Secondary Physics. Student book. Form Four

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Physics STUDENT BOOK Biology Writers 1. Abdikarim Hassan Muhumed (ina Calicaaruf) …………….(Chairman) 2. Ahmed Hussien Ali ( Ahmed dheere) 3. Mahamoud Mahamed Ali 4. Mohamed Osman Siciid (M.Somali) Cover Designer 1. Hamud Khaireh Yusuf 2. Liban Ali H.rabi Editors 1. Abdilahi Yusuf Warsame 2. Abdikarim Hassan Muhumed (ina Calicaaruf)

Biology Textbook © Ministry of Education and Higher Studies Republic of Somaliland Curriculum Development Institute is the department of the Somaliland Ministry of Education and Higher Studies that is responsible for the curriculum development and textbook production.

Design and layout © HEMA Books 2016

First Edition 2016

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 prior permission in writing of Ministry of Education and Higher Studies of the Republic of Somaliland, or are expressly permitted by law.

ISBN:

Printed by HEMA Books Hargeisa, Somaliland

PREFACE This book is designed for form two Somaliland secondary schools and contains all of the topics they require. Each topic is supported theories, concepts, explanations, analysis and exercises and it’s based on newly revised curriculum. The objective of this book is to satisfy educational needs of the learners and to help them to develop self-reliance and complete confidence in their abilities to understand and solve daily life examples. In the context of this objective, the book contains: 

Detailed notes, solved examples, content-concept, clear mathematical formulas,



Exercises which address all topics covered in the textbook.



Chapter review questions.

This book is an excellent teaching learning tool for both teachers and students. The language is simple, to improve accessibility for all students, Care is taken to introduce and use all the special terms that students need to gain a complete understanding of the physics concepts introduced. In the text, key terms are highlighted in bold. The depth and breadth of each topic is pitched at the appropriate GCSE level students. The key objective of this book is to improve the quality of secondary Physics education; all of it has been reviewed and revised, ensuring that the new specification is fully covered. Chapter Exercise questions in each chapter provide opportunities to check understanding. They often address misunderstandings that commonly appear in examination answers, and will help students to avoid such errors. The key objective of Modern Education is to give learners the skills, knowledge and attitudes they will need to succeed in a rapidly evolving world. In most developing countries learning resources are scarce. It is therefore necessary that the teacher uses alternative methods such as: Collection from the environment and Improvisation. The teacher should have the capacity and attitude to improvise resources from locally available materials which are often considered waste or valueless. Improvisation helps reduce the cost of teaching and learning since improvised resources cost very little or have no cost at all. In addition improvisation helps demystify science and bring it home to the learner as part and parcel of everyday life.

Contents Chapter One: Forces and motion ........................................................................................................ 4 1.0-Introduction .................................................................................................................................. 5 1.1-Kinematics .................................................................................................................................... 5 Distance and displacement .................................................................................................................. 5 Speed and velocity .............................................................................................................................. 6 Acceleration ...................................................................................................................................... 11 Motion Graphs .................................................................................................................................. 14 Equations for uniform acceleration ................................................................................................... 25 1.2-Dynamics .................................................................................................................................... 28 Balanced and unbalanced forces ....................................................................................................... 28 Newton’s law of motion.................................................................................................................... 29 Newton’s First law of motion ........................................................................................................... 29 Inertia and Mass ................................................................................................................................ 30 Newton’s Second law of motion ....................................................................................................... 30 Acceleration caused by gravity ......................................................................................................... 33 Air resistance and Terminal velocity ................................................................................................ 34 Free falling motion ............................................................................................................................ 36 Momentum ........................................................................................................................................ 37 Newton’s third law............................................................................................................................ 37 Impulse: ............................................................................................................................................ 46 1.3 Using vectors and projectile motion ........................................................................................... 50 Using vectors .................................................................................................................................... 50 Combining vectors ........................................................................................................................ 50 Resolution of vectors ........................................................................................................................ 61 Projectiles.......................................................................................................................................... 63 1.4 Circular motion ........................................................................................................................... 70 Introduction ....................................................................................................................................... 70 Centripetal acceleration .................................................................................................................... 70 Centripetal force................................................................................................................................ 71 Angular velocity................................................................................................................................ 73 Application of uniform circular motion ............................................................................................ 76 Newton’s Law of universal gravitation ............................................................................................. 77

1.5 Work, Energy and Power ............................................................................................................ 79 Doing work ....................................................................................................................................... 79 Kinetic energy: .............................................................................................................................. 84 Gravitational potential energy and kinetic energy Transformation .............................................. 87 1.6 Deforming solids ......................................................................................................................... 94 Introduction ....................................................................................................................................... 94 Springs in series and springs in parallel ............................................................................................ 97 Measuring young’s modulus ....................................................................................................... 105 Chapter two: Waves ......................................................................................................................... 109 2.1 Behaviour of waves................................................................................................................... 110 Introduction ..................................................................................................................................... 110 Wave motion ................................................................................................................................... 110 Characteristics of waves motion ................................................................................................. 111 Type of Wave Motion ..................................................................................................................... 111 Properties of longitudinal wave ...................................................................................................... 112 Properties of transverse wave ......................................................................................................... 114 Characteristics of waves ................................................................................................................. 115 Properties of waves ......................................................................................................................... 121 2.2 Sound ........................................................................................................................................ 129 2.21-Properties of sound ................................................................................................................. 134 Doppler Effect................................................................................................................................. 142 Musical note .................................................................................................................................... 145 Chapter three: Telecommunication ................................................................................................ 150 3.0 Introduction ............................................................................................................................... 151 3.1 Early history .............................................................................................................................. 151 Representing Information ............................................................................................................... 152 Basic communication System ......................................................................................................... 153 3.2-Radio waves .................................................................................................................................. 154 Radio system ................................................................................................................................... 155 Electrical oscillations ...................................................................................................................... 157 Simple Radio Receiver................................................................................................................ 159 3.3 Television.................................................................................................................................. 160 Chapter four: Radioactivity ............................................................................................................. 165 4.0-Introduction .............................................................................................................................. 166

4.1 Characteristics of radioactive materials .................................................................................... 166 4.2 Types of radiations .................................................................................................................... 166 Alpha, beta and gamma radiation ................................................................................................... 167 Alpha radiation (α) .......................................................................................................................... 168 Beta radiation (β) ............................................................................................................................ 168 Gamma radiation (γ) ....................................................................................................................... 168 4.3-Nuclear Equations..................................................................................................................... 169 4.4 Effects on the Nucleus .............................................................................................................. 171 4.5 Isotopes, Isobars and Nuclides .................................................................................................. 173 4.6 Rate of decay and half – life ..................................................................................................... 174 Nuclear energy ................................................................................................................................ 180 4.7 Applications and Hazards of Radioactivity .............................................................................. 181 Applications of radioactivity ........................................................................................................... 181 Hazards of radioactivity .................................................................................................................. 184 Chapter five: Electromagnetic induction ........................................................................................ 189 5.0-Introduction .............................................................................................................................. 190 5.1 Definition of electromagnetic induction ................................................................................... 190 5.3-Faraday’s law............................................................................................................................ 191 An induced Emf in a coil (Lenz’s law) ........................................................................................... 195 Lenz’s law ....................................................................................................................................... 195 Applications of electromagnetic induction (generator effect) ........................................................ 200 5.4 Mutual induction (Transformer effect....................................................................................... 205 Applications of mutual induction (transformer effect) ................................................................... 208 Types of Current in a Transformer ................................................................................................. 209 Types of Transformer ..................................................................................................................... 210 Glossary ............................................................................................................................................... 221 Index.................................................................................................................................................... 228 References .......................................................................................................................................... 230

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Chapter One 1.0-Introduction Motion is the most fundamental and common physical phenomenon. It consists in the change in position of bodies or their pails relative to one another. The motion of a body is influenced by the bodies surrounding it, i.e., by its interactions with them. Mechanics is a fundamental area of physics, which studies the universal principles of motion. The aim of mechanics is to analyse and predict the motion of bodies resulting from the different interactions. The nature of the particular interactions is considered by other branches of physics. Mechanics consists of two branches, kinematics and dynamics.

1.1-Kinematics

Kinematics describes the motion without considering the causes of motion, the word “Kinematic” comes from the Greek word Kinema, meaning motion. Kinematics is also defined as the study of motion in terms of distance, displacement, speed, velocity and acceleration. Dynamics studies the relation between the motion and the causes of the motion. Sometimes a study of the equilibrium of bodies is considered to be a separate branch of mechanics, called statics.

Distance and displacement Distance When an object is in motion its distance increases, the total length of the path traveled by the object from one location to another is called distance. As distance does not take into account the direction traveled by the object, distance is therefore scalar quantity (this means it has only magnitude but no direction). The SI unit of distance is meter (m). For example, the little girl shown on the figure 1.1 starts her motion from the red spot, she moved a total distance of 12m (1m+2m+1m+2m+3m+3m), and so distance travelled is 12m. Long distances can be measured in Km or mile, while short distances can be measured in meters, Centimeters and even millimeters, Figure 1.2 For example distance between Hargeisa and Borama is 118 kms or 73.3 miles , Shown on figure1.2.

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5

Displacement Displacement is the distance moved in specific direction. In another words, it is the shortest distance of the moving body from its starting position. Displacement describes an object's change in position. It shows the object's final position with respect to its initial position and it’s a vector quantity, because it has both magnitude and direction. The SI unit of displacement is meter (m). as shows figure 1.1 above, the total displacement of the little girls from its starting point is 3m.

Speed and velocity Speed One of the most obvious aspects of an object in motion is how fast it is moving. Speed (v) is defined as the distance travelled by an object per unit of time. Speed is a scalar quantity. The SI unit for speed is meters per second or m/s. other units are km/h or mile/hour

Speed =

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡𝑡𝑡𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡𝑡𝑡𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

or, In symbols, 𝑣𝑣𝑣𝑣

=

𝑆𝑆𝑆𝑆

𝑇𝑇𝑇𝑇

 Speed is a scalar quantity, because it has only magnitude (or size)  Speed of a car is measured by an instrument called speedometer, as we learned in form one

There four classifications of speed exist: constant, non-uniform, average, and instantaneous speed. It is important to understand each and how all four speeds are related. a) Constant speed Constant speed occurs when the object travels the same distance in equal periods of time. It is an example of uniform motion. b) Non-uniform Speed If an object changes its speed while moving, the object is said to be in non uniform, speed. In the above figure, speed of the car varies at different interval of time. Therefore, speed of the car is non-uniform. c) Average speed Average speed is defined as the total distance traveled by an object per unit time. Since speed is derived from two scalar quantities, distance and time, it is also a scalar quantity. The SI unit for speed is meters per second or m/s. d) Instantaneous speed Instantaneous speed is the speed at a specific instant in time. It can be thought of as the speed that the speedometer reads at any given moment.

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Chapter One Example 1 Yahye went to visit his parents for the long weekend. He traveled a total distance of 346 km. His total trip took him 4.5 hours. What was the average speed for his drive home?

Yahye’s average speed throughout the trip was 77 km/hr. he probably traveled faster than this for part of his trip and at some point he probably traveled slower. He may even have stopped the car for a junk food break Velocity Velocity ( ) is an objects' displacement (change in position) per unit time. Since velocity is derived from a vector quantity, displacement, it is also a vector quantity and therefore requires a direction as well as a magnitude. The SI unit for velocity is m/s. Average velocity= Example: 2

Displacement Time taken

or, in symbols

A man and a woman are walking with a speed of 5km/h in opposite directions, as show on the figure. What is the difference between their speed and velocity? They have same speed of 5km/hr, but different velocities because they have to travel opposite direction. Example .3 A bus traveled covers a distance of 160km from Hargiesa to Berbera towards east in 3 hours in the morning and returns to Hargiesa in the evening covering the same distance of 160km in the same time of 3 hours. a) Calculate the average speed of the car? b) Calculate the average velocity of the car?

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Solution a) Average speed = km/h

Total distance travelled

=

160km+160km

=

Total displscement

=

160km−160km

=

Totoal time taken

b) Average velocity =

Totoal time taken

3h+3h

3h+3h

320km 6h

0km 6h

= 53.3

= 0 km/h

You might say your car is going 60 mph, but the velocity would also need some type of directional description, such as north, south or whatever direction you are heading. The velocity of a body can be changed in two ways: i) By changing the magnitude (speed of the body), and ii) By changing the direction of the body. On the opposite figure (a), the velocity of the car is changing with direction, but the magnitude (speed) is the same at all points (A, B, C, D, E and F). In figure (b), the car is moving the same direction but the velocity is changing as magnitude no as direction.

8

Figure (a)

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Chapter One Determining speed In laboratory, speed of an object can be determined by different instruments such as; 1. Tickertape timer 2. Two light gates( photo-gate timer) 3. Motion sensor Ticker-timer experiments The ticker timer makes dots on a paper tape. So if a piece of tape is pulled through the timer for a second there will mark 50 dots on it. Since the spaces between the dots take such a short time (1/50 s) the ticker timer is a very useful instrument for measuring short intervals of time. Ticker timer marks dots on the tape at 1

regular intervals, usually 50 s (i.e. 0.02s) it is called ‘tick’.

Figure (b)

The distance between successive dots equal the average speed of whatever is pulling the tape in, Because there are 50 dots in each second, each 5-dot length is produced in 1/10th of a second (0.1s). You can calculate the speed of the moving object by using this expression; Length of Strip Velocity = 0.1s

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If the 5-dot lengths are stuck side by side in the right order, you can create a velocity- time graph. For a constant speed motion, the strips all have the same length and so the graph should look like as figure (a). For an accelerating object, each progressive strip gets longer and longer and the graph looks like as shown figure (b)

Tape charts are made by sticking successive strips of tape, usually ten tick lengths, side by side. That in Figure (a) represents a body moving with uniform speed since equal distances have been moved in each ten ticks interval. The chart in Figure 2.3b is for uniform acceleration: the ‘steps’ are of equal size showing that the speed increased by the same amount in every ten tick (1/5 s). The acceleration (average) can be found from the chart as follows. The speed during the first ten ticks is 2 cm for every 1/5s, or 10 cm/s. During the sixth ten ticks it is 12 cm per 1/5 s or 60 cm/s. And so during this interval of 5 ten ticks, i.e. 1 second, the change of speed is (60 − 10) cm/s = 50 cm/s. 𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜 𝐨𝐨𝐨𝐨𝐨𝐨𝐨𝐨 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐬𝐬𝐬𝐬 50𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐/𝑠𝑠𝑠𝑠 = =50cm/s2 acceleration 𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐜𝐜𝐜𝐜 𝐭𝐭𝐭𝐭𝐜𝐜𝐜𝐜𝐭𝐭𝐭𝐭𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜 1𝑠𝑠𝑠𝑠

The dots on the tape describes the motion of the object

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Chapter One Acceleration

Acceleration of a body is defined as the rate of change of its velocity with the

time.( or Acceleration is how quickly an object is moving). Its unit of acceleration is m/s2 When the velocity of the body is changing either as direction or magnitude, that body is said to be accelerating and if the velocity of the body is increasing, that body is accelerating while when the velocity of the body is decreasing that body is said to be decelerating. For  The acceleration of a body at any instant is called example, if the acceleration of a body is 3m/s2, it means instantaneous acceleration. that its velocity is increasing by 3m/s every second and if the deceleration of the body is -2m/s2, it means that the velocity of the body decreases 2m/s every second. From the definition of acceleration, if the velocity does not change with time, then the acceleration of the body is zero. Thus the acceleration of a body moving with uniform velocity is zero. Acceleration is a vector quantity, because it has both magnitude and direction. it changes with the change in the magnitude of velocity, change in 𝒗𝒗𝒗𝒗 − 𝒖𝒖𝒖𝒖 direction of the velocity, or change in both magnitude and direction of 𝒂𝒂𝒂𝒂 = 𝒕𝒕𝒕𝒕 velocity. Acceleration =

𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂 𝐓𝐓𝐓𝐓𝐂𝐂𝐂𝐂𝐓𝐓𝐓𝐓𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐓𝐓𝐓𝐓𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂

u= 0, if an object moves from rest. v= 0, if an object came to rest or stops

 Negative acceleration is called deceleration or retardation, so velocity decreases.

 If a velocity is a constant (in both magnitude and direction) acceleration of the body is said to be zero.

Acceleration

Uniform acceleration An object has uniform acceleration when it is traveling straight line and the velocity of the object is increasing at the same rate.

Non-uniform acceleration

An object has a non-uniform acceleration when either the velocity of the body does not increase at the same rate or the direction of the object changes.

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Example (1) A car is moving a long straight road increase its velocity uniformly from 30m/s to 70m/s in 5s, what its acceleration? Solution Given

𝑣𝑣𝑣𝑣 − 𝑢𝑢𝑢𝑢 70 − 30 40𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠 u= 30m/s 𝑎𝑎𝑎𝑎 = = = 𝑡𝑡𝑡𝑡 5 5𝑠𝑠𝑠𝑠 v=70m/s Example: 2 t =5 a = 8m/s2 A sprinter accelerates from rest with a uniform acceleration of 4m/s2for 15s, calculate second her final velocity Solution a=? v = u + at Given: v = 0 + (4m/s2) (15s) u= 0 v = 60m/s a= 4m/s2 t = 15s v=? Exercise 1. What is the average speed of a) a car that travels 400 m in 20 s, b) an athlete who runs 1500 m in 4 minutes? 2. A train increases its speed steadily from 10 m/s to 20 m/s in1 minute. a) What is its average speed during this time, in m/s? b) How far does it travel while increasing its speed? 3. A motorcyclist starts from rest and reaches a speed of 6 m/s after travelling with uniform acceleration for 3 s. What is his acceleration? 4. An aircraft travelling at 600 km/h accelerates steadily at 10 km/h per second. Taking the speed of sound as 1100 km/h at the aircraft’s altitude, how long will it take to reach the ‘sound barrier’? 5. A vehicle moving with a uniform acceleration of 2 m/s2 has a velocity of 4 m/s at a certain time. What will its velocity be a) 1 s later, b)5 s later? 6. If a bus travelling at 20 m/s is subject to a steady deceleration of 5 m/s2, how long will it take to come to rest? 7. The tape in Figure below was pulled through a timer by a trolley travelling down a runway. It was marked off in ten ticks lengths. a) What can you say about the trolley’s motion? b)Find its acceleration in cm/s2

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Chapter One

8. A car has an instantaneous speed of 4.5m/s. if he keeps this speed up for 15s. How far will he travel? 9. A man fires a bullet from a gun. The bullet flies with a velocity of 200m/s. how long will the bullet to hit with a target of 800m away? 10. An automobile travelling at 15m/s along a straight road accelerates to 80m/s in 5 second. What is the acceleration of the automobile? 11. A car driver brakes gently. Her car slows down from 23m/s to 11m/s in 20s. What is it deceleration? 12. A train slows down from 60m/s to 20m/s in 50s. a) What is its acceleration? b) Write your answer in words? 13. The acceleration of car is 10m/s2. If the car starts from rest, a) What will be its speed after 10s? b) How long it takes to reach a speed 1000m/s?

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Motion Graphs The motion graphs which are used to represent the motion of object, these graphs change when the motion of the object changes. In case of linear motion graphs, linear graphs are made between the varying quantities with respect to the time. Varying quantity whose graphs are made with respect to time could be distance, velocity, acceleration, so It can be distancetime graph ,this graph between distance and time which shows the variation in distance with time, speed - time graph, 1. Distance-time graph and displacement time graph The graph of distance/ displacement is plotted with respect to the time. Generally, time is plotted on the x -axis and distance/ displacement is taken on the y-axis of the coordinate system.

A graph of distance travelled against time taken is called distance-time graph. And the graph of displacement covered against time taken is called displacement-time graph. The slope of the distance-time graph tells us about the speed of the object while the slope of the displacement-time graph tells us about the velocity of the object. The area under the distance-time graph and the area under the displacement-time graph have no meaning.

The gradient of the distance-time graph represent the speed (or velocity) of the motion of the object

Gradient =

Vertical change

=

S2 −S1

Horizental change t2 −t1

𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺

The value of the speed is given by the gradient of the line

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Chapter One Types of Distance-Time graph Shape of the graph

Name of the graph 1. Constant speed or

Description  Every point (x, y and z) of the graph, speed is the same.

Uniform speed

 The body is traveling equal distance with equal time interval

2. Stationary object.

 Gradient of the graph is zero,so v =0 and hence object is stationary.



From the shape of the graph steeper the gradient, higher the speed.

4. Non-uniform



Gradient increases, so speed increases and it is accelerating.

5. Moving opposite



The gradient is negative so negative velocity means moving opposite direction.

3. The steeper, the gradient, the greater the speed.

speed

direction

Example: 1

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The distance-time graph below represents the journey of a cyclist from small village to the New-York city.

From the graph; a) Determine the distance between the village and New-York city? b) How long did he take to complete his journey? c) What was his average speed? d) How many stops did he make? e) How long did he stops for altogether? f) What was his average speed excluding stops? g) From the shape of the graph, which stage did he travel faster? Solution a) It is 60km b) He took 10 hours c) Given: 𝑆𝑆𝑆𝑆 60𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 V=? V= = 𝑇𝑇𝑇𝑇 10ℎ S = 60km V= 6km/h t= 10 h d) He made two stops.

e) At stage C, he was rest for 1 hour and stage F, he was rest for 2 hours. For altogether he was rest for 3 hours 𝑆𝑆𝑆𝑆 60𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 V= = f)Given: 𝑇𝑇𝑇𝑇 7ℎ v =? V=8.57km/h s = 60km t=7h g) It is stage E. because of the steeper the graph, the greater the speed.

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Chapter One Example: 2 The distance –time graph below represent the journey of a train from London to Paris and then return London.

From the graph: a) Determine the total distance Solution a) Total distance traveled by the train = 300m + 300m = 600m b) Train takes 30second to travel from London to Paris. c) Given. 600m s= 600m 𝑉𝑉𝑉𝑉 = 𝑆𝑆𝑆𝑆⁄𝑡𝑡𝑡𝑡= =10.9m/s 55s t=55s v=? Displacement 0m = = 0m/s d) Average velocity = Time taken 55s e) It was at stage C. f) The train was stationary for 7.5 seconds g) At stage B, the train moves faster.

traveled by the train? b) How long did the train take to travel from London to Paris? c) What was its average speed? d) Calculate average velocity of the train? e) Which stage did the train was stationary? f) How long the train was stationary?

g) From the shape of the graph, which stage did the train move faster?

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Example: 3 The displacement-time graph below shows the motion of a car.

a) What is the final displacement of the cyclist? b) What is the total distance of the cyclist? c) Convert the graph into distance-time graph? d) What is the cyclist’s speed at i. t = 4s ii. t = 8s iii. t = 12s

Solution a) The total displacement is -5m. The minus sign indicates that the cyclist is moving 5m back of its initial position. b) The total distance travelled is 5m forward and 10m backward. Therefore it is 15m c)

d) i) Speed = ∆S/∆t = 5m/5s = 1m/s. ii) Speed =∆S/∆t = 0/5s = 0m/s. therefore the car is stationary. iii) Speed = ∆S/∆t = 10m/10s = 1m/s. Velocity -Time Graph  The gradient of velocity - Time graph represent acceleration of the motion of the object.

Gradient =

changeofvelocity ∆𝑉𝑉𝑉𝑉 changeoftime

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 =

=

𝑐𝑐𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑣𝑣𝑣𝑣𝑎𝑎𝑎𝑎𝑣𝑣𝑣𝑣𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑣𝑣𝑣𝑣𝑡𝑡𝑡𝑡𝑣𝑣𝑣𝑣 ∆𝑉𝑉𝑉𝑉 𝑡𝑡𝑡𝑡𝑣𝑣𝑣𝑣𝑡𝑡𝑡𝑡𝑎𝑎𝑎𝑎 𝑡𝑡𝑡𝑡𝑎𝑎𝑎𝑎𝑡𝑡𝑡𝑡𝑎𝑎𝑎𝑎

𝐺𝐺𝐺𝐺𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐺𝐺𝐺𝐺𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴

18

∆𝑡𝑡𝑡𝑡

=

∆𝑡𝑡𝑡𝑡

.................

Chapter One Area under velocity-time graph and speed-time graph Velocity is a speed in a specific direction. Where there is no change in the direction of motion, a velocity-time graph looks the same as speed-time graph. It is easy to see why this is the case for an object moving at constant velocity. The displacement is simply velocity x time, which is the area of the shaded rectangle. Area under speed-Time graph equals

distance moves or covered.

See figure a. For changing velocity, again the area under the graph gives displacement (figure b).

Area=displacement/distance=length x Width

Area = distance traveled = l x w

1

Area= distance traveled = 𝑏𝑏𝑏𝑏ℎ 2

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Types of Velocity-Time graph

Shape of the graph

Name of the graph 1. Uniform

Acceleration

Description.  Every point (x, y and z) of the graph, accelerationis the same  When a body moves with uniform acceleration, its speed changes by equal amounts in equal intervals of time.

2. Constant

 Velocity does not change, so acceleration is zero.

3. The steeper, the



From the shape of the graph steeper the gradient, higher the acceleration.



Gradient increases, so velocity is changing in a different rate and it is accelerating.

velocity

gradient, the greater the acceleration.

4. Non-uniform acceleration

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Chapter One 5. Deceleration (or retardation)



The gradient is negative, so negative acceleration means deceleration.

Example: 1 The velocity-time graph below shows the motion of a car traveled from town A to town B From the graph: a) Name parts of the graph labeled A, B and C? b) Determine the maximum speed attained by the car? c) Calculate acceleration of the car for the first 4second? d) Calculate the total distance traveled by the car? e) Calculate the average velocity the car?

Solution

a) Line A = uniform acceleration d) Total distance traveled = total area Line B = constant velocity under the graph. Line C = deceleration ST = A1 + A2 +A3 b) Maximum speed attained by the car is 35m/s. A1 = 70m c) Given:𝑣𝑣𝑣𝑣−𝑢𝑢𝑢𝑢 35−0 35 𝑢𝑢𝑢𝑢 = 0 a= = = = 𝑡𝑡𝑡𝑡 4 4 A2 = l x w = 35 x 6= 210 m v= 35m/s 1 1 t= 4s a =8.8m/s2 A3= 2 𝑏𝑏𝑏𝑏ℎ= 2(4) (35) =70m 𝑎𝑎𝑎𝑎 =? e)

Given: V=? S= 350m T = 14s

Therefore, total distance traveled by the car is

V =S⁄T =

350m 14s

= 25m/s

ST = A1 + A2 +A3 ST = 70 m + 210m +70m = 350m.

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21

Example: 2 The velocity-time graph below shows the part of journey of a train

From the graph. a) Describe the motion of the train at stage A and stage C? b) Calculate the deceleration of the train at stage C? c) Calculate the distance traveled during deceleration? Solution

c) Given:

a) At stage A, the train moves at constant velocity. S =? At stage C, the train was decelerating. b = 4s b) Given:h =35m/s

u= 35 m/s v= 0 t= 4s a=?

22

𝑎𝑎𝑎𝑎 =

𝑣𝑣𝑣𝑣−𝑢𝑢𝑢𝑢 𝑡𝑡𝑡𝑡

=

0−35

a = 8.75 m/s2

4

=

−35 4

1

Distance =Area =2 𝑏𝑏𝑏𝑏ℎ 1

= 2(4s) (35m/s) Distance =70m

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Chapter One Exercise Question one The graph shows how the velocity of a motorbike changes when it is travelling along a straight road.

(i) What was the change in velocity of the motorbike in the first 5 seconds? ............................................................................................................................................. (ii) Write down the equation which links acceleration, change in velocity and time taken. ............................................................................................................................................. (iii) Calculate the acceleration of the motorbike during the first 5 seconds. Show clearly how you work out your answer and give the unit. ............................................................................................................................................. ............................................................................................................................................. (c) A car is travelling on an icy road. Describe and explain what might happen to the car when the brakes are applied. ............................................................................................................................................. ............................................................................................................................................. (d) Name three factors, other than weather conditions, which would increase the overall Stopping distance of a vehicle. ............................................................................................................................................. ............................................................................................................................................. .............................................................................................................................................

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23

Question two The velocity-time graph shows part of the journey of a train.

(a) (i) calculate the initial acceleration of the train. …………………………………………………………………………………………………… …………………………………………………………………………………………………… (ii) Calculate the distance travelled in the first 210 s. ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… (b) Between 250 s and 400 s the train travelled a distance of 3500 m. calculate the mean speed between 0 and 400 s. ……………………………………………………………………………………………………… …………………………………………………………………………………………………… Question three A car travelling with a constant speed of 20m/s for 10s, accelerated uniformly to a speed of 60m/s for 5s.the car maintained that speed for 10s, and then decelerated uniformly until it brought to a rest for 5s.

a) Plot a velocity time graph for the motion of the car b) From the graph find out i) What is the acceleration of the car while its speed was increasing ii) What is the total displacement of the car iii) What is the average velocity of the car Question four

A car is travelling at a constant speed of 72 km/h passes a stationary police car. The police car immediately gives chase, accelerating uniformly to reach a speed of 90 km/h in10h and continues at this speed until he overtakes the other car. Find: a) The time taken by the police car to catch up the car b) The distance travelled by the police car when this happens.

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Chapter One Equations for uniform acceleration

Problems involving bodies moving with uniform acceleration can often be solved quickly using the equations of motion. There are four equations of motion. Take care when you use them; they only apply (i) To motion in a straight line (ii) To an object moving with a constant acceleration

When you use these equations take the following steps: 1. Step one: write down the quantities which you know and the quantity which you want to know. 2. Choose the equation which links these quantities and substitute in the values. 3. Calculate the unknown quantity.

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25

Example: 1 A rocket lifts off from a rest with an acceleration of 20ms-2. How fast will it be travelling after 50sec? Solution u=O t = 50sec a = 20m/s v =? v = u + at = O + 20 x 50= 1000m/s. Example: 2 The car shown below is travelling along the road at 8m/s. It accelerates at 1m/s-2. For a distance of 18m how fast is it then travelled? Solution

Example: 3 A train travelling at 20ms-1 accelerates at 0.5ms-2for 30sec how far will it travel in this time. Solution

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Chapter One Example: 4

The cyclist shown figure right is traveling at 15m/s. She breaks, so that she does not collide with the wall, what deceleration must she have?

Solution

Example: 5 A sprint cyclist starts from rest and accelerates at 1 m/s2 for 20 seconds. He then travels at a constant speed for 1 minute and finally decelerates at 2 m/s2 until he stops. Find his maximum speed in km/h and the total distance covered in metres. Solution First stage u = 0,a = 1 m/s,2t = 20 s We have v = u + at = 0 + 1 m/s2 × 20 s = 20 m/s 20 𝑥𝑥𝑥𝑥 60 x60=72 km/h v = 1000 The distance s moved in the first stage is given by 1

s = ut + 2at2 = 0 × 20 s + 1/ 2 × 1 m/s2 × (20s) 2 s=1/2× 1 m/s2 × 400 s2 = 200 m

Exercise

A train starts from rest, and accelerates at 1 m·s−2 for 10 s. How far does it move? A bus is going 30 m·s−1 and stops in 5 s. What is its stopping distance for this speed? A racing car going at 20 m·s−1 stops in a distance of 20 m. What is its acceleration? A ball has a uniform acceleration of 4 m·s−1. Assume the ball starts from rest. Determine the velocity and displacement at the end of 10 s. 5. A motorcycle has a uniform acceleration of 4 m·s−1. Assume the motorcycle has an initial velocity of 20 m·s−1. Determine the velocity and displacement at the end of 12s. 6. An aeroplane accelerates uniformly such that it goes from rest to 144km·hr−1in 8 s. Calculate the acceleration required and the total distance that it has travelled in this time 7. A truck is travelling at a constant velocity of 10 m · s−1when the driver sees a child 50 m in front of him in the road. He hits the brakes to stop the truck. The truck accelerates at a rate of -1.25 m·s−2. His reaction time to hit the brakes is 0.5seconds. Will the truck hit the child? 1. 2. 3. 4.

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1.2-Dynamics Dynamics is a branch of mechanics concerned with the motion of bodies under the action of forces. The study of dynamics deals with such concepts as why an object moves, and what makes it start or stop moving. The main concept we are concerned with in the study of dynamics is force. A force is any kind of push or pull on an object. Thus, a force is required to push a crate or to throw a ball. Therefore describing motion in terms of force is called dynamics. Why does the motion of an object change? What might cause one object to remain at rest and another object to accelerate? Why is it generally easier to move a small object than a large object? The two main factors we need to consider are the forces acting on an object and the mass of the object. In this chapter, we begin our study of dynamics by discussing the three basic laws of motion, which deal with forces and masses and were formulated more than three centuries ago by Isaac Newton. Balanced and unbalanced forces If an object has two or more forces acting on it, we have to consider whether or not they are balanced. For example, when the two teams pulled with equal force in opposite direction, their forces were balanced.

Forces that are equal in size but opposite in direction are called balanced forces. They don’t cause a change in motion or the state of motion Unbalanced forces When two forces acting on an object are not equal in size, we say that they are unbalanced forces. If the forces on an object are unbalanced this is what happens:  an object that is not moving starts to move  an object that is moving changes speed or direction

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Chapter One Resultant forces The size of the overall force acting on an object is called the resultant force. If the forces are balanced, this is zero. The net force on the rope in the top of Figure is 0. When the net force on an object is 0 N, the forces acting on it are balanced forces. If the forces acting on an object are balanced, the object’s motion does not change. When the net force acting on an object is not 0, the forces acting on the object are unbalanced forces. In the example above, the resultant force is the difference between the two forces, which is 100 - 60 = 40 N. Newton’s law of motion Newton further studied Galileo’s ideas on force and motion and presented three fundamental laws that govern the motion of objects. These three laws are known as Newton’s laws of motion. The first law of motion is stated as: Newton’s First law of motion “A body will continue in its state of rest or uniform motion in straight line unless acted on by an external unbalanced force.” Example: The book on table will remain on table unless some force is applied on it. The ball moving on ground stops by itself because of friction (external force). If there were no frictional forces, the moving ball will continue to move unless we stop it. In other words, all objects resist a change in their state of motion. In a qualitative way, the tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. This is why, the first law of motion is also known as the law of inertia. The word ‘inertia’ is derived from the Latin word ‘iners’ meaning a state of idleness.

First law of motion is related to term “Inertia”. It’s the property of body by the virtue of which the body resists the external force. Inertia is the ability of an object to preserve its state of motion. Inertia depends on the mass and the speed of an object. The greater a mass and velocity of an object, the greater the inertia of the object, Also, the greater the inertia of an object, the greater the force needed to overcome Common examples of inertia in our day to day life: 1. The passengers fall forward when the bus suddenly stops. This is due to inertia of motion, the lower portion of body comes to rest but the upper portion of body continues to be in motion. 2. When we shake the branches, the fruits and leaves fall. The branches are in motion while the fruits and leaves are in rest so, they fall. 3. The dust particles get removed when we shake the carpet. This is, because the particles are at rest while the carpet is moving, so, the particles are removed. 4. When the person jumps from the moving bus, he runs through some distance due to inertia of motion. 5. Any moving body has momentum. Mathematically, the momentum is denoted by P. It’s the product of mass and its velocity.

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29

P = mass x velocity P=mxv Experiment about inertia  Set a coin on a stiff playing card covering an empty glass tumbler standing on a table as shown in Figure below.  Give the card a sharp horizontal flick with a finger. If we do it fast then the card shoots away, allowing the coin to fall vertically into the glass tumbler due to its inertia.  The inertia of the coin tries to maintain its state of rest even when the card flows off.

Figure-7 When the card is flicked with the finger the coin placed over it falls in the tumbler. Activity 2  Place a water-filled tumbler on a tray.  Hold the tray and turn around as fast as you can.  We observe that the water spills. Why? Observe that a groove is provided in a saucer for placing the tea cup. It prevents the cup from toppling over in case of sudden jerks.

Inertia and Mass All the examples and activities given so far illustrate that there is a resistance offered by an object to change its state of motion. If it is at rest it tends to remain at rest; if it is moving it tends to keep moving. This property of an object is called its inertia. Do all bodies have the same inertia? We know that it is easier to push an empty box than a box full of books. Similarly, if we kick a football it flies away. But if we kick a stone of the same size with equal force, it hardly moves. We may, in fact, get an injury in our foot while doing so! Similarly, in experiment, instead of a coin if we use a small plastic, we find that a lesser force is required to perform the activity. A force that is just enough to cause a small cart to pick up a large velocity will produce a negligible change in the motion of a train. This is because; in comparison to the cart the train has a much lesser tendency to change its state of motion. Accordingly, we say that the train has more inertia than the cart. Clearly, heavier or more massive objects offer larger inertia. Quantitatively, the inertia of an object is measured by its mass. We may thus relate inertia and mass as follows: Inertia is the natural tendency of an object to resist a change in its state of motion or of rest. The mass of an object is a measure of its inertia.

Newton’s Second law of motion

The first law of motion indicates that when an unbalanced external force acts on an object, its velocity changes, that is, the object gets acceleration. We would now like to study how the

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Chapter One acceleration of an object depends on the force applied to it and how we measure a force. The same is true for situations of uniform motion in straight line. If the forces are unbalanced the net force is not zero, then we have a new situation calling for the Newton’s second law, Newton’s Second Law investigate the relationship between Force, mass and acceleration And how the acceleration of an object is effected by the resultant force applied on the object and the mass of the object. Force, mass and acceleration If a force “F” causes an object of mass “m” to accelerate, the magnitude of the acceleration can be found by using Newton’s second law of motion. This law states that “the acceleration of an object is inversely proportional to the mass of the object and is directly proportional to the force that causes this acceleration.”

𝒇𝒇𝒇𝒇

Combining the results into one equation, we get a α 𝒎𝒎𝒎𝒎 or f α m a

Therefore F = kma Where K is the constant of proportionality and k =1 F = ma Where F = resultant (or unbalanced) force causing the acceleration a. So if m =1kg and a = 1ms-2 then F = 1N. Substituting in F = kma, we get k = 1 and so we can write F = 1kg x 1m/s2 = 1kg m/s2 The SI unit of the force is Newton “N”. One Newton can be defined as “a force that will give a mass of 1kg an acceleration of 1m/s2 in the direction of the force” 1N = 1kg x 1m/s2 or simply N = kg m/s2. Resultant force cause objects to accelerate, since it changes the velocity of the object. The amount by which an object accelerates depends on. a) The size and direction of the resultant force F b) And the mass of the object m

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31

Examples 1. A cyclist of mass 60 kg rides a bicycle of mass 20kg. When starting off the cyclist provides a force of 200N. Calculate the initial acceleration of the cyclist? Total mass = 60kg + 20kg = 80kg and Force = 200N. a = F/m = 200N/80kg = 2.5m/s2. 2. A car of mass 500kg is travelling at 20m/s. the driver sees red traffic light ahead and slows to halt in 10s. Calculate the breaking force provided by the car? a = (V – u) /t = (0 – 20m/s)/10s = -2m/s2. F = ma = (500kg) (-2m/s2) = -1000N (The minus sign indicates that the force is a braking force and acts opposite to the direction of motion). 3. A block of mass 2kg is pushed along constant velocity by a force of 5N, when the push is increased to 9N, what is a) The resultant force? b) The acceleration of the object? a) When the object is moving with a constant velocity, the resultant force is zero (this means the forward force and the backward force are balanced) so the force against motion is 5N. when the force is increased to 9N, the resultant force is 9N – 5N = 4N b) a = F/m = 4N/2kg = 2m/s2. Exercise 1. Calculate the force needed to give a car of mass 800kg an acceleration of 2m/s2? 2. A rocket has a mass of 5000kg. At a particular instant the net force acting on it is 200,000N. Calculate its acceleration? 3. A motorcyclist of mass 40kg rides a bike of mass 60kg. As she sets off from the light, the forward force on the bike is 200N. Assuming the net force on the bike remains constant, what is the cyclist’s velocity after 5s? 4. The figure below shows a model of car of mass 40kg.

a) What is the resultant force? b) What is the car’s acceleration? c) If the frictional force against motion is increased to 15N. What happen to the car’s motion? s 5. A wooden block of mass 5kg is pushed along a table with a constant velocity by a force of 10N. The force is then increased to 14N. calculate a) The resultant force? b) The acceleration of the block? 6. A car of mass 100kg travelling at 36km/hr is brought to rest over a distance of 40m. calculate

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Chapter One a) The acceleration of the car? b) The average braking force of the car? 7. A car of mass 500kg is travelling along a flat road. The forward force provided by the car is 300N and the force of the air resistance against the car 200N as shown below.

Calculate a) The acceleration of the car? b) What happens to the car’s motion if the force of air resistance increases to 300N? 8. A car engine can provide maximum forward force of 500N. The air resistance which the car experiences depends on its speed according to F = 0.2V2 (where F is the force of the air resistance). What is the car’s top speed?

Acceleration caused by gravity

Acceleration due to gravity is defined as the inherent property of earth to attract every object to its center. There is a force known as force of gravity and the acceleration generated by this force is known as the acceleration due to gravity. It is denoted by the letter 'g'. If you drop a ball or stone, it falls to the ground. It speeds up as it falls due to the pull of the earth’s gravity. Another name of this force is the weight of the object. Heavy objects have a greater weight than lighter objects, but if you drop over a cliff at a same time, they will fall at the same acceleration regardless of their weight. In fact, lighter objects fall more slowly than a heavy object because the force of the air resistance has a greater effect on the lighter objects which causes to reach a very low terminal velocity and the force of air resistance has almost no effect on the heavy objects, this cause to reach a very high terminal velocity. That is why heavy object fall more quickly than lighter objects.

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33

Acceleration or the rate of change of velocity is existing because of net force acting on a body. A permanent force is present on earth which is pulling downward to everybody to its center. If we consider gravity is the only force which is acting on a body, then we can find the body will accelerate at a rate of 9.8 m/s2 vertically downward direction to the center of the earth. Every object fall at this same rate if there is no air friction is present The force that caused the acceleration due to gravity (weight) can be found by using Newton’s second law of motion as follows: F = ma and W = mg Where, W = weight (downward force), m = mass of the object, g = acceleration due to gravity (gravitational field strength) On the earth g = 9.8m/s2 (approximately 10m/s2) and on the mood g = 1.6m/s2.

Air resistance and Terminal velocity When an object falls in air, first it accelerates freely. At the start of the fall, the only force acting on it is its weight as shown in the figure on the right, but the air resistance opposes its weight. The air resistance opposing its motion increases as its speed increases, so reducing its acceleration. Eventually, air resistance acting upwards equals the weight of the object acting downwards. The resulting force on the object is then zero since the weight of the object balances the air resistance. The object falls at a constant velocity called its terminal velocity, whose value depends on the size, shape, and weight of the object. A small dense object e.g. a parachute has a low terminal velocity. The graph below shows how the velocity of the parachutist varies during a descent. The idea of a parachute is to greatly increase the force of air resistance. Then the terminal velocity is reduced, and parachutist can land safely. Terminal velocity depends on how big you are. For insects, air resistance is much more important for human being, and so their terminal velocity is quite low. Causes of air resistance and its effects Air resistance is due to the colliding of an object with molecules of air. A falling object collides with air molecules during the downward fall. These air molecules create a force pushing upward which is opposite to the object's direction of travel. The amount of air resistance encountered by the object depends on three factors: 1: The speed of the object. The faster it goes the more air particles it collides with per second therefore the bigger the air resistance will be. If it is not moving the air resistance will be zero.

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Chapter One 2: The surface area of the object. The larger the surface area of the object the more air particles it will hit into per second and the higher the air resistance will be. If it is very streamlined the area of intercept will be small and there will be less air resistance acting on it.

Free falling motion A free falling object is an object that is falling under the influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects:  

Free-falling objects do not encounter air resistance. All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s2 (often approximated as 10 m/s2)

Because free-falling objects are accelerating downwards at a rate of 9.8 m/s2, a ticker tape trace or dot diagram of its motion would depict acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward.

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35

Free falling motion Objects released from rest or thrown upward that move under the influence of gravity alone is said to be in free fall. Free fall is a uniformly accelerated motion, so the equations of linear motion are applicable, except the acceleration is changed to acceleration due to gravity. Original form For free falling motion V = u + at

V = u + gt

V2 = u2 + 2as

V2 = u2 + 2gh

S = ut + ½ at2

h = ut + ½ gt2

S=(

𝑽𝑽𝑽𝑽+𝒖𝒖𝒖𝒖 𝟐𝟐𝟐𝟐

)t

h=(

𝑉𝑉𝑉𝑉+𝑢𝑢𝑢𝑢 2

)t

Examples: 1. A ball is thrown vertically upwards, with an initial velocity of 30m/s. a) How high will it rise, when its speed is 20m/s? b) How high will it rise? c) For how long will it be in the air? (Given that the acceleration due to gravity is 10m/s2) a) u = 30m/s, g = -10m/s2 V = 20m/s h =? 2 V = u + 2gh (20)2 = (30)2 + 2(-10) h 400 = 900 – 20h h = - 500/-20 = 25m V=0 h =? g = -10m/s2 2 V = u + 2gh (0)2 = (30)2 + 2(-10) h 0 = 900 – 20h h = - 900/-20 = 45m c) First we find the time taken to reach the maximum height, then the time that the ball in air is twice the time taken to reach the maximum height. u = 30m/s V=0 g = -10m/s2 t =? 2 2 t = (V – u)/g = (0 – 30m/s)/-10m/s = (-30m/s)/ (-10m/s ) = 3s. Therefore the time taken in air = 2 x 3s = 6s. Exercise 1. An egg falls off a table. The floor is 0.8m from the table-top. (Take g = 10m/s2) a) How long will take the egg to reach the floor? b) How fast will it be moving when it hits the floor? b) u = 30m/s,

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Chapter One 2. The astronaut on the moon drops a spanner. It takes 1.7s to fall a distance of 2.3m. Determine the acceleration due to gravity on the moon? 3. A climber, climbing a big cliff accidentally knocks loose a large rock. She sees it reaches the bottom of the cliff 8s later. (Take g = 10m/s2) a) What is the speed of the impact? b) How far did the rock fall? c) What is the average velocity of the rock?

Momentum

Momentum is a property of an object with mass. It is a measure of its instantaneous kinetic energy and is a measure of how easy it is to change its velocity, i.e. to change its direction or speed it up/slow it down. The higher the momentum, the harder it is to stop. Momentum is defined as the product of the mass of an object and its velocity. Momentum (p) = mass (m) x velocity (V) Momentum is a vector quantity. The direction of momentum is the direction of the velocity. The SI unit of momentum is kilogram meter per second (kgm/s) or Newton second (Ns). Example 1 Calculate the momentum of a car of mass 200kg travelling 5ms-1 Answer P = mv = 2000x5 = 1000kgms-1 Example 2 A bicycle and cyclist of total mass 100kg travelling 6ms-1 If the velocity of the body changes increased its velocity to 10ms-1 calculate the change in from u to v. the change in momentum of the cyclist? momentum ∆P equals:∆p = mv – mu = m (v – u) Solution Initial momentum = mu = 100x6 = 600Ns Final momentum = mv = 100x10 = 1000Ns Change in momentum = 1000– 600 = 400Ns.

Newton’s third law

The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. The force exerted on the first object is called action and the force exerted on the second object is called reaction. These forces act on different objects and never on the same object. So Newton’s third law states every action there is equal and opposite reaction. Suppose you are standing at rest and intend to start walking on a road. You must accelerate, and this requires a force in accordance with the second law of motion. Which is this force? Is it the muscular effort you exert on the road? Is it in the direction we intend to move? No, you push the road below backwards. The road exerts an equal and opposite reaction force on your feet to make you move forward. It is important to note that even though the action and reaction forces are always equal in magnitude, these forces may not produce accelerations of

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37

equal magnitudes. This is because each force acts on a different object that may have a different mass. When a gun is fired, it exerts a forward force on the bullet. The bullet exerts an equal and opposite reaction force on the gun. This results in the recoil of the gun. Since the gun has a much greater mass than the bullet, the acceleration of the gun is much less than the acceleration of the bullet. The third law of motion can also be illustrated when a sailor jumps out of a rowing boat. As the sailor jumps forward, the force on the boat moves it backwards (Fig. 12).

Conservation of Momentum Suppose two objects (two balls A and B, say) of masses mA and mB are traveling in the same direction along a straight line at different velocities uA and uB, respectivelyFigure(a). And there are no other external unbalanced forces acting on them. Let uA> uB and the two balls collide with each other as shown in Figure (b). During collision which lasts for a time t, the ball A exerts a force FAB on ball B and the ball B exerts a force FBA on ball A. Suppose vA and vB are the velocities of the two balls A and B after the collision, respectively

38

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Chapter One

Figure. Conservation of momentum in collision of two balls. From Equation, the momentums of ball A before and after the collision are mAuA and mAvA, respectively. The rate of change of its momentum (or FAB, action) during the collision will be / 𝑚𝑚𝑚𝑚𝐴𝐴𝐴𝐴 (𝑣𝑣𝑣𝑣𝐴𝐴𝐴𝐴 − 𝑢𝑢𝑢𝑢𝐴𝐴𝐴𝐴 ) 𝑡𝑡𝑡𝑡

Similarly, the rate of change of momentum of ball B (= FBA or reaction) during the collision will be 𝑚𝑚𝑚𝑚𝐵𝐵𝐵𝐵 (𝑣𝑣𝑣𝑣𝐵𝐵𝐵𝐵 − 𝑢𝑢𝑢𝑢𝐵𝐵𝐵𝐵 ) 𝑡𝑡𝑡𝑡

According to the third law of motion, the force FAB exerted by ball A on ball B (action) and the force FBA exerted by the ball B on ball A (reaction) must be equal and opposite to each other. Therefore, FAB = – FBA or, mA (vA-uA)/t = - mB (vB-uB)/t and multiple both sides by t This gives, mAuA + mBuB = mAvA + mBvB

Since (mAuA + mBuB) is the total momentum of the two balls A and B before the collision and (mAvA + mBvB) is their total momentum after the collision, from Equation we observe that the total momentum of the two balls remains unchanged or conserved provided no other external force acts Examples 1. Calculate the momentum of a truck of mass 2000kg travelling at 5m/s? p = mv = (2000kg)(5m/s) = 10,000kgm/s = 10,000Ns 2. A bicycle and a rider of total mass 100kg travelling at 6m/s increases its velocity to 10m/s. calculate a) The initial momentum of the bicycle and the rider? b) The final momentum of the bicycle and the rider? c) The change in momentum?

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39

a) p1 = mu = (100kg) (6m/s) = 600kgm/s. b) p2 = mv = (100kg) (10m/s) = 1000kgm/s. c) ∆p = p2 - p1 = 1000kgm/s – 600kg/s = 400kgm/s. 3. A trolley of mass 2kg travelling at 3m/s collides with stationary trolley of mass 1kg. The two trolleys stick together after the collision and move in the same direction as the direction of the first trolley. What is their common velocity?

Total momentum before impact = total momentum after impact 𝑚𝑚𝑚𝑚1 𝑢𝑢𝑢𝑢1 + 𝑚𝑚𝑚𝑚2 𝑢𝑢𝑢𝑢2 = 𝑚𝑚𝑚𝑚1 𝑣𝑣𝑣𝑣1 + 𝑚𝑚𝑚𝑚2 𝑣𝑣𝑣𝑣2 (2 kg x 3 m/s) + 0 = (2 kg x V) + (1 kg x V) 6 kgm/s = (3kg) V V = (6 kgm/s)/3 kg = 2 m/s Therefore the common velocity of the two trolleys is 2m/s. Example: 2 A bullet of mass 20 g is horizontally fired with a velocity 150 m s-1 from a pistol of mass 2 kg. What is the recoil velocity of the pistol? Solution We have the mass of bullet, m1 = 20 g (= 0.02 kg) and the mass of the pistol, m2 = 2 kg; initial velocities of the bullet (u1) and pistol (u2) = 0, respectively. The final velocity of the bullet, v1 = + 150 m s-1. The direction of bullet is taken from left to right (positive, by convention, Figure). Let v be the recoil velocity of the pistol.

Total momentum of the pistol and bullet before the fire, when the gun is at rest = (2 + 0.02) kg × 0 m s-1 = 0 kg m s-1 Total momentum of the pistol and bullet after it is fired = 0.02 kg × (+ 150 m s-1) + 2 kg × v m s-1 = (3 + 2v) kg m s-1 According to the law of conservation of momentum Total momentum after the fire = Total momentum before the fire

40

.................

Chapter One 3 + 2v = 0 v = − 1.5 m s-1. Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet, that is, right to left. Exercise:

1. A body has a mass of 5kg. calculate a) Its momentum when it has a velocity of 2m/s? b) Its velocity when its momentum is 20kgm/s? 2. Which one of the following bodies has a greater momentum, A 4g bullet moving at a velocity of 300m/s and a stone sliding at 20m/s? 3. What is the total momentum of a 2kg mass moving at 100m/s due east and 4 kg mass moving at 150m/s due west? (hint: take west as – ve and east as + ve) 4. The manufacturer of a family car gave the following information. Mass of car 950 kg. The car will accelerate from 0 to 33 m/s in 11 seconds. (a) Calculate the acceleration of the car during the 11 seconds ..................................................................................................................................... ..................................................................................................................................... (b) Calculate the force needed to produce this acceleration. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (c) The manufacturer of the car claims a top speed of 110 miles per hour. Explain why there must be a top speed for any car. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... 5. A crane is used to lift a steel girder to the top of a high building. steel girder (mass = 200kg)

massive crane body

25m

When it is lifted by the crane: • the girder accelerates from rest to a speed of 0.6 m/s in the first 3 seconds; • it then rises at a steady speed. a)

Calculate the acceleration of the girder. (Show your working.) ..................................................................................................................................... .....................................................................................................................................

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41

What is the weight of the steel girder? Calculate the power of the crane motor as it lifts the girder at a steady speed of 0.6 m/s. (Show your working. You can ignore the weight of the cable and hook which is small compared to the weight of the girder.) .......................................................................................................................... . .......................................................................................................................... . .......................................................................................................................... . (c) A new motor is fitted to the crane. This motor accelerates the girder at 0.3 m/s2. Calculate the force which the crane applies to the girder to produce this acceleration. (Show your working.) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... 6. A sky-diver steps out of an aeroplane. After 10 seconds she is falling at a steady speed of 50m/s. She then opens her parachute. (b)

(i) (ii)

After another 5 seconds she is once again falling at a steady speed. This speed is now only 10m/s. (a) Calculate the sky-diver’s average acceleration during the time from when she opens her parachute until she reaches her slower steady speed. (Show your working.) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (b) Explain, as fully as you can: (i) why the sky-diver eventually reaches a steady speed (with or without her parachute). .......................................................................................................................... . .......................................................................................................................... . .......................................................................................................................... . .......................................................................................................................... . (ii) why the sky-diver’s steady speed is lower when her parachute is open. .......................................................................................................................... . (c) The sky-diver and her equipment have a total mass of 75kg. Calculate the gravitational force acting on this mass. (Show your working.)

42

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Chapter One ..................................................................................................................................... ..................................................................................................................................... 7.

(a) The amount of damage caused when a car collides with a wall depends on the amount of energy transferred. If the speed of a car doubles, the amount of energy transferred in a collision increases four times. Explain, as fully as you can, why this is so. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (b) The diagram shows a car and a lorry about to collide. speed = 20km/h speed = 40km/h 9.5 tonne 0.5 tonne

When they collide, the two vehicles become tightly locked together. (i) Calculate the speed of the vehicles immediately after the collision. (Show your working. There is no need to change to standard units.) .......................................................................................................................... . .......................................................................................................................... . .......................................................................................................................... . .......................................................................................................................... . .......................................................................................................................... . (ii) The collision between the car and the lorry is inelastic. Explain, in terms of energy, what this means. ........................................................................................................................... 8. A cyclist accelerates from a set of traffic lights. The driving force of the back tyre on the ground is 250 N. (a) How much work is done by this force when the cyclist travels 5 metres? (Show your working.) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (b)

9.

What happens to the energy transferred by this force? ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... The graph shows the speed of a runner during an indoor 60 metres race.

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43

Speed (metres per second)

10

5

0

9.

0

2

4 6 Time (seconds)

8

10

(a) Calculate the acceleration of the runner during the first four seconds. (Show your working.) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (b) How far does the runner travel during the first four seconds? (Show your working.) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (c) At the finish, a thick wall of rubber foam slows the runner down at a rate of 25 m/s2. The runner has a mass of 75kg. Calculate the average force of the rubber foam on the runner. (Show your working.) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... The diagram shows a shuttlecock that is used for playing badminton.

The shuttlecock weighs very little. When you drop it from a height of a few metres, it accelerates at first but soon reaches a steady speed. Explain, as fully as you can: (a) why the shuttlecock accelerates at first, .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (b) why the shuttlecock reaches a steady speed. ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... 10. A child goes out to visit a friend.

44

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Chapter One The graph shows the child’s journey. 150

Distance travelled (metres)

100

50

0

0

10

20

30

40

50

60

70

Time in seconds Calculate the child’s average speed for the whole journey. [Show your working and give the units in your answer.] ..................................................................................................................................... .....................................................................................................................................

(a)

(b)

How many times faster is the child travelling in part A of the graph than in part C? [You should show how you obtained your answer.] ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... 11. A bouncy ball is dropped vertically from a height of 2.00 m onto the floor. The graph shows the height of the ball above the floor at different times during its fall until it hits the floor after 0.64 s. 2.0

1.6

1.2 Height in metres 0.8

0.4

0

(a)

0

0.2

0.4

0.6 0.8 Time in seconds

1.0

1.2

1.4

What is the average speed of the ball over the first 0.64 s? Show clearly how you

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45

(b)

(c)

(d)

work out your answer. .................................................................................................................................... .................................................................................................................................... After it hits the floor the ball bounces back to a height of 1.25 m. It reaches this height 1. 16 s after it was dropped. Plot this point on the grid above and sketch a graph to show the height of the ball above the floor between 0. 64 s and 1.16 s. (i) The ball bounces on the floor 0.64 s after being dropped. How long after being dropped will it be before it bounces a second time? .......................................................................................................................... . .......................................................................................................................... . (ii) What distance will the ball travel between its first and second bounce? .......................................................................................................................... . .......................................................................................................................... . The ball was held stationary before being dropped. On the graph and your sketch mark two other points X1 and X2, where the ball is stationary, and in each case explain why the ball is not moving. X1 ............................................................................................................................... ................................................................................................................................... X2 .............................................................................................................................. ...................................................................................................................................

12. An object “A” of mass 20kg moving with a velocity of 3m/s makes a head on collision with object “B” of mass 10kg moving with a velocity of 2m/s in the opposite direction. If “A” and “B” stick together and move in the direction of “A”, calculate their common velocity?

Impulse:

Newton’s second law motion states F = ma and acceleration is defined as the rate of change of velocity a = (V – u)/t. by combining these two equation gives 𝑚𝑚𝑚𝑚𝑣𝑣𝑣𝑣 − 𝑚𝑚𝑚𝑚𝑢𝑢𝑢𝑢 ∆𝑝𝑝𝑝𝑝 𝑚𝑚𝑚𝑚(𝑣𝑣𝑣𝑣 − 𝑢𝑢𝑢𝑢) = = 𝐹𝐹𝐹𝐹 = 𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡 Therefore the rate of change of momentum is equal to the force causing the change. Force (F) = change in momentum (∆p) / time (t) This formula tells that the reaction force (F) is inversely proportional to the interaction time. This relationship is very important and is used to design car safety devices and so many other things. For example, the front and the back of the cars are designed to crumple, in order to spread out the time of a collision and so reduce the force on you. When a car crashes, the passenger will tend to continue his forward motion at a high speed

46

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Chapter One due to inertia. The passenger will crash into the front part of the car and get injured seriously. A seat-belt can protect the passenger. A seat-belt slightly stretches in a collision, thus it can lengthen the time during which the passenger decelerates and greatly reduce the impact force. Air bags in some cars are designed in a same way to reduce the reaction force and safes the passengers in the cars. This is also used for sporting events, for example in a cricket match as shown in the figure below: In a cricket match a fielder moves his arms back while trying to catch a cricket ball because the backward of his arms increases the interaction time between the ball of the arm, so this reduces the reaction force. Otherwise the ball may hurt the player or bounces back. Goal keepers use similar principle. We can write Newton’s second law of motion in terms of momentum change. Ft = ∆p The product of force and the time for which it acts “Ft” is called the impulse of the force. Impulse measures how hard and for how long a force acted. The SI unit of the impulse of the force is same as that of momentum (kg m/s or Ns). Impulse can also be obtained from the area under the force-time graph. If two cars that are travelling very quickly collide they come to a stop in a short space of time.

This means that the cars and their occupants experience a large change of momentum very quickly. Why could this cause a very serious injury?

A very large change of momentum in a short space of time means the car occupants will experience a very large force. Many modern car safety features work by increasing the amount of time taken for the person to decelerate in a collision. A longer deceleration means that change in momentum occurs over a longer time. There is therefore smaller force acting on the person. The feature of cars that use this principle are:   

seatbelts airbags crumble zones

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47

Example A tennis ball of mass 0.07 kg is moving towards a racket at a speed of 30 . After collision, the ball moves at the same speed in the opposite direction. What is the magnitude of the force acting on the ball if the time of contact between the ball and the racket is (a) 0.01 s and (b) 0.05 s? Solution a. Force acted on the ball b. Force acted on the ball

Exercise: 1. A motorcycle of mass 500kg is brought to rest from 30m/s in 10s. What is the average braking force? 2. A force of 10N acts on a ball of mass 1 kg for 2s. Find the impulse given to the ball? 3. A car of mass 1000kg is travelling at a velocity of 15m/s. it collides head with a wall. Calculate the force of the impact if it stops in a) 0.5s? b) 0.01s? Which time interval is preferable from the view point of the passenger’s safety? Why?

4. The diagram below shows a force-time graph for a super market trolley of mass 60kg. The trolley is at rest prior to the force acting on it.

48

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Chapter One a) b) c) d)

What is the change in momentum of the trolley during this time interval? What is the impulse given to the trolley? What is the average force exerted on the trolley? What is the speed of the trolley immediately after it has been pushed

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1.3 Using vectors and projectile motion Using vectors Using vectors are much wider than using scalars because vectors involve directions. You learnt distinction between vectors and scalars earlier. You should recall that a scalar quantity has magnitude only while vector quantity has both magnitude and direction. Let us look this example of scalar/vector pair; speed and velocity. To define the velocity of a moving object, you have to say how fast it is moving and the direction it is moving in, but defining the speed of a moving object, you have to say how fast it is moving. Here are some more examples of scalar and vector quantities; vector quantities are usually represented by arrows on diagrams. Examples of scalar quantities: distance, speed, mass, energy and temperature. Examples of vector quantities: displacement, velocity, acceleration and force. When we combine or add two vectors, we need to add both magnitude and direction. For example, take 2 steps in the forward direction, stop and then take another 3 steps in the forward direction. The result is that we have taken 5 steps in the forward direction.

The final answer when adding vectors is resultant vector. The resultant displacement in this case will be 5 steps forward. Resultant of vectors The resultant of a number of vectors is the single vector whose effect is the same as the individual vectors acting together.

Combining vectors Sometimes there are several vectors acting on the same object, for example the figure on the right shows some forces acting on a car as it struggles up the steep hill. The forces are:  Its weight, W  The contact force, N of the road ( its normal reaction )  Air resistance, R

 The forward force, F caused by the engine of the car

50

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Chapter One The combining effect of such forces is called resultant force a. Vectors in a straight line When adding two or more vectors in straight line, we can find the resultant vector by using algebraic sum and the sing of your final answer will tell you the direction of resultant force. So first, it is important to take account in to their directions and use the sign convention correctly. Sing convections  

Upward vectors are negative and downward vectors are positive. Vectors to the right are positive and vectors to the left are negative.

Worked examples 1- Calculate the resultant force and its direction of each of the following figures.

Solution a- Resultant force = 10N + 5N = 15N to the right b- Resultant force = -30 N + 70 N = 40 N to the right 2- A cyclist travels 20 km to the north and 45 km to the south. How many kilometers he is away from his starting point (means what is the resultant displacement)?

Solution Resultant displacement is + 45km – 20 km = + 25 km (25 km towards south)

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Exercise 1- Two teams are pulling a rope against each other as shown in the figure below. Calculate the resultant force.

2- A car moves forward with a velocity of 40 m/s and then moves backward with a velocity of 57 m/s. Calculate the resultant velocity of the car. 3- The figure below shows a falling tennis ball. Calculate the resultant force.

4- A parachutist weighs 1000N. When she opens her parachute, it pulls upward on her with a force of 2000N. a- Draw a diagram to show the forces acting on the parachutist. b- Calculate the resultant force acting on her. c- What effect will this force have on her? b. Vectors at right angles (perpendicular vectors) When two vectors acting at right angles upon a given point, the resultant vector is equal to the square root of the sum of the two vectors. This means we use Pythagoras theorem to find resultant vector of two perpendicular vectors. For example, the figure below-left shows two perpendicular forces acting upon a given point. The direction of resultant vector can be found by using tan θ. Add vectors by placing them tip-totail and drawing the resultant vector from the tail of the first vector to the tip of the last vector to form rightangled triangle sea the figure on the right.

52

𝐹𝐹𝐹𝐹𝑦𝑦𝑦𝑦

Tanθ = 𝐹𝐹𝐹𝐹

................. 𝐹𝐹𝐹𝐹 = √𝐹𝐹𝐹𝐹𝑥𝑥𝑥𝑥 2 + 𝐹𝐹𝐹𝐹𝑦𝑦𝑦𝑦 2

𝑥𝑥𝑥𝑥

Chapter One F = resultant force, Fx = Horizontal force, Fy = vertical force Note: In the case above, we take force an example of vectors, similarly we use these formulae for other perpendicular vectors, and only the names will be changed. For example, if we dealing with velocity the formula will be𝑉𝑉𝑉𝑉 = √𝑉𝑉𝑉𝑉𝑥𝑥𝑥𝑥 2 + 𝑉𝑉𝑉𝑉𝑦𝑦𝑦𝑦 2

Worked examples

1- Two forces act on this shuttlecock as it travels through the air, calculate the resultant force and direction of travel.

Given First, we have to draw a vector triangle Fx = 6 N Fy= 8N F =?

Solution

To find the direction of travel (direction of resultant force)

R = √𝐹𝐹𝐹𝐹𝑥𝑥𝑥𝑥 2 + 𝐹𝐹𝐹𝐹𝑦𝑦𝑦𝑦 2

Tanθ =

R = √(6𝑁𝑁𝑁𝑁)2 + (8𝑁𝑁𝑁𝑁)2 R = √36𝑁𝑁𝑁𝑁 2 + 64𝑁𝑁𝑁𝑁 2

θ = 530

𝐹𝐹𝐹𝐹𝑦𝑦𝑦𝑦 𝐹𝐹𝐹𝐹𝑥𝑥𝑥𝑥

=

8𝑁𝑁𝑁𝑁 6𝑁𝑁𝑁𝑁

= 4/3

the direction of travel is 530 below horizontal force.

R = √100𝑁𝑁𝑁𝑁 2 = 10N

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53

2- An aircraft is flying due north with a velocity of 200 m/s. A side wind of velocity 50 m/s is blowing due east. What is the magnitude and direction of the aircraft’s resultant velocity? Given

Solution

To find the direction of resultant velocity

R = √𝑉𝑉𝑉𝑉𝑥𝑥𝑥𝑥 2 + 𝑉𝑉𝑉𝑉𝑦𝑦𝑦𝑦 2

R=

√(50𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠)2

+ (200

𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠)2

R = √2500 𝑚𝑚𝑚𝑚2 /𝑠𝑠𝑠𝑠 2 + 40 000 𝑚𝑚𝑚𝑚2 /𝑠𝑠𝑠𝑠 2

𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜

𝑉𝑉𝑉𝑉𝑦𝑦𝑦𝑦

50 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠

Tanθ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑉𝑉𝑉𝑉 = 200 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠 = 0.25

θ = 140

𝑥𝑥𝑥𝑥

the direction of resultant velocity is 140north of east

R = √42500 𝑚𝑚𝑚𝑚2 /𝑠𝑠𝑠𝑠 2 = 206 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠

Exercise

1- A 100N force and 50N force act on a point P. the 100N force acts due north, and the 50N force acts due east. What is the magnitude and direction of the resultant force? 2- A spider runs two sides of a table as shown in the figure below. Calculate the magnitude and direction of its final displacemen c. Non-perpendicular vectors To determine the magnitude of the resultant vector of non-perpendicular vectors, use the cosine rule, and use the sine rule to find the direction of the resultant vector. For example, if two non-perpendicular forces act upon a point as shown in the figure below.

Complete by drawing a parallelogram then, the diagonal divides the parallelogram in to two congruent triangles. The longest side of one triangle (the diagonal of the parallelogram)

54

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Chapter One represents the resultant force. The two other sides of each of the two triangles represent the two non-

Non-perpendicular component forces. Magnitude of resultant force using F= √𝑭𝑭𝑭𝑭𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 + 𝑭𝑭𝑭𝑭𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 − 𝟐𝟐𝟐𝟐𝑭𝑭𝑭𝑭𝟏𝟏𝟏𝟏 𝑭𝑭𝑭𝑭𝟐𝟐𝟐𝟐 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 𝛉𝛉𝛉𝛉𝟑𝟑𝟑𝟑    



𝑭𝑭𝑭𝑭𝟐𝟐𝟐𝟐 𝒄𝒄𝒄𝒄𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝛉𝛉𝛉𝛉𝟑𝟑𝟑𝟑 𝐅𝐅𝐅𝐅

or

𝒄𝒄𝒄𝒄𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝛉𝛉𝛉𝛉𝟐𝟐𝟐𝟐 =

𝑭𝑭𝑭𝑭𝟏𝟏𝟏𝟏 𝒄𝒄𝒄𝒄𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝛉𝛉𝛉𝛉𝟑𝟑𝟑𝟑 𝐅𝐅𝐅𝐅

𝐹𝐹𝐹𝐹

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ3

=

𝐹𝐹𝐹𝐹2

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ1

, then 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ1 =

𝐹𝐹𝐹𝐹2 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ3 F

When expressing direction of resultant force from f2, we take Δ ACD applying the sine rule we have

  

𝒄𝒄𝒄𝒄𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝛉𝛉𝛉𝛉𝟏𝟏𝟏𝟏 =

θ is the angle between the two component forces θ1 is the angle between resultant force and f1 θ2 is the angle between resultant force and f2 F is the resultant force, f1 and f2 are component forces When expressing direction of resultant force from f1, we take Δ ABD applying the sine rule we have



direction of resultant force can be found

𝐹𝐹𝐹𝐹

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ3

=

𝐹𝐹𝐹𝐹1

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ2

, then 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ2 =

𝐹𝐹𝐹𝐹1 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ3 F

θ3 is the angle which opposes the resultant force θ = θ1 + θ2 θ3 = 1800 − θ [θ and θ3 are two adjacent angles of the parallelogram and they add up to 1800 ]

Worked examples 1- A force of 60 N and a force of 40 N act concurrently on a point P find the magnitude of their resultant force when the angle between them is 600.

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55

Given f1 = 60 N f2 = 40 N θ= 600 θ3 = 1800 – θ = 1800 – 600 = 1200 F =? Solution F= √𝐹𝐹𝐹𝐹 21 + 𝐹𝐹𝐹𝐹 2 2 − 2𝐹𝐹𝐹𝐹1 𝐹𝐹𝐹𝐹2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐θ3

= √602 + 402 − 2(60)(40) (cos 1200 ) =√5200 − 4800 cos 1200 =√5200 − 4800(−0.5)

=√5200 + 2400

=√7600

To find the direction of the resultant force, use Δ ACD 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝛉𝛉𝛉𝛉𝟐𝟐𝟐𝟐 =

𝑐𝑐𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ2 =

𝒇𝒇𝒇𝒇𝟏𝟏𝟏𝟏 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝛉𝛉𝛉𝛉𝟑𝟑𝟑𝟑 𝑭𝑭𝑭𝑭

60 𝑁𝑁𝑁𝑁(𝑐𝑐𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠1200 ) 60 (0.8660) 51.96 = = = 0.5959 87.2 N 87.2 87.18

θ2 = 36.60

The resultant force is 36.60 to the 40 N-force

Similarly, we can take Δ ACD and calculate 𝑐𝑐𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ1 as follows 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝛉𝛉𝛉𝛉𝟏𝟏𝟏𝟏 =

𝑐𝑐𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ1 =

𝒇𝒇𝒇𝒇𝟐𝟐𝟐𝟐 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝛉𝛉𝛉𝛉𝟑𝟑𝟑𝟑 𝑭𝑭𝑭𝑭

40 𝑁𝑁𝑁𝑁(𝑐𝑐𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠1200 ) 40 (0.8660) 34.64 = = = 0.3972 87.2 N 87.2 87.2

θ1 = 23.40

The resultant force is 23.40 to the 60 N-force

= 87.2 N The magnitude of the resultant force is 87.2N 2- An aircraft flies 80 km due east and then 20 km north-east as shown in the figure below. Calculate the final displacement (resultant displacement) of the aircraft.

Given S1 = 80 km S2 = 20 km θ3 = 1800 – 450 = 1350 S=?

Solution S= √𝑆𝑆𝑆𝑆 21 + 𝑆𝑆𝑆𝑆 2 2 − 2𝑆𝑆𝑆𝑆1 𝑆𝑆𝑆𝑆2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐θ3

56

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Chapter One = √802 + 202 − 2(80)(20) (cos 1350 ) =√6800 − 3200 cos 1350

=√6800 − 3200(−0.7071)

=√5200 + 2262.72 =√9062.72

= 95.2 km The magnitude of the resultant Displacement is 95.2 km

To find the direction of the magnitude, use sine rule 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ1 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ3 = S2 𝑆𝑆𝑆𝑆 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ1 =

𝑆𝑆𝑆𝑆2 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ3 𝑆𝑆𝑆𝑆

20 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘(𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠1350 ) 20 (0.7071) 14.1 = = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠θ1 = 95.2 km 95.2 95.2 = 0.1486 θ1 = 8.50

The final displacement is 8.50 North of East.

Exercise 1- The diagram shows the direction of two forces of 16 N and 12 N acting at an angle of 500 to each other.

Draw a vector diagram to determine the direction and magnitude of the resultant force.

2- A force of 20 N and 40 N act on a point and the angle between them is 300. Calculate the magnitude and direction of resultant force.

d- Three or more vectors The spider shown in the figure below is hanging by a thread. It is blown sideways by the wind. The diagram shows the three forces acting on it:  Weight acting downwards  The tension in the thread along the thread  The push of the wind

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57

In the figure above, blowing in the wind-this spider is hanging in equilibrium. The diagram also shows how these can be added together. In this case we arrive an interesting result. Arrows drawn to represent each of the three forces, end-to-end. The end of the third arrow coincides with the start of the first arrow, so the three arrows form a closed triangle. This tells us that the resultant force R on the spider is zero, that is R = 0. The closed triangle is known as triangle of forces. So, there is no resultant force. The forces on the spider balances each other out, and we say that the spider is in equilibrium. If the wind blew a little harder, there would be an unbalanced force on the spider, and it would move off to the right. We can use this idea in two ways:  If we work out the resultant force on an object and find that it is zero, this tells us that the object is equilibrium  If we know that an object is in equilibrium we know that the force on it must add up to zero we can use this to work out the value of one or more unknown forces. Worked examples 1- A stone was dropped in to fast flowing stream it does not fall vertically because of the side way push of the water.

a- Calculate the resultant force b- Is the stone is in equilibrium?

58

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Chapter One Solution First, we calculate the resultant force of the two straight line forces. R.F = - 0.5 N + 2.5 N = 2 N

Then, we find the resultant force of the two perpendicular forces

F= √1.52 + 22 F= √2.25 + 4

F= √6.25 F= 2.5N c- the resultant force of the three forces is 2.5 N d- The stone is not in equilibrium 2- Three forces are acting an object concurrently as shown in the figure below. a- Calculate the resultant force of the three forces b- Is the object is in equilibrium Given Tension in thread is 10 N Weight of the object is 8 N Bush of the win

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59

Solution First the resultant force of the two perpendicular forces F= √62 + 82 = √36 + 64 =

√100 = 10N

Then we calculate the resultant force of the three forces F = 10 N – 10 N = 0 N a- The resultant force of the forces is 0 N b- The object is in equilibrium

Exercise Find the resultant force of each of the following figures

60

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Chapter One Resolution of vectors

Resolution of vectors is the separation of single vector in to two or more component vectors acting in a given direction on the same point. This means we break down a vector quantity in to components which are dimensions at right angles to each other, which are horizontal component and vertical component. For example, the figure below shows a single force, F decomposed in to two component forces Fx and Fy (horizontal and vertical components respectively). Horizontal component is given by Fx = Fcosθ

Vertical component is given by Fy = Fsinθ

Worked examples 1- If an object is drawn with a force of 60 N over a level road, and the rope makes an angle of 300 with the horizontal road. Find: a- The horizontal component of the force b- The vertical component of the force Solution

a- Fx = Fcosθ = 60 N (cos300) = 60 N (0.866) = 51 .96 N b- Fy = Fsinθ = 60 N (sin300) = 60 N (0.5)= 30 N

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61

2- A person has covered a displacement of 70 km as shown in the figure below. Calculate the horizontal and vertical displacement’

a- Fx = Fcosθ = 70 (cos500) = 70 km (0.6428) = 45 km (nearest whole number) b- Fy = Fsinθ = 70 km (sin500) = 70 km (0.766) = 54 km (nearest whole number) Exercise 1- An object is pulled with a force of 200N. If the rope is inclined at 600 to the horizontal path, find: a- The horizontal component of the force b- The vertical component of the force. 2- An airplane flies a velocity of 150 m/s 300 above the ground. Calculate: a- Horizontal velocity b- Vertical velocity

62

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Chapter One Projectiles A projectile is a particle that is sent moving in to the air. The path followed by a projectile is called its trajectory. The horizontal distance traveled by a projectile is called its range. If the air resistance is negligible, projectiles all accelerate downwards at the same rate (acceleration due to gravity), and move horizontally at constant velocity. The motion of two dimensions of an object is called projectile motion, it has two components-Horizontal and vertical. The two components are independent; that is, they have no effect on each other. When the two motions are combined they form a curved path (trajectory). Here is an example to illustrate these ideas. In a toy, a ball-bearing is fired horizontally from a point 0.4 m above the ground. Its initial velocity is 25 m/s. Its position at equal intervals of time have been calculated and are shown in the table below these results are also shown in the figure below. Study the table and the graph. You should notice the following:  

The horizontal distance increases steadily. This is because the balls horizontal motion is un affected by the force of gravity. It travels at steady velocity horizontally. The vertical distance do not show the same pattern. The ball is accelerating downwards. (these figures has been calculated using g = 9.81 m/s)

The motion in two dimensions is explained by two main principles that are kinematic principles and Newton's laws of motion. Projectile thrown horizontally When an object is launched or thrown completely horizontally, such as a rock thrown horizontally off a cliff, the initial velocity of the object is equal to its initial horizontal velocity. Because horizontal velocity doesn’t change, this velocity is also the object’s final horizontal velocity, as well as its average horizontal velocity. Further, the initial vertical velocity of the projectile is zero. This means that you could hurl an object 1000 m/s

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63

horizontally off a cliff, and simultaneously drop an object off the cliff from the same height, and they will both reach the ground at the same time (even though the hurled object has traveled a greater distance).

Ux= Initial horizontal velocity Uy = initial vertical velocity Vx = final horizontal velocity Vy = final vertical velocity V = Velocity of motion of the projectile h = vertical displacement X = horizontal displacement

𝑋𝑋𝑋𝑋 = 𝑈𝑈𝑈𝑈𝑥𝑥𝑥𝑥 𝑡𝑡𝑡𝑡 +

X = 𝑈𝑈𝑈𝑈𝑥𝑥𝑥𝑥 𝑡𝑡𝑡𝑡

1 𝑎𝑎𝑎𝑎 𝑡𝑡𝑡𝑡 2 2

[a = 0 because of the constant speed, since air resistance is negligible]

1

h = 𝑈𝑈𝑈𝑈𝑦𝑦𝑦𝑦 𝑡𝑡𝑡𝑡 + 2 𝑔𝑔𝑔𝑔 𝑡𝑡𝑡𝑡 2 1

[Uy = 0 because there is no any vertical velocity at the beginning]

h = 2 𝑔𝑔𝑔𝑔 𝑡𝑡𝑡𝑡 2

2

V = √𝑉𝑉𝑉𝑉𝑥𝑥𝑥𝑥 + 𝑉𝑉𝑉𝑉𝑦𝑦𝑦𝑦

2

or V= √𝑈𝑈𝑈𝑈𝑥𝑥𝑥𝑥 2 + (𝑔𝑔𝑔𝑔𝑡𝑡𝑡𝑡)2

[𝑉𝑉𝑉𝑉𝑥𝑥𝑥𝑥 = 𝑈𝑈𝑈𝑈𝑥𝑥𝑥𝑥 ; 𝑉𝑉𝑉𝑉𝑦𝑦𝑦𝑦 = 𝑔𝑔𝑔𝑔𝑡𝑡𝑡𝑡]

Direction of motion of projectile can be found by using 𝑽𝑽𝑽𝑽𝒚𝒚𝒚𝒚

Tan θ= 𝑽𝑽𝑽𝑽

𝒙𝒙𝒙𝒙

Worked Examples 1- A stone thrown horizontally at 15m/s. It is thrown from the top of a cliff 44m high. a- How long does it take the stone to reach the bottom of the cliff? b- How far from the base of the cliff does the stone strike the ground? Given Ux = 15m/s, h = m44m, g = 10m/s Solution 1

a- When h = 2 𝑔𝑔𝑔𝑔 𝑡𝑡𝑡𝑡 2 2ℎ

2(44𝑚𝑚𝑚𝑚)

then 88 𝑚𝑚𝑚𝑚

t = √ 𝑔𝑔𝑔𝑔 = √10𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠2 = √10𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠2 = √8.8𝑠𝑠𝑠𝑠 2 = 2.97 s

64

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Chapter One b- X = 𝑈𝑈𝑈𝑈𝑥𝑥𝑥𝑥 𝑡𝑡𝑡𝑡 = 15m/s(2.97 s) = 44. 2- A particle is projected horizontally with a speed of 14m/s. Find the horizontal and vertical displacement of the particle from the point of projection after 2 seconds. Given Ux = 14m/s, t = 2 s. g = 10./s2 Solution a- Horizontal displacement X = 𝑈𝑈𝑈𝑈𝑥𝑥𝑥𝑥 𝑡𝑡𝑡𝑡 = 14m/s(2 s) = 28 m b- h =

1 2

1

𝑔𝑔𝑔𝑔 𝑡𝑡𝑡𝑡 2 = 2 10𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠 2 (2𝑠𝑠𝑠𝑠)2= 5m(4) = 20 m

3- A projectile is projected horizontally with a velocity of 40 m/s. 3 seconds after projection, find: a- The horizontal and vertical components of the velocity. b- The speed and direction of motion of the projectile. Given Ux = 40m/s, t = 3s a- Vx =?, Vy =?

b- V =?, direction of motion =?

Solution a- i-) Vx = Ux = 40 m/s ii-) Vy = gt = (10 m/s2) (3s) = 30 m/s b- i-) V = √𝑉𝑉𝑉𝑉𝑥𝑥𝑥𝑥 2 + 𝑉𝑉𝑉𝑉𝑦𝑦𝑦𝑦 2 =

√(1600 𝑚𝑚𝑚𝑚2 /𝑠𝑠𝑠𝑠 2 + (900 𝑚𝑚𝑚𝑚2 /𝑠𝑠𝑠𝑠 2 = √(2500 𝑚𝑚𝑚𝑚2 /𝑠𝑠𝑠𝑠 2 = 50m/s 𝑉𝑉𝑉𝑉𝑦𝑦𝑦𝑦

30𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠

ii-) Tan θ = 𝑉𝑉𝑉𝑉 = 40𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠 = 0.75 𝑥𝑥𝑥𝑥

θ = 36.90

The direction of motion is 36.90below the horizontal.

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Exercise 1- In the given picture below, Fatima throws the ball horizontally with an initial velocity of 10m/s. Time elapsed during the motion is 5s, calculate the height that object is thrown and vertical component of the velocity (Vy) after it hits the ground.

2- A bomb is released from a jet plane flying horizontally with a speed of 300 m/s at an altitude of 200m. a- Find the time to reach the ground b- Where will the bomb strike the ground 3- A particle is projected horizontally at 168 m/s. 5 seconds after projection, find: a- The horizontal and vertical components of the particle’s velocity b- the magnitude and direction of the velocity Projectile thrown at an angle When an object is projected from rest at an upward angle, its initial velocity can be resolved into two components. These two components operate independently of each other. The upward velocity undergoes constant downward acceleration which will result in it rising to a highest point and then falling backward to the ground. The horizontal motion is constant velocity motion and undergoes no changes due to gravity. The analysis of the motion involves dealing with the two motions independently. Suppose a particle is projected with a velocity of U m/s as shown in the figure

66

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Chapter One X = Ux t

Ux = Vx Ux = U Cos θ Uy = U Sin θ

ℎ = 𝑈𝑈𝑈𝑈𝑦𝑦𝑦𝑦 𝑡𝑡𝑡𝑡 +

Note that: a- Vertical upward motion Vy = 0 and g -10 m/s2 b- Vertical downward motion Uy = 0 and g 10 m/s2

1 𝑔𝑔𝑔𝑔 𝑡𝑡𝑡𝑡 2 2

Vy = Uy + 𝑔𝑔𝑔𝑔 t

The distance, d , of the projectile at any time is given by: d = √𝑋𝑋𝑋𝑋 2 + ℎ2 The projectile speed and direction are given by: 𝑉𝑉𝑉𝑉𝑦𝑦𝑦𝑦

Tan θ = 𝑉𝑉𝑉𝑉

V = √𝑉𝑉𝑉𝑉𝑥𝑥𝑥𝑥 2 + 𝑉𝑉𝑉𝑉𝑦𝑦𝑦𝑦 2

𝑥𝑥𝑥𝑥

Worked examples 1- A particle is projected upward with a velocity of 40 m/s at an angle of 300 to the horizontal. Find: a- Its time of flight b- The horizontal distance it travelled. c- Maximum height Given U = 40m/s, θ = 300, g = 10 m/s2 a- T =? b- X =? ch=? Solution Ux = U cos θ = 40 m/s (cos 300) = 40 m/s (0.866) = 34.64 m/s Uy = U sing θ = 40m/s (sin 300) = 40 m/s (0.5) = 20 m/s a- ℎ = 𝑈𝑈𝑈𝑈𝑦𝑦𝑦𝑦 𝑡𝑡𝑡𝑡 +

1 2

1

𝑔𝑔𝑔𝑔 𝑡𝑡𝑡𝑡 2

0 = 20 𝑡𝑡𝑡𝑡 + 2 (−10) 𝑡𝑡𝑡𝑡 2

[h = 0 when the projectile strike the ground]

0 = 20𝑡𝑡𝑡𝑡 – 5𝑡𝑡𝑡𝑡 2

0 = 5t (4 – t)

5t = 0 or u-t = 0 t = 0 or t = 4 t= 0 s at the start

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:. The time of flight is 4 seconds We can also find time of flight (total time) by using this formula Vy = Uy + 𝑔𝑔𝑔𝑔 t

and we double the

time we obtain.

First we find time for the vertical upward motion and remember that Vy is zero in this case.

Vy = Uy + 𝑔𝑔𝑔𝑔 t

0 = 20 m/s + (-10 m/s2)t 0 = 20 m/s –(10 m/s2)t

(10 m/s2)t = 20 m/s t=

(20 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠)/ 10 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠2

=2s

This time is the time required only for the half of the motion. Thus we multiply it with two to obtain time of flight. So, time of flight is 2 x2 s = 4 s

b-X = Ux t X = (34.64 m/s) (4s) X = 138.56 m 1

a- ℎ = 𝑈𝑈𝑈𝑈𝑦𝑦𝑦𝑦 𝑡𝑡𝑡𝑡 + 2 𝑔𝑔𝑔𝑔 𝑡𝑡𝑡𝑡 2

1 𝑔𝑔𝑔𝑔 𝑡𝑡𝑡𝑡 2 2 When Uy = 0 [we just consider downward vertical motion so we take t for 2 seconds and g for 10m/s2 1 ℎ = (10m/s2)(2 𝑠𝑠𝑠𝑠)2 2 h = 5m (4) h = 20 m ℎ=

2-An object is projected from point O with a speed of 40 m/s at 600 to the horizontal, 3 seconds after projection, find: a- The speed of the object b- The horizontal and vertical displacements of the object from O. c- The distance of P from O

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Chapter One Given U = 50 m/s, g = -10 m/s, t = 3s Solution Ux = U cos θ = 50 m/s (cos 600) = 50 m/s (0.5) = 25 m/s Uy = U sing θ = 50m/s (sin 600) = 50 m/s (0.866) = 43.3 m/s Vx = Ux = 25 m/s Vy = Uy +gt = 43.3 m/s + (-10 m/s2)(3s) = 43.3 m/s - 30 m/s = 13.3 m/s a- V = √𝑉𝑉𝑉𝑉𝑥𝑥𝑥𝑥 2 + 𝑉𝑉𝑉𝑉𝑦𝑦𝑦𝑦 2 = √(25 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠)2 + (13.3 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠)2

= √801.89𝑚𝑚𝑚𝑚2 /𝑠𝑠𝑠𝑠 2 = 28.3 m/s

= √625 𝑚𝑚𝑚𝑚2 /𝑠𝑠𝑠𝑠 2 + 176.89𝑚𝑚𝑚𝑚2 /𝑠𝑠𝑠𝑠 2

b- (i) X = Ux t = 25 m/s (3s) = 75 m

(𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 )

ℎ = 𝑈𝑈𝑈𝑈𝑦𝑦𝑦𝑦 𝑡𝑡𝑡𝑡 +

1 𝑔𝑔𝑔𝑔 𝑡𝑡𝑡𝑡 2 2

= 43.3(3) + 0.5(-10)(32) = 129.9 – 45 = 84.9 m c- d = √𝑋𝑋𝑋𝑋 2 + ℎ2

= √(75𝑚𝑚𝑚𝑚)2 + (84.9𝑚𝑚𝑚𝑚)2

= √5625𝑚𝑚𝑚𝑚2 + 7208.01𝑚𝑚𝑚𝑚2 = 113.28m

Exercise 1- John kicks a ball and the ball does projectile motion with an angle of 53º to horizontal. Its initial velocity is 10 m/s, find:

abcde-

Velocity of the projectile Horizontal and vertical velocity of the projectile The maximum height it can reach, Horizontal displacement of the projectile Distance of P from

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69

1.4 Circular motion Introduction

Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. At all instances, the object is moving tangent to the circle. Since the direction of the velocity vector is the same as the direction of the object's motion, the velocity vector is directed tangent to the circle as well. An object moving in a circle is accelerating. Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed. However, it is accelerating due to its change in direction. The direction of the acceleration is inwards. The final motion characteristic for an object undergoing uniform circular motion is the net force. The net force acting upon such an object is directed towards the center of the circle. The net force is said to be an inward or centripetal force. Without such an inward force, an object would continue in a straight line, never deviating from its direction. Yet, with the inward net force directed perpendicular to the velocity vector, the object is always changing its direction and undergoing an inward acceleration.

Figure 1: a) picture of rotating stone with string, b) free body diagram of the picture.

There are many examples of bodies moving in circular paths – rides at a funfair, clothes being spun dry in a washing machine, the planets going round the Sun and the Moon circling the Earth. When a car turns a corner it may follow an arc of a circle.

Centripetal acceleration An object is said to be moving in uniform circular motion when it maintains a constant speed while traveling in a circle. Remember that since acceleration is a vector quantity comprised of both magnitude and direction, objects can accelerate in any of these three ways: 1. Changing direction of velocity (linear acceleration); 2. Changing magnitude of velocity (centripetal acceleration); 3. Change in both magnitude and direction (angular acceleration).

70

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Chapter One In this lesson, we will be investigating centripetal acceleration and uniform circular motion that is, objects moving in circular paths at constant speeds. Centripetal acceleration is obtained by:

a=

𝒗𝒗𝒗𝒗𝟐𝟐𝟐𝟐 𝒓𝒓𝒓𝒓

Worked examples 1. A car is rounding a curve of radius 100m at a velocity of 2m/s. find the centripetal acceleration. Given V = 2m/s, s= 100m, m=1000kg, a =? Solution 𝒂𝒂𝒂𝒂 =

𝒗𝒗𝒗𝒗𝟐𝟐𝟐𝟐 𝒓𝒓𝒓𝒓

=

(𝟐𝟐𝟐𝟐𝒎𝒎𝒎𝒎/𝒔𝒔𝒔𝒔)𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝒎𝒎𝒎𝒎

= 0.04m/s2.

2. A car is moving with a constant velocity around a circular path. If the radius of the circular path is 50 m and the centripetal acceleration is 8 m/s2, what is the tangential speed of the car? Solution: Given that, r=50 m, a=8 m/s2 𝒂𝒂𝒂𝒂 =

𝒗𝒗𝒗𝒗𝟐𝟐𝟐𝟐 𝒓𝒓𝒓𝒓

.

𝑣𝑣𝑣𝑣 = √𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎.

V= √400𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚2 = 20m/s.

Centripetal force To make a body move in a circle requires a centripetal force. This force produces a centripetal acceleration which changes the velocity of the body continuously (even though its speed may be constant) as it moves round a circle because its direction is continuously changing. Centripetal force means force seeking center. Centripetal force increases if:    

The mass of the body increases The velocity of the object increases The radius of the circle decreases The radius of the circle decreases

................

71

Practical work: Investigating circular motion Use the apparatus in Figure 2 to investigate the various factors that affect circular motion. Make sure the rubber bung is tied securely to the string and that the area around you is clear to other students. The paper clip acts as an indicator to aid keeping the radius of the circular motion constant. Spin the rubber bung at a constant speed while adding more weights to the holder; it will be found that the radius of the orbit decreases. Show that if the rubber bung is spun faster, more weights must be added to the holder to keep the radius constant. Are these findings in agreement with the formula of centripetal force, F = mv2/r? Centripetal force is obtained by: F= ma from newton’s second law of motion F=

Figure 2:

where a = 𝒎𝒎𝒎𝒎𝒗𝒗𝒗𝒗𝟐𝟐𝟐𝟐

𝒗𝒗𝒗𝒗𝟐𝟐𝟐𝟐 𝒓𝒓𝒓𝒓

then force is given by

𝒓𝒓𝒓𝒓

Worked examples 1. Calculate the centripetal force of a 3.0 Kg mass tied in a rope and swung in a circle of radius 0.75 m, which is moving in a velocity of 4.0 m/s? Solution: The given parameters are m=3.0 Kg, v=4.0 m/s, r =0.75 m F=

𝒎𝒎𝒎𝒎𝒗𝒗𝒗𝒗𝟐𝟐𝟐𝟐 𝒓𝒓𝒓𝒓

=

𝟑𝟑𝟑𝟑.𝟎𝟎𝟎𝟎×𝟒𝟒𝟒𝟒.𝟎𝟎𝟎𝟎𝟐𝟐𝟐𝟐 𝟎𝟎𝟎𝟎.𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕

= 64N

2. A car of mass 1000 Kg moves in a circular path of radius 50 m. The constant speed of car is given by 12 m/s. Determine the centripetal force of the car? Solution From the question it is given that, m=1000 Kg, r=50 m, v=12 m/s F=

𝒎𝒎𝒎𝒎𝒗𝒗𝒗𝒗𝟐𝟐𝟐𝟐 𝒓𝒓𝒓𝒓

=

𝟏𝟏𝟏𝟏𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎×𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕

= 2880N

3. A car of mass 1000 Kg moves in a circular path of radius 50 m. The constant speed of car is given by 12 m/s. Determine the centripetal force of the car? Solution

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Chapter One From the question it is given that, m=1000 Kg, r=50 m, v=12 m/s F=

𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝟐𝟐𝟐𝟐 𝒓𝒓𝒓𝒓

=

𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏×𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝟓𝟓𝟓𝟓𝟏𝟏𝟏𝟏

= 2880N

4. A centripetal force of a bicycle and cyclist moving in a curve path with a radius of 200m is 60N, if the total mass is 120kg calculate the speed of the bicycle? Solution From the question it is given that, m=120 Kg, r=200 m, F=60N F=

𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝟐𝟐𝟐𝟐 𝒓𝒓𝒓𝒓

then V=√

Angular velocity

Fr 𝒎𝒎𝒎𝒎

=√

60×200 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏

= 10m/s.

The body moving from point A to B in a straight line (figure 3) has linear velocity. Linear velocity (v) is defined as the rate of change of linear displacement.

Figure 3

Linear velocity (v) =

𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥 𝐝𝐝𝐝𝐝𝐥𝐥𝐥𝐥𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐝𝐝𝐝𝐝𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐝𝐝𝐝𝐝 𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐭𝐭𝐭𝐭𝐥𝐥𝐥𝐥

=

𝐒𝐒𝐒𝐒 𝐝𝐝𝐝𝐝

A body moving from point A to B in circular motion (figure 4) has angular velocity (𝛚𝛚𝛚𝛚). Angular velocity is defined as the rate of change of angular displacement. Consider a particle moving along a circular path covering an arc of length AB in a time (t), (figure 4) the angular displacement of radius OA in 𝜃𝜃𝜃𝜃 in the same time (t). Hence Angular velocity (𝛚𝛚𝛚𝛚).=

𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥 𝐝𝐝𝐝𝐝𝐥𝐥𝐥𝐥𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐝𝐝𝐝𝐝𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐝𝐝𝐝𝐝

=

𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐭𝐭𝐭𝐭𝐥𝐥𝐥𝐥

𝜽𝜽𝜽𝜽 𝒕𝒕𝒕𝒕

Angular velocity is expressed in radian per second (rad/sec). Relationship between angular velocity and frequency

Figure 4:

For one complete circular motion 𝜽𝜽𝜽𝜽= 360o=2𝝅𝝅𝝅𝝅 radians and time taken t=T,(referred to as the periodic time). 𝜽𝜽𝜽𝜽

Hence 𝝎𝝎𝝎𝝎 = 𝒕𝒕𝒕𝒕 =

𝟐𝟐𝟐𝟐𝝅𝝅𝝅𝝅 𝑻𝑻𝑻𝑻

𝟏𝟏𝟏𝟏

Sin the frequency of revolution (f) = 𝑻𝑻𝑻𝑻, we have 𝜔𝜔𝜔𝜔= 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋. There fore Angular velocity (𝝎𝝎𝝎𝝎) = 2𝝅𝝅𝝅𝝅f

................

73

Relationship between the angular velocity and the linear speed We have seen that the arc length l =r 𝜽𝜽𝜽𝜽.

Where r = radius of curved path and 𝛉𝛉𝛉𝛉= angular displacement in radian.

Dividing both sides by t, we have, 𝐿𝐿𝐿𝐿 𝑟𝑟𝑟𝑟𝜽𝜽𝜽𝜽

𝐿𝐿𝐿𝐿

𝜃𝜃𝜃𝜃

= . But 𝑡𝑡𝑡𝑡 is the linear speed v of the rotating particles and 𝑡𝑡𝑡𝑡 is its angular velocity (𝝎𝝎𝝎𝝎), 𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡

therefore

Linear speed (v) = radius (r) × angular velocity (𝝎𝝎𝝎𝝎)

V=r 𝝎𝝎𝝎𝝎

Worked examples

1. Figure below shows the motion of second hand of clock. Calculate is angular velocity. Solution 𝛉𝛉𝛉𝛉

Angular velocity (𝛚𝛚𝛚𝛚) = 𝐭𝐭𝐭𝐭

The tip of a second hand takes 60 seconds to make one complete revolution i.e 2πrad. Therefore θ

2π

𝛚𝛚𝛚𝛚 = t = 60 = 0.105rad/s

2. A bicycle wheel makes 300 revolutions per minute (rpm). Calculate the angular velocity of the wheel. Solution The wheel makes 300 revolutions in 1 minute. There in each second the wheel makes

300 60

= 5 revoultions

1 revolution= 2πrad 5 revoultions = 5x2πrad = 10πrad θ

𝛚𝛚𝛚𝛚 = t =

10π 1

= 31.4rad/min or 0.523rad/s

 The angular velocity is 31.4rad/m or 0.523rad/s 3. Calculate the angular velocity of the earth when it is rotating about its own axis.(time period of the earth about its own axis =24hours). Solution The earth takes 24 hours to rotate once about its axis. Hence the angular velocity @ is given by, θ

2π

2π

𝛚𝛚𝛚𝛚 = = = = 86400 = 7.3×10-5 rad/s t (24×60×60)

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Chapter One 4. A ball is tied to a string is rotated at uniform speed in a circle of radius 10cm. it takes 1.5 second to describe an arc of length 6cm. calculate a) Linear speed b) Angular velocity c) Periodic time Solution a) Linear speed (v) =

arc lenght (l) time (t)

=

6cm 1.5s

= 4.0cm/s 𝐕𝐕𝐕𝐕

b) Since linear speed (v) =r 𝛚𝛚𝛚𝛚then 𝛚𝛚𝛚𝛚 = = c) 𝛚𝛚𝛚𝛚 =

𝟐𝟐𝟐𝟐𝛑𝛑𝛑𝛑 𝐓𝐓𝐓𝐓

then 𝐓𝐓𝐓𝐓=

𝟐𝟐𝟐𝟐𝛑𝛑𝛑𝛑 𝛚𝛚𝛚𝛚

=

𝟐𝟐𝟐𝟐𝛑𝛑𝛑𝛑

𝟎𝟎𝟎𝟎.𝟒𝟒𝟒𝟒

= 15.7s

𝐫𝐫𝐫𝐫

𝟒𝟒𝟒𝟒.𝟎𝟎𝟎𝟎 𝟏𝟏𝟏𝟏𝟎𝟎𝟎𝟎

= 𝟎𝟎𝟎𝟎. 𝟒𝟒𝟒𝟒𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫/𝐬𝐬𝐬𝐬

Centripetal force in terms of angular velocity We learnt that centripetal force is given by F=

𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝟐𝟐𝟐𝟐 𝒓𝒓𝒓𝒓

and also V=r 𝛚𝛚𝛚𝛚therefore

F=

𝟐𝟐𝟐𝟐

𝒎𝒎𝒎𝒎(𝐫𝐫𝐫𝐫 𝛚𝛚𝛚𝛚) 𝒓𝒓𝒓𝒓

Worked example

𝒎𝒎𝒎𝒎𝐫𝐫𝐫𝐫𝟐𝟐𝟐𝟐 𝛚𝛚𝛚𝛚𝟐𝟐𝟐𝟐

=

𝒓𝒓𝒓𝒓

= mr ω2

F = mr 𝛚𝛚𝛚𝛚2

1. A 0.5kg ball is tied to a string is rotated at uniform speed in a circle of radius 20cm. it takes 2 second to describe a linear velocity of 50cm/s. calculate a) Angular velocity b) Centripetal force

Solution

Given: m=0.5kg, r = 20cm, v=50m/s 𝑽𝑽𝑽𝑽

a) V=r 𝛚𝛚𝛚𝛚 then 𝛚𝛚𝛚𝛚= 𝒓𝒓𝒓𝒓 = 𝟓𝟓𝟓𝟓𝟎𝟎𝟎𝟎 𝟐𝟐𝟐𝟐𝟎𝟎𝟎𝟎

=2.5rad/s

b) F = mr ω2 = 0.5× 0.2 × 2. = 0.625

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75

Application of uniform circular motion A car negotiating a circular path on a level of horizontal road When a car rounds a bend, a frictional force is exerted inwards by the road on the car’s tyres, so providing the centripetal force needed to keep it in its curved path (Figure 5 a). Here friction acts as an accelerating force (towards the center of the circle) rather than a retarding force. The successful negotiation of a bend on a flat road, therefore, depends on the tyres and the road surface being in a condition that enables them to provide a sufficiently large frictional force – otherwise skidding occurs. Safe cornering that does not rely entirely on friction is achieved by ‘banking’ the road as in Figure 5 Figure 5 b. Some of the centripetal force is then supplied by the part of the contact force N, from the road surface on the car, that acts horizontally. A bend in a railway track is banked, so that the outer rail is not strained by having to supply the centripetal force by pushing inwards on the wheel flanges. 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝟐𝟐𝟐𝟐

F= from this formula it can be calculated form the 𝒓𝒓𝒓𝒓 maximum safe speed for the motorists not to skis off the track. 𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟

𝐕𝐕𝐕𝐕𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 = √𝐦𝐦𝐦𝐦 . i.

ii.

where 𝑽𝑽𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 = maximum safe speed.

Banked tracks In order that a motorists does not fully depend on the friction force between the tyres and the road, circular paths are given a small banking angle, 𝜃𝜃𝜃𝜃 i.e. The outer edge of the road raised a little above the Figure 6 inner side so that the track is slopping towards the center of the curve. An air craft in a circular turn When an air craft takes a turn in horizontal plane, it must make a correct banking angle in midair as shown in figure 7, in order successfully negotiate the curved path.

Figure 7

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Chapter One Newton’s Law of universal gravitation Newton’s law of universal gravitation states:

“The force of attraction between two objects is directly proportional to the product of the two masses of the objects and inversely proportional to the square of the distance between them"

𝐅𝐅𝐅𝐅 =

𝐆𝐆𝐆𝐆𝐦𝐦𝐦𝐦𝟏𝟏𝟏𝟏 𝐦𝐦𝐦𝐦𝟐𝟐𝟐𝟐 𝐝𝐝𝐝𝐝𝟐𝟐𝟐𝟐

Where: G = Universal gravitation constant (6.67x10-11Nm2/kg2) F = force of gravity m1 and m2= the two masses d = distance between the centers of the two masses Worked examples 1- Calculate the gravitational attraction of two cars 5m apart if their masses are 1000kg and 1200kg

Given

m1, and m2 = 11kg and 12000kg respectively S = 5m, G = 6.67x10-11Nm2/kg2, F =?

Solution

Gm1 m2 S2 (6.67x10−11 NM2 /𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘2 )(1000𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘)(1200𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘) F= (5m)2 F= F=

(8004000x10−11 N 25

F= 3.2016 x10−6 N

= 320160x10−11 N

2- The mass of the moon is about 7.3x1022 kg and the mass of earth is 6x1024 kg. if the center of the two are 3.9 x108 m apart, what is the gravitational force between them? Given m1 = 7.3x1022 kg m2 = 6x1024 kg d = 3.9 x108 m Solution Gm1 m2 F= S2 F= F=

(6.67x10−11 )(7.3x1022 )(6x1024 ) (3.9x108 )2

292.146𝑥𝑥𝑥𝑥1035 15.21x1016

F= 19.2 x1019 N

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F= 1.92 x1020

Summary  Angular displacement of a particle is the angle swept through by the radius joining the particles to the center of circle.  Angular displacement is measured in radians.  2𝜋𝜋𝜋𝜋 radian = 360o.  The length of an arc of a circle is equal to r𝜃𝜃𝜃𝜃, where 𝜃𝜃𝜃𝜃 is in the radian measure.  Angular velocity is the rate of change of angular displacement, it is measured in radian per second.    

For a body in circular motion angular velocity 𝜔𝜔𝜔𝜔 =

𝟐𝟐𝟐𝟐𝝅𝝅𝝅𝝅 𝑻𝑻𝑻𝑻

= 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋.

Linear speed (V) = radius (r) × angular velocity (𝜔𝜔𝜔𝜔) For a body to under go a circular motion, there should be a force acting towards the center along the radius of the circle. Centripetal force is the force which acts towards the center of the circle an keeps body in a circular path. 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝟐𝟐𝟐𝟐

= mr ω2.



Centripetal force (F) =



Centripetal acceleration (a) =

Exercise

𝒓𝒓𝒓𝒓

𝒎𝒎𝒎𝒎𝟐𝟐𝟐𝟐 𝒓𝒓𝒓𝒓

= r ω2.

1. Define centripetal force and write the equation for centripetal force. 2. A student makes the following statement “a body moving with uniform speed is acceleration” justify the statement. 3. A satellite orbits the earth every 4 hours. Calculate a) The angular velocity of the satellite b) The centripetal acceleration of the satellite, if the radius of the satellite’s orbiting is 12800km, c) The linear speed of the satellite. 4. A 5kg mass moves at uniform speed of 18m/s in a circular path of radius 0.5m. Calculate the centripetal force acting on the mass. 5. A satellite moves round the earth in a circular orbit of radius 8000km and takes 84 minutes to revolve around the earth ones. Calculate a) The angular velocity b) The centripetal acceleration 6. A particle revolves at 2Hz in a circle of radius 2m. calculate a) Linear speed b) Angular velocity 7. Calculate the gravitation attraction between two masses of 400 kg and 800 kg separated by a 8m.

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Chapter One 1.5 Work, Energy and Power Doing work Work is done on an object when an applied force moves it through a distance. To do work 

A force must be exerted on to an object.



The object must move as a result of this force.

Work done (W) = force (F) x distance moved in the direction of the force (S). The unit of work is joule (J). 1J = 1Nm. Work is a scalar quantity. When the applied force and the direction of motion are in a same direction as shown below,

Work done can be found by:

Figure 2.5.1: You have to start the car moving

W = FS Work done by force acting at an angle to the direction of motion. When there is an angle “Ø” between the applied force and the direction of motion the component of the force which is same direction as the direction of motion is doing the work. Therefore we must calculate the components of the force in the direction of motion and then use that to work out the work done. For a force F acting at angle θ to the direction of motion: W = Fcosθs The object travels at distances parallel to the ground. Force in the direction of travel = Fcosθ. So the work done = FScosθ.

Figure 1.5.2: Force acting on an object at angle

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Work done against or by the gravity 

If a body falls from the top of a tower, work is done by the force of the gravity (weight).



Also for an object of mass “m” lifted vertically to a height “h” with a uniform velocity the minimum upward force used equals to its weight.



Applied force = gravitational force = weight of the object. Therefore W = FS where F = weight = mg And s = vertical height h

Therefore W = mgh Work done and Energy transferred Doing work is a way of transferring energy. The amount of work done, calculated using W = Fs tells us the amountof energy transferred: Work done = energy transferred = Fs Since work done = energy transferred: A joule is also the amount of energy transferred when a force of 1Newton moves a distance of 1m in the direction of force, Example: 1) A car of mass 600kg traveling at 20m/s is brought to a rest in 2 seconds. If the engine of a car produces a force of 400N. Calculate the work done by the engine? Answer m = 600kg W = FS W = Fs -1 S =u + v t = 400 x 20 u = 20m/s -1 v = 0m/s 2 = 8000j t = 2sec = (20 + 0) 2 =20m F = 400N 2 2) If a force of 50N is used to pull a box along the ground a distance of 8m and the box moves in the same direction as the force calculate the work done by the force Solution W = Fs = 50N x 8m = 400Nm = 400j 3) A horizontal force of 80N is needed to push an object across a horizontal floor through a distance of 40m. How much work is done? W = Fs = 80N x 40m = 3200Nm = 3200J 4) An automobile is pushed with a horizontal force of 60N at a velocity of 3m/s on a level road. How much work is done in 2 seconds? W = Fs = F (V x t) = 60 N x 3 m/s x 2s = 360Nm = 360J

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Chapter One 5) A force of 50 N acts on the block at the angle shown in the figure 1.5.3. The block moves a horizontal distance of 3.0 m. Calculate the work being done by the force?

Figure 1.5.3

W = FS cos = 60 N x 3m cos300= 155.88Nm = 155.88J 6) A stone of mass 0.5kg rolls 50m down a slope of height 30m as shown in the figure. What is the work done by the force of the gravity? (g = 10m/s2).

Figure 1.5.4: A stone sliding down an inclined plane

Note: the distance moved by the stone down the slope is 50m, but distance moved in the direction of the force is 30m. Therefore, W = mgh = (0.5kg) (10m/s2) (30m) = 150Nm = 150J Exercise 1.5.1 1. Farah runs up a staircase of 39 steps. Each step is 9 cm in height. Given that student’s

mass is 60 kg the acceleration due to gravity is 10m/s2, find the work done by student to reach the top of the staircase? 2. A cart load of sand is pulled 12 m across the ground as shown figure 1.5.5 below. The

tension in the rope is 500 N and is directed 30 degrees above the horizontal. How much work is done in pulling the load?

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Figure 1.5.5.

3. A man of mass 84 kg climbs stairs of height 4.5m how much work does he do against

the force of gravity? (take g = 9.8ms-2) 4. A stone of weight 20 N falls from the top of a 25 m high cliff a) How much work is done by the force of gravity b) How much energy is transferred to the stone

Gravitational potential energy If you lift a heavy object, you do work. You are providing an upward force to overcome the down ward force of gravity on the object: your force moves the object upwards, so the force is doing work. In this way energy is transferred from you to the object. You lose energy, and the object gains energy. We say that its gravitational potential energy has increased. The work done is equal to the gravitational potential energy gained by the object; Work done and gravitational potential energy Imagine an object of mass “m” which is raised from a height “h” above the ground to a greater height “H” as shown in the figure below:

Figure 1.5.6: Work done and GPE

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Chapter One At the initial position, GPE = mgh

and at the final position GPE =

mgH Therefore ∆GPE = mgH – mgh = mg (H – h) The value of the “mg” is the force needed to raise the object, because “F = ma” and a = g ∆GPE = F (H – h) “H – h” is the distance moved by the object, therefore we can say Work done = change in gravitational potential energy. For example: when you lift a 10 N weight ( a mass of 1 kg) from the floor to a high shelf, a height difference of 2 m, you have done Figure 1.5.7: Doing work increases GPE work on the weight. The work done = 10 N x 2 m = 20 J and this is equal to the increase in GPE of the weight as shown in figure 1.5.7.

Worked examples 1) A weight lifter raises weight weights with a mass o f200kg form the ground to a height of 1.5m as figure below shows: how much work does he do? By how much does the GPE of the weight s increase? Solution Figure  The down ward force on the weight is their weight W = mg. An equal, upward force F is required to lift them. W = F = mg = 200kg x 9.8N/kg = 1960N Now we can calculate the work done by the force F Work done = Force x distance moved = 1960N x 1.5m = 240j (Note that the distance moved is in the same direction ass the force) so the work done on the weights = GPE increase in the weights GPE increase of the weights = 2940j A Formula for GPE The change in the GPE of an object ∆GPE depends on the Change in its height ∆h. We can calculate ∆GPE as illustrated above using the equation.

∆GPE = mg∆h

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Note! The force needed to lift an object is equal to its weight mg. work done by this force is given by: Force x distance moved or weight x change in height. Work done against gravity = mgh = (40kg x 10m/s2 12 m = 4800J

Figure 1.5.9

Exercise 1. A man pushes an object of mass 10kg with a constant force of 80 N at an angle of 450 to the horizontal. So 800 J of work is done. How far does he push the object? 2. A man applied a constant horizontal force of 65 N to a block initially on a horizontal surface. The block reaches a velocity of 12 m/s after being pushed for 3 seconds. Calculate the work done by the man? Kinetic energy: Kinetic energy is the energy of a moving body. It is the ability that an object has to do work as a result of its motion. Kinetic energy can be found by: KE = ½ mv2 KE = kinetic energy, m = mass and V = velocity. Therefore the kinetic energy of an object depends on the velocity of the object and it’s mass. The greater the mass and the velocity of an object, the higher its kinetic energy 

A body moving with speed V possesses kinetic energy. The value of KE is equal to the amount of work needed to bring that body from rest to a speed V. and expression for its value can be found as follows.



Consider a body of mass m which starts from rest and reaches a speed V after moving through the distance S under the action of a constant force F in figure 1.5.10.

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Chapter One

Figure 1.5.10

Kinetic energy is the energy a body has because of its motion. Suppose a body of mass m starts from rest and is acted on by a steady force F which gives it a uniform acceleration a. if the velocity of the body is v when it has travelled a distance s, then using V2= u2 + 2as we 𝑣𝑣𝑣𝑣 2

get since u=0 v2 = 2as or 𝑠𝑠𝑠𝑠 = 2𝑎𝑎𝑎𝑎

F= ma the work done by this force is w W= fs by substituting f= ma and s = v2/2a 𝐯𝐯𝐯𝐯 𝟐𝟐𝟐𝟐

W = 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 ( ) 𝟐𝟐𝟐𝟐𝐚𝐚𝐚𝐚

w = ½ mv2

1

but this work done equals kinetic energy ∴ 𝑘𝑘𝑘𝑘. 𝑒𝑒𝑒𝑒. = 2 𝑚𝑚𝑚𝑚𝑣𝑣𝑣𝑣 2

 You can do work on a body to change its KE. In fact the work done on a body is equal to its change in KE. KE gained by an object of mass m = work done the force F giving an acceleration. 

Generally, the force acting on a body either increases or decreases its KE, depending on the direction of the force. Note: - Work done by the force = KE gained or lost by the body = Final KE – initial KE = 1/2 mv2 - 1/2mu2

Examples 1. What is the kinetic energy of 1200kg car moving with constant velocity of 20m/s? KE = ½ mv2 = ½ 1200 x 20m/s2 = 240000J 2. A satellite with a mass of 1000kg orbits the earth. If the kinetic energy of the satellite is 9.8 x109J. Calculate the speed of the satellite? 2(9.8 𝑥𝑥𝑥𝑥 109 )𝐽𝐽𝐽𝐽 2𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾 = √ = 4.4 𝑥𝑥𝑥𝑥103 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠 𝑉𝑉𝑉𝑉 = √ 1000𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚 Exercise 1) Calculate the change in KE of a ball of mass 800g when it bounces. Assume that it hits the ground with a speed of 20 m/s and leaves it at 18.5 m/s? 2) A body of mass 14kg starts from rest. Find its KE after travelling through a distance of 8 m with an uniform acceleration of 3 m/s2.

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3) A body of mass 10 kg is pulled from rest with a constant force of 40 N. The force is applied for 2 sec. calculate (a) The distance travelled (b) Work done on the body (c) The KE of the body (d) The final velocity of the body 4) Calculate the kinetic energy of car of mass 50 kg that has a velocity of 108 km/h? 5) What is the velocity of an object of mass 2.5 kg that has 850 J of kinetic energy? 6) A particle of mass 100 g is thrown straight up with a velocity of 8 m/s. what potential energy did the object gain at its maximum height? 7) Calculate the KE. of a a) 2 kg trolley travelling at 1.5 m/s, 8) Calculate the PE of a 5 kg mass when it is a) 13 m, above the ground. (g = 10 N/kg) b) 6.8 m, above the ground. (g = 10 N/kg) 9) Calculate the increase in kinetic energy of a car of mass 500 kg when it accelerated from 20m/s to 30m/s? Further Examples 1. A car is accelerating from rest at constant acceleration of 2m/s2 for 20s. If the car has a mass of 1000kg. a) Find the work done by the engine of the car? b) What is the kinetic energy of the car after 20s? c) What is the velocity of the car after 20s? a) F = ma = (1000kg)(2m/s2) = 2000N and S = ut + ½ at2 = 0 + ½ (2m/s2)(20)2 = 400m Therefore W = FS = (2000N) (400m) = 800,000J = 800kJ b) Work done = energy transferred = 800,000J = 800kJ c) V = u + at = 0 + (2m/s2)(20s) = 40m/s Another way of finding the velocity is by using the formula of the kinetic energy. 2𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾 2(800000)𝐽𝐽𝐽𝐽 𝑉𝑉𝑉𝑉 = √ = √ = 40 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚 1000𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘

2. A stone of weight 10N falls from the top of 250m high cliff.

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Chapter One a) How much work is done by the force of gravity by pulling the stone to the foot of the cliff? b) How much energy is transferred? a) W = FS = (10N)(250m) = 2500J b) Energy transferred = work done = 2500J. Exercise 1. A force of 15N acts on brick of mass 2 kg moving it 10 m horizontally from rest. Find a) The work done by the force? b) The kinetic energy gained by the object? c) The velocity of the object 2. A 20 kg object is raised 60 m to the platform shown. (take g = 10m/s2)

Figure 1.5.11

a) Calculate the gravitational potential energy of the object at the top of the plat form?

Figure 1.5.12: A ball falling from a certain height

b) If the object slides down the inclination of the platform, calculate the kinetic energy of the object at the bottom? c) What is the speed of the object as it reaches to the bottom of the platform? Gravitational potential energy and kinetic energy Transformation When an object falls, it speeds up. Its GPE decreases and its KE increases. Energy is being transformed from GPE to the KE. If no energy lost in the process, Decrease in GPE = gain in KE

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For example Let as consider a ball that is allowed to fall freely towards the ground. Before it is dropped, it has GPE equal to mgh. As it falls, it’s GPE decreases but its KE increases. Just before hitting the ground, it has only KE. Infect, its KE at the bottom is exactly equal to the GPE it had at the top. We can prove this by using this equation: since v = 0 and a = g

we have v2 = 2gh

whereh is the distance fallen. Thus the KE after falling a distance h is ½ mv2= ½ m (2gh) = mgh Another example: Suppose a ball of mass ‘m’ is dropped from a height ‘h’. Total mechanical energy (ET) = Kinetic energy (K.E.) + Gravitational Potential energy (P.E.) At point A: The body is at rest so, P.E. = mgh and K.E. = 0 because velocity at A is zero. So, Total mechanical energy (ET) at A = 0+mgh = mgh At point B: The body hits the ground with velocity, v so, P.E = 0 as height is zero and K.E. = ½mv2. Also initial velocity, u =0, ad g is positive. Using the equation v2 – u2 =2gh we get v2 = 2gh So, Total mechanical energy (ET) at B=½mv2 +0 = ½m (2gh) = mgh This example is an illustration of the law of conservation of energy: The total energy is neither increased nor decreased in any process. Also if the ball is thrown upwards, it slows down, its KE decreases while its GPE increases. Energy is being transformed from KE into a GPE Decrease in KE = Increase in GPE The GPE on the top point A equals the KE which is thrown first from point B But the total mechanical energy is constant. The law of conservation of energy states: “Energy cannot be created or destroyed; it may only be transferred from one form to another but the total

................. Figure 1.5.13: The total mechanical energy is constant

88

Chapter One amount of energy never changes.” For example, there is a continuous transformation of kinetic energy and potential energy as shown in the figure 1.5.14 below:

Figure 1.5.14: GPE and KE transformation

In a same way, when an abject falls it speeds up, so its kinetic energy increases and its potential energy decreases since its height decreases. Energy is transformed from a gravitational potential energy to kinetic energy, so if there is no energy lost Decrease in GPE = increase in KE Worked Examples  A pendulum consists of a brass sphere of mass, 5kg hanging along string as shown left. The sphere is pulled to the side so that it is 0.15m above its lowest position. It is then released. How fast will it be moving when it passes through the lowest point along its path see figure 1.5.15. ?

Solution Step1:- calculate the loss in GPE as the sphere falls from its height position ∆GPE = mg∆h = 5 x 9.8-0.15 = 7.35j Step2: - this tells us that the gain in the sphere’s KE is 7.35j We can use this to calculate the sphere’s velocity KE = ½ mv2 V2 = 2 x KE = 2 x 7.35 = 2.94 m 5 V= 2.94 = 1.71m/s

Figure 1.5.15

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3. A cyclist of mass 60kg is moving from the top of Small Mountain of 10m high to the foot of the mountain as shown in the figure 1.5.16 shown below. (Take g = 10m/s2).

Figure

1.5.16: A cyclist rides over a bicycle

Calculate a) The potential energy of the cyclist at top of the mountain? b) The kinetic energy of the cyclist at the foot of the mountain? c) The speed of the cyclist at the foot of the mountain? a) PE = mgh = 60 kg x 10m/s2 x10 m = 6000J b) Increase in KE = decrease in PE = 6000J c) 𝑉𝑉𝑉𝑉 = √

2𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾 𝑚𝑚𝑚𝑚

= √

2(6000)𝐽𝐽𝐽𝐽 60𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘

= 14.14 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠

4. A boulder of mass 4kg rolls over a cliff and reaches the beach below with a velocity of 20m/s. (take g = 10m/s2) a) What is kinetic of the boulder just before it lands? b) What is its potential energy on the cliff? c) How high is the cliff? a) KE = ½ mv2 = ½ (4kg) (20m/s) 2 = 800J b) ∆PE = ∆KE therefore potential energy at the top of the cliff is 800J. c) h = PE/mg = 800J/(4kg x 10m/s2) = 20m

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Chapter One Exercise 1. A block of mass 2 kg is dropped from a height of 25 m given that acceleration due to gravity is 10m/s2. Calculate a) Potential energy at this height? b) The kinetic energy as it hits the ground? c) The speed of the object when it hits the ground? d) Work done by the object as it falls from the ground? 2. A 200g steel ball falls from a height of 1.8m onto a metal plate and rebounds to a height of 1.5 m. find a) The PE of the ball before the fall (g = 10m/s2)? b) Its KE as it hits the plate? c) Its velocity on hitting the plate? d) Its velocity of the rebound

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Power Power is defined as the rate of doing work or the rate of energy transfer. Power measures how fast work is done or how fast energy is transferred. Power (P) = work done (W)/time taken (t) Power is a scalar quantity. The SI unit of power is watt (W).

1W = 1 J/s

When a constant force “F” moves an object through a distance “S” with an average velocity “Vav” P = W/t = FS/t We know that S/t = Vav therefore P = Fav At first, power is measured in horse power. One horse power = 750W. Examples: 1- Muna weighs 500N runs a flight of stairs in 5s.

Figure 1.5.17: Muna runs a flight of stair

What is the rate of doing work against gravity if the total height of the stairs is 4m? P = W/t = FS/t =

500N x 4m 5s

= 400W

2- A motor of a lifter provides a force of 20kN (20,000N). How high the motor moves in 10 seconds if the power of the motor is 40kW (40,000W)? W = P x t = (40,000W) (10s) = 400,000J Distance moved = W/F = 400,000J/20,000N = 20m.

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Chapter One Exercise 1. Calculate how much work is done by a 50kW car engine in a time of 1.0minute? 2. A 250.-kilogram car is initially at rest at point A on a roller coaster track. The car carries a 75-kilogram passenger and is 20. meters above the ground at point A. [Neglect friction.]

Figure 1.5.18

a) Calculate the total gravitational potential energy, relative to the ground, of the car and the passenger at point A. b) Calculate the speed of the car and passenger at point B c) Compare the total mechanical energy of the car and passenger at points A, B,and C. 3. A climber of mass 60 kg is attached by a rope to point A on a rock face. She climbs up to point B in 20 seconds. Point B is 3.2 m vertically above point A.

Figure 1.5.19.

(a)(i) Calculate the average speed of the climber between A and B. (ii) Calculate the weight of the climber. (iii) Calculate her gain in potential energy. (b) She then loses her footing and free falls from point B. After passing point A she is held safely by the rope. Calculate her speed as she passes point A

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1.6 Deforming solids Introduction

One of the effects of the force is the deformation of solids. Deformation is a change in the shape or size of an object due to an applied force. There are several different deformations of solids and some of them are:  Elastic deformation This type of deformation is reversible. Once the forces are no longer applied, the object returns to its original shape. E.g rubber exhibit elastic deformation.  Plastic deformation This type of deformation is irreversible and mostly done beyond the elastic deformations. E.g. plastic bags.  Fracture This type of deformation is also irreversible. A break occurs after the material has reached the end of the elastic. The forces that cause the deformation of solids may be compressive or tensile. If the forces applied to an object caused squeezing, the forces are said to be compressive and if the forces cause an object to stretch, they are said to be tensile forces.

Stiffness is the property that solids resist the forces that cause to change their shape and size. In other words, Stiffness is the rigidity of an object. Stretching a spring A spring is one of elastic materials. If a small mass is hung on a vertically suspended spring it will stretch down. When you remove the mass the spring regain its shape. But if a large mass is hung on the spring, the spring may lose its elasticity. As a result when you remove the big mass the spring will not regain its shape, because it is extended beyond its elastic limit. Verifying Hook’s law The experiment below made by candlelight secondary school students verifies hook’s law. Apparatus: clamp, retort stand, spring, different masses and ruler Procedure 1. Setup apparatus as shown diagram 1.6.2 2. Measure natural length of spring (with no load) using mm ruler 3. Add one mass at a time, allow spring to come to rest and measure extension (CHANGE IN LENGTH from original to stretched) 4. Do this six times for enough measurements 5. Repeat experiment for reliability and calculate average value for each weight

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Chapter One

Figure 3.6.2

(a)

(b)

It is important to record all of the readings taken and we plot the result a graph of load against extension. On plotting a graph between the load and extension, one gets a straight line as shown figure 1.6.2 (b). The first section of this graph OE is a straight line through the origin. The extension x is directly proportional to the applied force (load) F. the behavior of the spring in the linear region OE of the graph can be expressed by the following equation: X∝F F = kx Where k is the force constant of the spring (sometimes called either the stiffness or the spring constant of the spring). The spring constant k is the force per unit extension. Or it is a force which extends the spring by one meter, The spring constant k of the spring is given by the equation F

k=x

The SI unit for the force constant(or spring constant) k is Newton per meter or N/m. we can find the spring constant k from the gradient of section OE of the graph. 1

k = gradient OE

The spring constant ‘K’ is a measure of how stiff the spring is or how much force would stretch the spring by one meter. A stiffer spring will have a large value for the spring constant k. Beyond A the spring is no longer straight line. This is because the spring has become permanently deformed. It has been stretched beyond its elastic limit. This is summarized by Hooks law Hook’s law states that extension of spring is directly proportional to the stretching force provided the elastic limit is not exceeded. Notice thatOE the spring has elastic behavior if a force is removed it springs back its original form. If the load is increased further, it stretches beyond A and changes its behavior, now it has a plastic behavior. Figure 1.6.3

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Factors affecting spring constant  Types of material used to make the spring:Different materials have different spring constant and stretch differently, for example steel is stiffer than copper.  length of the spring The shorter the spring the stiffer it is and therefore the greater the spring constant.  Number of turns per unit length The more the number of turns per unit length the stiffer the spring  Diameter of the spring A wider spring is less stiff than a spring with a small diameter.  Thickness of the wire used to make the spring A thicker wire is stiffer than a thin wire. Example: 1 When a load of 12 N is applied to a steel spring it produces an extension of 80 mm without exceeding the elastic limit of the spring. Calculate: a) The spring constant of the spring b) The weight of the object that produces an extension of 60 mm. Solution a) F = 12 N b) W = F = kx K = 30 mm =0.08 m = 150N/m x 0.06 m K =? = 9N K = F/X = 12 N/0.08 mm = 150 N/m 2. A spring extends 50mm when applied by a force of 5N. if the spring obeys Hooke’s law a) Calculate the spring constant? b) What is the extension when a force of 7N is applied to the spring? c) What force is needed to get an extension of 100mm? a. k = F/x = 5N/0.05m = 100N/m. b.

x = F/k = 7N/100Nm-1 = 0.07m = 70mm

c. F = kx = (100N/m)(0.1m) = 10N 3. A 2.5kg mass is hung on a vertical spring with k = 50N/m.

Figure 1.6.4

By how much is the spring extended? (g = 10N/kg)

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Chapter One Applied force = weight of the object W = mg = (2.5kg) (10N/kg) = 25N Extension = F/k = 25N/50Nm-1 = 0.5m.

Springs in series and springs in parallel A system of springs can be arranged in series or in parallel.

Figure 1.6.5

When the springs are arranged end to end, they are called spring in series. Springs in series are easy to stretch, this means when the springs are arranged in series their stiffness decreases and can be stretched easily. The combined spring constant of number of springs arranged in series can be found by 1 1 1 1 1 = + + ………………………… 𝑘𝑘𝑘𝑘1 𝑘𝑘𝑘𝑘2 𝑘𝑘𝑘𝑘3 𝑘𝑘𝑘𝑘𝑛𝑛𝑛𝑛 𝑘𝑘𝑘𝑘𝑇𝑇𝑇𝑇

If the springs are arranged side by side, they are said to be in parallel. Springs in parallel are difficult to stretch, this means their stiffness increases and it cannot be stretched easily. The combined spring constant of a number of springs in parallel can be found by Examples

𝑘𝑘𝑘𝑘𝑇𝑇𝑇𝑇 = 𝑘𝑘𝑘𝑘1 + 𝑘𝑘𝑘𝑘2 + 𝑘𝑘𝑘𝑘3 + … … … … … … … 𝑘𝑘𝑘𝑘𝑛𝑛𝑛𝑛

1) a spring extends 10mm when applied by a force of 1N. a) Calculate the spring constant of the spring?

b) What will be the extension a force of 4N is applied to it? c) If the spring and another identical spring are arranged in series i)

What is their combined spring constant?

ii)

What will be the extension when a force of 4N is applied to it?

d) If the spring and other identical spring is arranged in parallel i)

What is their combined spring constant?

ii)

What will be the extension when a force of 4N is applied to it?

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Answers k = F/x = 1N/0.01m = 100N/m a) x = F/k = 4N/100Nm-1 = 0.04m = 40mm. b) i)

1

𝑘𝑘𝑘𝑘𝑇𝑇𝑇𝑇

=

1

𝑘𝑘𝑘𝑘1

+

1

𝑘𝑘𝑘𝑘2

therefore KT = (k1k2)/(k1 +k2) = (100 x 100)/(100 + 100) = 10,000/200 = 50N/m

-1

ii) x = F/k = 4N/50Nm = 0.08m = 80mm. i) kT = k1 + k2 = 100N/m +100N/m = 200N/m ii) x = F/k = 4N/200Nm-1 = 0.02m = 20mm. 2) Two springs each with a force constant of 300N/m set up in parallel with an applied force of 30N. What would be the extension? X = F/kT = (30N)/ (300N/m + 300N/m) = 30N/ (600N/m) = 0.05m = 50mm. Exercise

1. The figure below the part of the force against extension graph for a spring. The spring obeys Hooke’s Law for forces up to 5.0N.

Figure 1.6.6. Force-extension graph

a) Use your graph to answer in the following question?

98

i.

What is the extension when the force is 2N?

ii.

Calculate the spring constant of the spring by using your answers in “a”?

iii.

Calculate the extension produced by a force of 5 N?

.................

Chapter One b) The spring that its force extension graph is shown above and another identical spring are arranged as follows and a force of 5N is applied together.

Figure 1.6.7 Springs in parallel

i.

Calculate their combined spring constant?

ii.

Calculate the combined extension caused by the force of 5N?

2. a)Fill in : Hooke’s law states that the …………………… is directly proportional to the ………………………….. unless the elastic limit is exceeded. b) Label the following setup used to investigate Hooke’s law.

Figure 1.6.8

c) After carrying out the investigation, the following results are obtained:

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99

(i) The length of the unloaded spring is ……………………………. mm. (ii) Complete the missing values by calculating the extension in each case. (iii) Continue plotting the graph of Extension (mm) on the y-axis against Load (N) on the axis. Write the title of the graph and label each axis.

Figure 1.6.9

3. a) Hooke’s law states that a force acting on a material produces an extension which is ............................................................. to the force. (b) A student attaches a load to the end of a spring.

Figure 1.6.10

(i)

100

Name the type of force acting in the stretched spring.

.................

Chapter One (ii) The student measured the length of the spring for different loads. The table shows her data.

1. Deduce the load in newtons that would produce a length of 130 mm. Load = ................................. N 2. Estimate the maximum load in newtons at which the spring obeys Hooke’s Law. Maximum load = .................................. N (c) A force–extension graph for a material is shown in figure 1.6.11. Three regions A, B and C are labelled.

Figure 1.6.11

(i) In which region is Hooke’s law obeyed? (ii) In which region is the material easiest to extend? (iii) Explain your answer to (ii).

4. The flowing results were obtained when a spring was stretched.

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Load N Length of spring cm

1.0 12.0

3.0 15.5

4.5 19.0

6.0 22.0

7.5 25.0

a) Use the result to plot a graph of length of spring against load b) Use the graph and find out i) The unload length of the spring ii) Extension produced of 7.0 N load iii) Load required increasing the length of the spring by 5.0 cm Stress and Strain Wires obey Hooke's Law just like a spring. This is because bonds between atoms stretch just like springs: If we stretch a wire, the amount it stretches by depends on:  its length  its diameter  the material it’s made of. If we have two of the same material and length but of different thicknesses, it is clear that the thicker wire will stretch less for a given load. We make this a fair test by using the term tensile stress which is defined as the tension per unit area normal to that area. The term normalmeans at 90o to the area. We can also talk of the compression force per unit area, i.e. the pressure. Stress = Load (N) = F area (m2) A In some text books you may see stress given (sigma, a Greek letter 's'). Therefore:

You will have met the expression F/A before. It is, of course, pressure, which implies a squashing force. A stretching force gives an expression of the same kind. Units are newtons per square metre (N/m2) or Pascals (Pa). 1 Pa = 1 N m-2 You must always convert areas to square metres for this equation to work. Remember that radii will often be given in mm or cm. This is a common bear trap.  1 mm2 = 1 x 10-6 m2  Therefore if you get an area of 10-2 m2, you probably have forgotten to do the conversion to square metres. If we have a wire of the same material and the same diameter, it doesn’t take a genius to see that the wire will stretch more for a given load if it is longer. To take this into account, we

102

.................

Chapter One express the extension as a ratio of the original length. We call this the tensile strain which we define as the extension per unit length. Strain = extension (m) original length (m) There are no units for strain; it’s just a number. It can sometimes be expressed as a percentage.



You will find that the same is true for when we compress a material.

The Young Modulus The Young Modulus is defined as the ratio of the tensile stress and the tensile strain. So we can write: Young modulus = tensile stress tensile strain We know that Tensile stress = force = F area A And that Tensile strain = ___extension __ = ∆l original length l Young Modulus has the physics code E, so we can write:

This becomes:

The Young Modulus is named after Thomas Young (1773 to 1829) who was a British polymath, contributing to our understanding of optics, physiology, and Egyptology, among other fields The Young Modulus is also known as Young's Modulus or the elastic modulus or tensile modulus. It is a measure of the stiffness of a material that is independent of the particular sample of a substance. That means a generic value can be given for a material without its dimensions being known (like the values given for resistivity). It is the ratio of tensile stress (force/cross sectional area) to tensile strain (extension/ original length)

Units for the Young Modulus are Pascals (Pa) or newtons per square metre (Nm-2). The Young Modulus describes pulling forces.

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103

Stress-Strain graphs Stress-strain graphs are really a development of force-extension graphs, simply taking into account the factors needed to ensure a fair test. A typical stress-strain graph looks like this in figure 1.6.12:

Figure 1.6.12

We can describe the details of the graph as:  P is the limit of proportionality, where the linear relationship between stress and strain finishes.  E is the elastic limit. Below the elastic limit, the wire will return to its original shape.  Y is the yield point, where plastic deformation begins. A large increase in strain is seen for a small increase in stress.  UTS is the ultimate tensile stress, the maximum stress that is applied to a wire without its snapping. It is sometimes called the breaking stress. Notice that beyond the UTS, the force required to snap the wire is less.  S is the point where the wire snaps. We can draw stress-strain graphs of materials that show other properties.

................. Figure 1.6.13

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Chapter One 



 

Curve A shows a brittle material. This material is also strong because there is little strain for a high stress. The fracture of a brittle material is sudden and catastrophic, with little or no plastic deformation. Brittle materials crack under tension and the stress increases around the cracks. Cracks propagate less under compression. Curve B is a strong material which is not ductile. Steel wires stretch very little, and break suddenly. There can be a lot of elastic strain energy in a steel wire under tension and it will “whiplash” if it breaks. The ends are razor sharp and such a failure is very dangerous indeed. Curve C is a ductile material Curve D is a plastic material. Notice a very large strain for a small stress. The material will not go back to its original length.

Measuring young’s modulus We can measure the Young Modulus by doing a simple experiment below:

Figure 1.6.14

We need to measure the diameter of the wire using a micrometer. Then we measure the extension as we increase the load, recording the observations. We need to convert the force into stress, and extension into strain. From that we plot a stress-strain graph, using the gradient to calculate the Young Modulus. The value we get for E is often rather low. This is because the wire might suffer the following defects:  It might not be of uniform diameter. A smaller diameter will cause a greater stress.  The crystal structure of the wire may be degraded by the wire drawing process. The wire is made by pulling the metal through a small hole called a die. There are also many uncertainties. The greatest uncertainty comes in measuring the diameter.

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105

Deducing young modulus from stress- strain graph

The Young Modulus is the gradient of the stressstrain graph for the region that obeys Hooke’s Law. This is why we have the stress on the vertical axis when we would expect the stress to be on the horizontal axis.  The area under the stress strain graph is the strain energy per unit volume (joules per metre3). Strain energy per unit volume = 1/2 stress x strain. The units arise because stress is in N m-2 and strain is m m-1 (NOTE: This unit here is not "millimetres to the Figure 1.6.15 minus one", but metres per metrewhich mean no units). N m-2 x m m-1 = N m m-3. N m is joules, hence Jm-3 Area is the strain energy per unit volume. So we can write this equation: Area = ½ × stress × strain Worked examples 1. A tow rope for a car has a diameter of 5cm and has a pull on it of 400N. What is the 

tensile stress in the rope? Cross-sectional area = 𝜋𝜋𝜋𝜋𝑟𝑟𝑟𝑟 2 = 3.14 x (2.5 x10-2)2 = 19.94 x 10-4 m2 Stress = F/A = (400N)/ (19.94 x 10-4 m2) = 2.04 x 105Nm-2.

2. Find the longitudinal strain produced when steel 2.4m long is increased in length by 0.6mm? Strain = ∆L/L= 0.6 mm/2400 mm = 2.5 x 10-4. 3. A steel bar of cross-sectional area 16cm2 is 2m long. When a tensile force of 8 x104 N is applied to it, the length increases 0.5mm. Find the young’s modulus of this material? F = 8 x104 N, L = 2m, ∆L = 0.5mm = 5 x 10-4m, A = 16cm2 = 1.6 x 10-3 m2. E = (FL)/ (A x ∆L) = (2m x 8 x 104N)/ (1.6 x 10-3 m2 x 5 x 10-4m) = 2.0 x 1011Nm-2. 4. A steel bar in a roof truss has a diameter of 2 cm and is 4 m long. It supports a tensile load of 8 x104 N. given the young’s modulus for this material is 2.0 x 1011Nm-2, calculate the elongation under this load? F = 8 x104 N, L = 4m, 𝐴𝐴𝐴𝐴 = 𝜋𝜋𝜋𝜋𝑟𝑟𝑟𝑟 2 = 3.14 x 10-4m2, E = 2.0 x 1011Nm-2, ∆L = ? ∆L = (FL)/ (AE) = (4m x 8 x 104N)/ (3.14 x 10-4 m2 x 2.0 x 1011Nm-2) = 5.1 x 10-3.

5. What force will break a wire of 2.5 x 10-7m2 cross-sectional area if it has an ultimate tensile strength of 109Nm-2? U = maximum load/original area,

106

.................

Chapter One Therefore; maximum load = U x original area = (109Nm-2) (2.5 x 10-7m2) = 2.5 x 102N. Any force greater than (or equal to) this value will break the wire. Exercise 1. A wire made of a particular material is loaded with a load of 500 N. The diameter of the wire is 1.0 mm. The length of the wire is 2.5 m, and it stretches 8 mm when under load .What is the Young Modulus of this material? 2. What is the elastic strain energy per unit volume for the wire in question 1? Storing (strain) Energy Some elastic materials are intended to absorb energy. The more a material is extended, the greater the force required.

Figure 1.6.16

 The energy stored

as a material is deformed is represented by the area between the curve and the extension axis.  If a force F produces an extension e below the limit of proportionality, then the energy 1 stored = 2Fx, as shown in the diagram above.  Since F = ke,

1

energy stored = 2kx 2.

If seat belts and climbing ropes did not stretch, the forces exerted in stopping a person in an emergency would be too great for the body to withstand. Example This figure 1.6.17 shows a simplified version of force against extension graph for a piece of metal.

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107

Figure 1.6.17

Find the elastic potential energy when the metal is stretched to its elastic limit and the total work that must do to break the wire? Elastic potential when the object is stretched to its elastic limit is found by: Strain energy = ½ (20) (5 x 10-3) = 50 x 10-3 J = 0.05J Total work = total area under the graph = 0.05J + (25 x 10-3 x 20) J = 0.05J + 0.5J = 0.55J

108

.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .

Chapter Two

.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..

109

2.1 Behaviour of waves Introduction Sound waves, electromagnetic waves, mechanical waves etc are the waves which are used by us in our daily life. When we are watching Television or enjoying Radio programmes, everything, we see and hear is transmitted to us from the station through Radio or Electromagnetic waves produced by the stations The telephones are also based on the concept of waves. In a telephone we can hear the sound, as the energy moves from one point to another point, but there is no movement of the object and the message is transferred in the form of sound waves. Another example of waves is, when we drop a stone in a pond of water, the water ripples, means the waves spread out on water surface around the point where the stone hits the water surface.

Wave motion Wave motion is a form of disturbance, which travels through a medium due to the repeated periodic motion of the particles of the medium about their mean position. The motion being handed over from one particle to another is known as wave motion. A medium is a substance or material that carries the disturbance (either a pulse, wave or periodic wave) from sound waves use air as a medium.

A waves is a travelling disturbance in a medium that transfers energy and information from its source to another points without transferring the particles of the medium one place to another. For example,

Sound waves cannot be transmitted through a vacuum. The particles of the medium vibrate around their fixed positions. The moving waves or (waves that transfer energy) are called progressive waves. All particle in progressive waves have same amplitude, Progressive wave is a result of repetitive pulses in a medium or space. The waves which can be propagated only in a material medium are called Mechanical waves e.g. waves on water surface, sound waves etc. As you learned in form three, there are waves which do not require material medium to travel through which are called electromagnetic wave e.g. light wave, X-rays, gamma-rays, Radio waves, Microwaves etc. All the electromagnetic waves traveling through vacuum with the speed v of light which is 3 x 108m/sec. Sometimes the wave motions are also known as progressive waves. Explanation of wave When we throw a stone in a pond of still water, the water surface oscillates up and down, while the energy caused by the dropping stone is carried across the surface of water. The water ripples or wave spreads out on the water surface around the point where the stone hits the water surface. As shown in figure, the water ripples, spread away in the form of crests and

110

................

Chapter Two troughs. If we place leaf near the centre of disturbance at certain distance, we find that when the ripple reaches the leaf, it will simply move the leaf up and down and no movement of the leaf takes places in the direction of propagation of the disturbance. This shows that it is only the disturbance which travels outwards, while the water particles keep on moving up and down and continue to transfer the motion and energy to the neighbouring particles. Such a disturbance is called wave motion (or progressive waves).

Characteristics of waves motion

The Characteristics of wave motion are given below 1. Wave motion is a disturbance travelling through a medium. 1. There is no bodily motion of the particles of the medium from one part to another. The particles of the medium execute vibrating about their mean positions. 2. The displacement of a vibrating particle of the medium is zero over one complete vibration. 3. For the propagation of wave motion, a material medium having properties of elasticity and inertia is essential. 4.There is a regular phase difference between the various particles of the medium. 5. The energy is propagated form one part of the medium to another, without any net transport of the material medium. 6. The disturbance from one particle reaches its next neighbouring particle a little later and as the disturbance reaches the neighbouring particle, it also starts executing vibration about its mean position

Type of Wave Motion There are two types of wave motion namely transverse waves and longitudinal waves. That can be distinguished by how the particles of medium vibrate with the direction of wave motion. 1. Longitudinal waves When the particles of a medium vibrate about their mean position in the directions of propagations of disturbance, the wave motion is called the longitudinal wave motion in other

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111

words; Waves that the particles of the medium vibrate parallel to the direction of wave motion are called longitudinal waves. In a longitudinal wave the particles of the medium vibrate to and fro to the direction of wave motion. The vibrations of particles in a longitudinal wave compress in one region and form compression and stretch next region and form rarefaction

Modeling a longitudinal wave Sound is one of the most common longitudinal waves. You can't see sound, although you can feel it as a vibration by touching a speaker or talking into an in inflated balloon and feeling the other side of the balloon. Modeling a longitudinal wave can also be done with a slinky spring. To do this, tie one end of the slinky to a post with a string. Stretch the slinky out and hunch up five or six coils from the ends of the slinky and press them together with your fingers. You are storing energy in the compressed spring coils. If you let the compressed coils go, you will see a pulse of compressed coils travel along the length of the slinky as shown in figure. The pulse produced in the spring acts just like a longitudinal wave. The compression travels along the length of the spring. On each side of the compressed pulse there is a zone where the slinky coils are spread apart. These are as arc known as rarefaction. The motion of the particles produced by the moving energy of the wave is back and forth in the direction of wave propagation. In the case of sound waves, the pulses are compressed air particles (compressions) surrounded on either side by air particles that have been spread apart (rarefactions). Compressions arc zones of high air pressure and rarefactions arc zones of low air pressure. A longitudinal wave travels in the form of Compressions and Rarefactions  A compression is a region of the medium, in The wavelength, λ, of a longitudinal have is which the particles of medium come closer equivalent to the distance between adjacent zones to each other than the normal distance of compression or rarefaction (see figure). The between them; it has high pressure and small amplitude of a longitudinal wave is equal to the volume. size of the maximum displacement of the panicles  A rarefaction is a region of the medium, in which particles of the medium move from their equilibrium position. apart from each other than the normal distance between them, it has low pressure and large volume

112

................

Chapter Two Properties of longitudinal wave The following are the characteristic properties of longitudinal wave motion; i) In a longitudinal wave motion the particles of the medium vibrate about their mean

positions in the directions of the propagation of the waves. ii) The waves travel in the form of compressions and rarefactions. iii) There is a phase change from one particle to the next. iv) Particles move forward in a compression and backward in a rarefaction. v) There is a transfer of energy by the longitudinal waves. vi) Each particle starts vibrating a little later than its predecessor. 2. Transverse wave When the particles of the medium vibrate about their mean position in a direction perpendicular to the direction of propagation of disturbance, the wave motion is called transverse wave motion. In other words, Waves that the displacement of the particles is at right angles to the direction of wave motion are called transverse waves. In a transverse wave the particles of the medium vibrates up and downward to the direction of wave.

A pulse is a single disturbance that transfers energy from one place to another. For example, the figure below shows a single pulse that is travelling along a string.

Modeling of transverse waves Transverse waves transfer energy in a direction perpendicular to the direction of the disturbance in the medium. For example, if we fix a string at one end, a disturbance in the form of the pulse travels along the length of the string; it means that particles of the string vibrate along a direction perpendicular to the direction of the disturbance. A vibrating string is an example of a transverse wave. Although all points on the string itself are constrained to move only up and down, wave pulses move perpendicularly along the length of the string. A transverse wave may consist of more than one pulse. Transverse waves consisting of many pulses result when the wave source oscillates about some equilibrium position for long periods of time. Under such conditions an initial pulse is followed immediately by another pulse of opposite displacement. A series of such alternating pulses is known as a wave train (transverse wave).

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113

Transverse waves travel in the form of crests and troughs. A crest is a position of the medium, which is highly raised above the normal position of rest of the particles of the medium when a transverse wave passes through it. A trough is a portion of the medium, which is highly depressed below the normal positions of rest of the particles of the medium when a transverse wave passes through it

Properties of transverse wave

 Wave pulse is a single disturbance to the medium.  Wave train is a series of wave pulse to the medium.

The following are the characteristic properties of transverse wave i) In transverse waves, the particles vibrate about their mean position at right angles to the directions of propagation of the wave. ii) The amplitude and time period of the wave motion remains the same. iii) Every particle begins to vibrate a little later than its proceeding particles. iv) The transverse waves are in the form of crest and trough. Example of transverse waves 1. Water waves 2. Waves on a rope a string 3. Electromagnetic waves 4. Secondary (s) seismic waves

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Chapter Two The table below shows Difference between longitudinal wave and transverse wave LONGITUDNAL WAVE

TRANSVERSE WAVE

1. a longitudinal wave, the particles of the medium vibrate about their mean position in the same direction to the direction of the propagation.

1. In a transverse wave, the particles of the medium vibrate about their mean position at right angle to the direction of propagation.

2. Longitudinal waves travel in the form of compressions and Rarefaction.

2. Transverse wave travel in the form of crests and troughs.

3. One compression and one rarefaction constitute one wave.

3. One crest and one trough constitute one wave.

4. Longitudinal waves cannot be polarised.

4. Transverse waves can be polarised.

5. There is a change of density of the medium which is higher in compression as compared to rarefaction.

5. There is no change of density of the waves.

6. Sound waves are longitudinal waves In nature.

6. Light waves or all electromagnetic waves are transverse in nature

Exercise 1. How is a wave produced? Give some examples of different ways of producing waves. 2. Waves carry energy where does this energy come from? 3. A wave caused a cork to move up and down in water. What type of energy does the moving leaf has? 4. What is the difference between a longitudinal wave on a spring and transverse wave on a spring? 5. What is meant by compression and rarefaction in a spring? Show these two features on a diagram.

Characteristics of waves 1) Wave length

The distance between the two successive crests or troughs is known as the wavelength. (Or) It can also be defined as the distance travelled by the waves during a time when the vibrating particle of medium completes one cycle. In other words, it is the shortest distance between two points on a wave that are in phase.

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115

Inlongitudinal waves ,wavelength is the distance between two successive compressions

 symbol of wavelength is a Greek letter called lambda(λ)  the unit of wavelength is meter(m)  for example: BBC can be listen 41mbond (41m is wavelength) In symbol,

Wave phases Points on a wave which are a whole number of wavelengths apart are said to be in phase. Points A and A1 in the diagram shown are in phase

Points which are an odd number of half wavelengths apart are said to be out of phase. Points A and B in the diagram shown right are out of phase. Two points of a wave are said to be in phase if: i) They are moving with same speed. ii) They are moving in the same direction. iii) They are having same displacement from the rest position

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Chapter Two Example From the diagram below determine its wavelength?

Solution 3λ=60 λ=

60

Amplitude

3

= 20m

wavelength = 20m

The amplitude of a wave refers to the maximum displacement of a particle on the medium from its rest position. In a sense, the amplitude is the displacement from rest to crest. Similarly, the amplitude can be measured from the rest position to the trough position. On the diagram below at the points K and B, the particles of the medium make the maximum displacement, but at the position F, the particle make minimum displacement also, Amplitude is a measure of how much the wave vibrates the medium it passes through. It could be height of water wave or pressure of sound wave. The energy of the wave is carried by the amplitude of the wave, The higher the amplitude, the more energy a wave contains.

 

Unit of amplitude is meter (m) The energy of the wave is carried by the amplitude of the waves.

 

The large the amplitude, the more energy the wave carry. The amplitude decreases as the distance from the source increase. Because energy of the wave decreases.

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117

For example The crest of water waves decreases as the distance from source increases. A) longitudinal waves  Frequency is the number of compressions produced per second.  The unit of frequency is Hertz (Hz) 1𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 = (1 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) ⁄ 𝑠𝑠𝑠𝑠𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 Formula of frequency In symbols

 Frequency and wavelength are inversely proportional.

Frequency determine the type of sound F Example:1 Calculate the frequency and the amplitude each of the following diagrams.

λ

Solution a) Given f =? t = 2s # Cycle = 2.5 b) Given

f =?

118

t =8s # Cycles= 5

i. Frequency

=

= 0.8 Hz

ii.

i. Frequency

=

= 0.5625Hz

ii. Amplitude = 4cm

................

Chapter Two Example: 2 A longitudinal sound wave has a speed of 340m/s in air, if its wavelength is 0.85m, what is its frequency? Given v = 340m/s λ= 0.85m f =?

Solution Frequency

=

= 400Hz

Example: 3 The diagram below shows a water wave in a small triple tank which has speed of 16cm/s From the diagram above a) Determine the wavelength of the wave? b) Determine amplitude of the wave? c) Calculate the frequency of the wave?

Solution a) 3λ =75cm

λ=

= 25 cm

b) Amplitude = 20cm c) Given v = 16cm/s Frequency λ =25cm f =?

λ = 25 cm

=

= 0.64Hz

1. Period (T)  Period of the wave is the time taken for one complete cycle (wave).or is the time taken for a wave to move a distance of one wavelength.  Its unit is second(s).

 Period is the inverse of the frequency.

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119

2. Wave speed  The speed of the wave is the distance traveled by a crest per unit time taken.  Its is measured in meter per second (m/s) S λ V= = T T

Substitute by F

and

V=fλ

Example: 1 If a wave with a frequency of 400Hz and wavelength of 2m is produced from a source a) Find the speed of the wave? b) Find the period of the wave? Solution Given f= 400Hz λ= 2m a) V=? b) T=?

a) V =? V=fλ V = (400Hz)(2m) V = 800m/s

b) Period =? Period =

=

Period = 0.0025second

Example: 2 If the distance between the point A and point B is 60cm, as shown the diagram below From the diagram, calculate the:a) b) c) d) e) Solution a) Period =? Period = 0.8s b) Frequency =? Frequency =

e) Amplitude =? Amplitude = 3cm

Number of cycles Time taken

c) Wavelength =? 2λ =60cm λ = 60⁄2 λ = 30cm d) Speed =? Speed = Frequency x Wavelength Speed = (1.25Hz) (0.3m) Speed = 0.375m/s

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Period of the wave? Frequency of the wave? Wavelength of the wave? Speed of the wave? Amplitude of the wave?

=

2

1.6𝑠𝑠𝑠𝑠

= 1.25Hz

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Chapter Two Properties of waves Ripple tank A ripple tank for showing the properties of water waves comprises a shallow transparent tray of water with a point light source above it and a white screen on the floor below as shown the figure. Before adding the water the tray is leveled with a spirit-level to ensure a uniform water depth of rather less than 1 cm. straight parallel waves may be produced by a horizontal metal strip, or circular waves by a vertical ball-ended rod. When either of these is dipped into the water To study water waves with a ripple tank pulse of ripples is sent across the surface. Alternatively, continuous ripples may be obtained by fixing the dipper to a horizontal bar suspended by rubber bands. The bar is moved up and down by the vibrations of a small electric motor having an eccentric (off-centre) metal disc on its rotating spindle. A rheostat in the motor

Wave front: A wave front is a line or surface on which the disturbance has the same phase at all points. If the source is periodic, it produces a succession of wave front, all of the same shape. Ripples on a pond are the example wave front can be obtained by joining all the Crests of a wave. Wave fronts can either be spherical or plane.

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Reflection of waves Reflection of a wave occurs when a wave strikes an obstacle such as barrier, plane reflector, mirror and wall. The reflection of waves obeys the law of reflection: (a) The angle of incidence is equal to the angle of reflection. (b) The incident wave, the reflected wave and the normal lie in the same plane.

When the reflection of a wave happened, the wavelength (λ), the frequency and the speed (v) do not change but the direction of propagation of the wave changes, but after reflection every crest changes into trough

Reflection of water waves To investigate the reflection of water waves a metallic plane reflector is placed at the centre of a ripple tank. The motor with a wooden bar attached is switched on to produce plane waves which propagate towards the reflector. The reflector repositioned to produce different angles of incidence. Some analyses of water wave reflection with a different reflector are shown below. The shape and the direction of the reflected waves depend on:1. The shape of the incident waves. 2. Shape of the reflecting surface

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Chapter Two

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Refraction of wave Refraction is the changing of the direction of wave when it travels from one medium into another of different densities. When wave is refracted, the refraction changes 1. Wavelength of the wave 2. Direction of the wave , But the frequency does not change (or remains constant) 3. It is caused by a change of speed of the wave

Refraction of Waves A ripple tank can also be used to study the behavior of waves as they move from one medium into another. In general, when a wave strikes a boundary, some of its energy is reflected and some is transmitted or absorbed. When a wave crosses a boundary into another medium, its velocity changes because the new material has different characteristics. Entering the medium obliquely (at an angle), the transmitted wave moves in a direction different from that of the incident wave. This phenomenon is called refraction

The figures above show a glass plate placed in a ripple tank. The water above the plate is shallower than in the rest of the tank. The velocity of water waves depends on the water depth, so the water above the plate acts like a different medium. The left diagram shown above has the glass plate on the bottom and parallel to the edge of the picture. The diagram on the right has the glass plate, again on the bottom, but this time angled towards the top right. How does the velocity of waves depend on water depth? In the figure shown below on the left, a wave train passes over a glass plate, changing the depth of the water. As the waves move from deep to shallow water, their wave length decreases and the direction of the waves changes. Because the waves in the shallow water are generated by the waves in deep water, their frequency is not changed. From the equation v = f, the decrease in the wavelength of the waves means that the velocity is lower in the shallower water.

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Chapter Two

Diffraction Diffraction refers to the bending of waves around an edge of an object. The spreading of waves is called diffraction when it passes through gap or go around obstacle. For example, if you stand along an outside wall of a building near the corner, you can hear people talking around the corner. Assuming there are no reflections or air motion (wind), this would not be possible if the sound waves traveled in a straight line. In general, the effects of diffraction are evident only when the size of the diffracting object or opening is about the same as or smaller than the wavelength waves. This relationship can be readily observed in water waves. For example, water waves in a pond pass by blades of grass or reeds with little noticeable effect because the widths of these objects are much smaller than the wavelength of the dependence of diffraction on wavelength and size of the object is illustrated in the figures shown below.

Diffraction usually changes the direction and shape of the wave. Diffraction turns a plane wave into a circular wave when the wave passes through a narrow opening. Factors that affect the amount of diffraction are wavelength and size of the widths Maximum diffraction occurs when the gap is about equal to the wavelength of the waves Also when the waves meet the edge of an obstacle, the waves start to diffract around the edge. This diffraction is greatest when the waves have a longer wavelength, as shown in the diagrams shown below.

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However, it is important to stress again, that diffraction will only occur when the size of the object is of the same order as the size of the wavelength of the waves. Interference Interference is said to occur when waves from two or more coherent sources superpose with one another producing a resultant wave. Interference of water waves can be demonstrated using a ripple tank with two vibrating rod dippers. (see Figures below) The two dippers send our circular waves that are in phase and of the same frequency. An interference pattern consisting of easily observed lines of constructive (maximum) and destructive interference (minimum) is seen.

The lines of constructive interference (maxima) are places where the resultant amplitude is double the amplitude of one wave. The lines of destructive interference (minima) are places where the waves cancel out and the resultant amplitude is zero. a. Constructive Interference Let the two sources (dippers) be S1 and S2 respectively. Assume that the two sources generate waves in phase.

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Chapter Two For constructive interference to occur at a point P, the two waves must arrive in phase at P. This means that the wave crest from one source always meet the wave crest of the other. This can only occur if the path difference of the waves from the two sources is zero or they differ by an integral multiple of wavelength, i.e. nλ.

The condition for constructive interference is Path difference S2 P – S1 P = nλ where n = 0,1,2,3…..is the order of the maxima. b. Destructive Interference Destructive interference occurs if the waves from the two sources arrive exactly out of phase at point P, i.e. the wave crest always meets the trough. This happens when the path difference of the waves is (n – ½)λ.

The condition for destructive interference is Path difference S2 P – S1 P = (n-½)λwhere n = 1,2,3…..is the order of the minima. The Figure below illustrates the production of an interference pattern by two sources.

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127

(a) Constructive interference occurs at P when the waves meet in phase because their path difference is zero. (b) Constructive interference also occurs at Q because the path difference is 1 wavelength. (c) Destructive interference occurs at R because the path difference is half a wavelength, In general, for two sources which generate waves in phase, the conditions for 1. Constructive Interference: Path Difference of the 2 waves = nλ 2. Destructive Interference: Path Difference of the 2 waves = (n – ½)λ Standing wave A standing wave is an interference pattern that is set up when two waves of the same frequency, same speed, travel in opposite directions in the same medium. The interference of such waves produce points where the displacement of the standing wave

is always zero (and the standing wave has zero amplitude), these points are called nodes. Between two consecutive nodes there are regions of maximum displacement (where the amplitude of standing wave is larger) which are called antinodes. As shown in the diagram below A standing wave may be set up when a wave reflects back from a surface and the reflected wave interferes with the wave still travelling in the original direction as illustrated in the diagram below. In standing waves, the wave disturbance does not seem to travel at all. It seems to be standing still.

Conditions for formation of stationary waves When two progressive waves of equal amplitude and equal frequency travelling with the same speed in opposite directions are superposed, a stationary or standing wave is formed. StationaryWave No net transfer of energy from one point to Energy another. Energy is confined within the wave and there is interchange of K.E. and P.E. All particles between two adjacent nodes are in Phase phase. Particles on opposite sides of a node will be in antiphase. Amplitude Varies from zero at nodes to maximum at antinode. Wavelength 2 x distance between adjacent nodes or antinodes. Frequency

128

All particles vibrate in SHM with same frequency except at nodes.

Progressive Wave Energy is transported in the direction of travel of the wave. All particles within one wavelength vibrate with different phase. Same amplitude for all particles in the wave. Distance between adjacent particles which are in phase. All particles vibrate in SHM with same frequency.

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Chapter Two Waveform

Does not advance.

Advances in the direction of velocity of wave.

Characteristics of stationary waves      



The waveform remains stationary. Nodes and antinodes are formed alternately. The points where displacement is zero are called nodes and the points where the displacement is maximum are called antinodes. Pressure changes are maximum at nodes and minimum at antinodes. Amplitude of each particle is not the same; it is maximum at antinodes decreases gradually and is zero at the nodes. The velocity of the particles at the nodes is zero. It increases gradually and is maximum at the antinodes. 1

Distance between any two consecutive nodes or antinodes are equal to 2 λ, where as 1

the distance between a node and its adjacent antinodes is equal to 4

2.2 Sound

Sound is a form of energy propagated by vibrating body which can travel from place to place. We hear sounds such as clocks ticking, airplanes flying, cars moving, people talking, dogs barking and music being played all around us. But what causes these different sounds? If you clamp one end of a ruler to the bench and flick the overhanging end, you can hear the ruler vibrating. A sound is produced.  Likewise, when a guitar string is plucked, it vibrates and produces sound. These simple experiments show that sound is caused by a vibration. Because vibrations can be produced in different ways, so can sounds. Speech sounds from our bodies are the result of vibrations of the focal cords in the throat. 

Sound wave is a mechanical longitudinal wave. This means it needs a material medium to propagate and the particles of the medium vibrate in the direction of wave travel. Different sounds have different frequencies.

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The human ear can hear sound waves of frequencies between 20Hz to 20000Hz, this is called audible range. Human speech contains sounds between 600Hz and 4800Hz.  Sound waves that have frequencies below 20Hz are called infrasonic. (or simply infrasound) while sound waves that have frequencies above 20 000Hz are called ultrasonic (or simply ultrasound).  Human speech contains sound of frequencies between 600Hz to 4800Hz. Examples of sources of sound include:



 Vocal cord.  Tuning fork.  Car moving.  Vibrating string.  Sound transmission and speed of sound Sound does not travel through a vacuum (empty space), it is transmitted only a material medium. Liquids are better transmitters of sound than gases, and solids are better transmitters of sound than either liquids or gases. Sound waves have a definite velocity of propagation in a given medium.. The speed of sound waves in a medium is determined by the type of medium and the temperature of that medium. The table below shows speed of sound in different media.

Speed of sound in various media Medium

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Speed

Air (0C0 )

332 m/s

Air (20C0 )

344 m/s

Helium (0C0)

972 m/s

Water (25C0)

1493 m/s

Seawater (25C0)

1533 m/s

Copper (25C0)

3560 m/s

Iron (25C0)

5130 m/s

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Chapter Two Sound transmission needs a material medium Demonstration We can demonstrate that sound cannot travel through a vacuum by using bell jar experiment. The bell jar experiment is a common experiment used to demonstrate that sound needs a medium to travel. What is a bell jar? A bell jar is laboratory equipment used for creating a vacuum. It is so named as its shape is similar to that of a bell. A bell jar is placed on a base which is vented to a hose fitting that can be connected via a hose to a vacuum pump. By pumping the air out of the bell jar, the air pressure inside the jar can be varied.

How does the experiment work? The experiment is done by placing an electrical bell in the bell jar. As the air is pumped out of the sealed bell jar, the sound from the bell jar fades. At a particular vacuum, no more sound is heard from the bell, but we can see that the hammer continues hitting the gong and sound is produced. However, the sound is not audible to our ears because of the vacuum inside the jar. This demonstrates that the sound wave cannot travel through vacuum. That is, a sound wave needs a material medium for its propagation.

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Sound can travel through solids, liquids and gases We obviously know that the sound we normally hear reaches us through air, but it can also be transmitted through liquids and solids as demonstrated the activities below. Activity-1:sound travel through liquid Take a bucket or a bathtub. Fill it with clean water. Take a small bell in one hand. Shake this bell inside the water to produce sound. Make sure that the bell does not touch the body of the bucket or the tub.

Place your ear gently on the water surface (Fig.2.2). (Be careful: water should not enter in your ear). Can you hear the sound of the bell? Does it indicate that sound can travel through liquids? Yes, it indicates that sound can travel through liquids, and that is how whales and dolphins might be communicating under water.

Activity-2: sound travel through solids Take two match box trays, cotton thread of length 15m, two wooden splints. Hold the two tray boxes so that the string is taut. Let one student speak at one end, A, as another listens at end B. What does happen? When the student speaks at one end softly, the other student hears clearly. This shows that sound travel through solids.

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Chapter Two Speed of sound in air The speed of sound in air is 331.5 m/s at 0C0 and changes by 0.6m/s for each degree Celsius. Where, temperature is in degree Celsius.

V = (331 + 0.6T) m/s

Example-1 What is the speed of sound in air at 350C? Solution V = (331 + 0.6T) m/s V = (331+ 0.6(35) m/s V = (331+ 21) m/s V = 352m/s Example-2

A lightning flash is seen 10 seconds before thunder is heard. Find the distance to the lightning flash if the temperature is 20 C0. Solution V =(331+0.6T)m/s

Distance = speed x time

V = (331 +0.6(20)m/s

= 343m/s x 10s

V= (331 +12)m/s

= 3430m

V= 343m/s V = 343 m/s

Exercise 1. When a sound wave passes through air which of these occurs? a. The particles of air oscillate up and down b. The particles of air oscillate parallel to the direction of wave travel c. The particles of air oscillate perpendicularly to the direction of wave travel d. The particles of air do not move at all. 2. What condition is necessary for sound to travel from one place to another? 3. Sound waves cannot pass through: a. Fresh water b. Concrete c. Wood d. Outer space

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e. Earth 4. A sound source produces 160 compressions in 10 seconds. The distance between successive compressions is 20 m. Calculate: a. Frequency of the wave. b. Speed of the wave. 5. What is the speed of a sound in air at? a- 300C

b- 40 Factors affecting velocity of sound in air

a- Temperature of air: Sound waves travel faster in hot air than in cold air b- Humidity: Humidity refers to the moisture content of the air in the atmosphere. The density of moist air is more than that of dry air, so sound travels faster in moist air than in dry air c- Wind:If wind blows in the same direction as sound, then the speed of sound increases and vice versa. d-Density:Sound travels faster in gases of lower density

2.21-Properties of sound 1- Reflection Sound waves are reflected well from hard flat surfaces such as walls or cliffs and obey the same laws of reflection as light. The reflected sound is called an echo.

In some halls, sound waves are reflected from the walls, floor and ceiling. Since the echo time is short, the echo overlaps with the original sound. The original sound thus seems to be prolonged, an effect called reverberation. Surfaces of materials such as cotton, wool and foam rubber absorb most of the energy of incident sound waves. Because of this property, such materials are used in places where echo effects are not desirable. The walls of broadcasting studios and concert halls are thus made of absorbent materials.

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Chapter Two Experiment: To demonstrate reflection of sound Apparatus Two plastic tubes, a ticking clock, smooth hard wall.

Procedure   

Place the clock near the end of one of the tubes as shown in the figure above Point the open end of the tube towards a hard wall at an angle of incidence, i. With the ear close to the end of the sound open tube, listen to the reflection of the sound from the wall at different angles of reflection, r , and note the angle at which the reflected sound is loudest.

Observation it is observed that maximum loudness of the reflected sound occurs when:  The angle of reflection, r, is equal to the angle of incidence, i,  Both tubes and the normal to the reflecting surface lie in the same plane. Conclusion Sound waves obey the laws of reflection.

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Application of reflection of sound a) Determination of speed of sound Experiment: To determine the speed of sound in air by echo Apparatus Hard smooth wall, two wooden pieces, and stop-watch Procedure Stand some distance away from and directly in front of a large high wall. Tap two pieces of wood together and listen to the echo. Now tap repeatedly and try to arrange the next tap to coincide with the echo heard. Time a number of successive taps and determine the average time between successive taps.  Measure the perpendicular distance between the observer and the wall as in the figure above. Observation

   

Since the echo from one tap coincides with the sound from next tap, it means that the time taken to make a tap after the preceding one equals the time taken by sound to travel from the observer to the wall and back. If the distance between the observer and the wall is d meters, the number of tap intervals n 𝐭𝐭𝐭𝐭

and the time t seconds, the sound travels 2d meters in 𝐧𝐧𝐧𝐧seconds.

Hence, speed =

𝐭𝐭𝐭𝐭

𝐭𝐭𝐭𝐭 = 𝟐𝟐𝟐𝟐𝐝𝐝𝐝𝐝 ÷ 𝐧𝐧𝐧𝐧

Example-1

𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐭𝐭𝐭𝐭𝐝𝐝𝐝𝐝𝐧𝐧𝐧𝐧𝐝𝐝𝐝𝐝 𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐝𝐝𝐝𝐝𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐝𝐝𝐝𝐝

𝐭𝐭𝐭𝐭 =

𝐭𝐭𝐭𝐭𝐝𝐝𝐝𝐝𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭 𝐭𝐭𝐭𝐭𝐝𝐝𝐝𝐝𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝

Then

𝐭𝐭𝐭𝐭 =

𝟐𝟐𝟐𝟐𝐝𝐝𝐝𝐝 𝐭𝐭𝐭𝐭

𝟐𝟐𝟐𝟐𝐧𝐧𝐧𝐧𝐝𝐝𝐝𝐝 𝐭𝐭𝐭𝐭

Two boys stand 200m from a wall. One bangs two pieces of wood together while the second starts a stop-watch and stops it when he hears the echo. If the time shown on the stop-watch is 1.2 seconds, calculate the speed of the sound. Given d = 200m t = 1.2 s, v= ?

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Solution V=

𝟐𝟐𝟐𝟐𝐝𝐝𝐝𝐝 𝐭𝐭𝐭𝐭

=

𝟐𝟐𝟐𝟐(𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝐭𝐭𝐭𝐭) 𝟏𝟏𝟏𝟏.𝟐𝟐𝟐𝟐𝐝𝐝𝐝𝐝

=

𝟒𝟒𝟒𝟒𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝐭𝐭𝐭𝐭 𝟏𝟏𝟏𝟏.𝟐𝟐𝟐𝟐 𝐝𝐝𝐝𝐝

= 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑. 𝟑𝟑𝟑𝟑 𝐭𝐭𝐭𝐭/𝐝𝐝𝐝𝐝

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Chapter Two Example: 2 The speed of sound in air is 340m/s. A loudspeaker placed between two walls A and B, but nearer to wall A than wall B, is sending out constant sound pulses. How far is the speaker from wall B if it is 200 m from wall A and the time between the two echoes received is 0.176 seconds? Given

Solution

V = 340 m/s

𝑣𝑣𝑣𝑣 =

t1 = time taken to hear echo from wall A t2 = time taken to hear echo from wall B Then, t2 - t1 = 0.176 s. d1 = distance between loudspeaker and wall A = 200 m d2 = distance between loudspeaker and wall B =?

2𝑁𝑁𝑁𝑁

t1 =

2𝑁𝑁𝑁𝑁2 𝑣𝑣𝑣𝑣

𝑡𝑡𝑡𝑡 2𝑁𝑁𝑁𝑁1

-

2𝑁𝑁𝑁𝑁2

and t =

𝑣𝑣𝑣𝑣 2𝑁𝑁𝑁𝑁1

340 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠

𝑣𝑣𝑣𝑣

-

and

2𝑁𝑁𝑁𝑁 𝑣𝑣𝑣𝑣

t2=

= 0.176 s 2(200 𝑚𝑚𝑚𝑚) 340 𝑚𝑚𝑚𝑚/𝑠𝑠𝑠𝑠

2𝑁𝑁𝑁𝑁2 𝑣𝑣𝑣𝑣

= 0.176s

2𝑑𝑑𝑑𝑑2 − 400 𝑚𝑚𝑚𝑚 = 0.176 s(340 m/s) 2𝑑𝑑𝑑𝑑2 -400 𝑚𝑚𝑚𝑚 = 59.84 m 2𝑑𝑑𝑑𝑑2 = 59.84 m + 400 m = 459.84 𝑚𝑚𝑚𝑚 𝑑𝑑𝑑𝑑2 = 459.84 𝑚𝑚𝑚𝑚 ÷ 2 = 229.9 𝑚𝑚𝑚𝑚

Example: 3 A student stands at a distance of 100m from a wall and claps two pieces of wood together. After the first clap, the student claps whenever an echo is heard from the wall. Another student starts a stopwatch at the fifth clap and stops it at the fifty-fifth clap. The stopwatch records a total time of 25 s. Find the speed of the sound.

Given

Solution

d =10m

d=

N = 50 t = 25s v=?

=

𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽 𝟐𝟐𝟐𝟐𝑵𝑵𝑵𝑵

2𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑡𝑡𝑡𝑡

→v=

2𝑥𝑥𝑥𝑥50𝑥𝑥𝑥𝑥100 = 400 𝑚𝑚𝑚𝑚 25

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b) Echolocation Echolocation is the process finding distances by sending ultrasonic pulses and detecting it after reflection. Echolocation allows animals such as bats and dolphins to locate the object’s exact position.

Ships with sonar (sound navigation and ranging) can measure the depth of a seabed by sending ultra-sound pulses and measuring the echo time.

Example-1 A ship sends sound wave and receives an echo after 3 seconds. How deep is the water if the speed of sound in water is 1500m/s? Solution 𝒔𝒔𝒔𝒔 =

138

𝟒𝟒𝟒𝟒𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝒎𝒎𝒎𝒎 𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎/𝒔𝒔𝒔𝒔( 𝟑𝟑𝟑𝟑 𝒔𝒔𝒔𝒔) = = = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐

................

Chapter Two Exercise: 2 1. A boy is 60m from a large wall. He is hammering a block of wood. Every time he hits the block of wood. He hears an echo 0.5 seconds later. Calculate the speed of the sound waves. 2.

A man standing in a gorge between two large cliffs claps his hands at a steady rate and hears two echoes. The first comes after 2 seconds and the other after 3 seconds. If the speed of sound is 340 m/s what is the distance between the two cliffs?

3. A submarine emits ultrasound, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff? 4. What it is echolocation? 2- Refraction Sound can be refracted like light. Just as for light, refraction takes place at the boundary between two media in which the speed of travel is different. In the open air, there is no sharp boundary. However, air temperature decreases as the altitude increases. Furthermore, sound travels faster in air at higher temperatures. Combining these two effects, the different layers of air at different altitudes serve to refract sound gradually. This explains why sound is usually more difficult to hear in the day time than at night. You can notice this on a hot day or a cold night. On a hot day the air near the ground is hot so the sound wave bends upwards from the hot air into the cold air.

On a cold night the air near the ground is cold and so the sound wave bends downwards. This is why you can sometimes hear sounds from a long way away if the night air is cold.

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3: Interference Interference of sound is a phenomenon in which two sound waves superpose to form a resultant wave of greater or lower amplitude. If the two waves reinforce with each other and become louder, the effect is called constructive interference. But if the two sound waves cancel each other, the effect is called destructive interference. Interference usually refers to the interaction of waves that are correlated or coherent with each other, either because they come from the same source or because they have the same or nearly the same frequency. Experiment: Interference of sound waves Apparatus Take two loudspeakers and signal generator Procedure The two loudspeakers are connected to the same signal generator so that the sound waves from each are in phase. The signal generator is set at a frequency of about 1000Hz. In order to produce an observable interference pattern, the two loudspeakers must be separated a distance of about 0.5m apart.

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Chapter Two Observation When walking along a straight line in front of the loudspeakers, the observer will hear alternating loud and soft sounds. The loud sounds are due to constructive interference whereas the soft sounds are due to destructive interference. Example In the figure L1 and L2 are two loudspeakers emitting identical sound waves simultaneously. The frequency of the sound waves is 3400Hz. a. Calculate the path difference L2A – L1A. b. Find the wavelength if the velocity of the sound in air is 340m/s c. Would a loud or soft sound be heard at A? Explain.

Solution a. L2A = √1.22 + 0.52 = 1.3𝑚𝑚𝑚𝑚 , L1A = 1.2m, so L2A – L1A = 0.1m

b. 𝜆𝜆𝜆𝜆 =

𝑣𝑣𝑣𝑣 𝑓𝑓𝑓𝑓

=

340 3400

= 0.1

c. Loud, since the path difference is a whole multiple of the wavelength.

4- Diffraction Diffraction of sound is the spreading of sound waves when they passé edges and go through small openings. Sound from your mouth spreads out in all directions because of diffraction by the sides of the mouth. Therefore, a person behind you can hear what you say even when you are facing away from him. Waves with longer wave lengths are diffracted more than waves with shorter wave lengths.

Diffraction is a property that is demonstrated by all waves. The effect is most noticeable

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when the wave length of the wave is approximately equal to the size of the hole through which they are moving

Examples of diffraction of sound include sound waves that are diffracted as they pass through doorways. Diffraction also happens when waves pass a single edge, you can think an edge as just one side of a wall. Examples of diffraction around edge include sound waves that are diffracted as they pass along the edge of wall.

This student can hear the sound of the radio even though it is behind an obstacle. This is because; the sound of the radio spreads around the corner of the wall due to diffraction of sound wave.

Doppler Effect

The Doppler effect (or the Doppler shift) is the change in frequency of a wave (or other periodic event) for an observer moving relative to its source. It is named after the Austrian physicist Christian Doppler, who proposed it in 1842 in Prague. We are most familiar with the Doppler Effect because of our experience with sound waves. Perhaps you recall an instance in which a police car or emergency vehicle was traveling towards you on the highway. As the car approached with its siren blasting, the pitch of the siren sound (a measure of the siren's frequency) was high; and then suddenly after the car passed by, the pitch of the siren sound was low. That was the Doppler Effect - an apparent shift in frequency for a sound wave produced by a moving source.

Application of Doppler Effect The handheld radar guns used by police to check for speeding vehicles use the Doppler effect. Here is how they work: a - A police officer takes a position on the side of the road.

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Chapter Two b - The officer aims his radar gun at an approaching vehicle. The gun sends out a burst of radio waves at a particular frequency. c - The radio waves strike the vehicle and bounce back toward the radar gun. d - The radar gun measures the frequency of the returning waves. Because the car is moving toward the gun, the frequency of the returning waves will be higher than the frequency of the waves initially transmitted by the gun. The faster the car's speed, the higher the frequency of the returning wave. e - The difference between the emitted frequency and the reflected frequency is used to determine the speed of the vehicle. A computer inside the gun performs the calculation instantly and displays a speed to the officer. Resonance All objects have a natural frequency of vibrations. The vibration can be started and increased by another object vibrating at the same frequency. The effect is called resonance. You can find the natural frequency of a large glass (thin one works best) by wetting your finger and rubbing it gently around the top edge of the glass. As your finger dries it vibrates on the glass and a little experimentation with speed of rubbing will mean that this vibration resonance with the glass. Energy is then transmitted from your finger to the glass and it sings. Resonance increases the loudness of the sounds produced by many musical instruments. For example, wind instruments standing waves in the tube resonate with the vibration in the mouthpiece, amplifying the sound of the instrument. Ultra sound Sounds above the range of human hearing are called ultrasonic sounds or ultrasound. It is above 20 kHz. It can be detected electronically and displayed on a CRO. Some animals like dogs can detect it.

Application of ultrasound a- Ultrasound is used in SONAR to measure the depth of the sea and to locate the underwater objects like shipwrecks, submarines and sea rocks. Some animals like dogs, bats and dolphins can send ultrasound and detect the reflection of the ultrasound to find the position of an object b- Ultrasound is used to investigate inside the human body. Echocardiography helps the doctors to study the condition of the hurt of a patient. Stones in the gall-bladder and kidneys can be located with ultrasound. c- Ultrasound is used in industry for detecting flaws in metal blocks or sheets without damaging them. It is used to clean delicate machine parts or parts that are located hard-to-reach places. These are placed inside the cleaning solution which is then subjected to high frequency waves. These dislodge the dirt and dust particles and the parts get thoroughly cleaned. d- Ultrasonic guidance device for the blind person use ultrasound. The blind person carries a transmitter of ultrasonic waves. The reflected waves are detected by an

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electronic device and heard through an earpiece. A trained user can learn to see the obstacles. e- Ultrasound is used to scan the worm to monitor the health and sometimes to determine the

sex of an unborn baby, an ultrasonic transmitter-receiver is scanned over the mother’s abdomen and a detailed image of the fetus is built up. Reflection of ultrasonic pulses occurs from boundaries of soft tissue.

We use ultrasound for scanning over mother’s abdomen rather than X-rays because  Ultrasound is safer than X-rays  Ultrasound can distinguish between different layers of soft tissue.

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Chapter Two Musical note Musical notes originate from a source vibrating in a uniform manner Musical notes can be described in terms of; pitch, loudness and quality and these are effects on the ear caused by intensity, frequency and harmonic content respectively. We will consider these three effects one by one. a) Intensity and loudness The intensity of a sound is the time rate at which the sound energy flows through a unit area. More briefly, the intensity is the average power transported by unit area. Intensity is measured in watts per square meter. 𝐼𝐼𝐼𝐼 =

𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴

The intensity of a sound wave is dependent on its amplitude. It inversely proportional to the square of its distance from the source. The loudness of sound is the effect of the intensity of sound on the ears. While loudness increases with intensity, there is no simple linear relationship. It is more nearly logarithmic. Example Sound energy is radiated uniformly in all directions from a small source at rate of 1.2 watts. Find the intensity of the sound at a point 25 m from the source. Given P = 1.2 w, A is a sphere with radius 25 m 𝐴𝐴𝐴𝐴 = 4𝜋𝜋𝜋𝜋𝑟𝑟𝑟𝑟 2 = 4(3,14)(25𝑚𝑚𝑚𝑚)2 = (12.56)(625)𝑚𝑚𝑚𝑚2 = 7850𝑚𝑚𝑚𝑚2 Solution 𝑃𝑃𝑃𝑃

𝐼𝐼𝐼𝐼 = = 𝐴𝐴𝐴𝐴

1.2 𝑤𝑤𝑤𝑤

7850 𝑚𝑚𝑚𝑚2

= 0.0001529 = 1.529 𝑥𝑥𝑥𝑥10−4 w/m2

b) Frequency and pitch

Pitch is a property of sound determined by its frequency. High-pitched sounds are associated with high frequencies. Low-pitched sounds are associated with low frequencies. A man’s speaking voice produces sound with frequencies around 130 Hz; a woman’s speaking voice produces sound with an average frequency of 250 Hz. c) Harmonics and qualities A taut wire or string which vibrates as a single unit produces its loose frequency called its fundamental. Fundamental (f0) and tones having frequencies which are whole number multiples of the fundamental are called harmonics, such as 2f1, 3f1, 41 and so on. Thus the fundamental frequency f0 is the first harmonic, 2f0 (the first overtone) is the second harmonic, 3f0 (the second overtone) is the third harmonic and so on. The table below shows an example of harmonic overtones

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Frequency

Order

Name 1

Name 2

1 · f = 440 Hz

n=1

fundamental tone

1st harmonic

2 · f = 880 Hz

n=2

1st overtone

2nd harmonic

3 · f = 1320 Hz

n=3

2nd overtone

3rd harmonic

4 · f = 1760 Hz

n=4

3rd overtone

4th harmonic

The quality of a sound is a result of the number of harmonics which are present. Musical instruments A musical instrument is an instrument created or adapted to make musical sounds (note) that originates from a source vibrating in a uniform manner. Irregular vibrations cause noise. Musical instruments can be classified in to three main categories, and they are:  Stringed instrument  Wind instrument  Precaution instrument a) Stringed instrument Stringed instruments have strings that produce sound when they vibrate e.g. guitar.

Sound is produced by a guitar when a string vibrates. When the string is plucked a standing wave is formed between the two ends of the string. The standing wave contains nodes and antinodes. If the standing note has one loop a fundamental note is produced. If there are more than one loop overtones are produced. The distance between two successive nodes is one half of the wavelength.

Experiments have shown that the frequency (in hertz) of a sound produced by a string depends on the following factors:  The length (L) of the string in meters. Shorter strings give higher notes while longer strings give lower notes.

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The tension (T) of the string in Newton. Tighter strings give higher notes while loose strings give lower notes. The mass (M) per unit length in kilogram per meter (kg/m). Less dense strings (thinner strings) give higher notes while more dense strings (thicker strings) give lower notes. 1

𝑇𝑇𝑇𝑇

𝑓𝑓𝑓𝑓 = 2𝐿𝐿𝐿𝐿 √𝑀𝑀𝑀𝑀

or

𝑓𝑓𝑓𝑓 =

𝑉𝑉𝑉𝑉 𝐿𝐿𝐿𝐿

Where: - f is the fundamental frequency. - V is the speed of sound. The frequency of the n-th overtone of a stringed instrument can be found by: 𝑓𝑓𝑓𝑓𝑛𝑛𝑛𝑛 = (𝑛𝑛𝑛𝑛 + 1)𝑓𝑓𝑓𝑓 Example: 1 Calculate the fundamental frequency for a 0.56 m wire with mass per unit length 0.002 kg m-1 under a tension of 77.4 N? What is the frequency of the emitted sound wave if the third overtone is produced? Solution 1 𝑇𝑇𝑇𝑇 1 77.4 √ = √ = 175.6𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 2𝐿𝐿𝐿𝐿 𝑀𝑀𝑀𝑀 2(0.56) 0.002 𝑓𝑓𝑓𝑓𝑛𝑛𝑛𝑛 = (𝑛𝑛𝑛𝑛 + 1)𝑓𝑓𝑓𝑓 = 𝑓𝑓𝑓𝑓3 = (3 + 1)175.6 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 = 4 × 175.6 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 = 702.4 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑓𝑓𝑓𝑓 =

b) Wind instruments The instruments in this category, standing waves are produced by longitudinal stationary waves in air columns e.g. organ pipes. Air columns can vibrate in a number of different ways and the fundamental note is produced when the mode of vibration is simplest. Air columns can be tubes open at one end or both ends.

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For closed pipes:

𝑉𝑉𝑉𝑉

The fundamental frequency can be found by: 𝑓𝑓𝑓𝑓 = 4𝐿𝐿𝐿𝐿 where: - f is the fundamental frequency - V is the speed of sound - L is the length of the tube. The frequency of the n-th overtone can be found by: 𝑓𝑓𝑓𝑓𝑛𝑛𝑛𝑛 = (2𝑛𝑛𝑛𝑛 + 1)𝑓𝑓𝑓𝑓 For open pipes:

𝑉𝑉𝑉𝑉

The fundamental frequency can be found by: 𝑓𝑓𝑓𝑓 = 2𝐿𝐿𝐿𝐿 and the frequency of the n-th overtone can be found by: 𝑓𝑓𝑓𝑓𝑛𝑛𝑛𝑛 = (𝑛𝑛𝑛𝑛 + 1)𝑓𝑓𝑓𝑓 Example -1 A person blows hard across the mouth of a closed tube of length 0.3 m. a) What is the frequency of the fundamental note? b) What is the frequency of the first and the fourth overtones? c) If the tube is an open tube find: i. The fundamental frequency. ii. The frequency of the first and fourth overtones. (Take the speed of sound = 340 m/s)

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Chapter Two Solution 𝑉𝑉𝑉𝑉 340 340 a) 𝑓𝑓𝑓𝑓 = 4𝐿𝐿𝐿𝐿 = 4(0.3) = 1.2 = 283.3 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻

b) 𝑓𝑓𝑓𝑓𝑛𝑛𝑛𝑛 = (2𝑛𝑛𝑛𝑛 + 1)𝑓𝑓𝑓𝑓 then: 𝑓𝑓𝑓𝑓1 = (2(1) + 1)𝑓𝑓𝑓𝑓 = 3𝑓𝑓𝑓𝑓 = 3(283.3 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻) = 849.9 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑓𝑓𝑓𝑓4 = (2(4) + 1)𝑓𝑓𝑓𝑓 = 9𝑓𝑓𝑓𝑓 = 9(283.3 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻) = 2549.7 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 c) 𝑉𝑉𝑉𝑉 340 340 i. 𝑓𝑓𝑓𝑓 = 2𝐿𝐿𝐿𝐿 = 2(0.3) = 0.6 = 566.7 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 ii.

𝑓𝑓𝑓𝑓𝑛𝑛𝑛𝑛 = (𝑛𝑛𝑛𝑛 + 1)𝑓𝑓𝑓𝑓 then: 𝑓𝑓𝑓𝑓1 = (1 + 1)𝑓𝑓𝑓𝑓 = 2𝑓𝑓𝑓𝑓 = 2(566.7 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻) = 1133.4 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑓𝑓𝑓𝑓4 = (4 + 1)𝑓𝑓𝑓𝑓 = 5𝑓𝑓𝑓𝑓 = 5(566.7 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻) = 28433.5 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻

c) Percussion instruments

Percussion instruments are made out of just about any material; wood, metal, glass, ceramic, or animal bones and teeth. To make percussion instruments sound, you usually strike them with a stick, mallet, or your hand to make them vibrate. In percussion instruments such as drums or the xylophone, two dimensional standing waves are produced in the skin or the metal that is struck.

Exercise 1. Sound energy is radiated uniformly in all directions from a small source at rate of 2.4 watts. Find the intensity of the sound at a point 30 m from the source. 2. A monochord is an instrument that consists of one string or wire that can be plucked to produce a note. The monochord contains a 20 cm long wire with mass 0.1 g under a tension of 100 N. Calculate: a- Fundamental frequency of the monochord b- Frequency of the first and third overt

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150

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Chapter Three 3.0 Introduction Telecommunication is concerned with sending and receiving information over a distance. In the broadest sense, information can be in the form written or spoken words, numbers, diagrams, pictures, music or computer data.

3.1 Early history The earliest line-of-sight methods involved sending smoke signals from hilltop. It was the fore-runner of flag signaling by semaphore and ship-to-ship signaling by Aldis lamp. The mid-19th century saw the invention of the electric telegraph by Wheatstone, Figure 3.1a, in which messages were sent in Morse code as currents in cables. Shortly afterwards Bell developed the telephone, Figure 3.1b, making it possible to transmit speech electronically. The use of electrical signals increased the speed of communications substantially.

Figure: 3. 1 a

Figure: 3.1c

Fig ure: 3.1b

Figure: 3.1d

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In 1895, after Hertzhas shown how to produce the electromagnetic waves predicted mathematically by Maxwell, Marconi succeeded in transmitting them over a mile. In 1901 he established that wireless telegraphy signals could be sent and received across the Atlantic; shortly afterwards radio telephony became possible and the ‘wireless’, Figure 3.1c, transformed home life. Picture transmission, i.e. television, Figure 3.1d, came in the 1930s after much of the early development work had been done by Baird. Since the 1990s light has been used to transmit information at an even greater rate along optical fibers. Rapid communication has been established worldwide via fax, e-mail and the Internet.

Representing Information Information can be represented electrically in two ways. a) Digital Method In this method electricity is switched on and off and the information is in the form of electrical pulses. For example, in the simple circuit of Figure 3.2 a, data can be sent by the ‘dots’ and ‘dashes’ of Morse code by closing the switch for a short or a longer time. In Figure 3.2 b, the letter A (∙ −) is shown. Computers use the simpler binary codewith 1 and 0 represented by ‘high’ or ‘low’ voltages respectively. They can only handle numbers (i.e. 1 and 0) and so there is a pattern of 1s and 0s for each of the 26 letters of the alphabet and for other symbols. Figure 3.2crepresents capital S in this code.

Figure: 3. 2

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Chapter Three b) Analogue method In this case the information is changed to a voltage or current that varies continuously and smoothly over a range of values. The waveform of the voltage or current, i.e. its variation with time, represents the information and is an analogue of it, Figure 3.3. For example, the loudness and frequency (pitch) of a sound determines the amplitude and frequency of the waveform produced by a microphone on which the sound falls.

Figure: 3. 3 Analogue information

Basic communication System Electrical signals representing ‘information’ from a microphone, a television camera, a computer, etc., can be sent from a place to place using either cables or radio waves. While some signals can be sent directly through cables, in general, and certainly in radio and television, as we will see later, a ‘carrier’ wave is required to transport them. The basic building blocks of any communication system are shown in Figure 3.4. Signals from the input transducer are added to the carrier in the modulator or encoder by the process of modulation. This involves impressing one wave system (the signal) upon another, higher frequency, system (the carrier). The modulated signal is then sent by the transmitter into the propagating medium (i.e. cable or radio wave).

Figure: 3. 4 Communication system

At the receiving end, the receiver may have to select and perhaps amplify the modulated signal before the demodulator or decoder extracts from it the information signal for delivery to the output transducer. Some form of storage by a tape or disc may also be required.

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Basically the transmitting and receiving ends each consists of an electronic system comprising an input transducer, a processor consisting of several subsystems, and an output transducer.

3.2-Radio waves Radio waves are members of the family of electromagnetic radiation considered in Chapter 4 that you saw in form 3. They are energy carriers which travel at the speed of light c, their frequency f and wavelength 𝝀𝝀𝝀𝝀being related, as for any wave motion, by the equation

v=fλ

Where v = c = 3.0 × 108 m/s in a vacuum (or air). Radio waves can be classified either by their frequency or their wavelength; if one is large the other is small. Frequency is more fundamental since, unlike λ(and v), f does not change when the waves travel from one medium to another. The table 3.1 below shows how they are grouped into different frequency bands. Table: 3. 1 Classification of radio waves

Figure: 3. 5 Transmission of radio waves

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Chapter Three Radio waves can travel from a transmitting aerial in one or more of three different ways, Figure 3.5.

a) Surface or ground waves These follow the earth’s surface and have a limited range, being greatest (1500 km) for long waves but much less for VHF. b) Sky waves These travel skywards and if they are below a certain critical frequency (typically 30 MHz) are returned to Earth by bouncing off the ionosphere. This consists of layers of air molecules (the D, E and F layers), stretching from about 80 km above the Earth to 500 km, which have become positively charged (i.e. ionized) by the removal of electrons due to the Sun’s ultraviolet radiation. The critical frequency varies with the time of day and the seasons. Sky waves of low, medium and high frequencies can travel thousands of kilometers. c) Space waves These give straight-line transmission over 100 km or so on Earth in the absence of intervening obstacles such as hills and buildings, and are the way VHF, UHF and microwaves travel. They can also penetrate the ionosphere so satellite communication uses this type of transmission. Radio system Radio waves are emitted by aerials when a.c. flows in them but the length of the aerial must be comparable with the wavelength of the wave produced for the radiation to be appreciable. A 50 Hz a.c. corresponds to a wavelength of 6 × 106 m (since v = f λ) where v = 3 × 108 m/s and f = 50 Hz). Alternating currents with frequencies below about 20 kHz are called audio frequency (a.f.) currents; those with frequencies greater than this are radio frequency (r.f.) currents. Therefore, so that aerials are no too large, they are supplied with r.f. currents. However, speech and music generate a.f. currents and so some way of combining a.f. with r.f. is required if they are to be sent over a distance. a) Transmitter In this an oscillator produces an r.f. current which would cause an aerial connected to it to send out an electromagnetic wave, called a carrier wave, of constant amplitude and having the same frequency as the r.f. current. If a normal receiver picked up such a signal nothing would be heard. The r.f. signal has to be modified or modulated so that it ‘carries’ the a.f. This is done in various ways. In amplitude modulation (AM) the amplitude of the r.f. is varied so that it depends on the a.f. current from the microphone, the process occurring in a modulator. This type of transmitter is used for medium and long-wave broadcasting in Britain; Figure 3.7a shown below is a block diagram for one. In frequency modulation (FM), which is used in VHF broadcast, the frequency of the carrier is altered at a rate equal to the frequency of the a.f. but the amplitude

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remains constant, Figure 3.6. This kind of signal is fairly free from electrical interference.

Figure: 3. 6 Frequency modulation (FM)

b) Receiver A block diagram of a simple AM receiver is shown in Figure 3.7b. This tuning circuit selects the wanted signal from the aerial. The detector (or demodulator) separates the a.f. (speech or music) from the r.f. carrier. The amplifier then boosts the a.f. which produces sound in the loudspeaker.

Figure: 3. 7 AM radio system

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Chapter Three Electrical oscillations The production of r.f. currents is impossible with mechanical generators but is readily achieved electrically by using an oscillatory circuit. This contains a capacitor C and a coil L, Figure 3.8a. If C is charged and discharges through L, currents flows backwards and forwards round the circuit, alternately discharging and charging C, first one way then the other. The frequency of the a.c. produced depends on C and L; the smaller they are the higher is the frequency and each circuit has a natural frequency of oscillation. Due to the resistance of the coil, the current eventually stops and a decaying or damped oscillation is obtained, Figure 3.8b.

Figure: 3. 8

Slow damped oscillations (about 2 Hz) can be shown using the circuit of Figure 3.9 with the CRO on its slowest time base speed.

Figure: 3. 9 Demonstration circuit

Tuning circuit Different transmitting stations send out radio waves of different frequencies. Each induces a signal in the aerial of a radio receiver but if the aerial is connected to an L-C circuit, Figure 3.10, only the signal whose frequency equals the natural frequency of the L-C circuit is

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selected. If the value of C is varied different stations can be ‘tuned in’. This is an example of electrical resonance.

Figure: 3. 10 Turning circuit

Detection or demodulation In detection or demodulation the a.f. which was ‘added’ to the r.f. in the transmitter is recovered in the receiver. Suppose the amplitude modulated r.f. signal V of Figure 3.11a is applied to the detector circuit of Figure 3.11d. The diode produces rectified pulses of r.f. current I, Figure 3.11b. These charge up C during the positive half-cycles as well as flowing through the earphone. During the negative half-cycles when the diode is non-conducting, C partly discharges through the earphone. The p.d. Vcacross C (and the earphone) varies as in Figure 3.11c if C and R have the correct values. The voltage has been smoothed and apart from the slight r.f ripple Vc has the same frequency and shape as the modulating a.f. and reproduces the original sound in the earphone. Germanium diodes are used as detectors. If a loudspeaker is to be operated, the detector is followed by several stages of a.f. amplifications. A suitable resistor then replaces the earphone.

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Figure: 3. 11 Detection of an AM signals

Simple Radio Receiver

Figure: 3. 12 Simple radio receiving circuit

Connect the circuit of Figure 3.12 on an S-DeC, Figure 3.13. The transistor leads can be lengthened and connections made to the ‘tags’ on the variable capacitor. For an aerial, support a length of wire (e.g. 10 m) as high as you can; making an earth connection to a water tap will improve reception, Figure 3.14. You should be able to tune in to one or two stations (depending on your location) by altering the variable capacitor.

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Figure: 3. 13 Radio receiving built on an S-DeC

Figure: 3. 14 Connect the aerial to earth via a tap

3.3 Television a) Black and white A television receiver is basically a CRO with two time bases. The horizontal or line time base acts as in the CRO. The vertical or frame time base operates at the same time and draws the spot at a much slower rate down to the bottom of the screen and then returns it almost at once to the top. The spot thus draws a series of parallel lines of light (625) which covers the screen, Figure 3.15, and is called a raster.

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Figure: 3. 15 Pattern of lines (raster) on a TV screen

A picture is produced by the incoming signal altering the number of electrons which travel from the electron gun to the screen. The greater the number the brighter the spot. The brightness of the spot varies from white through grey to black as it sweeps across the screen. A complete picture appears every 1/25 s, but because of the persistence of vision we see the picture as continuous. If each picture is just slightly different from its predecessor, the resultant effect is that of a ‘movie’ and not a sequence of ‘stills’. b) Color One type of color television has three electron guns and the screen is coated with about a million tiny light emitting dots arranged in triangles. One ‘dot’ in each triangle emits red light when hit by electrons, another green light and the third blue light. As the three electron beams scan the screen, an accurately placed ‘shadow mask’ consisting of a perforated metal plate with about one- third of a million holes ensures that each beam strikes only dots of one ‘color’, e.g. electrons from the ‘red’ gun strike only ‘red’ dots, Figure 3.16.

Figure: 3. 16 Colour TV screen

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When a triangle of dots is struck it may be that the red and green electron beams are intense but not the blue. The triangle will emit red and green light strongly and appear yellowish. The triangles of dots are struck in turn, and since the dots are so small and the scanning so fast, we see a continuous color picture. c) Energy transfers Electrons emitted by an electron gun in the cathode ray tube of a television receiver or a CRO are accelerated towards the screen by the high p.d. which is applied across the tube. If this V and an electron has charge e and a massm, then an amount of electrical energy eV (from W = QV) is transferred to kinetic energy of the electron. Therefore 𝟏𝟏𝟏𝟏 𝐤𝐤𝐤𝐤. 𝐞𝐞𝐞𝐞. = 𝐦𝐦𝐦𝐦𝐯𝐯𝐯𝐯 𝟐𝟐𝟐𝟐 = 𝐞𝐞𝐞𝐞𝐞𝐞𝐞𝐞 𝟐𝟐𝟐𝟐

Where V is the final velocity of the electron. This k.e. is transferred to heat and light when the electron hits the screen We can calculate the k.e. if we take 𝑒𝑒𝑒𝑒 = 1.6 × 10−19 𝐶𝐶𝐶𝐶 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑉𝑉𝑉𝑉 = 25 𝑘𝑘𝑘𝑘𝑉𝑉𝑉𝑉 𝑡𝑡𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑎𝑎𝑎𝑎 𝑘𝑘𝑘𝑘. 𝑒𝑒𝑒𝑒. = 𝑒𝑒𝑒𝑒𝑉𝑉𝑉𝑉 = 1.6 × 10−19 𝐶𝐶𝐶𝐶 × 25 × 103 = 4.0 × 10−15 J

Exercise 1. List the limitations of sending information by: a) a line-of-sight method, b)

telegraphy,

c) Telephony. 2. a) The Figure below represents a radio.

Figure: 3. 17

Which of the blocks in the diagram: (i)

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detects radio waves;

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Chapter Three (ii)

increases the voltage of the sound signal;

(iii)

Produces sound when there is a varying current in it?

b) The Figure below represents a radio wave carrying a sound signal.

Figure: 3. 18

(i)

What is happening to the amplitude of the radio wave?

(ii)

Use similar axes to sketch the sound signal that the radio wave is carrying.

3. In the Figure below shows the circuit of a simple radio receiver, a) State the names of the parts labelled A, B, C. b) What is the purpose of the part labelled D?

Figure: 3. 19

4. Ayan is listening to a radio as shown in the figure 3.20

Figure: 3. 20

a) Complete the passage below using words from the table 3.2 below

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Table: 3. 2

Sound

Amplifier

Light

Microphone

Aerial

Battery

Tuner

Decoder

Electrical

The ................................................. of a radio receives detect signals from many different stations and convert them into electrical signals. The ................................................. selects one particular station from many. The .................................................... increase the amplitude of these electrical signals. The energy required to do this is supplied by the............................................................. The loudspeaker in a radio receiver converts ..................................................energy into...................................energy. 5. a) The names of five basic blocks used in communications systems are shown in the Figure below. The uses of these blocks are shown on the right. The uses are not in the correct order. Copy and complete the diagram by drawing a line between each block and its correct use. One has been done for you.

Figure: 3. 21

b) People can communicate by talking and listening. Different parts of the body have different functions in this method of communication. Complete the table 3.3 using words from the box.

Table: 3. 3

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Chapter Four

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4.0-Introduction In 1896, Henri Becquerel discovered almost by accident that some substance (like uranium and radium) can blacken a photographic film even in the dark. These substances are radioactive material. In 1898, Marie Curie and her husband discovered two other radioactive elements namely polonium and radium. Radioactivity is the emission of sub-atomic particles or rays from the nucleus of an atom. Elements which emit radiation from their nuclei are called radioactive elements. Radioactivity arises from unstable nuclei which may occur naturally (natural radioactivity) or be produced in artificially in nuclear reactors (induced radioactivity). Most of the elements having atomic number greater than 82 emit invisible radiation naturally. Induced radioactivity can be done by forcing an extra neutron to the nucleus which disturbs the delicate balance of the neutrons and protons in the nucleus. The nucleus becomes unstable and the result is radioactive change.

4.1 Characteristics of radioactive materials  They are continually decaying (breaking down) into simpler atoms as a result of emitting radiations. The decay of radioactive materials is a random process.  Radiations from radioactive elements produce bright flashes when they strike fluorescent compounds such as zinc sulphate.  Radiations from radioactive materials cause the ionization of air molecules.  Radiations from radioactive element penetrate the photographic film and blacken it.  Radiations from radioactive materials can destroy the germinating power of plant seeds and can kill bacteria.

4.2 Types of radiations Whenever an unstable nuclide disintegrates it emit one or more of these three radiations. 1. Alpha (α) radiation 2. Beta (β) radiation. 3. Gamma radiation. Most radioactive substances emit two kinds of radiations simultaneously (usually alpha and gamma or beta and gamma)

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Chapter Four Alpha, beta and gamma radiation There are three main types of nuclear radiation alpha, beta and gamma which can be identified by their penetrating power, ionizing ability, speed, mass, charge, and effects in an electric and magnetic fields. The figure 4.1 below shows the penetrating power of these three types of radiations:

Figure: 4. 1

The passage of alpha, beta and gamma through a magnetic field and electric field is as shown figure 4.2 below:

Figure: 4. 2

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167

Alpha radiation (α)  It is positively charged particle from the nucleus of the radioactive elements.  It is helium nucleus, this means its charge is +2 and its mass is 4. It can be denoted as      

𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐𝜶𝜶𝜶𝜶

or 𝟒𝟒𝟒𝟒𝟐𝟐𝟐𝟐𝑯𝑯𝑯𝑯𝑯𝑯𝑯𝑯. It travels at about 10% of speed of light. It has a high power of ionization of air molecules since it has +2charge it attracts electrons from nearby atoms in the gas. Alpha particles have a least penetrating power. It travels only a few centimeters in air and can be stopped by a thin sheet paper or the outer layer of the skin. Its ionization is concentrated in a small volume due to its low speed and low penetrating power. Alpha can be detected on a photographic film, in a cloud chamber (it is a straight thick tracks, all of about the same length). Examples of alpha emitters are radium, uranium, thorium and the good laboratory source of alpha radiations is americium-241.

Beta radiation (β) The term beta is referred to an electron from the nucleus of the atom. Electron do not exist in the nucleus but one neutron is changed into proto-electron pair as shown below then the electron can be lost from the nucleus. 𝟏𝟏𝟏𝟏 𝟎𝟎𝟎𝟎𝒏𝒏𝒏𝒏

= 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝑷𝑷𝑷𝑷 + −𝟏𝟏𝟏𝟏𝟎𝟎𝟎𝟎𝑯𝑯𝑯𝑯

 Bata particles are fast moving electrons and relatively charge of -1.  They are much lighter than alpha and are usually more penetrating having ranges of several meters in air or they can be stopped by few centimeter of aluminum  They travel about 90% of speed of light.  They are much less ionizing power than alpha but are more easily deviated by the field.  Beta particles can be detected by Geiger-Muller (Gm) counters (it is very thin and twisted tracks), cloud chamber or photographic film.  Examples of some pure beta emitters: carbon-14, tritium, sulfur-35 and its good laboratory source is strontium-90.

Gamma radiation (γ)  A gamma radiation is high-energy electromagnetic radiation emitted from the nucleus of an atom. It is not a stream of particles.  It travels at about speed of light.  It has no net charge.  Gamma radiations have low ionizing power and have high penetrating power. They are never completely absorbed but their intensity can be reduced by several centimeters of lead.  Gamma emission generally occurs after an alpha or beta.

168

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Chapter Four  Gamma rays can be detected by a photographic film, cloud chamber and GeigerMuller counter.  Examples of some gamma emitters: iodine-131, cesium-137, radium-226, and technetium-99m. Its main laboratory source is cobalt-60. Table: 4. 1 below summarizes the properties of the three types of radiations:

Type of radiation Nature of radiation

Alpha (α)

Beta

(β)

Gamma

()

Nucleus of Helium

High speed of electron

Electromagnetic wave

+2

-1

0

Up to 10% of speed of light

Up to 90% of the speed of light

Speed of light

Ionizing effect

Strong

Weak

Very weak

Effects of fields

Deflected by fields

Deflected by fields

Not deflected

Not very penetrating , stopped a thick sheet of paper

Penetrating, but stopped by few centimeters of aluminium

Very penetrating, never completely absorbed, through lead may reduce its density.

Relative charge Speed

Penetrating power

4.3-Nuclear Equations In nuclear changes, just as a chemical change, nothing is very lost or gained. The total mass number and total atomic number remain the same. The mass number and atomic number of an atom can be written in the form of: 𝐒𝐒𝐒𝐒𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦 𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒

𝐀𝐀𝐀𝐀 𝒁𝒁𝒁𝒁𝐗𝐗𝐗𝐗

Where A= N + P

And Z = n (P) .  E.g. Carbon -14 is written as 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟔𝟔𝟔𝟔𝐂𝐂𝐂𝐂; and uranium -328 is written as 𝟗𝟗𝟗𝟗𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝐔𝐔𝐔𝐔 𝐀𝐀𝐀𝐀 A general atom X is written as where A= N + P and Z = n(P) 𝒁𝒁𝒁𝒁𝐗𝐗𝐗𝐗  Neutron … 𝟏𝟏𝟏𝟏𝟎𝟎𝟎𝟎𝐦𝐦𝐦𝐦 ;  Proton…… 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝐏𝐏𝐏𝐏;  Electron... −𝟏𝟏𝟏𝟏𝟎𝟎𝟎𝟎𝐦𝐦𝐦𝐦

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 Alpha particle  Beta particle

𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐∝ 𝟎𝟎𝟎𝟎 −𝟏𝟏𝟏𝟏𝛃𝛃𝛃𝛃

𝟒𝟒𝟒𝟒 𝟐𝟐𝟐𝟐𝐇𝐇𝐇𝐇𝐇𝐇𝐇𝐇

or

Remember: In any nuclear equation, the total mass number and atomic on the LHs must equal to the total mass number and atomic on the RHS.

Examples 1) When a neutron bombards a nitrogen atom, carbon -14 and other nuclide result. Identify the unknown nuclide? 𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒 𝟕𝟕𝟕𝟕𝐍𝐍𝐍𝐍

+ 𝟏𝟏𝟏𝟏𝟎𝟎𝟎𝟎𝐧𝐧𝐧𝐧

𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒 𝟔𝟔𝟔𝟔𝐂𝐂𝐂𝐂

A + 14 = 15 and therefore A= 1

+ 𝐀𝐀𝐀𝐀𝒁𝒁𝒁𝒁𝐗𝐗𝐗𝐗

Z+ 6 = 7 and therefore Z = 1

So the unknown particle is hydrogen since it has a mass number of 1 and an atomic number of 1 𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒 𝟕𝟕𝟕𝟕𝐍𝐍𝐍𝐍

+ 𝟏𝟏𝟏𝟏𝟎𝟎𝟎𝟎𝐧𝐧𝐧𝐧

𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒 𝟔𝟔𝟔𝟔𝐂𝐂𝐂𝐂

+ 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝐇𝐇𝐇𝐇

2) What other product is formed when polonium -218 decay to Astatine? 𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟖𝟖𝟖𝟖𝟒𝟒𝟒𝟒𝐏𝐏𝐏𝐏

𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀

218 = 218 + A

So

84 = 85+Z

+ 𝐀𝐀𝐀𝐀𝒁𝒁𝒁𝒁𝐗𝐗𝐗𝐗

A= 218 – 218 = 0 So

Z = -1

The unknown particle is an electron 𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀

𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝟖𝟖𝟖𝟖 𝟖𝟖𝟖𝟖𝟒𝟒𝟒𝟒𝐏𝐏𝐏𝐏

Exercise

+ −𝟏𝟏𝟏𝟏𝟎𝟎𝟎𝟎𝐇𝐇𝐇𝐇

1. When two atoms of hydrogen 𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝐇𝐇𝐇𝐇 are fused to gather they form a neutron and one new atom. What is that new atom? 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝐇𝐇𝐇𝐇

+ 𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝐇𝐇𝐇𝐇

2. Which of these three types of radiation (i) Is form of electromagnetic radiation? (ii) Has no charge? (iii)Carries positive charge? (iv) Travel at the speed of light? (v) Is made up of electrons?

170

𝟏𝟏𝟏𝟏 𝟎𝟎𝟎𝟎𝐧𝐧𝐧𝐧

+ 𝐀𝐀𝐀𝐀𝒁𝒁𝒁𝒁𝐗𝐗𝐗𝐗

................

Chapter Four (vi) Has the most ionizing effect? (vii) Can be stopped few centimeters of aluminum? (viii) Can be penetrating a thick sheet of lead?

4.4 Effects on the Nucleus Alpha decay In an alpha decay, the atomic number Z of the nucleus goes down by 2 and the mass number A, goes down by 4, such a change is called a nuclear transmutation. 𝐀𝐀𝐀𝐀−𝟒𝟒𝟒𝟒 𝟒𝟒𝟒𝟒 𝐀𝐀𝐀𝐀 𝒁𝒁𝒁𝒁𝐗𝐗𝐗𝐗 𝒁𝒁𝒁𝒁−𝟐𝟐𝟐𝟐𝐗𝐗𝐗𝐗 + 𝟐𝟐𝟐𝟐∝ For example, uranium-238 producing thorium-234 238 234 + 42𝛼𝛼𝛼𝛼 92𝑈𝑈𝑈𝑈 90𝑇𝑇𝑇𝑇ℎ Uranium thorium alpha

Figure: 4. 3

Beta radiation A beta particle is an electron. When an electron is ejected from the nucleus, the atomic number “Z” is increased by 1 (because a new proton has been created), but the mass number “A” remains unchanged. 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 + −10𝑒𝑒𝑒𝑒 𝑍𝑍𝑍𝑍𝑋𝑋𝑋𝑋 𝑍𝑍𝑍𝑍+1𝑌𝑌𝑌𝑌 Unstable nucleus daughter nucleus beta particle  A beta particle is an electron. Electrons do not exist inside the nucleus but can be produced if a neutron changes into electron-proton pair For example, when strontium-90 undergoes a beta decay the daughter nucleus becomes Yttrium-90. 90 90 + −10𝑒𝑒𝑒𝑒 38𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 39𝑌𝑌𝑌𝑌 Strontium Yttrium beta particle

Figure: 4. 4

Gamma emission Gamma rays are e.m radiation, the emission of an alpha or beta particle from a nucleus leaves the protons and neutron in an “exited” arrangement. As the protons and neutrons

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171

rearrange to become more stable, they lose energy. This is carried as burst of gamma radiation. When a nucleus emits gamma rays, the nucleus keeps the same atomic and mass number. For example: cobalt-60 is a common gamma-emitting nuclide ∗60 60 γ 27𝐶𝐶𝐶𝐶 27𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 + Unstable cobalt stable cobalt Gamma

Figure: 4. 5

Radioactive decay Radioactive decay is the process by which unstable nucleus breaks down or disintegrate releasing radiation. Disintegration Series (Decay Series) When an unstable nuclide disintegrates, it forms a new nuclide. The new nuclide may also be unstable; hence it also decays to give a further nuclide. This process continuous until a stable nuclide is formed this process is called decay series. For example uranium decays: “Heavy” atoms (greater than Bismuth, #83) naturally decay to smaller atoms along a consistent path, or series, of decays. Radioactive U-238 → Th-234 + 𝛼𝛼𝛼𝛼 Th-234 → Pa-234 + 𝛽𝛽𝛽𝛽 Pa-234 → U-234 + 𝛽𝛽𝛽𝛽

U-234 → Th-230 + 𝛼𝛼𝛼𝛼

Figure: 4. 6

Th-230 → Ra-226 + 𝛼𝛼𝛼𝛼

Ra-226 → Rn-222 + 𝛼𝛼𝛼𝛼 Rn-222 → Po-218 + 𝛼𝛼𝛼𝛼 Po-218 → Pb-214 + 𝛼𝛼𝛼𝛼

172

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Chapter Four Pb-214 → Bi-214 + 𝛽𝛽𝛽𝛽

Bi-214 → Po-214 + 𝛽𝛽𝛽𝛽

Po-214 → Pb-210 + 𝛼𝛼𝛼𝛼 Pb-210 → Bi-210 + 𝛽𝛽𝛽𝛽

Bi-210 → Po-210 + 𝛽𝛽𝛽𝛽

Po-210 → Stable Pb-206 + 𝛼𝛼𝛼𝛼

Exercise 1. How many alpha particles are emitted when uranium 234 decays to lead 214 as shown below? 𝟗𝟗𝟗𝟗𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝟗𝟗𝟗𝟗𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝐔𝐔𝐔𝐔 → 𝟖𝟖𝟖𝟖𝟗𝟗𝟗𝟗𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏+ X( 𝟗𝟗𝟗𝟗𝛼𝛼𝛼𝛼) 2. An isotope of uranium -235 can be written U. what is the significance of number of 235 and 92. a) Complete the following radioactive decay equation 𝒚𝒚𝒚𝒚 𝟗𝟗𝟗𝟗𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝟎𝟎𝟎𝟎 𝟖𝟖𝟖𝟖𝟗𝟗𝟗𝟗𝐗𝐗𝐗𝐗 → 𝒙𝒙𝒙𝒙𝐘𝐘𝐘𝐘+ −𝟐𝟐𝟐𝟐𝛃𝛃𝛃𝛃 + 𝟗𝟗𝟗𝟗𝛼𝛼𝛼𝛼 b) Find the values of P and q in the equations. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 𝐏𝐏𝐏𝐏 i. 𝒒𝒒𝒒𝒒𝐗𝐗𝐗𝐗 → 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝐘𝐘𝐘𝐘 +2𝛼𝛼𝛼𝛼 𝒑𝒑𝒑𝒑

𝟗𝟗𝟗𝟗𝟐𝟐𝟐𝟐𝟖𝟖𝟖𝟖 ii. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝐗𝐗𝐗𝐗 → 𝒒𝒒𝒒𝒒𝐘𝐘𝐘𝐘 + 𝟓𝟓𝟓𝟓𝛼𝛼𝛼𝛼 3. A radioactive nuclide emits an alpha particle followed by beta particle. The mass number and atomic number of the resulted nucleus respectively are. a) A + 4, Z + 1 b) A-4, Z-1 c) A, Z+1 d) A-4, Z-2

4.5 Isotopes, Isobars and Nuclides

Isotopes are atoms of same element which have different mass number because they have different number of neutrons in their nuclei. E.g. Cl-35 Cl-37 and H-1, H-2 H-3. Isobars are atoms of different elements which have same mass number. E.g. Ra-288, Ac-288 and thanum-288. Nuclides are atoms which have different nuclei; whether the difference is the number of protons, the number of neutrons or both. Background radiation Nuclear radiations can damage or destroy living cells. It may upset the chemical instructions in cells so that cell may grow up abnormally and can cause cancer. However there is small amount of radiation around us all the time because of radioactive material in the

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environment. This is called background radiation. It mainly comes from natural source such as soil, rocks building materials, air, food, Drinks and cosmic rays from space it. Nuclear power station and nuclear bombs also increase back ground radiation. The pie chart below shows the distribution of nuclear radiations in the world. Figure: 4. 7

4.6 Rate of decay and half – life

A very unstable nuclide decays more frequently than with greater stability, but in every case the decay follows the same rule. “The rate of radioactive decay is proportional to the number of nuclei present” Rate of decay α N or rate of decay = λN Where λ = constant of proportionality and each nuclide have different value for λ. The slope of the graph of the number of atoms “N” against time “t” tells us the rate of decay. The curve of number of atoms against time (decay curve) is an exponential curve. This means it does not touch the time axis, so the number of atoms do not reach zero.

Figure: 4. 8

In the exponential decay the time taken to halve the number of active nuclei is constant. Every radioactive element has its own definite decay rate, expressed by its half-life. “The average time for half of the atoms in a given sample to decay is called half-life. In other wards it is the time taken for the activity of a given sample to fall to half the original value. It can be denoted as t½. The decay curve of the figure 4.9 below shows the activity of sample of radioactive substance against time.

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Figure: 4. 9

The activity falls by the same fraction in successive time intervals. It falls from 80Bq to 40 Bq in 10 minutes; from 40Bq to 20Bq in the next 10 minutes, from 20Bq to 10Bq in the third minutes and so on. The half life of the sample is 10 minutes. Note:  Radioactivity is measured in becquerels (Bq). 1 Bq = 1 nuclear disintegration per second The bequerel is a very small unit. In elementary work, quantities of kBq to MBq range are common. Quantities of radioactive materials are commonly estimated in bequarels rather than mass, because equal masses of radioactive materials with different half lives have different activities.  An older unit of radioactivity was the currie (Ci). currie is a very large unit (1Ci = 3.7x1011Bq) so quantities of µCi and mCi range are used. Different nuclides have different half-lives as shown in the table below: Table: 4. 2 shows some materials and their half-lives

Element Uranium Plutonium Carbon Iodine Sodium Polonium-218 Radon Polonium-214

Half life 4500 x106 years 24000 years 5600 years 8 days 15 hours 3 minutes 54 second 1.6 x10-4 seconds

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Half-life of a nuclide is a characteristic of that nuclide; it is governed by the structure of the nucleus. It is not affected by the temperature, pressure and any other physical condition. The number of nuclides that is not decayed “N” after a time “T” is found by: N = No (½) T/t

where No = original number of atoms,

N = number of un-decayed atom or remain atom t = half life

Or the activity of the nuclide “A” after a time “T” can be found by: A = Ao(½ )T/t

where Ao = original activity, t = half life.

Examples 1. In a laboratory there are 12g of radioactive nuclide, if the half live of the radioactive nuclide is 5 days, how many atoms remain after 20 days? Answer: N = No (½) T/t = (12g) (½)20/5 = (12g) (½)4 = 12g/16 = 0.75g will remain. 2. A radioactive nuclide has a half-life of 8 day a) What fraction of the original number of atoms of the isotope will be left after 32 days? b) If there are 10,000 atoms in the sample at present, how many atoms will remain after 32 days? Answer a) N = No (½) T/t = (1) (½)32/8 = (½)4 =1/16 of the original atom will be left after 32 days b) N = No (½) T/t = (10, 0000) (½)32/8 = (10,000) (½)4 = 625 atoms will remain 3. Carbon-14 has a half life of 5700 years. A 10g sample of wood cut recently from a living tree has an activity of 160 counts per minute. A piece of charcoal taken by from a prehistoric campsite also weighs 10g but has an activity of 40counts per minute. Estimate the age of the charcoal? Answer: A = Ao(½ )T/t

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Chapter Four 40 = 160(½) T/5700 1 4

……… by dividing both sides by 160, we get

= (½) T/5700. Solving for T, we get T = 2 x 5700years = 11400 years

Exercise 1. A radioactive substance has decayed to 1/128 of its original activity after 49 days. What is its half life? 2. 800g of radioactive nuclide has a half life of two hours. Calculate How many grams will remain; i. After 6 hours ii. After 10 hours 3. In an experiment to find the half-life of radioactive nuclide the count rate falls from 120Bq to 15 Bq in 30 days. What it half-life? 4. 16000 of radioactive nuclide has a half-life of 20 days , how many grams will remain a) After 60 days b) After 100 days 5. A radioactive nuclide has a half- life of 20 minutes. a) What fraction from the original nuclide will remain one hour? b) If there are 8000 atoms in the sample at present how many un-decayed atoms will remain after one hour. 6. A bone of living human being contains 160 units of carbon 14 an identical bone was found ancient burial around, which contains 10 units of carbon 14. The half-life of carbon 14, is 5730 yrs. calculate the age of skeleton? 7. A radioactive isotope has a half-life of 2.5h a) At what fraction from the original nuclide will remain after 20 hours? b) If there are 4000g in the sample at present how many grams will remain after 20 hours? Radiation detectors 1. Photographic film Photographic plates can be used to detect radioactive rays, because such plates are blackened by the rays. 2. Ionization detector The ionization property of the radiations is used to detect and measure the radiations.

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3. Spark counter This contains a positively charged grid and a negatively charged plate.

Figure: 4. 10

When radioactive rays pass between a positively charged wire and a negatively charged metal plate, ionization occurs. A large current is produced and pass suddenly between the grid and the plate, sparks can be seen and heard. 4. Cloud chamber

Figure: 4. 11

The base of the chamber is cooled by dry ice to a bout -800C. A felt ring inside the top of the chamber is moistened with alcohol. When radiations are emitted from a radioactive substance, it produces ions along its path and alcohol vapor from felt ring condenses around these ions that can be seen as narrow white lines.

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Figure: 4. 12

5. Scintillation Detectors This works by the radiation striking a suitable material (such as Sodium Iodide), and producing a tiny flash of light. This is amplified by a "photomultiplier tube" which results in a burst of electrons large enough to be detected. 6. Geiger-Muller tube (GM tube)

Figure: 4. 13

When radioactive rays enter the GM tube, the ions which are produced allow a sudden large current pulse to pass through the tube. This pulse can be detected by a scalar or rate meter. A scalar records a total number of counts, whereas a rate meter records the number of pulses or counts per second. 7. The gold leaf electroscope (alpha only) Dry air is normally a good insulator, so a charged electroscope will stay with as it is. Since the charge cannot escape. When an electroscope is charged, the gold leaf sticks out, because the charges on the gold repel the charges on the metal stem. When a radioactive source comes near, the air is ionized, and starts to conduct electricity. This means that the charge can "leak" away, the electroscope discharges and the gold leaf falls.

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Nuclear energy Mass and energy Einstein predicated that if the energy of body changes by an amount E, its mass changes by an amount m given by the equation. E = MC 2 The energy changes in physical and chemical changes are very small, but those in nuclear changes are millions of times greater. Nuclear reactions In nuclear reactor, the chain reaction is steady and controlled so that an average only on neutron from each fission. The reaction rate is controlled by neutron absorbing rods called moderators. The energy released in nuclear reactor is used to produce a super heated steam at high pressure which turns turbo –generators and result powerful electricity is obtained

Figure: 4. 14

There are two types of nuclear reactions and they are: 1. Nuclear fusion The combination of two light nuclei to form a heavier nucleus is called nuclear fusion.

Figure: 4. 15

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Chapter Four For example: two light nuclei of hydrogen (deuterium H-2 & Tritium H-3) can be smashed together to form a nucleus of helium and a neutron.

2. Nuclear fission The splitting of a heavy nucleus into nuclei with smaller masses is called nuclear fission. The fragments weigh very slightly than the original nucleus. The loss mass appears as energy, so large amount of energy is released.

Fission can occur when a nucleus of a heavy atom captures a neutron as shown below: For example: 235 92𝑈𝑈𝑈𝑈

95 1 -11 + 10𝑛𝑛𝑛𝑛 138 56𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵+ 36𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾+ 3 0𝑛𝑛𝑛𝑛+ (4.8 x 10 J)

Nuclear fission differs from early nuclear reaction in three aspects:  The nucleus is deeply divided into larger fragments of equal masses.  The mass decreases or Q value is appreciable  Other neutrons called fission neutrons are emitted in the process which in turn split other nuclei

4.7 Applications and Hazards of Radioactivity Applications of radioactivity Radiations from radioactive substances can be used in several different ways such as: 1. As a thickness gauge If a radioisotope is placed on one side of the moving sheet of material and GM tube on the other, the count-rate decreases if the thickness increases. This technique is used to control the thickness of paper, plastic and metal sheets during manufacture. The manufacture of aluminum foil (for cooking) is a good example. A radioactive source is placed above the foil and a detector below it as shown below.

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Figure: 4. 16

Some of the radioactivity is absorbed by the foil and some passes through to the detector. The thicker the foil, the fewer radiations passes through it to the detector. 2. As a Tracers Radioisotopes can be detected in very small (and safe) quantities, so they can be used as tracers- their movements can be tracked, Example include: a. Checking the function of body organs. For example, to check thyroid function, a patient drinks a liquid containing iodine-132, a gamma emitter. Over the next 24 hours, a detector measures the activity of the tracer to find out how quickly it becomes concentrated in the thyroid gland. b. Tracing the plants uptake of fertilizers from roots to leaves by adding a tracer to the soil water. c. Detecting leaks in underground pipes by adding a tracer to the fluid in the pipes. For tests like those above, artificial radioisotopes with short half-lives are used so that there is no detectable radiation after a few days. 3. Radiotherapy Cobalt-60 is strong gamma emitter. So a high concentrated beam from a cobalt-60 source can be used to kill cancer cells. Treatment like this radiotherapy

Figure: 4. 17

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Chapter Four 4. Radio dating (Archaeology) Some of Radioisotopes have very long half lives and archaeologists use to pinpoint the age of ancient materials.

Figure: 4. 18

5. As an energy source Radioactive materials can be used as the energy source of artificial pacemaker for the heart. Radioisotopes are also used in nuclear electric power systems. Such systems are used in space stations and remote stations on earth. 6. As a sterilizing agent Radiations from radioactive elements can be used to sterilize medical equipment and

food. 7. Smoke detector

Figure: 4. 19

Many homes are fitted with smoke alarm. This contains a weak source made of americium-241. This emits alpha particles which ionize the air, so that it conducts electricity and small current flows. If smoke enters the alarm, it absorbs the alpha particles, the current reduces, and the alarm sounds.

Figure: 4. 20

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Hazards of radioactivity Radiation can affect the people in a number of ways:  It can destroy cells and tissues  It can also change the DNA, causing mutations and effecting the future generation if the sex cells are effected  Radiation can cause skin burns and cancer To reduce hazards from radioactive material People should be shielded from radioactive source by suitable absorber. There should be as large a distance as possible between person and radioactive source.  People who work with radioactive material wear a badge containing photographic film that measure the amount of exposure to radiation.  Radioactive substances should be kept in thick lead boxes.  Minimize time spent near radiation source.  Never eat or drink in the lab. Exercise:  

1. Natural uranium is mainly 238 92𝑈𝑈𝑈𝑈. About 0.7% of natural uranium is atoms are known as isotopes. (a) Explain what is meant by the term isotope. (b) Complete the following table for U-238.

235 92𝑈𝑈𝑈𝑈

. These two

(c) Plutonium is a fuel sometimes used in nuclear reactors. One possible nuclear fission reaction for plutonium is:

(i)

Calculate the value of x and state the conservation law used to calculate its value.

(ii)

Calculate the value of yand state the conservation law used to calculate its value.

(d) Stars produce their energy by nuclear fusion. In stars larger than our sun this is achieved by the Carbon–Nitrogen–Oxygen cycle. One possible reaction for this cycle is:

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(i)

Complete the following equation by writing the correct symbols for proton and X; and the correct numbers for m and n in the given square brackets.

2. The following equation represents a nuclear reaction.

(a) Calculate the numbers represented by the letters X and Y in the above nuclear equation. i) Number X ii) Number Y (b) There are more neutrons produced than used in this fission reaction. Explain the importance of these neutrons in a nuclear reactor. (c) The above reaction is a fission reaction. State the meaning of the term fission reaction. (d) In the sun a nuclear reaction takes place between a deuterium 12𝐻𝐻𝐻𝐻 nucleus and a tritium 3 1𝐻𝐻𝐻𝐻 nucleus. The products of this reaction are a helium nucleus and a neutron. (i) Complete a balanced equation for this nuclear equation. Give the correct symbols for the products of the reaction.

(ii)

Name this type of reaction.

3. An equation for a nuclear reaction is:

(a) Calculate the value of x. (b) The following equations represent three different nuclear reactions.

Choose from the list above the reaction that represents a nuclear fusion reaction. Give your reason. (c) State what “235” stands for in the symbol 235 92𝑈𝑈𝑈𝑈

(i)

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4.

(a) The following equation represents a nuclear reaction.

(i) State what type of reaction this is. Explain your answer. a) Reaction type: b) Explanation: (ii) Calculate the value of X in the above reaction. (b) Moderators are an essential part of a nuclear reactor. Name a material used as a moderator in a nuclear reactor and explain its purpose. (i) Name: (ii) Explanation: (c) The diagram below shows the components inside a power station with a nuclear reactor.

(i)

With the aid of the above diagram, explain how electricity is produced continuously in a nuclear power station. (d) In a plastic manufacturing factory a beta radiation source is used to measure the thickness of the plastic sheets produced. The setup is shown in the diagram below.

Explain what happens to the nucleus of a radioactive atom when beta decay occurs. (ii) Discuss why beta particles are more suitable than alpha particles or gamma rays for determining the thickness of plastic sheets. (e) Strontium-90 can be used as a beta particle source. Use the table below to complete the equation for the decay of strontium-90. (i)

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(f) The half-life of strontium-90 is 28 years. The radioactive source contains 2g of strontium-90. Draw a decay graph on the grid below to show how the mass of strontium-90 changes over three half-lives.

(g) Gamma rays, another form of radioactive decay, have several uses. Write down one use of gamma rays. 5. Carbon Dating, Fossils are dated using the radioactive decay of carbon-14. Carbon-14 decays with a half-life of 5 730 years. (a) Describe what is meant by the term half-life. (b) The following graph shows how the radioactivity of 1.0 g of processed carbon decreases.

(c)

1.0 g of a similarly processed sample of carbon obtained from a fossil had an activity of 5.0 counts per minute. (i) Use the graph to estimate the age of the fossil. Carbon-14 emits beta particles when it decays.

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(i) Describe the changes in the nucleus of a radioactive atom when Alpha decay occurs. (ii) Complete the equation for the beta decay of carbon-14, using the information in the boxes below. Show all atomic and mass numbers.

(d)

188

The half-life of an unknown material was investigated using a Geiger counter. The initial count rate was 156 per second. After 7 minutes the count rate fell to 39 per second. Calculate the half-life of the material.

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5.0-Introduction 





In 1819, Hans’s Aorist discovered that whenever, an electric current flows through conductor, magnetic field is produced around it, if such conductor is placed in magnetic field, it experiences magnetic force (motor-effect). But by moving a coil in a magnetic field, can an electric current be generated? This question was answered by Michael faraday and Joseph Henry in the year 1831. The two scientists, although working independently, were able to show that an electric current can be generated, this method of producing electricity called electromagnetic induction. For current to flow in a conductor there must be deriving force called electromotive force (emf) so that an emf is first induced in a conductor by magnetic fields in this process, and this induced emf cause an induced current to flow in conductor, if the conductor forms part of complete circuit).

5.1 Definition of electromagnetic induction 



Electromagnetic induction is the phenomenon in which an e.m.f. is induced in the conductor when there is a change in the magnetic flux linked with the conductor. In other word, the processes which in electric currents are induced in a conductor by magnetic field are called electromagnetic induction. In electromagnetic induction, the mechanical energy (i.e. the work done in moving the coil or the magnet to change the magnetic flux) is transformed into electrical energy (in the form of current in the conductor)

Conditions necessary for electromagnetic induction 

Induced electromagnetic force or current in a conductor can be generated in two ways: 1. By relative movement (generator effect) 2. By changing magnetic field (mutual induction or transformer effect)

1) Relative movement (generator effect) a) Moving a magnet through a coil



190

If the north pole of a magnet is pushed into the coil, the needle of galvanometer deflects in one direction (to the right). When the bar magnet is stationary the needle returns to its zero position. When the magnet is being removed out of the coil the needle deflects in the opposite direction (to the left). Thus: whenever, a magnet moves inside a coil a current is induced in the coil. This is also valid if a coil of wire is moved over stationary magnet, only relative motion is needed.

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Chapter Five  

The relative motion between the coil and a magnet produces a current. Faraday made the following three conclusions: 1. A current flows in the coil only when there is a relative motion between the coil and the magnet due to which the galvanometer connected with the coil shows deflection. 2. The direction of deflection in galvanometer is reversed if the direction of motion (or polarity of the magnet) is reversed. 3. The current in the coil (deflection in the galvanometer) is increased  By the rapid motion of the magnet (or coil)  By the use of strong magnet  By increasing the area and number of turns of the coil. b) Moving wire in a magnetic field

A current can also be induced in a wire, when the wire moves in a magnetic field. The figure below shows a copper wire connected to a galvanometer. Part of the wire AB is in a magnetic field. If AB moves at right angles across the magnetic field lines, a current is induced in the wire. The deflection of the needle is opposite to the direction of the movement. There is no current induced if the wire is stationary or moves parallel to the field.

5.3-Faraday’s law Faraday explained that whenever a voltage is induced in a conductor whenever it cuts magnetic field lines, but not when the conductor is parallel with the field or is at rest. Based on faraday’s law: the induced voltage in a wire can be increased by increasing: the speed of motion of the wire of magnet The length of a wire in a magnetic field The strength of the magnet Mathematically those three points can be summarized as an equation. 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵

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Where; EMF = electromotive force or voltage B = magnetic field strength L = length of the wire V = speed of the motion If the wire makes an angle “θ” to the magnetic field, the equation becomes as follows: 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 Induced EMF is maximum when θ =900 “i.e. when the wire moves at right angle to the magnetic field lines” and it is zero when θ = 00 “i.e. when the wire moves parallel to the magnetic field lines.” For the coil of wire of N turns 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = 𝑁𝑁𝑁𝑁𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 Where; N = number of turns of the wire. 

Faraday’s law relates the size of the induced e.m.f. to the change in flux linkage. The law is summarized as the following:

Faraday’s law states that: the size of the induced e.m.f. is proportional to the rate of change of flux linkage. As the proportionality constant is equal to 1, for a uniform rate of change of flux linkage this can be written as: Magnitude of induced e.m.f. N =

𝜟𝜟𝜟𝜟𝜟𝜟𝜟𝜟 𝜟𝜟𝜟𝜟𝜟𝜟𝜟𝜟

where ΔΦ is the change of flux in time Δt.. Direction of induced current in a wire If a straight, wire is moving at right angles to a magnetic field the direction of induced current can be found by using Fleming’s right hand rule as shown below The law says “if the first finger of your right hand indicates the direction of the magnetic field and the thumb shows the direction in which the wire is moved in the field, then your second finger will point the direction of the induced current.:”

 Either of the following will reverse the direction of induced emf or current in a wire. 1. Reversing the direction of motion 2. Reversing the direction of magnetic field lines.

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Chapter Five Examples 1. A wire 15cm long is moving right at a speed of 1m/s at right angle to a magnetic field of strength 0.01T which is directed into the page as shown figure 5.1.5:

a) Find the induced EMF? b) Illustrate an arrow points the direction of the current on the diagram? a) EMF = BLv = (0.01T)(0.15m)(1m/s) = 0.0015V = 1.5 x 10-3V b) By using Fleming’s right hand rule, the direction of the current is from B to A 2. A copper rod “AB” 1.3 m long is pushed along two smooth rails “CD” and “EF” at constant speed of 3 m/s by a constant force of “F”. A 2Ω resistor is connected between “C” and “F” and a magnetic field of 0.35T is directed into the page.

a) Calculate the induced EMF on the rod AB? b) Work out the magnitude and the direction of the current passing through the 2Ω resistor? c) Calculate the magnitude of the constant force? a) EMF = Blv = (0.35T)(1.3m)(3m/s) = 1.365V b) I = V/R = (1.365V)/(2Ω) = 0.682A and by using Fleming’s right hand rule its direction is from “B” to “A”

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c) F = BIL = (0.35T)(0.682A)(1.3m) = 0.31N 3. A wire 15 cm long is moved upwards at speed of 4 m/s at right angles to a magnetic of 3 T. The field goes from left to right as shown in the figure 5.1.9. Find the EMF induced and illustrate its direction in the diagram Use

ʘ out the page Into the page

Ans EMF  LV  3(0.15)(4)  1.8V

Using FRHR we find the induced current goes out of the page along the wire. Exercise 1. The figure 5.1.7 below shows a conductor AB in a magnetic field.

a) Mark in the direction of the magnetic field? b) Which direction will a current flows in the conductor AB when it is i.

Moved into the page?

ii.

Moved out of the page?

2. A 5cm length of wire passes through a magnetic field at right angles the field with speed 12m/s. the field is 1.5T. Calculate the induced EMF?

3. A conducting rod of length 0.30 m and resistance 10.0 Ω moves with a speed of 2.0 m/s through a magnetic field of 0.20 T which is directed out of the page.

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a) Calculate the EMF induced in the rod? b) Find the magnitude of the current in the rod and the direction it flows. c) Calculate the force acting on the rod as it moves through the coil? An induced Emf in a coil (Lenz’s law) This law is named after the German physicist Heinrich Friedrich Lenz (1804–1865), who announced it in 1833. Lenz's law enables us to determine the direction of the induced current in the coil.

Lenz’s law When a magnet is moved into and out of a coil, the induced current that flows through the coil can be determined from Lenz's Law. Lenz's Law states that the induced current always flows in the direction that opposes the change producing it. Lenz's Law obeys the principle of conservation of energy. Work is done to move the magnet against the repulsive force. This work done is converted to electric energy which manifests as an induced current. 

Combining Faraday’s and Lenz’s laws gives the equation for induced e.m.f.:

Where ε is the induced e.m.f. The negative sign shows that the induced e.m.f. is in opposition to the change of flux causing it.

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In figure 5.1.10 (a) a north pole of a bar magnet is pushed into the coil. The induced current must be such that it turns into an electromagnet with North Pole at end facing the north pole of the magnet. The magnet is repelled and its motion is opposed.  In figure 5.1.10(b), the north pole of the magnet is pulled out of the coil. The induced current turns the end of the coil facing the north pole of the magnet into south-pole. The magnet is attracted and its motion is opposed.  In figure 5.1.10 (c) when the S pole of a magnet moves towards end of a coil, the end will become a magnetic S pole to oppose the motion of the magnet.  In figure 5.1.10 (d) when the S pole is pulled away from end of the solenoid, the end will become a magnetic N pole so as to oppose the motion of the magnet.  Lenz’s law is based on the principle of conservation of energy. The work done in moving the magnet is converted to electrical energy. o The pole-type (north or south) is controlled by the direction in which the current is induced. The direction of the current is given by the right-hand grip rule as shown figure 5.1.11. 

The fingers point in the conventional current direction and the thumb gives the North Pole.

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Chapter Five ACTIVITY 5.1. Title: Induced voltage in a conductor Aim: To investigate the factors that affects the size of an induced voltage in a conductor. Apparatus: 2 magnadur magnets, centre zero voltmeter, wire, a bar magnet, a 120 turn coil, a bicycle dynamo and a C.R.O.

Instructions • Move the wire to and fro along the line a) AB and b) CD. • Note the effect on the meter when a) the number of wires cutting the field is increased b) the speed of movement is increased c) the strength of magnetic field is increased. • Note the kind of voltage produced across the wire. • Repeat the procedure with the magnet moved in and out of the coil as shown figure 5.1.14 below.

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Make a list of your conclusions from the above experiments.

Exercise 1. The figure 5.1.14 below shows a magnet being pushed into a coil of wire, which is connected to a galvanometer.

Which of the following statement is/are correct? a) The induced current will flow from A to B through the coil. b) The induced current will flow from B to A through the coil. c) No induced current will flow. d) End B will become a North Pole.

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Chapter Five 2. When the magnet in the figure 5.1.15 shown below is moved towards side B of the solenoid, the pointer of the galvanometer deflects momentarily to the right.

(a) Explain why this happens. (b) What is the process called? (c) Without moving the magnet, how can you produce another momentary deflection of the pointer? (d) State two ways of increasing the size of the deflection of the pointer. (e) Name the main energy change taking place when the magnet is moved into the coil. 4. Electricity can be generated by moving a magnet in a coil of wire. The diagram 5.1.16 below shows a magnet held above a coil of wire.

Experiments with this apparatus can show how the electricity is generated. (a) Draw a straight line from each experiment to what happens on the meter.

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The first line has been done for you.

Applications of electromagnetic induction (generator effect) 1. Generators A generator is a device that converts mechanical energy into electrical energy. There are two types of generators and they are: a) AC generator or AC dynamo (Alternator) AC generator is used to produce an alternating current and its main parts are shown in the figure 5.2.1 shown below:

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In an AC generator the coil is made of insulating copper wire and is rotated in the magnetic field. The slip rings are fixed to the coil and rotate it. The brushed are to two contacts which rub against the slip rings and keep the coil connected to the outside circuit. They are usually made of carbon. ‘



When the coil rotates, it cuts the magnetic field lines, so an EMF is induced. This makes a current to flow. As the coil rotates each side rotates upward, downward, upward, downward and so on, through the magnetic field. So the current flows backward, forward, backward and so on. In other words an AC current is induced.



This is the case as the motion of each side of the coil will change direction relatively to the magnetic field.



The graph above in figure 5.2.1 (b) shows how the current varies in one cycle (rotation).it is maximum when the coil is horizontal and cutting the magnetic field line at the fastest rate. It is zero when the coil is vertical and cutting no field lines.

b) DC generator ( DC dynamo) 

Direct current (DC) is “one-way” current like that from a battery. DC generators are similar in construction to DC motors, with a fixed magnet, rotating coil, brushes and a commutated to reverse the connection to the outside circuit every half-turn as shown in the figure 5.2.2 (a)



When the coil is rotated, alternating current is generated. However, the action of the commutator reverses the connection and this means that the current in the outside circuit always flows in the same way. In other words, it is d.c. The graph in figure 5.2.2 (b) shows how the coil rotates in the magnetic field. Initially the coil is vertical. No cutting of

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magnetic flux occurs and hence induced current does not exist. When the coil rotates, the change in flux increases and the induced current correspondingly increases in magnitude. After rotating by 90°, the coil is in the horizontal position. The change in magnetic flux is maximum and hence the maximum induced EMF is produced. When the coil continues to rotate, the change in flux decreases. At the 180° position, there is no change in flux hence no induced current exists. The induced current is achieves its maximum value again when the coil is at 270°. After rotating 360°, the coil returns to its original position. Remember The only difference in construction between generator and motor is that, there is an electrical appliance (such as bulb or resistor) in the external circuit of the generator while there is a source of current (such as a battery) in the external circuit of the motor. 

In both AC and DC generators, the induced EMF (and current) can be increased by  Increasing the number of turns on the coil.  Increasing the area of the coil.  Using stronger magnet.  Rotating the coil faster.

Table 5.2.1 below shows the difference between an a.c. generator and a d.c generator.

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Chapter Five Bicycle dynamo

Figure 4.2.3: bicycle dynamo

The diagram 5.2.3 above shows a bicycle dynamo. It works in similar way as generators that mentioned above. A magnet is rotated near a coil of wire so that the lines of the flux are cut by the wire and a current is induced that is used to light the lamp of the bicycle. The coil is wound on soft iron core so that the magnetic field is strong. 2. Record player pick-up In moving coil pick up the needle is connected to a pair of coils placed in a magnetic field of permanent magnet. The grooves of the record cause the needle to vibrate. This vibration is transmitted to the coil and the induced a current in the two coils. This induced current is used for the pick-up of the stereo sound production.

Figure 5.2.4: Record player pick-up

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3. A moving coil microphone A moving coil microphone is built like the loudspeaker in reverse. Sound waves from your throat makes the thin diaphragm of the microphone to vibrate, this makes the coil to vibrate. Because the coil is in a magnetic field a current is induced. The current is varied since sound waves are varied. This current is amplified and sent to the loudspeaker where it is changed into sound waves that is matched the original.

4. A play-back of cassette recorder When a recorder audio cassette is played a motor pulls the tape past a play back head. This time the varying magnetic field of the particles in the tape induces a varying current in the wire round the head. The signal is amplified and then sends to the loudspeaker where it is converted into sound waves which matched the original.

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Chapter Five 5.4 Mutual induction (Transformer effect 

A mutual induction is the change in magnetic field of one coil which induces a current in the other coil.



If two coils are arranged side by side as shown figure 5.3.1 below, when a current flows through the first coil magnetic field is set-up around it. If this field is changing (by varying the current or turning it on and off) an emf is induced in the second coil. If the current in the first coil is steady there is no current in the second coil because there is no field change.

The coil that has the input current or a source of voltage is called primary coil and the other coil is called secondary coil. When an iron core is inserted through the two coils, the effect of mutual inductance increases because the field lines become very concentrated. In summary mutual induction occurs if and only if there is a field change in one of the coils. An easy way to do this is by using AC. When the magnetic flux linking with second coil changes, an emf is induced in it causing current to flow, hence a deflection in the galvanometer 



The current in the first coil can be changed 1. By switching the current on and off 2. By varying the current causing a rheostat 3. By applying an a.c. in the first coil. Emf induced in the second coil can increasedby 1. Winding in the primary and the secondary coils on a soft iron core 2. By having more turns on secondary coil 3. Winding the secondary coil on the primary coil.

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

With a steady current through the primary coil no emf is induced b/c the magnetic field is not changing. As the switch is opened an emf is induced in the opposite direction to the secondary coil



The direction of the induced current can be worked out using Lenz’s law.

Example A primary coil “AB” and a secondary coil “CD” are arranged side by side a shown in the figure below. The current in the AB can be changed by sliding rheostat “R” to the left and right.

a) Use an arrow to indicate the direction of the current in the primary coil when switch “S” is closed? b) When the rheostat slides to the left in which direction is the current in the second coil flows? By using right hand grip rule, end “A” is South Pole as the rheostat slides to the left, the resistance decreases and so the current increases. The field strength of the coil increases. Coil “CD” produces a magnetic field to oppose this change. A current will flow in “CD” so that end “D” becomes South Pole. The induced current in “CD” will thus go from “D” to “C” by using right hand grip rule. c) When the current in the coil “AB” is steady (constant), in which direction is the current in the secondary coil “CD” will flow? No current is induced in the “CD” because there is no field change since the current is steady. d) Will the coil “CD” be repelled or attracted by the coil “AB”? “CD” will repel “AB”, because end “A” and end “D” are both South Poles.

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Chapter Five Exercise 1. In the experiment shown below two coils are arranged side by side;

What happens when? a) When the switch is closed (turned on)? b) The switch is then opened (turned off)? 2. In the experiment shown below two coils are arranged side by side.

What would be the effect of a) Extending the iron core so that it goes through both coils? b) Replacing the battery and switch by an AC supply?

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Applications of mutual induction (transformer effect) Transformers

What is a transformer? A transformer is a device which makes use of mutual induction to change voltage. It is a device that changes an alternating voltage for different levels needed. Function of transformer The function of a transformer is to increase or decrease the potential difference of an alternating current supply. Every mains-operated television and record players has one. Battery chargers and model train sets depend on transformer. The whole system of distribution of electricity across the country also depends on them. The transformer transfers electrical energy from one circuit o another, by mutual induction, b/w two coils. Structure and Technical Terms A transformer consist of 3 parts, namely 1. The primary circuit 2. The core 3. The secondary Circuit 1. Primary Circuit The primary circuit is the circuit that connected to the input energy source. The current, potential difference and coil (winding) in the primary circuit are called the primary current (Ip), primary potential difference (Vp) and primary coil respectively.

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Chapter Five 2. Core The core is the ferromagnetic metal wound by the primary and secondary coil. The function of the core is to transfer the changing magnetic flux from the primary coil to the secondary coil. 3. Secondary Circuit The secondary circuit is the circuit that connected to the output of the transformer. The current, potential difference and coil (winding) in the secondary circuit are called the secondary current (Is), secondary potential difference (Vs) and secondary coil respectively.

Working Principle of a Transformer A transformer consists of a primary coil and a secondary coil wound on a soft iron core. When an alternating current flows in the primary coil, a changing magnetic flux is generated around the primary coil. The changing magnetic flux is transferred to the secondary coil through the iron core. The changing magnetic flux is cut by the secondary coil, hence induces an emf. In the secondary coil. The magnitude of the output voltage can be controlled by the ratio of the number of primary coil and secondary coil. Types of Current in a Transformer The current in the primary circuit must be alternating current because alternating current can produce changing magnetic flux. A changing magnetic flux is needed to induce emf. In secondary coil. The induced current in secondary is also an alternating current. The frequency of the alternating current in secondary coil is same as the frequency of the primary current.

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The alternating in the secondary circuit can be converted into direct current by using a pair of diode. The figure on……….. The left shows the symbol of a transformer. The 2 lines in between the coil denote the core.

Types of Transformer There are 2 types of transformer, namely a. the step up transformer b. the step down transformer a) Step-up transformer:

In this kind of transformer 

Input voltage (primary voltage) is less than output voltage (secondary voltage) Vp < Vs



Number of the turns in the primary coil is less than the number of turns in the secondary coil. Np < Ns



The current in the primary coil is greater than the current in the secondary coil. I P> IS

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b) Step-down transformer:



Conversely, a step-down transformer is one where the e.m.f. in the secondary coil is less than the e.m.f. in the primary coil. It is used to reduce the potential difference. Vp > Vs



The number of windings in the primary winding is greater than the number of windings in the secondary coil.



Np > Ns

The current in the primary coil is lesser than the current in the secondary coil. I P< I S

Calculation of Potential Difference Change For a transformer, in which all field lines in primary coil cuts the secondary coil turns ratio is equal to the ratio of the input voltage and the output voltage. 𝑁𝑁𝑁𝑁𝑝𝑝𝑝𝑝 𝑉𝑉𝑉𝑉𝑝𝑝𝑝𝑝 = 𝑁𝑁𝑁𝑁𝑠𝑠𝑠𝑠 𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠

Vp = input (primary) potential difference Vs = output (secondary) potential difference Ip = input (primary) current Is = output (secondary) current

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Power loss in a transformer 

In practice the output power of a transformer is always smaller than the input power of the transformer due to the energy loss in the transformer as heat. If a transformer is 100% efficient (this means there is no energy loss), then; Input power = output power

This can be written as:

Dividing throughout the

equation of IPVP =

ISVS by “IPVS” gives: 𝑉𝑉𝑉𝑉𝑃𝑃𝑃𝑃 𝐼𝐼𝐼𝐼𝑆𝑆𝑆𝑆 = 𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠 𝐼𝐼𝐼𝐼𝑃𝑃𝑃𝑃

In summary

𝑉𝑉𝑉𝑉𝑝𝑝𝑝𝑝 𝑁𝑁𝑁𝑁𝑝𝑝𝑝𝑝 𝐼𝐼𝐼𝐼𝑆𝑆𝑆𝑆 = = 𝑁𝑁𝑁𝑁𝑠𝑠𝑠𝑠 𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠 𝐼𝐼𝐼𝐼𝑃𝑃𝑃𝑃 Examples:

1. A transformer has a primary coil of 1200 turns and a secondary coil of 60 turns. When the primary coil is connected to 240V AC supply the current through it is 0.1A. a) Is the transformer being step-up or step-down transformer? Explain your answer? The transformer is step-down transformer, because the number of turns in the secondary coil is less than the number of turns in t primary coil. b) What is the voltage across the secondary coil? 𝑁𝑁𝑁𝑁𝑝𝑝𝑝𝑝 𝑉𝑉𝑉𝑉𝑝𝑝𝑝𝑝 = 𝑁𝑁𝑁𝑁𝑠𝑠𝑠𝑠 𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠

Therefore, the secondary voltage can be found by: 𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠 =

𝑉𝑉𝑉𝑉𝑝𝑝𝑝𝑝 𝑥𝑥𝑥𝑥𝑁𝑁𝑁𝑁𝑠𝑠𝑠𝑠 240𝑉𝑉𝑉𝑉𝑥𝑥𝑥𝑥 60 = = 12𝑉𝑉𝑉𝑉 𝑁𝑁𝑁𝑁𝑝𝑝𝑝𝑝 1200

𝐼𝐼𝐼𝐼𝑠𝑠𝑠𝑠 =

𝑉𝑉𝑉𝑉𝑝𝑝𝑝𝑝 𝑥𝑥𝑥𝑥𝐼𝐼𝐼𝐼𝑝𝑝𝑝𝑝 240𝑉𝑉𝑉𝑉𝑥𝑥𝑥𝑥 0.1𝐴𝐴𝐴𝐴 = = 2𝐴𝐴𝐴𝐴 𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠 12𝑉𝑉𝑉𝑉

c) What current will flow in the secondary coil?

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2. Calculate the number of turns in the secondary coil for the transformer shown figure 5.3.8 below.

Ns =

Np x Vs 100 x 120V = Vp 240V = 50 turns

3. A step-down transformer has a primary coil of 1200 turns and secondary coil of 40 turns. When the primary coil is connected to 200V AC supply what is the voltage across the secondary coil? 𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠 =

𝑉𝑉𝑉𝑉𝑝𝑝𝑝𝑝 𝑥𝑥𝑥𝑥𝑁𝑁𝑁𝑁𝑠𝑠𝑠𝑠 200𝑉𝑉𝑉𝑉𝑥𝑥𝑥𝑥 40 = = 6.67𝑉𝑉𝑉𝑉 𝑁𝑁𝑁𝑁𝑝𝑝𝑝𝑝 1200

4. A transformer steps down the mains supply from 240V to 6V to operate a machine. a) What is the turn’s ratio? 𝑉𝑉𝑉𝑉𝑝𝑝𝑝𝑝 𝑁𝑁𝑁𝑁𝑝𝑝𝑝𝑝 240𝑉𝑉𝑉𝑉 40 = = = 𝑁𝑁𝑁𝑁𝑠𝑠𝑠𝑠 𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠 6𝑉𝑉𝑉𝑉 1

Therefore the turn’s ratio is 40: 1

b) How many turns are there in the primary coil, if the secondary coil has 50 turns? NP =

Vp x Ns 240V x 50 = = 2000 turns Vs 6V

5. A 240V mains transformer has 1000 turns in the primary and N turns in the secondary. It is used to supply energy to 12V/24W lamp.

a) How many turns in the secondary coil? 𝑁𝑁𝑁𝑁𝑠𝑠𝑠𝑠 =

𝑁𝑁𝑁𝑁𝑝𝑝𝑝𝑝 𝑥𝑥𝑥𝑥𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠 1000𝑁𝑁𝑁𝑁𝑥𝑥𝑥𝑥 12𝑉𝑉𝑉𝑉 = = 50 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑉𝑉𝑉𝑉𝑝𝑝𝑝𝑝 240𝑉𝑉𝑉𝑉

b) What is the efficiency in the transformer if the current drawn from 240V supply is 125mA? Power input = IpVp = (0.125A) (240V) = 30W

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Power output = power of the lamp in the output circuit = 24W Efficiency = (Pout /Pin ) x 100% = (24W/30W)x 100% = 80% 6. The figure 5.3.9 below shows a step-down transformer used to light a bulb.

Calculate a) The readings of the voltmeter “v” when the switch “K” is opened? 𝑉𝑉𝑉𝑉𝑠𝑠𝑠𝑠 =

𝑉𝑉𝑉𝑉𝑝𝑝𝑝𝑝 𝑥𝑥𝑥𝑥𝑁𝑁𝑁𝑁𝑠𝑠𝑠𝑠 240𝑉𝑉𝑉𝑉𝑥𝑥𝑥𝑥 50 = = 12𝑉𝑉𝑉𝑉 1000 𝑁𝑁𝑁𝑁𝑝𝑝𝑝𝑝

b) The readings of ammeter “A” when the switch “K” is closed? I = V/R = (12V)/4Ω = 3A c) The power dissipated in the lamp? P = IV = (3A)(12V) = 36W d) The primary current if the efficiency of the transformer is 90%? Efficiency = (Pout/Pin) x 100% Pin = (Pout x 100) /efficiency = (36W x 100)/90 = 40W Therefore; I = P/V = 40W/240W= 0.167A = 167mA ACTIVITY 5.3

Teacher Demonstration Title: Transformers Aim: To find out how transformers work. Apparatus: One magnadur magnet, 2 iron C cores, 2 coils (60 turns), one 1.5 V cell, one centre-zero voltmeter, one C.R.O. and one signal generator.

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Instructions • State the condition required to produce a voltage across the coil of wire in demonstration 1. • In demonstration 2, explain the meter movement when a) the switch is closed b) the switch remains closed c) the switch is opened. • Describe what happens to the meter when a low frequency a.c. is applied to the primary coil as shown in demonstration 3. • Draw the output CRO trace from the transformer when an alternating voltage is supplied to the input coil as in demonstration 4. • Explain why a transformer will only work continuously with a.c. ACTIVITY 5.4 Title: Transformers Aim: To find the relationship between the number of turns in the primary and secondary coils and the voltage across the primary and secondary coils. Apparatus: One voltmeter (a.c.), low voltage supply, 2 iron C-cores and 60, 120, 250 and 500 turn coils.

Instructions • Set up the apparatus as shown above. • Supply an input voltage of 2 volts to the primary coil.

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• Vary the number of turns on the primary and secondary coils using the 60, 120, 250 and 500 turn coils provided. • For each set up measure the primary and secondary voltages. • Take at least five different sets of results. • Present your results in a table. • Find the relationship between the quantities. ACTIVITY 5.5 Title: Transformers Aim: To compare the input and output power of a transformer. Apparatus: One mains ammeter, 12 V ac supply (transformer), one voltmeter, one ammeter, two 12 V bulbs (36 W, 24 W).

Instructions • Note the values of the input and output currents and voltages. • Calculate the input and output powers. • Repeat with a 24 W bulb. • Tabulate your results. • In which case is the transformer more efficient? • Is this an ideal transformer? Exercise 1. A transformer has a primary coil of 1600 turns and a secondary coil of 40 turns. When the primary coil is connected to a 240V a.c supply the current through it is 2A. a) Is it step-down or step-up transformer? Explain? b) What is the voltage a cross the secondary coil? c) What current will flow the secondary coil? 2. A step-down transformer gives a current of 6A at 12V. If the primary voltage is 240V. find a. primary current b. the power input c. The power output, assuming no power loss.

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Chapter Five 3. A current of 2A is passed through the primary coil (50 turns) of a transformer. The secondary coil has 400 turns. What current would be obtained from this transformer? What causes the power lost in transformers? Most of the energy lost in transformers is lost as heat and is created by one of the following ways: 1. Resistance of the winding: the coil of the wires has some resistance. Some energy is used to overcome this resistance and then heat is produced. This can be reduced by: 

Use thick copper wires of low resistance.



Use coolant to decrease the temperature of the transformer.

2. Eddy current: In the iron core. When the magnetic field in the iron core fluctuates, currents called eddy currents are generated in the iron core which causes heating. This can be reduced: 

Use a laminated iron core whereby each layer is insulated with enamel paint to prevent the flow of eddy currents.

3. Leakage of field lined: Some of the induced magnetic flux from the primary coil is not transmitted to the secondary coil; therefore the EMF in the secondary coil is decreased. This can be reduced by: 

The iron core should form a closed loop in order to minimize the leakage of the field lines.

4. Hysteresis lost: The energy used in the magnetization and de-magnetization of the iron core each time current changes its direction is known as hysteresis. This energy is lost as heat which subsequently heats up the iron core. This can be reduced by: 

Use a soft iron core that is easily magnetized and de-magnetized.

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Electrical Power Transmission: Electrical power is transmitted by power lines overlarge distance from power stations to the consumers.

Power transmission is always done by using AC, because: 

AC generators are simple, cheaper, more reliable and more efficient than DC ones.



AC voltage can easily be stepped up or stepped down by using transformers.

Because of the enormous distance the current has to flow, the resistance of the wire cannot be neglected. When the current passes through the wires power is lost because of the heating effect of the current. For same amount of power (constant amount of power) delivered by the 1

power line current is is small for higher voltages (P = IV, if P is constant𝑉𝑉𝑉𝑉 ∝ 𝐼𝐼𝐼𝐼 ).

The amount of electrical energy converted into heat energy is directly proportional to the square of the current (i.e. P = I2R) so high voltage is used in power lines to reduce power lost in transformers. Examples: 1. Cables of resistance 2Ω supply 2 kW of power. Calculate the power lost in the transmission lines when the voltage is

(a) 200V

(b) 2000V?

a) First calculate the current through the transmission line. I = P/V = 2000W/200V = 10A Then calculate the power lost. P = I2R = (10A)2 (2Ω) = 200W b) First calculate the current through the transmission line. I = P/V = 2000W/2000V = 1A Then calculate the power lost. P = I2R = (1A)2 (2Ω) = 2W

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Chapter Five 2. The figure 5.3.12 shown below shows an electrical power transmission system. The resistance of the transmission wire on each side is 4Ω. A 6V, 24W lamp is operated at its correct rating. Assuming that there is no energy lost in transformers, calculate

a) The current through the lamps? b) The current through the transmission line? c) The power lost in the transmission line? a. I = P/V = 24W/6V = 4A b. The c current through the transmission line is the primary current of the step-down transformer. 𝐼𝐼𝐼𝐼𝑃𝑃𝑃𝑃 =

𝐼𝐼𝐼𝐼𝑠𝑠𝑠𝑠 𝑥𝑥𝑥𝑥𝑁𝑁𝑁𝑁𝑠𝑠𝑠𝑠 4𝐴𝐴𝐴𝐴𝑥𝑥𝑥𝑥 1 = = 0.2𝐴𝐴𝐴𝐴 𝑁𝑁𝑁𝑁𝑝𝑝𝑝𝑝 20

c. P = I2R = (0.2)2(8Ω) = 0.32W

Problems Involving Electricity Transmission.

[ 1. Power Loss during Transmission. 2. The high voltage transmission cable is very dangerous. 3. The cost of the cables is high. 4. Charge leakage may happen between cables and earth.

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5. Pylons may be struck by lightning. 6. Pylons and cables may be struck by light aircraft. What is a national Grid Network? A national Grid Network is a network of cable that connects all the power stations in a country to transmit electricity to the consumers throughout the nation. The advantages of the National Grid Network 1. Reduces power lost during transmission. The potential difference is increased before Transmission. This can reduce the current and hence reduces the energy lost during transmission. 2. Electricity supply is more stable and reliable. This ensures a continuous supply of electrical energy to the whole country. 3. Electric current can be distributed to different users according to the voltage requirement. Transformer is used to step down the voltage to certain level according to the needs of the consumers. 4. Maintenance and repair work can be done at any time. This is because any power stations can be shut down without affecting users in other areas

Exercise 1. What is the advantage of high-voltage power transmission? 2. What is the advantage of AC power transmission over DC transmission? 3. Briefly describe how power is transmitted from a power station to your home? 4. What is the function of the transformer in domestic power transmission? 5. Cables of resistance 8Ω supply 4kW of power. Calculate the power loss in the cable if power is transmitted at a) 200V?

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Glossary A. Air resistance: friction due to the motion of an object through air; proportional to the object’s velocity

Ah, Ampere-hour(s): A measure of battery capacity. A 4Ah battery could, for instance, deliver 1A for 4 hours, 1/2A for 8 hours, etc.

AC generator: an instrument that converts mechanical energy into electrical energy and produces alternating current (current that oscillates back and forth) Acceleration: the rate of change of velocity of an object Acceleration due to gravity: the acceleration of an object towards the centre of a celestial body when the gravitational attraction of the mass of the body is the only force acting on the object

Alternator: An electromechanical device that converts mechanical power into AC electrical power.

Alpha particle: one or more helium nuclei ejected from a radioactive nucleus Amplitude: the distance from the rest position to the maximum displacement for an object in periodic motion; or, for a wave, the distance from the rest position to the maximum point of the crest or minimum point of the trough Anode: the electrode that accepts electrons; in a voltaic cell, the negative electrode; in an electrolytic cell, the positive electrode Antinodal line a stationary line of points caused by constructive interference of individual waves Antinode: positions of maximum amplitude of a standing wave caused by the constructive interference of two individual waves travelling in opposite directions Armature: in a motor or generator, the coil with an iron core that rotates in a magnetic field (same as rotor) Audible: sound frequencies in the range 20 to 20 000 hertz (Hz) Average acceleration: rate of change of velocity depending only on initial and final values Average velocity the quotient of the displacement and the time interval depending only on initial and final values

AM, Amplitude Modulation: A modulation method in which the carrier amplitude changes with the input signal amplitude. Amplifier: An electrical circuit that produces an output that is a replica of the input. The output may be scaled or have increased drive, or it may provide Analog: A system in which an electrical value (usually voltage or current, but sometimes frequency, phase, etc.) represents something in the physical world. The electrical signal can then be processed, transmitted, amplified, and finally, transformed back into a physical quality. AND: Combining two signals so that the output is on if both signals are present. This can be accomplished by an AND logic gate (two inputs, one output which is high if both inputs are high). B. Beat frequency: frequency of envelope wave produced by the superposition of two waves of similar but not identical frequencies Beats: periodic variations in amplitude of a wave caused by superimposing two waves of nearly the same frequency Bel: a unit of sound intensity level; 1 bel 10 pW/m2 Beta particle: high-speed electrons or positrons ejected from a radioactive nucleus C. Centripetal acceleration: the centre-directed acceleration of a body moving continuously along a circular path; the quotient of the square of the object’s velocity and the radius of the circle Centripetal force: the centre-directed force required for an object to move in a circular path

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Circular orbit: an orbit produced by a centripetal force Closed air column: an air column that is closed at one end and open at the other Component of a vector: part of a vector that is parallel to one of the axes of the coordinate system Coherent: light that is in phase (the maxima and minima occur at the same time and place) component wave: a wave that combines with another wave to produce a resultant wave Compression: a region of higher pressure compared to the surrounding medium; longitudinal waves have both compressions and rarefactions Constant acceleration: acceleration that is not changing over a certain interval of time Constant velocity: the velocity that is unchanging in a given time interval Constructive interference: resultant wave has a larger amplitude than its component wave Coherent light that is in phase (the maxima and minima occur at the same time and place) Conservation of mechanical energy: the change in the total mechanical energy (kinetic plus potential) of an isolated system is zero Conservation of momentum: the total momentum of two objects before a collision is the same as the total momentum of the same two objects after they collide Crest: the highest point on a wave Cycle: one complete repeat of a pattern of periodic motion, such as the crest of a wave to the next crest D. Daughter nucleus: the nucleus remaining after a transmutation reaction DC generator: a device that converts mechanical energy into electrical energy and produces a direct current Decibel: the most common unit to describe sound intensity level; 1 decibel 0.1 bel Destructive interference: the situation when a

combined or resultant wave has a smaller amplitude than at least one of its component waves Diffraction: the bending of waves around a barrier Displacement: the change in the position of an object; the difference of the final and initial positions Doppler effect: change in the observed frequency (or wavelength) of a sound due to motion of the source or the observer Dynamics: the study of the motions of bodies while considering their masses and the responsible forces; simply, the study of why objects move the way they do E. Eardrum: a thin layer of skin stretched completely across the interior end of the auditory canal that vibrates in response to sound waves Echo: reflected sound Echolocation: determining position of obstacles and prey by emitting sound pulses and detecting time interval for sound to be reflected Eddy currents: electric currents induced within the body of a conductor when that conductor is subjected to a changing magnetic field Efficiency: the ratio of useful energy or work to the total energy or work input; describes how well a machine or device converts input energy or work into output energy or work Elastic collision: a collision in which both momentum and kinetic energy are conserved Electric generator: a device that converts mechanical energy into electric energy Electromagnetic induction: the generation of a current due to the relative motion of a conductor and a magnetic field Equations of motion: set of mathematical equations describing the motion of an object undergoing

uniform acceleration that relate velocity, displacement, acceleration, and time Exhaust velocity: the backward velocity of the gas ejected from the combustion chamber of a rocket relative to the combustion chamber External force: any force exerted by an object that is not part of the system on an object within the system F. Force of gravity: mutual force between any two masses Frequency: number of cycles of periodic motion completed in a unit of time; frequency is the inverse of the period and is measured in s−1 or hertz Free fall: a situation in which gravity is the only force acting on an object Fundamental frequency: the lowest natural frequency able to produce resonance in a standing wave pattern G. Gamma ray: high-frequency radiation emitted from a radioactive nucleus Generator: a device that converts mechanical energy into electrical energy generator effect: the generation of a current in a coil due to the relative motion of a magnet and a conductor Gravitational potential energy: the potential energy an object has due to its location in a gravitational field; objects at higher altitudes have greater gravitational potential energy than objects at lower altitudes Gravitational field strength: the quotient of the gravitational force and the magnitude of the test mass at a given point in a field; the product of the universal gravitation constant and mass, divided by the square of the distance of a given location from the centre of the object

Gravitational force: infinite range force that operates between all massive particles H. Half-life: the time in which the amount of a radioactive nuclide decays to half its original amount Harmonic: the fundamental frequency and any overtone Hertz (Hz): unit used to measure frequency, defined as s−1 Hooke’s law: states that the applied force is directly proportional to the amount of extension or compression of a spring I. Impulse: the product of the force exerted on an object and the time interval over which the force acts Impulse-momentum theorem: states that the impulse is equal to the change in momentum of an object involved in an interaction Inelastic collision: a collision in which momentum is conserved, but kinetic energy is not conserved Induced current: a current produced in a conductor by the motion of the conductor in a magnetic field induced emf: the potential difference induced in a circuit by a wire moving through a magnetic field or sitting in a changing magnetic field inertia: the natural tendency of an object to stay at rest or uniform motion in the absence of outside forces; proportional to an object’s mass Inertial frame of reference: a frame of reference in which the law of inertia is valid; it is is a nonaccelerating frame of reference Infrasonic: sound frequencies lower than 20 Hz In phase: the periodic motion of two individual systems vibrating with the same frequency and always in the same stage of the cycle

Instantaneous acceleration: the acceleration of an object at a particular moment in time Instantaneous velocity: the velocity of an object at a particular instant in time

Longitudinal wave: a wave in which the particles of a medium vibrate parallel to the direction of motion of the wave; for example, sound waves

Ionizing radiation: radiation of sufficient energy to liberate the electrons from the atoms or molecules

Loudness: the perceived strength of a sound

J. Joule (J): the SI unit of energy or work; equivalent to applying one newton of force on an object over a distance of one meter K. Kinematics: the study of the motions of bodies without reference to mass or force; the study of how objects move in terms of displacement, velocity and acceleration Kinetic energy: the energy of an object due to its motion

L.

M. Mechanical energy: the sum of the kinetic and potential energy Mechanical wave: a wave that travels through a medium as a disturbance in that medium Mechanics: the branch of physics comprising kinematics and dynamics; simply, the how and the why of simple motion Modulation: a process of adding data to an electromagnetic wave by changing the amplitude or frequency Momentum: the product of an object’s mass and velocity Musical instrument: an object that can be used to create sounds through vibration of one or more of its parts; classified as brass, woodwind, string, or percussion N.

Law of conservation of energy: the total energy of an isolated system remains constant; energy may be converted from one form to another, but the total energy does not change Law of inertia: an object remains at rest or continues in straight-line motion unless acted on by an outside force; also known as Newton’s First Law Law of universal gravitation: the force of gravity between any two objects is proportional to the product of their masses and inversely proportional to the square of the distance between their centers Lenz’s law: the direction of the force the magnetic field exerts on the induced current opposes the direction of the motion of the conductor

Newton: the unit of force required to accelerate 1 kilogram of mass at a rate of 1.0 m/s2 Newton’s laws of motion: three fundamental laws of motion which are the basis of Newtonian mechanics Nodal line: a stationary line in a medium caused by destructive interference of individual waves Node: stationary points in a medium produced by destructive interference of two waves travelling in opposite directions Noise: mixture of sound frequencies with no recognizable relationship to one another Non-uniform acceleration: acceleration that is changing with time Non-uniform motion: the velocity is changing, either in magnitude or in direction, or both Nuclear fission: the splitting of a large nucleus into two or more lighter nuclei; usually caused by the

impact of a neutron and accompanied by the release of energy Nuclear fusion: the formation of a larger nucleus from two or more lighter nuclei, accompanied by the release of energy Nuclide: the nucleus of a particular atom, as characterized by its atomic number and atomic mass number

Primary coil: the coil in a transformer that is connected to the initial fluctuating or alternating current (i.e., the power supply) Principle of superposition: a combined or resultant wave is the sum of its component waves R.

Out of phase: the periodic motion of two individual systems vibrating with the same frequency is said to be out of phase if they both don’t reach the same amplitude at the same instant

Radioactive isotope (radioisotope): an isotope of an element that has an unstable nucleus and therefore disintegrates, emitting alpha, beta, or gamma radiation Radioactive material: material that contains radioactive nuclei Radioactivity: the spontaneous disintegration of the nuclei of certain elements, accompanied by the emission of alpha, beta, or gamma radiation Range: the horizontal distance a projectile travels

Overtone: all natural frequencies higher than the fundamental frequency in a standing wave pattern

Recoil: the interaction that occurs when two stationary objects push against each other and then move apart

O. Open air column: an air column that is open at both ends

P. Parent nucleus: the initial nucleus involved in a transmutation reaction Period: the time required for an object to complete one cycle of its repeated pattern of motion Phase difference: the angular difference between two systems in periodic motion that are not in phase Projectile an object that is given an initial thrust and allowed to move through space under the force of gravity only Pitch: an attribute of a sound that determines its position in a musical scale; pitch is measured in frequency Power: the rate at which work is done, measured in watts (W), or joules per second; also defined as the rate at which energy is transferred or transformed Power output: the rate at which an appliance can transform electric energy into a desired form such as light, heat, or sound

Restoring force: the force exerted by a spring on an object; proportional to the amount of extension or compression of the spring Rarefaction: a region of lower air pressure compared to the surrounding medium; longitudinal waves have both compressions and rarefactions Resolved vectors: components of a vector that are at right angles to each other; the components lie parallel to the axes of the coordinate system Resonance: phenomena that occurs when energy is added to a vibrating system at the same frequency as its natural frequency Resultant vector: a vector obtained by adding or subtracting two or more vectors Resultant wave: a wave produced by combining or superimposing two or more individual waves Right-hand rules: the rules which help to visualize the directions of vectors describing properties involved in electromagnetism by using the fingers and thumb of your right hand.

S. Scalar: a physical quantity that has only a magnitude or size Secondary coil: the coil in a transformer in which currents are induced Slip-ring commutator: a circular conductor attached to the coil of a motor or AC generator that is in contact with brushes that allows the continuous electric connection to the rest of the circuit Solenoid: a closely wound helix of wire that acts as a magnet when current runs through the wire Sonic boom: an acoustic pressure wave caused by an object moving faster than the speed of sound Sound intensity level: the rate of energy flow across a unit area; measured in an exponential scale in units of bels (B) Speed: the distance an object travels divided by the time the object was travelling; speed is a scalar quantity Spring constant: the amount of force a spring can exert per unit distance of extension or compression Standing wave: a stationary wave consisting of nodes and antinodes, formed when two equal travelling waves pass through one another in opposite directions Step-down transformer: a transformer that has fewer windings on its secondary coil, and acts to decrease voltages Step-up transformer: a transformer that has more windings on its secondary coil, and acts to increase voltages T. Terminal velocity: the velocity of a falling object at which the force of friction is equal in magnitude to the force of gravity

Thrust: the force with which gases ejected from a rocket push back on the rocket Trajectory: the path described by an object moving due to a force or forces Transmutation: the conversion of one element into another, usually as a result of radioactive decay Transmitter: A circuit that accepts signals or data in and translates them into a form that can be sent across a medium (transmitted), usually over a distance. The medium can be wireless or wired. Transformer: a device used to convert power from the high voltages used in transmission to low voltages safe for use in homes; increases or decreases AC voltages with little loss of energy Transverse wave: a wave in which the particles of a medium vibrate at right angles to the direction of motion; for example, water waves Trough: the lowest point on a wave U. Ultrasonic: sound frequencies higher than 20 000 Hz Uniform acceleration: acceleration that is constant throughout a particular time interval Uniform motion: moving at constant velocity V. Vector: a quantity that has a magnitude and a direction; vectors must be defined in terms of a frame of reference Vector components: parts of a vector that are parallel to the axes of a coordinate system, into which a vector can be resolved; they are scalar quantities Vector diagram: a diagram, with a coordinate system, in which all quantities are represented by vectors Velocity the rate of change of displacement of an object; a vector quantity. W.

Watt (W): a unit of power, equivalent to 1 joule per second Wave: a disturbance that transfers energy through a medium

the wave to its frequency and wavelength Wavelength: the shortest distance between any two points in a medium that are in phase; commonly measured from one trough to the next trough, or one crest to the next crest

Wave front: a group of adjacent points in a wave that all have the same phase, usually indicated by a line drawn along the crests of a wave Wave equation: the fundamental equation governing the motion of waves that relates the velocity of

Weight: the force that gravity exerts on an object due to its mass Work: the transfer of mechanical energy; equivalent to a force acting through a dista

Index Air resistance, 32 Alpha, 171 amplitude, 122 Angular velocity, 72 -Beta, 173 binary code, 157 Centripetal force, 70 Characteristics of waves, 120 Circular motion, 69 Cloud chamber, 184 Color, 165 Combining vectors, 48 Deformation, 96 Diffraction, 130 Displacement, 6 Doppler effect, 148 Dynamics, 26 Echolocation, 143 Electromagnetic induction, 195 energy, 85 Energy transfers, 166 equations of motion, 23 Faraday’s law, 196 Free falling motion, 34 frequency, 129 Gamma, 173 Geiger-Muller tube (GM tube):, 185 Generators, 205 gold leaf electroscope, 185 half – life, 179 harmonics, 151 Hertz, 156 Hook’s law states, 98 Impulse, 45 inertia, 27 intensity, 150 Interference, 131 Ionization, 183 Isobars, 178 Isotopes, 178 Kinematics, 5 Lenz’s law, 200 Longitudinal waves, 116 Momentum, 35 Motion, 5

motion graphs, 12 moving coil microphone, 209 musical instrument, 151 mutual induction, 210 natural frequency, 161 Newton’s First law of motion, 27 Newton’s law of universal gravitation, 76 Newton’s Second law of motion, 29 Newton’s third law, 35 Nuclear energy, 185 Nuclear Equations, 174 Nuclear fission, 186 Nuclear fusion, 186 nuclear reactor, 185 Nuclides, 178 Period, 124 Photographic film, 183 Pitch, 150 Power, 1, 79, 93, 216, 219, 223, 225, 230 projectile, 60 Radio system, 160 Radioactivity, 171 Receiver, 160 Reflection, 127 Refraction, 129 Resolution of vectors, 59 Resonance, 148 resultant force, 27 ripple tank, 129 Sound, 136 standing wave, 133 Stiffness, 96 Telecommunication, 156 television, 164 terminal velocity, 32 Transformers, 213 Transmitter, 160 Transverse wave, 118 uniform acceleration, 10 uniform speed, 10 vectors, 47 Wave speed, 125 waves, 115 Work, 79 Young Modulus, 107

Air resistance, 32 Angular velocity, 72

Centripetal force, 70 Circular motion, 69

Combining vectors, 48 Deformation, 96 Displacement, 6 Dynamics, 26 energy, 85 equations of motion, 23 Free falling motion, 34 Impulse, 45 inertia, 27 Kinematics, 5 Momentum, 35 Motion, 5 motion graphs, 12

Newton’s First law of motion, 27 Newton’s law of universal gravitation, 76 Newton’s Second law of motion, 29 Newton’s third law, 35 Power, 93 projectile, 60 Resolution of vectors, 59 resultant force, 27 terminal velocity, 32 uniform acceleration, 10 uniform speed, 10 vectors, 47 Work, 79

References 1. George Bethel and David cop pock (1999), physics first. New York: Oxford. 2. Tom Duncan and Heather Kennett (2001), GCSE physics, London: NW1 3BH. 3. Stephen pople (2007), complete physics IGCSE, New York: Oxford 4. Ken Dobson (1999). The physical world. London: WC1E7DP. 5. Jim Breithaupt and Ken Dun (1995). A-level physics. London: W128AW. 6. Dr Michael Taylor. (1999).MACMILLAN PHYSICS. Nairobi: Longman publishing printer. 7. Stephen pople (2007), complete physics IGCSE, Singapore: (KHL printing co, pte Ltd.) 8. Keith Johnson (2006), physics for you. CHELTENHAM: Nelson thorns Ltd. 9. Tom Duncan and Heather Kennett (2001), GCSE physics, London: NW1 3BH. 10. David Sang and Keith Gibbs and Robert Hitchings (2006), physics 1. Cambridge: Cambridge university press 11. K. Balaraman, C. kariuki, P. (Kanga 2005), longhorn secondary physics 1,2,3.Nairobi: printpack, Likoni Road, 12. George Bethel and David coppock (1999), physics first. New York: Oxford. 13. http://www.seai.ie/Schools/Post_Primary/Subjects/Physics/Unit_5_-_waves /Experiment_1/ 14. http://www.hk-phy.org/contextual/heat/tep/temch02_e.html 15. http://sciencecity.oupchina.com.hk/everybody/ch01e.htm 16. http://equilibriumchemistry11.blogspot.com/2010/10/heatingcooling-curve-of-puresubstance.html 17. http://www.cyberphysics.co.uk/topics/heat/latentheat/changingState.html 18. http://www.spaceflight.esa.int/impress/text/education/Solidification/index.html 19. http://ygraph.com/chart/1535