0  c = _ 8G

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D 

coSmic DenSity parameter The cosmic density parameter,  0 is the ratio o the average density o matter and energy in the Universe, , to the critical density, c   0 = ___ C

I  0 > 1 , the universe is closed. I  0 < 1 , the universe is open. I  0 = 1 , the universe is fat.

Dark energy Gravitational attraction between masses means that the rate o expansion o the Universe would be expected to decrease with time. Measurements using type Ia supernovae as standard candles have provided strong evidence that the expansion has not, in act, been slowing down over time (see page 204) . Observations currently indicate that the Universes rate o expansion has been increasing.

 Dark matter is hypothesized to explain the missing matter that must exist within galaxies or the known laws o gravitational attraction to be able to explain a galaxys rate o rotation. Dark matter adds to the attractive force of gravity acting within galaxies implying more unseen mass than had been previously expected, hence the name dark mass.

Currently there is no single accepted explanation or this observation and, o course, it is possible that our theories o gravity and general relativity need to be modied. Perhaps we are on the brink o discovery o new physics. Whatever the cause, the reason or the Universes accelerating expansion has been given the general name dark energy.

 The observation that expansion o the Universe is accelerating means that then there must be a orce that is counteracting the attractive orce o gravity. Dark energy opposes the attractive force of gravity between galaxies. The resulting increase in energy implies an unseen source o energy, hence the name dark energy.

Dark energy and dark matter are two dierent concepts. In both cases experimental evidence implies their existence but physicists have yet to agree a theoretical basis that explains the existence o either concept.

effect of Dark energy on the coSmic Scale factor The existence o dark energy counteracts the attractive orce o gravity. This will cause the cosmic scale actor to increase over time. The graph below compares how a fat Universe is predicted to develop with and without dark energy.

cosmic scale factor, R

at Universe with dark energy (accelerating expansion)

at Universe without dark energy (approaches maximum size)

now

212

O p t i O n D  A s t r O p h ys i c s

time

HL

asss s

aStrophySicS reSearch Much o the current undamental research that is being undertaken in astrophysics involves close international collaboration and the sharing o resources. Scientists can be proud o their record o international collaboration. For example, at the time that the previous edition o this book was being published, the Cassini spacecrat had been in orbit around Saturn or several years sending inormation about the planet back to Earth and is currently (201 4) continuing to produce data. The CassiniHuygens spacecrat was unded jointly by ESA (the European Space Agency) , NASA (the National Aeronautics and Space Administration o the United States o America) and ASI (Agenzia Spaziale Italiana  the Italian Space Agency) . As well as general inormation about Saturn, an important ocus o the mission was a moon o Saturn called Titan. The Huygens probe was released and sent back inormation as it descended towards the surace. The inormation discovered is shared among the entire scientic community. Many current projects, or example the Dark Energy Survey (involving more than 1 20 scientists or 23 institutions worldwide) , continue this process. All countries have a limited budget available or the scientic research that they can undertake. There are arguments both or, and against, investing signicant resources into researching the nature o the Universe.

 It is one o the most undamental, interesting and important areas or humankind as a whole and it thereore deserves to be properly researched.  All undamental research will give rise to technology that may eventually improve the quality o lie or many people.  Lie on Earth will, at some time in the distant uture, become an impossibility. I humankinds descendents are to exist in this uture, we must be able to travel to distant stars and colonize new planets. Arguments against:  The money could be more useully spent providing ood, shelter and medical care to the many millions o people who are suering rom hunger, homelessness and disease around the world.  I money is to be allocated on research, it is much more worthwhile to invest limited resources into medical research. This oers the immediate possibility o saving lives and improving the quality o lie or some suerers.  It is better to und a great deal o small diverse research rather than concentrating all unding into one expensive area. Sending a rocket into space is expensive, thus unding space research should not be a priority.  Is the inormation gained really worth the cost?

Future research, such as the Euclid mission to map the geometry o the dark Universe continues to be planned. Arguments or:  Understanding the nature o the Universe sheds light on undamental philosophical questions like:  Why are we here?  Is there (intelligent) lie elsewhere in the Universe?

current obSerVationS Three recent scientic experiments that have studied the CMB in detail have together added a great deal to our understanding o the Universe. Particular experiments o note include:

 calculated the Hubble constant to be 67.1 5 km s - 1 Mpc- 1

 NASAs Cosmic Background Explorer (COBE)

 showed the Universe to be fat,  0 = 1

 NASAs Wilkinson Microwave Anisotropy Probe (WMAP)

 calculated the Universe to be composed o 4.6% atoms, 23% dark matter and 71 .4% dark energy.

 ESAs Planck space observatory. Together these experiments have:  mapped the anisotropies o the CMB in great detail and with precision  discovered that the rst generation o stars to shine did so 200 million years ater the Big Bang, much earlier than many scientists had previously expected  calculated the age o the Universe as 1 3.75  0.1 4 billion years old

 showed that their results were consistent with the Big Bang and specic infation theories

In summary, current scientic evidence suggests that, when dark matter and dark energy are taken into consideration, the Universe:  is fat  has a density that is, within experimental error, very close to the critical density  has an accelerating expansion  is composed mainly o dark matter and dark energy.

O p t i O n D  A s t r O p h ys i c s

213

ib questons  astrophyscs This question is about determining some properties o the star Wol 359.

3.

a) The star Wol 359 has a parallax angle o 0.41 9 seconds. (i) Describe how this parallax angle is measured.

[4]

(ii) Calculate the distance in light-years rom Earth to Wol 359.

[2]

(iii) State why the method o parallax can only be used or stars at a distance o less than a ew hundred parsecs rom Earth.

[1 ]

b) The ratio

a) The spectrum o light rom the Sun is shown below.

relative intensity

1.

0.4

[4]

0.2 0 0

Show that the ratio luminosity o Wol 359 _____________________ is 8.9  1 0 4 . (1 ly = 6.3  1 0 4 AU) luminosity o the Sun c) The surace temperature o Wol 359 is 2800 K and its luminosity is 3.5  1 0 2 3 W. Calculate the radius o Wol 359.

4.

B

A

1.0  10 3 luminosity L/L s 1.0  10 1

a)

3.0  10 4 1.2  10 4 3.0  10 3 surface temperature T/K

5.

(i) Draw a circle around the stars that are red giants. Label this circle R. [1 ] (ii) Draw a circle around the stars that are white dwars. Label this circle W. (iii) Draw a line through the stars that are main sequence stars.

b) Explain, without doing any calculation, how astronomers can deduce that star B has a larger diameter than star A.

[2] [2]

a) Explain how Hubbles law supports the Big Bang model o the Universe.

[2]

b) Outline one other piece o evidence or the model, saying how it supports the Big Bang.

[3]

c) The Andromeda galaxy is a relatively close galaxy, about 700 kpc rom the Milky Way, whereas the Virgo nebula is 2.3 Mpc away. I Virgo is moving away at 1 200 km s 1 , show that Hubbles law predicts that Andromeda should be moving away at roughly 400 km s 1 .

[1 ]

Apparent brightness o the Sun Apparent brightness o star A Mean distance o Sun rom Earth 1 pc

= = = =

1 .4 4.9 1 .0 2.1

[3]

A quasar has a redshit o 6.4. Calculate the ratio o the current size o the universe to its size when the quasar emitted the light that is being detected.

[3]

HL 6. [1 ]

d) Explain why the distance o star A rom Earth cannot be determined by the method o stellar parallax. [1 ]

Explain the ollowing: a) Why more massive stars have shorter lietimes

[2]

b) The jeans criterion

[2]

c) How elements heavier than iron are produced by stars [2]

[3]

 1 0 3 W m2  1 0 9 W m2 AU  1 0 5 AU [4]

i B Q u E s t i O n s  A s t r O p h ys i c s

e) I light o wavelength 500 nm is emitted rom Andromeda, what would be the wavelength observed rom Earth?

[1 ]

d) How type 1 a supernovae can be used as standard candles [2]

c) Using the ollowing data and inormation rom the HR diagram, show that star A is at a distance o about 800 pc rom Earth.

214

(i) The elements present in its outer layers. (ii) Its speed relative to the Earth.

d) Andromeda is in act moving towards the Milky Way, with a speed o about 1 00 km s 1 . How can this discrepancy rom the prediction, in both magnitude and direction, be explained? [3]

1.0  10 -1 1.0  10 -3

[2]

b) Outline how the ollowing quantities can, in principle, be determined rom the spectrum o a star.

[2]

The diagram below shows the grid o an HR diagram, on which the positions o selected stars are shown. (L S = luminosity o the Sun.)

500 1000 1500 2000 2500 3000 wavelength / nm

Use this spectrum to estimate the surace temperature o the Sun.

d) By reerence to the data in (c) , suggest why Wol 359 is neither a white dwar nor a red giant. [2]

1.0  10 5

0.8 0.6

apparent brightness o Wol 359 _____________________________ is 3.7  1 0 1 5 . apparent brightness o the Sun

2.

1.0

7.

e) The signifcance o observed anisotropies in the Cosmic Microwave background

[2]

) The signifcance o the critical density o universe

[2]

g) The evidence or dark matter

[2]

h) What is meant by dark energy

[2]

Calculate the critical density or o the universe using the Hubble constant o 71 km s - 1 Mpc- 1 [3]

17 a P P e n d i x gp Plotting graPhs  axes and best fit

 All the ints have been ltted crrectly.

The reasn r ltting a grah in the frst lace is that it allws us t identiy trends. T be recise, it allws us a visual way  reresenting the variatin  ne quantity with resect t anther. When ltting grahs, yu need t make sure that all  the llwing ints have been remembered:

 Errr bars are included i arriate.

 The grah shuld have a title. Smetimes they als need a key.  The scales  the axes shuld be suitable  there shuld nt,  curse, be any sudden r uneven jums in the numbers.  The inclusin  the rigin has been thught abut. Mst grahs shuld have the rigin included  it is rare r a grah t be imrved by this being missed ut. I in dubt include it. Yu can always draw a secnd grah withut it i necessary.  The fnal grah shuld, i ssible, cver mre than hal the aer in either directin.  The axes are labelled with bth the quantity (e.g. current) AND the units (e.g. ams) .  The ints are clear. Vertical and hrizntal lines t make crsses are better than 45 degree crsses r dts.

 A best-ft trend line is added. This line NEVER just jins the dts  it is there t shw the verall trend.  I the best-ft line is a curve, this has been drawn as a single smth line.  I the best-ft line is a straight line, this has been added WITH A RULER.  As a general rule, there shuld be rughly the same number  ints abve the line as belw the line.  Check that the ints are randmly abve and belw the line. Smetimes ele try t ft a best-ft straight line t ints that shuld be reresented by a gentle curve. I this was dne then ints belw the line wuld be at the beginning  the curve and all the ints abve the line wuld be at the end, r vice versa.  Any ints that d nt agree with the best-ft line have been identifed.

Measuring intercePt, gradient and area under the graPh

The gradient  a curve at any articular int is the gradient  the tangent t the curve at that int.

Grahs can be used t analyse the data. This is articularly easy r straight-line grahs, thugh many  the same rinciles can be used r curves as well. Three things are articularly useul: the intercept, the gradient and the area under the graph.

y

1. Intercept In general, a grah can intercet (cut) either axis any number  times. A straight-line grah can nly cut each axis nce and ten it is the y-intercept that has articular imrtance. (Smetimes the y-intercet is reerred t as simly the intercet.) I a grah has an intercet  zer it ges thrugh the rigin. Proportional  nte that tw quantities are rrtinal i the grah is a straight line THAT pASSES THRoUGH THE oRIGIN. Smetimes a grah has t be cntinued n (utside the range  the readings) in rder r the intercet t be und. This rcess is knwn as extrapolation. The rcess  assuming that the trend line alies between tw ints is knwn as interpolation.

extrapolation

y P

y

x x

x x at point P on the curve, gradient = y x

gradient of straight line = y x

3. Area under a graph The area under a straight-line grah is the rduct  multilying the average quantity n the y-axis by the quantity n the x-axis. This des nt always reresent a useul hysical quantity. When wrking ut the area under the grah:  I the grah cnsists  straight-line sectins, the area can be wrked ut by dividing the shae u int simle shaes.  I the grah is a curve, the area can be calculated by cunting the squares and wrking ut what ne square reresents.

y-intercept The extrapolation of the graph continues the trend line.

y

The line is interpolated between the points.

 The units r the area under the grah are the units n the y-axis multilied by the units n the x-axis.  I the mathematical equatin  the line is knwn, the area  the grah can be calculated using a rcess called integration.

y

y

2. Gradient The gradient  a straight-line grah is the increase in the y-axis value divided by the increase in the x-axis value. The llwing ints shuld be remembered:  A straight-line grah has a cnstant gradient.  The triangle used t calculate the gradient shuld be as large as ssible.

x area under graph

x area under graph

 The gradient has units. They are the units n the y-axis divided by the units n the x-axis.  only i the x-axis is a measurement  time des the gradient reresent the RATE at which the quantity n the y-axis increases.

APPEN D I X

215

gp y d dm  p equation of a straight-line graPh All straight-line graphs can be described using one general equation

You should be able to see that the physics equation has exactly the same orm as the mathematical equation. The order has been changed below so as to emphasize the link. v = u + at

m and c are both constants  they have one fxed value.  c represents the intercept on the y-axis (the value y takes when x = 0)  m is the gradient o the graph. In some situations, a direct plot o the measured variable will give a straight line. In some other situations we have to choose careully what to plot in order to get a straight line. In either case, once we have a straight line, we then use the gradient and the intercept to calculate other values. For example, a simple experiment might measure the velocity o a trolley as it rolls down a slope. The equation that describes the motion is v = u + at where u is the initial velocity o the object. In this situation v and t are our variables, a and u are the constants.

y = c + mx By comparing these two equations, you should be able to see that i we plot the velocity on the y-axis and the time on the x-axis we are going to get a straight-line graph.

velocity v / m s -1

y = mx + c y and x are the two variables (to match with the y-axis and the x-axis) .

20 15 10

gradient = 20 = 4 m s -2 5

5 1

intercept = 0

2

3

4

5 time t / s

The comparison also works or the constants.  c (the y-intercept) must be equal to the initial velocity u  m (the gradient) must be equal to the acceleration a

 Identiy which symbols represent variables and which symbols represent constants.  The symbols that correspond to x and y must be variables and the symbols that correspond to m and c must be constants.  I you take a variable reading and square it (or cube, square root, reciprocal etc.)  the result is still a variable and you could choose to plot this on one o the axes.  You can plot any mathematical combination o your original readings on one axis  this is still a variable.  Sometimes the physical quantities involved use the symbols m (e.g. mass) or c (e.g. speed o light) . Be careul not to conuse these with the symbols or gradient or intercept. Example 1

Example 2 I an object is placed in ront o a lens we get an image. The image distance v is related to the object distance u and the ocal length o the lens f by the ollowing equation. 1 +_ 1 =_ 1 _ u v f There are many possible ways to rearrange this in order to get it into straightline orm. You should check that all these are algebraically the same. uv v + u = _ f

The gravitational orce F that acts on an object at a distance r away rom the centre o a planet is given by the equation GMm where M is the mass o the planet and the F = _ r2 m is mass o the object. I we plot orce against distance we get a curve (graph 1 ) . GMm We can restate the equation as F = ____ + 0 and i we plot F r 1 on the y-axis and __ on the x-axis we will get a straight-line r (graph 2) .

or

v v _ _ u = f - 1

or

1 = _ 1 - _ 1 _ u v f

v u

With a little rearrangement we can oten end up with the physics equation in the same orm as the mathematical equation o a straight line. Important points include

In this example the graph tells us that the trolley must have started rom rest (intercept zero) and it had a constant acceleration o 4.0 m s - 2 .

(v + u)

choosing what to Plot to get a straight line

gradient = 1 f

gradient = 1 f

uv

intercept = 0

v intercept = 1

1

A

2

F

force F

2

A gradient = GMm

B C distance r

C

B

intercept = 0

216

APPEN D I X

I r2

1 u

2

intercept = 1 f

gradient = 1

1 v

gpc y  mc fuc

Then lg (a) = b [t be abslutely recise lg1 0 (a) = b] Mst calculatrs have a lg buttn n them. But we dnt have t use 1 0 as the base. We can use any number that we like. Fr examle we culd use 2.0, 563.2, 1 7.5, 42 r even 2.71 8281 828459045235360287471 4. Fr cmlex reasns this last number IS the mst useful number t use! It is given the symbl e and lgarithms t this base are called natural logarithms. The symbl fr natural lgarithms is ln (x) . This is als n mst calculatrs. If p = e q Then ln (p) = q

When an exerimental situatin invlves a wer law it is ften nly ssible t transfrm it int straight-line frm by taking lgs. Fr examle, the time erid f a simle endulum, T, is related t its length, l, by the fllwing equatin.

A lt f the variables will give a curve, but it is nt clear frm this curve what the values f k and p wrk ut t be. on t f this, if we d nt knw what the value f p is, we can nt calculate the values t lt a straight-line grah.

The int f lgarithms is that they can be used t exress sme relatinshis (articularly wer laws and exnentials) in straight-line frm. This means that we will be ltting grahs with lgarithmic scales.

6

7

8

9

10

11

A lgarithmic scale increases by the same rati all the time. 10 1

0

10 10

1

10

2

1 00

10

Bth the gravity frce and the electrstatic frce are inverse-square relatinshis. This means that the frce  (distance aart) - 2 . The same technique can be used t generate a straight-line grah.

l / metres

force =

k (distance apart) 2

ln (T) = ln (k l p )

3

1 000

distance apart log (force)

The trick is t take lgs f bth sides f the equatin. The equatins belw have used natural lgarithms, but wuld wrk fr all lgarithms whatever the base.

intercept = log (k) gradient = -2

ln (T) = ln (k) + ln (l p )

log (distance apart)

ln (T) = ln (k) + p ln (l) This is nw in the same frm as the equatin fr a straight line y = c + mx

A nrmal scale increases by the same amunt each time. 4 5

In l plt f ln (time erid) versus ln (length) gives a straight-line grah

Fr examle, the cunt rate R at any given time t is given by the equatin

Inverse square relatinshi  direct lt and lg-lg lt

R

( )

1 = - ln (c) ln _ c These rules have been exressed fr natural lgarithms, but they wrk fr all lgarithms whatever the base.

3

intercept = ln (k)

Time erid versus length fr a simle endulum

ln (cn ) = n ln (c)

2

gradient = p

k and p are cnstants.

ln (c  d) = ln (c) + ln (d)

1

The gradient will be equal t p The intercet will be equal t ln (k) [s k = e ( in te rce  t) ]

T = k lp

The werful nature f lgarithms means that we have the fllwing rules

ln (c  d) = ln (c) - ln (d)

Thus if we lt ln (T) n the y-axis and ln (l) n the x-axis we will get a straightline grah.

In T

If a = 1 0 b

Power laws and logs (log  log)

force

logs  base ten and base  Mathematically,

T / seconds

hl

R0

R = R0 e -  t

R = R0 e -t

R0 and  are cnstants. If we take lgs, we get ln (R) = ln (R0 e -  t)

Natural lgarithms are very imrtant because many natural rcesses are exnential. Radiactive decay is an imrtant examle. In this case, nce again the taking f lgarithms allws the equatin t be cmared with the equatin fr a straight line.

ln (R) = ln (R0 ) - t [ln (e) = 1 ]

ln (R) = ln (R0 ) - t ln (e) This can be cmared with the equatin fr a straight-line grah

t ln (R)

exPonentials and logs (log  linear)

ln (R) = ln (R0 ) + ln (e -  t)

intercept = ln (R0 ) gradient = -

y = c + mx

t

Thus if we lt ln (R) n the y-axis and t n the x-axis, we will get a straight line. Gradient = -  Intercet = ln (R0 )

APPEN D I X

217

Answers Topic 1 (Page 8): Measurements and uncertainties 1. (a) (i) 0.5  acceleration down the slope (a) (iv) 0.36 ms - 2 2. C 3. D 4. D 5. (b) 2.4  0.1 s (c) 2.6  0.2 ms - 2 6. (b) (i) -3; (b) (ii) 2.6 1 0 - 4 Nm- 3

Topic 9 (Page 104): Wave phenomena ; 1. B 2. (b) 27.5 m s - 1 3. (a) 0.2; 4. (a) (i) zero; (ii)  or _ 2 -1 0 (iii) zero; (b) 1 1 0 nm; 5. (b) (i) 1 .5  1 0 m; (d) (ii) 5.0  1 0 1 9 m s - 2

Topic 2 (Page 24): Mechanics 1. C 2. D 3. B 4. B 5. (a) 520 N; (b) (i) 1 .2 MJ; (b) (ii) 270 W 6. (a) equal; (b) let; (c) 20 km hr- 1 ; (e) car driver; () No 7. (c) 3.50 N

Topic 10 (Page 111): Fields 1. A 2. C 3. C 4. (a) (i) 1 .9  1 0 1 1 J; (a) (ii) 7.7 km hr- 1 (a) (iii) 2.2  1 0 1 2 J; (c) 2.6 hr 5. (b) (i) 2.5 m s - 2

Topic 3 (Page 32): Thermal Physics 1. B 2. D 3. D 4. D 5. (a) (i) length = 20 m, depth = 2 m, width = 5m, temp = 25 C; (a) (ii) $464; (b) (i) 84 days 6. (a) (i) 7.8 J K mol- 1 Topic 4 (Page 50): Waves 1. C 2. C 3. (a) longitudinal (b) (i) 0.5 m; (ii) 0.5 mm; (iii) 330 m s-1 4. (c) (i) 2.0 Hz; (ii) 1.25 (1.3) cm; () (i) 4.73  10-7 m; (ii) 0.510 mm 5. 45 Topic 5 (Page 64): Electricity and magnetism 1. C 2. A 3. (c) (ii) 7.2  1 0 1 5 m s - 2 (c) (iv) 1 00 v 4. (d) B; (e) (i) Equal; (ii) approx. 0.4A; (iii) lamp A will have greater power dissipation; Topic 6 (Page 68): Circular motion and gravitation M; 1. A 2. A 3. C 4. (a) (ii) No; (b) 1 .4 m s - 1 5. (b) (i) g = G _ R2 M _ 27 24 (b) (ii) 1 .9  1 0 kg 6. (a) g = G 2 ; (b) 6.0  1 0 kg; R Topic 7 (Page 81): Atomic nuclear and particle physics 1. B 2. D 3. A 4. D 5. B 6. (a) (i) uud; (ii) electron is undamental; (iii) 3 quarks or 3 antiquarks; (iv) a quark and an __ 0 2 26 24 4  ] 8. (b) (i) 1 H + 1 2 Mg  1 1 Na + 2 He antiquark; 7. uu [ 12 12 9. (a) 6 C  7 N + -01  + ; (b) (ii) 1 1 600 years; 10. (a) (i) 3; (b) (i) 1 .72  1 0 1 9 11. A Topic 8 (Page 94): Energy production 1. (c) 1 5 MW (d) (i) 20% 4. (a) 1 000 MW; (b) 1 200 MW; (c) 1 7% ; (d) 43 kg s - 1 5. (c) 1 .8 MW

Topic 11 (Page 120): Electromagnetic induction 1. D 3. B 4. B 5. D 6. (b) 0.7 v 7. (a) 7.2  1 0 - 4 C; (b) 2.9  1 0 - 3 s; (c) (ii) 5CR = 3.5 s; (c) (iii) No Topic 12 (Page 130): Quantum and nuclear 1. C 2. B 3. (b) ln R & t; (c) Yes; (e) 0.375 hr- 1 ; (h) 1 .85 hr 5. (b) (i) 6.9  1 0 - 3 4 Js; (b) (ii) 3.3  1 0 - 1 9 J 6. 4.5  1 0 4 Bq Option A (Page 151): relativity 2. (a) (i) 1 .40c; (a) (ii) 0.95c; (c) 6.0  1 0 1 9 J 3. a) 2 yrs; b) 4 yrs ; c) x = 5 ly; d) 0.5 c 4. (c) ront; (d) T:1 00 m, S:87 m; (e) T:75m, S:87 m; 5. (a) (i) zero; (a) (ii) 2.7 m0 c2 ; (b) (i) 0.923 c; (b) (ii) 2.4 m0 c2 ; (b) (iii) 3.6 m0 c2 ; (c) agree. Option B (Page 170): Engineering physics 2. a) 0.95 N m; b) 25.2 J; c) 1 3.4 N 3. (a) No; (b) Equal; (c) 300 J; (d) -500 J; (e) 500 J; () 1 50 J; (g) 1 6% 4. (b) 990K; (c) (i) 1 ; (c) (ii) 2 & 3; (c) (iii) 3; 6. Laminar (R=1 200) 7. (a) 2Hz; (b) 21 mW Option C (Page 189): Imaging 1. (a) 1 4 cm behind mirror, virtual, upright, magnifed ( 2) ; (b) 24 cm behind diverging lens, real, inverted, magnifed ( 3) ; (c) 4.5 cm behind second lens, real, upright & diminished (0.25) 2. (d) upside down; (e) 60 cm; 4. (a)  1 0 dB; (b) 0.5 mW 5. (a) 1 20 MHz; (b) (iii) d = 38mm, l = 1 30mm 6. (b) (ii) 4 mm; (b) (iii) 9.3 mm; (b) (iv) 1 8.6 mm Option D (Page 214): Astrophysics 1. (ii) d = 7.78 ly, (c) r = 8.9  1 0 7 m; 3. (a) 5800 K 4. (e) 499.83 nm 5. 1 4% o current size 7. 9.5  1 0 - 2 7 kg m- 3

Origin of individual questions The questions detailed below are all taken rom past IB examination papers and are all  IB. Topic 1: Measurement and uncertainties 1 N99S2(S2) 2 M98H1 (5) 3 N98H1 (5) 4 M99H1 (3) 5 M98SpH2(A2) 6 N98H2(A1 )

Topic 8: Energy production 1 NO1 S3(C1 ) 2 M99S3(C1 ) 3 M98SpS3(C3) 5 M98S3(C2) 6 M1 22H2(B2.1 )

Topic 2: Mechanics 1 M98S1 (2) 2 M98S1 (4) 3 M98S1 (8) 6 N00H2(B2) 7 M091 S2(A2)

Topic 9: Wave phenomena 1 N01 H1 (24) 2 N98H2(A5) 3 M1 1 1 H3(G3)

5 M1 01 S2(A2)

Topic 3: Thermal Physics 1 N99H1 (1 5) 2 N99H1 (1 6) 3 N99H1 (1 7) 6 M1 1 2H2(A5) 7 M091 S2(A2) Topic 4: Waves 1 M01 H1 (1 4) 2 N1 0H1 (1 5)

3 N03S2(A3)

Topic 10: Fields 3 N1 0H1 (24) 4 N98H2(B4) 5 M98 Sp2(B2)

4 5 N04H2(B4.1 )

Topic 5: Electricity and magnetism 4 N03 HL2 Q2.2 Topic 6: Circular motion and gravitation 1 N1 0S1 (7) 2 M1 1 1 H1 (4) 3 M1 01 S1 (8) 4 M1 1 1 2(A5) 5 M08 SpS3(A3) 6 N05H2(B2.1 )  part question  sections (d) to (g) Topic 7: Atomic, nuclear and particle physics 1 N98S1 (29) 2 M99S1 (29) 3 M99S1 (30 4 M98SpS1 (29) 5 M98SpS1 (30) 8 M98S2(A3) 9 M99S2(A3) 10 M99H2(B4) 11 M1 22H1 (32)

218

An s we r s to q u e s ti o n s

4 M98SpS3(C2)

5 N09 HL3 G4

5 N01 H2(A3)

Topic 11: Electromagnetic induction 1 N00H1 (31 ) 3 M98H1 (33) 4 M1 1 2H1 (24) 6 N98H2(A4)

5 N99H1 (34)

Topic 12: Quantum and nuclear physics 1 N1 0H1 (34) 2 M01 H1 (35) 3 N00H2(A1 ) Option A relativity 2 M1 1 1 H3(2) 4 M00H3(G1 )

5 N01 H3(G2)

6 M092H3(3)

Option B Engineering physics 3 N01 H2(B2) 4 N98H2(A2) Option C imaging 2 N00H3(H1 ) 5 M03H3(D2) Option D Astrophysics 1 M1 01 H3(E1 ) 2 M1 1 1 H3(E2)

3 N01 H3(F2)

4 N98H3(F2)

Index Page numbers in italics reer to question sections. A absolute magnitude 1 94 absolute uncertainties 5 absolute zero 29 absorption spectra 69 greenhouse gases 92 acceleration 9, 1 4, 1 08 acceleration-time graphs 1 0 acceleration, velocity and displacement during simple harmonic motion [SHM] 34, 95 equations o uniorm angular acceleration 1 52 experiment to determine ree-all acceleration 1 3 uniormly accelerated motion 1 1 achromatic doublets 1 78 acoustic impedance 1 87 addition 5 air resistance 1 6 albedo 90 alpha radiation 72 alpha decay 72, 1 27 alternating current 54 coil rotating in a magnetic eld  ac generator 1 1 4 diode bridges 1 1 5 losses in the transmission o power 115 RMS values 1 1 4 transormer operation 1 1 4 transmission o electrical power 1 1 5 ammeters 56 amperes 63 ampliers 1 83 amplitude 39, 48 angular impulse 1 57 angular magnication 1 77 angular momentum 1 57, 1 65 conservation o angular momentum 1 57 angular motion 1 24 angular size 1 77 anisotropies 21 0 antimatter 73 antineutrinos 1 29 antinodes 48 antiparticles 78 apparent magnitude 1 94 Archimedes principle 1 64 assumptions 3 asteroids 1 90 astronomical units (AUs) 1 91 , 1 93, 1 94 astrophysics 1 90, 21 4 accelerating Universe 204 astrophysics research 21 3 Big Bang model 201 Cepheid variables 1 98 cosmological principle and mathematical models 208 current observations 21 3 dark energy 21 2 uture o the Universe 21 1

galactic motion 202 HertzsprungRussell diagram 1 97 history o the Universe 21 0 Hubbles law and cosmic scale actor 203 luminosity 1 94 nature o stars 1 92 nuclear usion  the Jeans criterion 205 nucleosynthesis 1 96 nucleosynthesis o the main sequence 206 objects in the Universe 1 901 red giant stars 1 99 rotation curves and dark matter 209 stellar evolution 200 stellar parallax 1 93 stellar spectra 1 95 types o supernovae 207 atomic clock requency shits 1 48 atomic energy levels 1 29 atomic physics 69, 77, 81 atomic model 77 evidence or atomic model 77 explanation o atomic spectra 69 structure o matter 77 atomic spectra 69, 1 23 atoms 26 attenuation 1 83 attenuation coecient 1 85 mass attenuation coecient 1 85 Avogadro constant 30 B background radiation 73 background count 73 Balmer series 1 23 barium meals 1 86 barrel distortion 1 78 baryonic matter 209 baryons 78, 79 base units 2 batteries 60 becquerels 73 Bernoulli eect 1 65 applications o the Bernoulli equation 1 66 Bernoulli equation 1 65 beta radiation 72 beta decay 72, 1 29 Big Bang model 201 binding energy 75 binding energy per nucleon 76 binoculars 45 bireringence 41 black-body radiation 90, 1 94 black holes 1 50, 1 96, 200 Schwarzchild radius 1 50 blue shit 1 95 Bohr model o the atom 1 24 boundary conditions 49 Boyles law 31 Brackett series 1 23

Brewsters law 41 buoyancy 1 64 C cancer 72 capacitance 1 1 7 capacitors 1 1 6 capacitor [RC] discharge circuits 1 1 8 capacitor charging circuits 1 1 9 capacitors in series and parallel 1 1 7 energy stored in charged capacitor 119 carbon dioxide 92 carbon xation 92 Carnot cycles 1 63 Carnot engine 1 63 Carnot theorem 1 63 cells 60 Celsius scale 25 Cepheid variables 1 96 mathematics 1 98 principles 1 98 chain reaction 76 Chandrasekhar limit 200 charge 51 charge capacity 60 Coulombs law 51 current 54 particle acceleration and electric charge 1 45 Charless law 31 chemical energy 22 chlorofuorocarbons (CFCs) 92 circuits 55 capacitor [RC] discharge circuits 1 1 8 capacitor charging circuits 1 1 9 investigating diode-bridge rectication circuit experimentally 1 1 6 parallel circuits 56 potential divider circuits 57 rectication and smoothing circuits 116 sensor circuits 57 series circuits 56 circular motion 657, 68 angular velocity and time period 66 circular motion in a vertical plane 66 examples 65 mathematics o circular motion 65 mechanics o circular motion 65 Newtons law o universal gravitation 67, 68 radians 66 collisions 23 colour 79 comets 1 90 communications 1 84 coaxial cables 1 84 optical bres 1 84 wire pairs 1 84 complex numbers 1 25 composite particles 78 compression 1 4, 1 6

I n dex

219

compression waves 35 concentration o solutions 42 conduction 89 conduction electrons 54 conductors 51 conjugate quantities 1 26 conservation o energy 22 constant pressure 29, 1 60 constant temperature 29 constant volume 29 constellations 1 901 constructive intererence 40, 47 continuity equation 1 65 continuous spectrum 69 continuous waves 35 convection 89 conventional current 54 converging lenses 1 72, 1 73 convex lenses 1 73 Copenhagen interpretation 1 25 cosmic density parameter 21 2 cosmic microwave background (CMB) radiation 73 fuctuations in CMB 21 0 cosmic scale actor 203 cosmic scale actor and temperature 21 0 eect o dark energy on the cosmic scale actor 21 2 cosmological principle 208 Coulombs law 51 coulombs 53 couples 1 54 critical angle 45 critical density 21 1 CT (computed tomography) scans 1 86 current 54 D damping 1 68 dark energy 204, 21 2 eect o dark energy on the cosmic scale actor 21 2 dark matter 209 gravity 21 2 MACHOs, WIMPs and other theories 209 Davisson and Germer experiment 1 22 De Broglie hypothesis 1 22 deormation 1 4 derived units 2 destructive intererence 40, 47 dielectric material 1 1 7 diraction 46 basic observations 46, 97 Davisson and Germer experiment 1 22 diraction and resolution 1 01 diraction grating 99 electron diraction experiment 1 22 examples o diraction 46 explanation 97 multiple-slit diraction 99 practical signicance o diraction 46 resolvance o diraction gratings 1 01 single-slit diraction with white light 97 uses o diraction gratings 99 diode bridges 1 1 5

220

I n dex

investigating diode-bridge rectication circuit experimentally 1 1 6 direct current 54 discharge characteristics 60 dispersion 1 83 acceleration, velocity and displacement during simple harmonic motion [SHM] 34, 95 diverging lenses 1 75 denitions and important rays 1 75 images created by diverging lens 1 75 division 5 Doppler eect 1 02 Doppler broadening 1 03 examples and applications 1 03 mathematics o the Doppler eect 1 02 moving observer 1 02 moving source 1 02 drag 1 6 drit speed equation 54 drit velocity 54 dwar planets 1 90 dynamic riction 20 E Earth 1 90 day 1 91 year 1 91 earthquake waves 35 eddy currents 1 1 5 eciency 22 Einstein model o light 1 21 elastic collisions 23 elastic potential energy 22 electric elds 52, 61 comparison between electric and gravitational elds 1 1 0 comparison between electric and magnetic elds 1 32 energy dierence in an electric eld 53 representation o electric elds 52 electric potential dierence 53, 1 09 electric potential energy 53, 1 09 electrical conduction in a metal 54 electrical energy 22 electrical meters 56 electricity 51 60, 64 electric charge and Coulombs law 51 electric circuits 55 electric current 54 electric elds 52 electric potential energy and electric potential dierence 53, 1 09 example o use o Kirchos laws 59 internal resistance and cells 60 potential divider circuits and sensors 57 resistivity 58 resistors in series and parallels 56 electromagnetic orce 71 electromagnetic induction 1 1 21 9, 1 20 alternating current 1 1 41 5 capacitance 1 1 7 capacitor charge 1 1 9 capacitor discharge 1 1 8 induced electromotive orce (em) 1 1 2

Lenzs law and Faradays law 1 1 3 rectication and smoothing circuits 116 electromagnetic waves 37, 89 electromotive orce (em) 60 induced em 1 1 2 production o induced em by relative motion 1 1 2 transormer-induced em 1 1 3 electron degeneracy pressure 200 electrons 77 nuclear scattering experiment involving electrons 1 28 orbital 1 25 electronvolts 53 electrostatic orce 1 4, 1 6, 51 electrostatic potential energy 22 elementary particles 78 emission spectra 69 emissivity 90 energies, range o 1 energy 22 concepts o energy and work 22 conservation o energy 22 energy fow or stars 1 92 energy types 22 mass and energy 1 43 relativistic momentum and energy 1 44 wave energy 48 energy degradation 1 62 energy production 8293, 94 electrical power production 82 energy conversions 82 ossil uel power production 84 global warming 93 greenhouse eect 92 hydroelectric power 87 new and developing technologies 88 nuclear power 856 primary energy sources 83 radiation 90 secondary energy sources 88 solar power 87, 91 thermal energy transer 89 wind power and other technologies 88 energy sources 83 comparison o energy sources 83 non-renewable energy sources 83 renewable energy sources 83 specic energy and energy density 83 energy transer 35, 89 energy transormations 22 wind power 88 engineering physics 1 52, 1 70 Bernoulli  examples 1 66 equilibrium examples 1 55 rst law o thermodynamics 1 61 fuids at rest 1 64 fuids in motion  Bernoulli eect 1 65 orced oscillations and resonance 1 689 heat engines and heat pumps 1 63 Newtons second law  moment o inertia 1 56 rotational dynamics 1 57 second law o thermodynamics and

entropy 1 62 solving rotational problems 1 58 thermodynamic systems and concepts 1 59 translational and rotational equilibrium 1 54 translational and rotational motion 1 523 viscosity 1 67 work done by an ideal gas 1 60 entropy 1 62 equilibrium 1 6 equilibrium examples 1 55 hydrostatic equilibrium 1 64 translational and rotational equilibrium 1 54 equipotentials 1 06 equipotential suraces 1 06 examples o equipotentials 1 06 relationship to eld lines 1 06 equivalence principle 1 46 bending o light 1 46 errors 4 error bars 6 estimation 3 evaporation 89 exchange bosons 78 exchange particles 78 excited state 72 exponential processes 73, 1 29 F ar point 1 77 Faradays law 1 1 3 application o Faradays law to moving and rotating coils 1 1 3 Feynman diagrams 80 bre optics 1 823 elds 1 05, 1 1 1 describing elds 1 05 electric and gravitational elds compared 1 1 0 electric potential energy and potential 1 09 equipotentials 1 06 eld lines 1 05, 1 06 gravitational potential energy and potential 1 07 orbital motion 1 08 potential [gravitational and electric] 1 05 potential and eld strength 1 09 propagation 1 32 uniorm elds 1 1 0 rst harmonic 49 ssion 76 fuid riction 1 6 fuid resistance 1 3 fuids at rest 1 64 buoyancy and Archimedes principle 1 64 denitions o density and pressure 1 64 hydrostatic equilibrium 1 64 Pascals principle 1 64 variation o fuid pressure 1 64 fuids in motion 1 656 Bernoulli eect 1 656

ideal fuid 1 65 laminar fow, streamlines and the continuity equation 1 65 fux density 1 1 2 fux linkage 1 1 3 fux losses 1 1 5 orces 1 4 couples 1 54 dierent types o orces 1 4, 1 6 orces as vectors 1 4 undamental orces 71 , 78 magnitude 1 05 measuring orces 1 4 moment o orce (torque) 1 54 particles that experience and mediate the undamental orces 71 ossil uels 84 advantages and disadvantages 84 eciency o ossil uel power stations 84 energy transormations 84 ractional uncertainties 5 rames o reerence 9, 1 31 inertial rame o reerence 1 33, 1 42 ree-body diagrams 1 4 ree-all 1 1 experiment to determine ree-all acceleration 1 3 requency 48 atomic clock requency shits 1 48 driving requency 1 68 Larmor requency 1 88 natural requency and resonance 1 68 threshold requency 1 21 ultrasound 1 87 riction 1 4, 1 6 coecient o riction 20 static and dynamic actors aecting riction 20 undamental units 2 usion 28, 76 nuclear usion 205 G galaxies 1 91 distributions o galaxies 202 experimental observations 203 motion o galaxies 202 rotation curves 208, 209 Galilean transormations 1 31 ailure o Galilean transormation equations 1 31 gamma radiation 72 gases 25, 26 equation o state 30 experimental investigations 29 gas laws 2930 greenhouse gases 92 ideal gases and real gases 30 molecular model o an ideal gas 31 gauge bosons 78 Geiger counters 73 general relativity 1 46 applications o general relativity to the universe as a whole 1 49 geometric optics 43 geometry o mirrors and lenses 1 76 global positioning systems 1 48

global warming 93 evidence or 93 mechanisms 93 possible causes 93 gluons 79 GM tubes 73 graphs 1 0 acceleration-time graphs 1 0 choosing what to plot to get a straight line 21 6 displacement-time graphs 1 0 equation o a straight-line graph 21 6 exponentials and logs 21 7 graphical representation o uncertainty 4 intererence o waves 40 logs  base ten and base 21 7 measuring intercept, gradient and area under the graph 21 5 plotting graphs  axes and best t 21 5 power laws and logs 21 7 rotational motion 1 58 simple harmonic motion [SHM] 34 velocity-time graphs 1 0 gravitational elds 1 1 comparison between electric and gravitational elds 1 1 0 gravitational eld strength 67 gravitational orce 1 4, 1 6 gravitational lensing 1 48 gravitational potential 1 07 escape speed 1 07 gravitational potential energy 22, 1 07 gravitational potential gradient 1 08 gravitational red shit 1 47 evidence to support gravitational red shit 1 48 gravity 71 centre o gravity 1 55 dark energy 21 2 dark matter 21 2 eect o gravity on spacetime 1 49 greenhouse eect 92, 93 greenhouse gases 92 H hadrons 78 hal-lie 74 example 74 investigating hal-lie experimentally 74 simulation 74 hal-value thickness 1 85 harmonics 49 heat 26, 1 59 heat engines 1 63 heat fow 25 heat pumps 1 63 methods o measuring heat capacities and specic heat capacities 27 phases o matter and latent heat 28 specic heat capacity 27 Heisenberg uncertainty principle 1 26 estimates rom uncertainty principle 1 26 HertzsprungRussell diagram 1 97 interpretation 200 Higgs bosons 78, 79

I n dex

221

Hubble constant 203 Hubbles law 203 Huygens principle 97 hydraulic systems 1 64 hydroelectric power 87 advantages and disadvantages 87 hydrogen spectrum 1 23 hydrostatic equilibrium 1 64, 1 92 hysteresis 1 1 5 I ideal gases 30 ideal gas laws 29, 30 ideal gas processes 1 61 kinetic model o an ideal gas 31 work done during expansion at constant pressure 1 60 image ormation 1 71 image ormation in convex lenses 1 73 image ormation in mirrors 1 76 real and virtual images 1 71 stick in water 1 71 wave model o image ormation 1 72 imaging 1 71 88, 1 89 aberrations 1 78 astronomical refecting telescopes 1 80 channels o communication 1 84 compound microscope and astronomical telescope 1 79 converging and diverging mirrors 1 76 converging lenses 1 72 dispersion, attenuation and noise in optical bres 1 83 diverging lenses 1 75 bre optics 1 82 image ormation 1 71 image ormation in convex lenses 1 73 radio telescopes 1 81 simple magniying glass 1 77 thin lens equation 1 74 ultrasonic imaging 1 878 X-ray imaging techniques 1 86 X-rays 1 85 impulse 23 incompressibility 1 65 inelastic collisions 23 inra-red 89 insulators 51 , 89 intensity 39, 90, 1 85, 1 88 intererence o waves 40 thin parallel lms 1 00 two-source intererence 47, 98 intermolecular orces 26 internal energy 22, 26, 1 59 internal energy o an ideal monatomic gas 1 59 internal resistance 60 determining internal resistance experimentally 60 invariant quantities 1 36 inverse square law o radiation 39 isotopes 70 J Jeans criterion 205 K Kelvin scale 25

222

I n dex

kilograms 1 7 kinetic theory 26 kinetic energy 22, 33 kinetic riction 20 Kirchos circuit laws 55 example o use o Kirchos laws 59 L laminar fow 1 65 lamination 1 1 5 latent heat 28 methods o measuring latent heat 28 length contraction 1 38 calculation o time dilation and length contraction 1 40 derivation o length contraction rom Lorentz transormation 1 38 lengths, range o 1 lenses 1 72 centre o curvature 1 72 chromatic aberration 1 78 ocal length 1 72, 1 75 ocal point 1 72, 1 75 geometry o mirrors and lenses 1 76 linear magnication 1 72, 1 74 power 1 72 principal axis 1 72 spherical aberration 1 78 thin lens equation 1 74 Lenzs law 1 1 3 leptons 78 lepton amily number 78 lit 1 4, 1 6 light bending o light 1 46 bending o star light 1 48 circularly polarized light 41 light clock 1 37 light curves 207 light energy 22 light gates 1 1 light waves 35, 40 light years (lys) 1 91 light-dependent resistors (LDRs) 57 monochromatic light 47 partially plane-polarized light 41 plane-polarized light 41 polarized light 41 speed o light 1 32 unpolarized light 41 waveparticle duality 1 22 liquid-crystal displays [LCDs] 42 liquids 26 logarithmic unctions 21 7 natural logarithms 21 7 longitudinal waves 35 longitudinal sound waves in a pipe 49 Lorentz transormations 1 34 derivation o eect rom Lorentz transormation 1 37 derivation o length contraction rom Lorentz transormation 1 38 Lorentz actor 1 34 Lorentz transormation example 1 34 luminosity 1 94 Lyman series 1 23

M MACHOs (Massive Astronomical Compact Halo Objects) 209 magnetic elds 61 comparison between electric and magnetic elds 1 32 magnetic eld in a solenoid 63 straight wire 63 two parallel wires 63 magnetic orce 1 4, 1 6, 64 examples o the magnetic eld due to currents 62 magnetic eld lines 61 magnetic orce on a current 62 magnetic orce on a moving charge 62 magniying glass 1 77 angular magnication 1 77 angular size 1 77 near and ar point 1 77 magnitude 1 , 1 05, 1 94 orders o magnitude 1 , 3 Maluss law 41 mass 1 9 centre o mass 1 52 mass and energy 1 43 mass deect 75 point masses 67 range o masses 1 unied mass units 75 material dispersion 1 83 mathematics 5 Cepheid variables 1 98 Doppler eect 1 02 exponential decay 1 29 gravitational red shit 1 47 motion o galaxies 202 parabolic motion 1 2 stellar parallax 1 93 two-source intererence 47 wind power 88 matter structure 77 matter waves 1 22 Maxwells equations 1 32 mean position 33 measurement 1 7, 8 mechanics 923, 24 energy and power 22 equilibrium 1 6 fuid resistance and ree-all 1 3 orces and ree-body diagrams 1 4 mass and weight 1 9 momentum and impulse 23 motion 91 2 Newtons rst law o motion 1 5 Newtons second law o motion 1 7 Newtons third law o motion 1 8 solid riction 20 work 21 mesons 78, 79 metabolic pathways 72 meteorites 1 90 methane 92 micrometers 58 microscopes compound microscopes 1 79 scanning tunnelling microscopes 1 27 travelling microscopes 98 microscopic vs macroscopic 26

conduction 89 ideal gases 30 Milky Way galaxy 1 91 Minkowski diagrams 1 3942 mirrors 1 76 geometry o mirrors and lenses 1 76 molar gas constant 30 molar mass 30 mole 30 molecules 26 moment o inertia 1 56 example 1 56 moments o inertia or dierent objects 1 56 momentum 1 7 conservation o momentum 23 equations 1 44 linear momentum and impulse 23 relativistic momentum and energy 1 44 units 1 44 use o momentum in Newtons second law 23 motion 9 equations o uniorm motion 1 1 equilibrium 1 6 example o equation o uniorm motion 1 0 alling objects 1 1 fuid resistance and ree-all 1 3 orces and ree-body diagrams 1 4 rames o reerence 9, 1 31 graphical representation 1 0 instantaneous vs average 9 Newtons rst law o motion 1 5 Newtons second law o motion 1 7 Newtons third law o motion 1 8 practical calculations o uniormly accelerated motion 1 1 projectile motion 1 2 uniormly accelerated motion 1 1 multiplication 5 muons 78 muon experiment 1 38 N near point 1 77 nebulae 1 90 negative temperature coecient (NTC) 57 neutrinos 1 29 neutrons 77 neutron capture 206 Newtons rst law o motion 1 5 law o universal gravitation 67, 68 second law o motion 1 7 third law o motion 1 8 nitrous oxide 92 NMR (Nuclear Magnetic Resonance) 1 88 comparison between ultrasound and NMR 1 88 nodes 48 noise 1 83 normal reaction 1 4, 1 6 nuclear energy 22 nuclear energy levels 1 29 nuclear usion 205 nuclear physics 706, 81 , 1 30

ssion and usion 76, 205 hal-lie 74 nuclear energy levels and radioactive decay 1 29 nuclear reactions 75 nuclear stability 70 nucleus 1 28 nuclide notation 70 radioactivity 723 strong nuclear orce 71 weak nuclear orce 71 nuclear power 856 advantages and disadvantages 85 enrichment and reprocessing 86 usion reactors 86 health and saety risk 86 moderator, control rods and heat exchanger 85 nuclear weapons 86 thermal meltdown 86 nuclear reactions 75 articial transmutations 75 mass deect and binding energy 75 unied mass units 75 units 75 worked examples 75 nucleosynthesis 1 96 nuclear synthesis o heavy elements  neutron capture 206 nucleosynthesis o the main sequence 206 nucleus 1 28 deviations rom Rutherord scattering in high energy experiments 1 28 nuclear radii and nuclear densities 1 24 nuclear scattering experiment involving electrons 1 28 size 1 28 nuclides 70 O Ohms law 55 ohmic and non-ohmic devices 55 OppenheimerVolko limit 200 optic bre 1 82 attenuation 1 83 capacity 1 83 communications 1 84 material dispersion 1 83 noise, ampliers and reshapers 1 83 types o optic bre 1 82 waveguide dispersion 1 83 optically active substances 41 orbital motion 1 08 energy o an orbiting satellite 1 08 gravitational potential gradient 1 08 weightlessness 1 08 oscillations 33 damping 1 68 natural requency and resonance 1 68 phase o orced oscillations 1 69 Q actor and damping 1 68 undamped oscillations 96 output ripple 1 1 6 ozone 92 ozone layer 92

P pair production and pair annihilation 1 23 parabolas 1 2 parallax 1 93 parsecs (pcs) 1 91 , 1 93 particle acceleration and electric charge 1 45 particle physics 7880, 81 classication o particles 78 conservation laws 78 exchange particles 78 Feynman diagrams 80 leptons 78 particles that experience and mediate the undamental orces 71 quantum chromodynamics 79 quarks 79 standard model 78, 79 Pascals principle 1 64 Paschen series 1 23 path dierence 47 percentage uncertainties 5 periscopes 45 permittivity 1 1 7 Pund series 1 23 phase 48 in phase 33, 40 out o phase 33, 40 phase dierence 40 photoelectric eect 1 21 , 1 85 example 1 21 stopping potential experiment 1 21 photons 1 45 photovoltaic cells 87 piezoelectric crystals 1 87 pion decay 1 31 , 1 45 Plancks constant 1 24 plane o vibration 41 planetary nebula 200 planetary systems 1 90 planets 1 90 polarization 41 concentration o solutions 42 urther examples 42 liquid-crystal displays [LCDs] 42 optically active substances 41 polarizing angle 41 polaroid sunglasses 42 stress analysis 42 positive eedback 93 potential [electric or gravitational] 1 05, 1 07, 1 09 equipotentials 1 06 potential and eld strength 1 09 potential barrier 1 27 potential dierence [electric and gravitational] 1 05, 1 09 potential due to more than one charge 1 09 potential energy store 33 potential inside a charged sphere 1 09 potentiometers 57 PoundRebkaSnider experiment 1 48 power 22, 82 power dissipation 55 powers 5 precession 1 88 prexes 2

I n dex

223

pressure law 31 primary cells 60 prismatic refectors 45 projectile motion 1 2 horizontal component 1 2 mathematics o parabolic motion 1 2 vertical component 1 2 proper length 1 36 proper time 1 36 protonproton (pp) cycle 1 96 protons 77 pulsars 1 96, 200 pumped storage 87 Q Q actor and damping 1 68 quality 1 85 quantities 1 , 2 quantized energy 69 quantum chromodynamics 79 quantum physics 1 21 7, 1 30 atomic spectra and atomic energy states 1 23 Bohr model o the atom 1 24 Heisenberg uncertainty principle and loss o determinism 1 26 matter waves 1 22 photoelectric eect 1 21 Schrdinger model o the atom 1 25 tunnelling, potential barrier and actors aecting tunnelling probability 1 27 quarks 78, 79 quark connement 79 quasars 200 R r-process 206 radians 66 radiant energy 22 radiation 89 black-body radiation 90, 1 94 cosmic microwave background (CMB) radiation 73 equilibrium and emissivity 90 intensity 90 isotropic radiation 201 , 21 0 radioactive decay 72 mathematics o exponential decay 1 29 nature o alpha, beta and gamma decay 72 random decay 73 radioactivity 723 background radiation 73 eects o radiation 72 ionizing properties 72 properties o alpha, beta and gamma radiations 72 radiation saety 72 random errors 4 rays 35, 39 ray diagrams 43, 1 71 , 1 73 real gases 30 rectication 1 1 5, 1 1 6 red shit 1 03, 1 95 gravitational red shit 1 47 refection 43

224

I n dex

law o refection 43 refection and transmission 43 refection o two-dimensional plane waves 43 total internal refection and critical angle 45 types o refection 43 reraction 445 double reraction 41 methods or determining reractive index experimentally 45 reraction o plane waves 44 reractive index and Snells law 44 total internal refection and critical angle 45 relativity 1 31 50, 1 51 black holes 1 50 curvature o spacetime 1 49 equivalence principle 1 46 general relativity 1 46, 1 49 gravitational red shit 1 47 invariant quantities 1 36 length contraction and evidence to support special relativity 1 38 Lorentz transormations 1 34 mass and energy 1 43 Maxwells equations 1 32 reerence rames 1 31 relativistic mechanics examples 1 45 relativistic momentum and energy 1 44 spacetime diagrams [Minkowski diagrams} 1 3942 special relativity 1 33 supporting evidence 1 48 time dilation 1 37 twin paradox 1 401 velocity addition 1 35 relaxation time 1 88 reshapers 1 83 resistance 55 investigating resistance 58 resistivity 58 resistors in parallels 56 resistors in series 56 resolution 1 01 resonance 1 68, 1 88 examples o resonance 1 69 phase o orced oscillations 1 69 resonance tubes 49 rest mass 1 36 restoring orce 33 Reynolds number 1 67 root mean square (RMS) 1 1 4 rotation curves 209 mathematical models 208 rotational equilibrium 1 54 rotational motion 1 52 bicycle wheel 1 52 energy o rotational motion 1 57 problem solving and graphical work 1 58 relationship between linear and rotational quantities 1 53 summary comparison o equations o linear and rotational motion 1 58 Rydberg constant 1 23 Rydberg ormula 1 23

S s-process 206 Sankey diagram 82 scalars 7 scattering 1 28, 1 85 Schrdinger model o the atom 1 25 Schwarzchild radius 1 50 scientic notation 3 secondary cells 60 recharging secondary cells 60 sensors 57 sensor circuits 57 SI units 2, 1 44 signicant gures 3 signicant gures in uncertainties 4 simple harmonic motion [SHM] 33 acceleration, velocity and displacement during SHM 34, 95 energy changes during SHM 34, 96 equation 95 identication o SHM 95 mass between two springs 33 two examples o SHM 95 simultaneity 1 33 singularity 1 50 situation diagrams 65 smoothing circuits 1 1 6 Snells law 44 solar energy 22 solar power 87, 91 advantages and disadvantages 87 solar constant 91 Solar System 1 90 solid riction 20 solids 26 sound waves 35, 40 investigating speed o sound experimentally 38 spacetime curvature o spacetime 1 49 spacetime interval 1 36 spacetime diagrams 1 39 calculation o time dilation and length contraction 1 40 examples 1 39, 1 40 representing more than one inertial rame on same spacetime diagram 1 42 twin paradox 1 401 special relativity 1 33 length contraction and evidence to support special relativity 1 38 postulates o special relativity 1 33 simultaneity 1 33 specic heat capacity 27 spectral linewidth 1 26 spectrometers 99 standing waves 48 stars 1 901 bending o star light 1 48 binary stars 1 92 brown dwar stars 1 96 classication o stars 1 95 energy fow or stars 1 92 luminosity and apparent brightness 1 94 main sequence stars 1 96

mass-luminosity relation or main sequence stars 1 97 neutron stars 1 96, 200 red giant stars 1 96, 1 99, 200 red supergiant stars 1 96 stellar clusters 1 91 stellar evolution 200 stellar parallax 1 93 stellar types and black holes 1 96 time spent on the main sequence 205 white dwar stars 1 96, 200 static riction 20 steady fow 1 65 Stean-Boltzmann law 90, 1 95 stellar spectra 1 95 absorption lines 1 95 Stokes law 1 67 streamlines 1 65 stress analysis 42 strobe photography 1 1 strong interactions 79 strong nuclear orce 71 subtraction 5 Sun 1 91 equilibrium 1 92 supernovae 200 supernovae and the accelerating Universe 204 types o supernovae 207 superposition 40 superposition o wave pulses 40 systematic errors 4 T tangential stress 1 67 telescopes array telescopes 1 81 astronomical refecting telescopes 1 80 astronomical telescopes 1 79 cassegrain mounting 1 80 comparative perormance o Earthbound and satellite-borne telescopes 1 81 comparison o refecting and reracting telescopes 1 80 Newtonian mounting 1 80 radio intererometry telescopes 1 81 single dish radio telescopes 1 81 temperature 25, 26 cosmic scale actor and temperature 21 0 temperature dierence 27 tension 1 4, 1 6 thermal capacity 27 thermal energy 22 conduction 89 convection 89 radiation 89 thermal energy transer 89 thermal equilibrium 25, 90 thermal physics 25, 32 gas laws 2930 heat and internal energy 26 molecular model o an ideal gas 31 phases [states] o matter and latent heat 28 specic heat capacity 27 thermistors 57

thermodynamics 1 59 rst law o thermodynamics 1 61 heat 1 59 internal energy 1 59 internal energy o an ideal monatomic gas 1 59 second law o thermodynamics 1 62 surroundings 1 59 thermodynamic system 1 59 work 1 59 thin lens equation 1 74 real is positive 1 74 thin parallel lms 1 00 applications 1 00 conditions or intererence patterns 1 00 phase changes 1 00 ticker timers 1 1 time 201 time constant 1 1 8 velocity-time graphs 1 0 time dilation 1 37 calculation o time dilation and length contraction 1 40 derivation o eect rom rst principles 1 37 derivation o eect rom Lorentz transormation 1 37 time period 66 isochronous time period 33 range o times 1 tomography 1 86 torque 1 54 totally inelastic collisions 23 transormer operation 1 1 4 resistance o the windings (joule heating) 1 1 5 step-up and step-down transormers 114 turns ratio 1 1 4 transient oscillations 1 68 translational equilibrium 1 6, 1 54 translational motion 1 52 bicycle wheel 1 52 relationship between linear and rotational quantities 1 53 transverse waves 35 transverse waves on a string 49 travelling waves 35, 48 trigonometry 7 tubes o fow 1 65 tunnelling 1 27 alpha decay 1 27 scanning tunnelling microscopes 1 27 turbulent fow 1 65 Reynolds number 1 67 two-source intererence 47 double-slit intererence 98 Youngs double-slit experiment 47, 98 U ultrasound 1 87 A- and B-scans 1 87 acoustic impedance 1 87 choice o requency 1 87 comparison between ultrasound and NMR 1 88 piezoelectric crystals 1 87

relative intensity levels 1 88 uncertainties 47, 8 error bars 6 estimating the uncertainty range 4 graphical representation o uncertainty 4 Heisenberg uncertainty principle 1 26 mathematical representation o uncertainties 5 signicant gures in uncertainties 4 uncertainty in intercepts 6 uncertainty in slopes 6 uniormly accelerated motion 1 1 universal gravitation 67, 68 Universe 1 91 Big Bang 201 closed Universe 21 1 cosmic scale actor 203 expansion o the Universe 201 fat Universe 21 1 uture o the Universe 21 1 history o the Universe 21 0 history o the Universe 203 open Universe 21 1 supernovae and the accelerating Universe 204 upthrust 1 4, 1 6 V vaporization 28 vectors 7 addition/subtraction o vectors 7 components o vectors 7 orces as vectors 1 4 representing vectors 7 vector diagrams 65 velocity 9 acceleration, velocity and displacement during simple harmonic motion [SHM] 34, 95 angular velocity and time period 66 change in velocity 1 4 Galilean equation 1 35 relative velocities 9, 1 401 velocity addition 1 35 velocity gradient 1 67 velocity-time graphs 1 0 wave equations 36 vernier callipers 58 Very Long Baseline Intererometry 1 81 viscosity 1 67 non-viscosity 1 65 voltmeters 56 volts 53 W water 92 water ripples 35, 40 waveparticle duality 1 22 wave phenomena 951 03, 1 04 diraction 97 Doppler eect 1 023 multiple-slit diraction 99 resolution 1 01 simple harmonic motion [SHM] 956 two-source intererence 98 waveunction 1 25

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waveguide dispersion 1 83 wavelength 48 waves 3349, 50 boundary conditions 49 crests and troughs 35 diraction 46 electromagnetic spectrum 37, 89 graphs o simple harmonic motion 34 intensity 39 investigating speed o sound experimentally 38 nature and production o standing [stationary] waves 48 oscillations 33 polarization 41 2 refection 43 reraction 445 superposition 40 travelling waves 35 two-source intererence o waves 47 wave characteristics 36 wave energy 48 wave equations 36 wave model o image ormation 1 72 wave pulses 35 waveronts 35, 39 waves along a stretched rope 35 weak interactions 79 weak nuclear orce 71 weight 1 6, 1 9 weightlessness 1 08 Wiens law 90 WIMPs (Weakly Interacting Massive Particles) 209 wind power 88 advantages and disadvantages 88 work 21 , 1 59 concepts o energy and work 22 denition 21 examples 21 heat and work 26 pV diagrams and work done 1 60 when is work done? 21 work done by an ideal gas 1 60 work unction 1 21 X X-rays 1 85 basic X-ray display techniques 1 85 imaging techniques 1 86 intensity, quality and attenuation 1 85 Y Youngs double-slit experiment 47, 98

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OXFORD IB STUDY GUIDES

Physics

2014 edition

 o r T h e I B d I p lo m a

Author Ti Kik

Csy suting t pysics Cus Bk, tis cnsiv stuy gui efectively reinorces  t ky cncts  t tst sybus t Sl n hl (fst xin 2016) . pck wit ti assessment guidance, it suts t igst civnt in xs. Oxord IB study guides build unrivalled assessment potential. Yu cn tust t t: 

Comprehensively cv t sybus, tcing IB scifctins

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rinc all t ky tics in  cncis, us-iny t, cementing understanding

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Efectively prepare stunts  ssssnt wit visin sut n exam strategies

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mti is snt in cncis cunks, helping students ocus

Suting Cus Bk, v wit t IB 978 0 19 839213 2

digtic t bks wn cnging cncts, building understanding

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How to get in contact: web www.oxfordsecondary.co.uk/ib email [email protected] tel +44 (0)1536 452620 fax +44 (0)1865 313472