A Level Further Mathematics for OCR A: Mechanics Student Book (AS/A Level) 1316644278, 9781316644416, 9781316644270, 9781316644515

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Brighter Thinking

A Level Further Mathematics for OCR A

Mechanics Student Book (AS/A Level) Jess Barker, Nathan Barker, Michele Conway, Janet Such Course consultant: Vesna Kadelburg

Contents Introduction How to use this resource 1 Work, energy and power 1 Section 1: The work done by a force Section 2: Kinetic energy and the work–energy principle Section 3: Potential energy, mechanical energy and conservation of mechanical energy Section 4: Work done by a force at an angle to the direction of motion Section 5: Power Mixed practice 1 2 Dimensional analysis Section 1: Defining and calculating dimensions Section 2: Units and dimensions of sums, differences and angles Section 3: Finding dimensions from units and derivatives and predicting formulae Section 4: Summary of dimensions and units Mixed practice 2 3 Momentum and collisions 1 Section 1: Momentum and impulse Section 2: Collisions and the principle of conservation of momentum Section 3: Restitution, kinetic energy and impulsive tension Mixed practice 3 4 Circular motion 1 Section 1: Linear speed vs angular speed Section 2: Acceleration in horizontal circular motion Section 3: Solving problems involving motion in a horizontal circle Mixed practice 4 5 Centres of mass 1 Section 1: Centre of mass of a system of point masses Section 2: Centres of mass of standard shapes Section 3: Centres of mass of composite bodies Mixed practice 5 Focus on … Proof 1 Focus on … Problem solving 1 Focus on … Modelling 1 Cross-topic review exercise 1 6 Work, energy and power 2 Section 1: Work done by a variable force Section 2: Hooke’s law, work done against elasticity and elastic potential energy Section 3: Problem solving involving work, energy and power Section 4: Using vectors to calculate work done, kinetic energy and power Mixed practice 6 7 Linear motion under variable force Section 1: Working with acceleration, velocity and displacement Section 2: Variable force Mixed practice 7 8 Momentum and collisions 2

Section 1: Variable force and vector notation Section 2: Oblique impacts and the impulse–momentum triangle Section 3: Oblique collisions of two spheres and impulsive tensions in strings Mixed practice 8 9 Circular motion 2 Section 1: Conservation of mechanical energy Section 2: Components of acceleration (a general model) Section 3: Problem solving situations Mixed practice 9 10 Centres of mass 2 Section 1: Centres of mass by integration Section 2: Equilibrium of a rigid body Mixed practice 10 Focus on … Proof 2 Focus on … Problem solving 2 Focus on … Modelling 2 Cross-topic review exercise 2 AS Level practice paper A Level practice paper Formulae Answers Worked solution for chapter exercises 1 Work, energy and power 1 2 Dimensional analysis 3 Momentum and collisions 1 4 Circular motion 1 5 Centres of mass 1 6 Work, energy and power 2 7 Linear motion under variable force 8 Momentum and collisions 2 9 Circular motion 2 10 Centres of mass 2 Worked solution for cross topic review exercises Cross-topic review exercise 1 Cross-topic review exercise 2 AS Level practice paper A Level practice paper Glossary Acknowledgements

Introduction You have probably been told that mathematics is very useful, yet it can often seem like a lot of techniques that just have to be learned to answer examination questions. You are now getting to the point where you will start to see where some of these techniques can be applied in solving real problems. However, as well as seeing how maths can be useful, we hope that anyone working through this book will realise that it can also be incredibly frustrating, surprising and ultimately beautiful. The book is woven around three key themes from the new curriculum:

Proof Maths is valued because it trains you to think logically and communicate precisely. At a high level, maths is far less concerned about answers and more about the clear communication of ideas. It is not about being neat – although that might help! It is about creating a coherent argument that other people can easily follow but find difficult to refute. Have you ever tried looking at your own work? If you cannot follow it yourself it is unlikely anybody else will be able to understand it. In maths we communicate using a variety of means – feel free to use combinations of diagrams, words and algebra to aid your argument. And once you have attempted a proof, try presenting it to your peers. Look critically (but positively) at some other people’s attempts. It is only through having your own attempts evaluated and trying to find flaws in other proofs that you will develop sophisticated mathematical thinking. This is why we have included lots of common errors in our ’work it out’ boxes – just in case your friends don’t make any mistakes!

Problem solving Maths is valued because it trains you to look at situations in unusual, creative ways, to persevere and to evaluate solutions along the way. We have been heavily influenced by a great mathematician and maths educator, George Polya, who believed that students were not just born with problem solving skills – these skills were developed by seeing problems being solved and reflecting on the solutions before trying similar problems. You may not realise it but good mathematicians spend most of their time being stuck. You need to spend some time on problems you can’t do, trying out different possibilities. If after a while you have not cracked it then look at the solution and try a similar problem. Don’t be disheartened if you cannot get it immediately – in fact, the longer you spend puzzling over a problem the more you will learn from the solution. You may, for example, never need to integrate a rational function in future, but we firmly believe that the problem solving skills you will develop by trying it can be applied to many other situations.

Modelling Maths is valued because it helps us solve real-world problems. However, maths describes ideal situations and the real world is messy! Modelling is about deciding on the important features needed to describe the essence of a situation and turning that into a mathematical form, then using it to make predictions, compare to reality and possibly improve the model. In many situations the technical maths is actually the easy part – especially with modern technology. Deciding which features of reality to include or ignore and anticipating the consequences of these decisions is the hard part. Yet some fairly drastic assumptions – such as pretending a car is a single point or that people’s votes are independent – can result in models that are surprisingly accurate. More than anything else, this book is about making links. Links between the different chapters, the topics covered and the themes just discussed, links to other subjects and links to the real world. We hope that you will grow to see maths as one great complex but beautiful web of interlinking ideas. Maths is about so much more than examinations, but we hope that if you take on board these ideas (and do plenty of practice!) you will find maths examinations a much more approachable and possibly even enjoyable experience. However, always remember that the results of what you write down in a few hours by yourself in silence under exam conditions is not the only measure you should consider when judging your mathematical ability – it is only one variable in a much more complicated mathematical model!

How to use this resource Throughout this resource you will notice particular features that are designed to aid your learning. This section provides a brief overview of these features. In this chapter you will learn how to: calculate the work done by a force calculate kinetic energy use the work–energy principle

Learning objectives A short summary of the content that you will learn in each chapter.

Before you start… GCSE

You should know how to convert units of distance, speed and time.

1 Convert kilometres.

metres to

A Level Mathematics Student Book 1

You should know how to calculate the weight of an object from its mass, and know the unit of weight.

2 Calculate the weight of a car of mass , stating the unit with your answer.

Before you start Points you should know from your previous learning and questions to check that you’re ready to start the chapter.

WORKED EXAMPLE

The left-hand side shows you how to set out your working. The right-hand side explains the more difficult steps and helps you understand why a particular method was chosen.

PROOF

Step-by-step walkthroughs of standard proofs and methods of proof.

WORK IT OUT Can you identify the correct solution and find the mistakes in the two incorrect solutions?

Key point A summary of the most important methods, facts and formulae.

Explore Ideas for activities and investigations to extend your understanding of the topic.

Tip Useful guidance, including on ways of calculating or checking and use of technology. Each chapter ends with a Checklist of learning and understanding and a Mixed practice exercise, which includes past paper questions marked with the icon

.

In between chapters, you will find extra sections that bring together topics in a more synoptic way.

Focus on… Unique sections relating to the preceding chapters that develop your skills in proof, problem solving and modelling.

CROSS-TOPIC REVIEW EXERCISE

Questions covering topics from across the preceding chapters, testing your ability to apply what you have learned. You will find AS Level and A Level practice questions towards the end of the resource, as well as a glossary of key terms (picked out in colour within the chapters), and answers to all questions. Full worked solutions can be found on the Cambridge Elevate digital platform, along with a digital version of this Student Resource. Maths is all about making links, which is why throughout this book you will find signposts emphasising connections between different topics, applications and suggestions for further research.

Rewind Reminders of where to find useful information from earlier in your study.

Fast forward Links to topics that you may cover in greater detail later in your study.

Focus on… Links to problem solving, modelling or proof exercises that relate to the topic currently being studied.

Did you know? Interesting or historical information and links with other subjects to improve your awareness about how mathematics contributes to society. Colour-coding of exercises The questions in the exercises are designed to provide careful progression, ranging from basic fluency to practice questions. They are uniquely colour-coded, as shown here.

1

A uniform rectangular lamina has vertices at centre of mass of the lamina.

and

. Find the coordinates of the

10 Find the increase of elastic potential energy when a light elastic string of natural length modulus of elasticity

is extended from

to

and

.

13 An object of mass is attached to one end of a light elastic string of natural length with its other end attached to a fixed point, . The modulus of elasticity of the string is . is dropped from . Find the extension of the string when the object reaches its maximum velocity. 18 A light elastic spring with natural length rests on a smooth horizontal table. One end is attached to a fixed point and a mass is attached at the other end , held from . The modulus of elasticity of the spring is . 10 Road surface conditions are being assessed for a horizontal bend in a road that is formed by an arc of a circle of radious . The road surface could be made of asphalt or concrete. The coefficient of friction between car tyres and asphalt is , and between car and concrete . 14 A vertical hollow cylinder of radius

is rotating about its axis. A particle rough inner surface of the

cylinder. The cylinder and rotate with the same The coefficient of friction between and the cylinder is .

Black – drill questions. Some of these come in several parts, each with subparts i and ii. You only need attempt subpart i at first; subpart ii is essentially the same question, which you can use for further practice if you got part i wrong, for homework, or when you revisit the exercise during revision. Green – practice questions at a basic level. Blue – practice questions at an intermediate level. Red – practice questions at an advanced level. Purple – challenging questions that apply the concept of the current chapter across other areas of maths. Yellow – designed to encourage reflection and discussion.

  – indicates content that is for A Level students only

1 Work, energy and power 1

In this chapter you will learn how to: calculate the work done by a force calculate kinetic energy use the work–energy principle equate gravitational potential energy to work done against gravity perform calculations using power.

Before you start… GCSE

You should know how to convert units of distance, speed and time.

1 Convert

metres to kilometres.

A Level Mathematics Student Book 1

You should know how to calculate the weight of an object from its mass, and know the unit of weight.

2 Calculate the weight of a car of mass , stating the unit with your answer.

A Level Mathematics Student Book 1

You should be able to use Newton’s second law of motion:

3 A resultant force of acts on an object of mass . Calculate the acceleration of the object.

A Level

You should be able to

4 A force of

Mathematics Student Book 2

resolve a force into components at right angles to each other.

acts on a particle at an

angle of to the positive horizontal direction. What are the horizontal and vertical components of the force?

The relationship between work and energy You have already studied the effect of a force or system of forces in A Level Mathematics. In this chapter, you will learn the definition of the work done by a force, which is a quantity that is measured in joules, the same units that are used for energy. You will learn about propulsive and resistive forces. You will learn about the relationship between work done and two different types of energy: kinetic energy and gravitational potential energy. You will also learn about power, which is the rate of doing work. Ideas of work, energy and power are crucial in engineering, enabling engineers to design machines to do useful work. Hydroelectric power stations work by converting the work done by falling water, first into kinetic energy as the hydroelectric turbines rotate and then into electricity.

Fast forward

In Chapter 6, you will learn about elastic potential energy and its conversion to kinetic energy.

Section 1: The work done by a force Work is done by a force when the object it is applied to moves. The amount of work done is the product of the force and the distance moved in the direction of the force. Some forces promote movement, while others resist it. For example, when you cycle into a breeze, your pedalling promotes movement but the breeze acts against your movement. Forces that promote movement are called propulsive forces and those that resist movement are known as resistive forces. Other propulsive forces include the tension in a rope being used to drag an object across the ground and the driving force of a vehicle engine. The driving force of an engine is often described as its tractive force. Other resistive forces include friction, vehicle braking and resistance by moving through still air or a liquid.

Key point 1.1 For a force acting in the direction of motion: Work done is measured in joules , i.e.

.

For example a force of acting on an object that moves in the direction of the force does of work. Doubling the force to over the same distance would double the amount of work done to . Likewise, doubling the distance moved to with an unchanged force of would double the amount of work done to . WORKED EXAMPLE 1.1

A box is pushed by the force.

across a horizontal floor by a horizontal force of

. Calculate the work done

Use the definition of work done. State units of work done

with your answer.

WORKED EXAMPLE 1.2

A truck driver driving along a horizontal road applies a braking force of work done by the brakes, giving your answer in . Convert Work done by brakes

to

for

. Calculate the

as you need to work in standard units.

Use the definition of work done.

Change to

.

WORKED EXAMPLE 1.3

A

crate is lifted

by a rope and pulley system. Calculate the work done against gravity. Apply the definition of work done to the gravitational force. The force needed to lift the crate is equal to the crate’s weight and

work done against gravity

the distance moved is height gained.

Calculate the weight of the crate, based on the usual approximation for the acceleration due to gravity of

.

Work done against gravity

Use the definition of work done.

Key point 1.2 When a mass, , is raised or lowered through a height : work done against or by

Fast Forward In Section 3 you will learn the equivalence of work done against gravity and gravitational potential energy.

WORKED EXAMPLE 1.4

A competitor of mass

dives from a

-metre-high diving board into a pool. Air resistance

averages as he descends through the air. Resistance from the water then averages as he descends further. Calculate: a the total work done by gravity as the diver descends

metres

b the total work done against air and water resistance during this descent. a

Use

b

Use to calculate the work done against each of the resistances.

to calculate the work done by gravity.

WORKED EXAMPLE 1.5

A van of mass travels along a straight road. The driving force of the vehicle engine is and resistance to motion is , on average. The van travels from one delivery to the next, descending in height. Find: a the work done by the vehicle engine b the work done by gravity c the work done against resistance. a

Convert distance to metres. Work done by vehicle

b

Use engine. Use

to calculate the work done by the vehicle to calculate the work done by gravity.

c Work done against Use

to calculate the work done against resistance.

EXERCISE 1A 1

A parcel is dragged metres across a horizontal floor by a horizontal rope. The tension in the rope is . Calculate the work done by the tension in the rope.

2

Susan climbs a vertical rock against gravity.

3

Sunil descends a vertical ladder. His mass is height Sunil descends.

4

A ball of mass is dropped from a window. Calculate the work done by gravity as the ball falls vertically to the ground below.

5

A puck slides

high. Susan’s mass is

. Calculate the work done by Susan

and the work done by gravity is

metres across an ice rink, against a resistive force of

. Find the

. Calculate the work done

against resistance. 6

A cyclist travelling on horizontal ground applies a driving force of a resistance from friction of . The cyclist travels . Find:

against a headwind of

and

a the work done by the cyclist b the total work done against wind and friction. 7

A fish basket is raised from the sea floor to a fishing boat at sea level, the basket is . The resistance to motion from the seawater is against gravity and water resistance, in raising the fish basket.

8

A driving force of does Resistance to motion averages from to .

metres above. The mass of

. Calculate the total work done,

of work moving a van along a horizontal road from to . . Calculate the work done against resistance as the van moves

Section 2: Kinetic energy and the work–energy principle Kinetic energy is the energy an object has because it is moving.

Key point 1.3 An object of mass moving with speed has kinetic energy If mass is measured in

and speed is measured in

.

, kinetic energy is measured in joules.

Tip If speed is not given in calculations.

, you should convert to

before you start the rest of your

WORKED EXAMPLE 1.6

A particle of mass particle.

is moving with kinetic energy

joules. Calculate the speed of the

Use the formula for kinetic energy.

Substitute and rearrange to find speed. As mass was given in and kinetic energy in joules, speed is in . Speed is a positive scalar, so the negative option of the root can be disregarded. WORKED EXAMPLE 1.7

A cyclist slows down from to is . Calculate the loss of kinetic energy. Let be the starting speed and be the final speed:

. The combined mass of the cyclist and her bicycle

To convert

to

conversion factor

you must multiply by the , which simplifies to division by

.

and

Loss of kinetic energy initial kinetic energy final kinetic energy

WORKED EXAMPLE 1.8

Calculate the increase in kinetic energy when a boat of mass tonnes changes velocity from to . Give your answer in .

Use Pythagoras’ theorem to convert the velocity vectors to speeds. You need the square of the speed, not the velocity vector, for the kinetic energy formula.

You can write

in factorised form.

Convert tonnes to Divide by

to convert joules to

.

The work–energy principle is an essential idea in Mechanics that enables us to calculate the work necessary to cause a change in kinetic energy.

Key point 1.4 The net work done by all the forces acting on a particle, including its own weight, is equal to the change in kinetic energy of the particle.

WORKED EXAMPLE 1.9

A particle of mass force of

at rest on a smooth horizontal plane is acted on by a constant horizontal

. Find the speed of the particle after it has travelled metres.

Work–energy principle: since the particle is starting from rest,

WORKED EXAMPLE 1.10

Stephen is driving his car along a horizontal road at when he notices a broken-down vehicle, just off the road, ahead. Stephen and his car have a mass of and the total resistance to motion is assumed constant at . Stephen believes he should slow down and that he can slow down sufficiently without applying the brakes. Calculate Stephen’s speed, in , as he reaches the broken-down vehicle, taking account of the resistance to motion. Assume that Stephen allows the resistance to motion to slow his car down over . There is no driving or braking force. Work done against resistance

Calculate the work done against resistance.

Convert the initial speed of

to

.

Write down the expression for loss of kinetic energy. Work–energy principle: work done against

Substitute for . Rearrange and solve for .

Convert back to

.

EXERCISE 1B 1

Calculate the kinetic energy of a cyclist and her bicycle having a combined mass of travelling at . Give your answer in .

2

Calculate the mass of an athlete who is running at

3

Calculate the speed of a bus of mass .

4

A box of mass of magnitude .

, with kinetic energy

tonnes with kinetic energy

, .

. Give your answer in

is pulled from to across a smooth horizontal floor by a horizontal force . At point , the box has speed and at point the box has speed

Ignoring all other resistive forces, find: a the increase in kinetic energy of the box b the work done by the force c the distance

.

5

Calculate the loss of kinetic energy when a boat of mass to .

6

A car driver brakes on a horizontal road and slows down from the car and its occupants is .

tonnes reduces in velocity from to

. The mass of

a Find the loss in kinetic energy. b Given that the work done against resistance to motion is

, find the work done by the

brakes. 7

A child of mass descends a smooth slide, after propelling herself from the top at Ignoring air resistance, calculate her speed at the bottom of the slide, which is metres

.

lower down than the top. 8

A bullet of mass grams passes horizontally through a target of thickness . The speed of the bullet is reduced from to . Calculate the magnitude of the average resistive force exerted on the bullet.

9

A train with mass tonnes is travelling at on horizontal tracks, when the driver sees a speed reduction sign. The train’s speed must be reduced to over . Resistance to motion is approximately

10 A package of mass

. Calculate the braking force required, in

grams slides down a parcel chute of length

.

metres, starting from

rest. The bottom of the chute is metres below the top. The speed of the package at the bottom of the chute is . Find the resistance to motion on the chute. 11 Use the equation of motion, relation:

, together with the formula,

, to derive the

12 Eddy cycles up a hill. His mass, together with his bicycle, is . His driving force is and resistance from friction is . Eddy travels metres along the road, which rises through a vertical height of

metres. His starting speed is

. Find his final speed.

Section 3: Potential energy, mechanical energy and conservation of mechanical energy Consider an object of mass falling freely under gravity from height

to height

, with starting speed

and final speed . Since the only external force acting on the object is gravity, the work–energy principle becomes: u h1 – h2 h1

v h2 ground level

Rearranging this gives:

Each side of this equation is the sum of two terms, one of which is kinetic energy. The other term is gravitational potential energy. Gravitational potential energy (GPE) is the energy an object has by virtue of its position. For an object of mass raised a distance , the increase in GPE is equal to the product of its weight, , and the distance .

Key point 1.5 Gravitational potential energy where is the height above ground (zero) level.

Tip You can choose any height as your ground (zero) level but it is usually best to choose the lowest height reached by the moving object. The principle of the conservation of mechanical energy states that, if there are no external forces other than gravity doing work on an object during its motion, then the sum of kinetic energy and gravitational potential energy remains constant.

Key point 1.6 If the only force acting on an object is its weight then mechanical energy is conserved: where is the vertical height above the zero level. This diagram may help you to understand the formula for conservation of mechanical energy more easily. As an object descends in height it speeds up, so gravitational potential energy is converted into kinetic energy. As an object ascends in height it slows down, so kinetic energy is converted into gravitational potential energy.

mechanical energy is conserved

1 mu2 2

mgh1

obj 1 mv2 a s e c t s it d p e e 2 esc ds pot e n d u p e s c o n n tia k in v e r te l e n e r e tic d t g y e n o erg mgh2 y

1 2 n 2 mw w o d w s s lo n d s t c e o b j t a s c e i gy s a ner e c ti to y mgh3 kine verted energ con ential pot ground level

WORKED EXAMPLE 1.11

Faisal throws a ball of mass

grams vertically upwards from ground level with a speed of

. Assuming no external forces apply: a calculate the speed of the ball after it has risen metres b calculate the maximum height gained by the ball. a

Use conservation of mechanical energy over the first the ascent.

of

Take the gravitational potential energy at ground level to be zero. Calculate the kinetic energy of the ball at

.

Use the formula for

equal to .

with the speed at

Calculate the speed of the ball. b

Use conservation of mechanical energy over the whole ascent (final kinetic energy is zero). At the maximum height, all the initial kinetic energy will have been converted into gravitational potential energy. Calculate the maximum height gained.

Using energy to solve problems The principle of conservation of mechanical energy applies to the situation where the only force acting on an object is its weight. You have already used the work–energy principle to solve problems involving external forces, such as friction and driving forces. You are now ready to combine the work–energy principle and the principle of conservation of energy. Any change in the mechanical energy of a system is the result of work done by external forces:

The formula can also be written as: The change in the total energy of an object the work done on the object by external forces.

Tip

External forces are any forces acting on an object other than its own weight.

WORKED EXAMPLE 1.12

A package is attached to one end of an inextensible string. The string and package are being raised by the action of a pulley. The tension in the string is . Find the height gained by the package as it increases in speed from to . Calculate the increase in kinetic energy.

Calculate the change in gravitational potential energy. Calculate the work done in raising the package. Work done on object change in total energy

WORKED EXAMPLE 1.13

Helen cycles from rest at the top of a sloping track, above the valley floor. She pedals downhill, then continues along a horizontal track, before ascending on an uphill track and stopping. The total distance she travels is , and the average resistance to motion is . The combined mass of Helen and her bicycle is . Calculate the total work done by Helen, and the average driving force she applies.

35m 10m valley floor

Use the work–energy principle and the principle of conservation of energy. Helen has no kinetic energy at the start of her ride and none at the end. You do not need to consider her motion throughout her ride: just at the start and at the end. Work done against resistance

Calculate the work done against resistance. The distance used is the total distance along the road.

Let both initial and final kinetic energy be zero and rearrange: work done against resistance is the loss of potential energy of Helen and her bicycle.

Helen’s average driving force

Use Helen’s work done, together with her distance travelled ( ) to calculate her average driving force.

Conservation of energy is an important principle throughout Physics. Work done by a moving object against resistance, which is lost mechanical energy, is converted to other forms of energy such as thermal energy and sound. This means that total energy is still conserved.

Did you know? The mechanical equivalent of heat was first proposed by James Joule and explains the relationship between mechanical energy and thermal energy.

EXERCISE 1C 1

Calculate the increase in potential energy when a mass of

2

Calculate the loss of potential energy when a mass of tonnes is lowered

3

A boy of mass he gains.

4

A toy train loses of potential energy when it descends a spiral track losing the mass of the toy train.

5

Richard strikes a golf ball off an elevated tee. The golf ball has mass initial speed of to the ball.

gains

is raised

. .

of potential energy when climbing a vertical rope. Calculate the height in height. Find

grams and Richard imparts an

a Find the initial kinetic energy of the golf ball. The ball lands on the green

metres below the tee. Assume there is no significant air resistance.

b Calculate the loss of potential energy of the ball when it lands on the green. c Calculate the kinetic energy of the ball when it lands on the green. d Calculate the speed of the golf ball when it lands on the green. 6

Anita dives off a highboard into a diving pool. When Anita leaves the board she has a speed of and she is metres above the water surface. Anita’s mass is . a Find Anita’s kinetic energy as she leaves the board. b Calculate Anita’s kinetic energy as she enters the water. c Calculate Anita’s speed as she enters the water. d What modelling assumptions have you made to simplify your calculations?

7

Wing serves a

gram tennis ball with a speed of

from a height metres above the level of

the tennis court. Assuming there are no resistive forces acting on the ball, calculate: a the kinetic energy of the ball as Wing serves it b the potential energy lost by the ball as it descends to the level of the court c the kinetic energy of the ball as it strikes the court d the speed of the tennis ball as it strikes the court. 8

Preeti descends a slide starting from rest. Her mass is and her speed at the bottom of the slide is during her descent.

9

A package of mass is projected down a smooth sloping parcel chute with a speed of . The bottom of the chute is vertically below the top. Assuming there are no external resistive forces, calculate: a the loss of potential energy of the package

. Overall, her change in vertical height is . Calculate the work done against resistance

b the speed of the package at the bottom of the chute. 10 Karol slides on his sledge down a straight track of length , descending . The combined mass of Karol and his sledge is . Karol’s starting speed is and his speed at the end of his descent is . Calculate the average resistance to motion, , during Karol’s descent. 11 Loretta and her bicycle have a combined mass of . Loretta cycles up a straight hill , accelerating from rest at to at . The level of point is below the level of . Find: a the increase in kinetic energy of Loretta and her bicycle as she cycles from to b the increase in potential energy of Loretta and her bicycle. During her ride, the resistance to motion is constant at does

of work.

c Calculate the distance from to .

parallel to the road surface and Loretta

Section 4: Work done by a force at an angle to the direction of motion The problems you have worked with so far have had all forces in the direction of movement, either promoting motion or resisting it directly. But in many cases the forces causing motion are not in the direction of motion. Examples are: a man dragging a sledge along horizontal ground by pulling on a rope that is angled upwards a child descending a slide under gravity; the child’s weight acts vertically downwards but she travels down the slide at an angle to the vertical. force, F

θ object moving this way

F cos θ resolved component of F in the direction of motion

If a force is applied at an angle to the direction of motion as shown, the resolved component of the force that does work is . The resolved component that is perpendicular to the direction of motion,

, does no work.

Key point 1.7 If a force is acting at an angle to the direction of movement:

Rewind Resolving forces is covered in A Level Mathematics Student Book 2, Chapter 21.

Fast Forward In Chapter 6 you will learn that the formula for work done by a force at an angle to movement is the scalar product of the force vector and the displacement vector.

WORKED EXAMPLE 1.14

Jamal is dragging his son on a sledge on horizontal ground. He is pulling a rope, attached to the sledge, at an angle of to the horizontal. The tension in the rope is . Find the work Jamal does dragging the sledge . Use the definition of work done by a force acting at an angle to the direction of motion.

WORKED EXAMPLE 1.15

A girl of mass descends a straight smooth slide, starting from rest. The slide is in length and inclined at to the horizontal. Use work and energy to calculate the speed of the girl at the bottom of the slide. Let be the angle between the

The slope makes an angle of with the vertical.

with the horizontal but

vertical and the slope. Calculate the work done by gravity. Use the component of the weight acting down the slope.

Work done by girl’s weight

Use the work–energy principle: work done by gravity increase of kinetic energy The girl is travelling at the bottom of the slide.

at

WORKED EXAMPLE 1.16

A box of mass is projected with speed up a smooth inclined plane. The plane slopes at to the horizontal. Calculate how far the box travels up the plane by considering conservation of energy. The slope makes an angle of with the horizontal but with the vertical.

The component of the weight that does work acts down the plane, at an angle of to the weight.

Let gravitational potential energy at start of movement be zero ( ). At the end of the movement:

Let be the distance the box travels up the plane.As the box travels metres up the plane, its gravitational potential energy increases. This is equivalent to the work done against the weight of the box:

Compare mechanical energy when the box is projected with mechanical energy when it comes to rest: (when the box comes to rest). The box travels plane.

EXERCISE 1D

up the

EXERCISE 1D 1

A force of newtons is acting at a constant metres. Calculate the work done by the force.

2

A force of

is acting at a constant

to the line of movement of a particle that moves

to the line of movement of a particle that moves

metres. Calculate the work done by the force. Give your answer in

.

3

A particle is acted on by a force of newtons acting at a constant angle of to its direction of movement. The force does joules of work. Find the distance moved by the particle.

4

A particle is acted on by a force of newtons acting at a constant angle of to its direction of movement. The force does joules of work moving the particle metres. Find the value of .

5

A particle is acted on by a force of movement. The force does constant angle.

newtons acting at a constant angle to its direction of

joules of work moving the particle through

metres. Find the

6

Calculate the increase in potential energy when a mass of at to the horizontal.

7

Calculate the loss of potential energy when a mass of to the horizontal.

8

A car is towed at constant speed along a horizontal straight road. The tow rope is at horizontal and the tension in the tow rope is . The work done by the force is distance moved by the car.

9

A block of mass descends

grams is moved descends

is released from rest on a smooth plane inclined at

up a plane inclined

along a plane inclined at to the . Calculate the

to the horizontal, and

down the plane. Calculate:

a the loss of potential energy of the block b the gain in kinetic energy of the block. 10 A block of mass is dragged up a smooth slope from rest at to . The distance is and the slope is inclined at to the horizontal. The rope used to drag the block is parallel to the slope and has a tension of . Find: a the work done by the tension in the rope b the change in potential energy of the block c the speed of the block at .

Section 5: Power Power is the rate of doing work. Average power is defined as the total work done by a force divided by the time taken.

Key point 1.8 When the force applied is constant:

Power is measured in watts

. joule per second is equal to watt.

Often you consider power in relation to a driving force but it applies equally to any force acting on an object.

Did you know? James Watt (1736–1819) was a Scottish engineer and scientist. The unit of power is named after him.

WORKED EXAMPLE 1.17

A crane lifts a tonne concrete block in seconds. Calculate the average power rating of the crane during the lift, giving your answer in kW. Calculate the work done by the crane lifting the block against gravity.

Use the definition of power

.

WORKED EXAMPLE 1.18

The engine brakes on a truck have a power rating of . Calculate the total work done in seconds by the braking force at this average power rating, giving your answer in . Rearrange the definition of power to make work done the subject. Convert your answer to

.

WORKED EXAMPLE 1.19

A pump is used to raise water from a well. In one minute,

litres of water is raised metres

before being ejected into a tank at a speed of

. The density of water is

.

a Calculate the gain of potential energy of the water per second. b Calculate the gain of kinetic energy of the water ejected per second. c Calculate the power of the pump, in watts. There are

litres in

, with a mass of

Work out the mass of

litres of water.

. 1500 litres of water has a mass of

. Use to calculate the gain in potential energy of the water per second.

a

Use

b

to calculate the gain of kinetic energy of

the water per second.

c Work done by pump per second

Use conservation of mechanical energy: work done by the The gain of total mechanical energy per second is the power of the pump.

Therefore power

You can make use of an alternative formula for power when solving problems. From the definitions of power and work done:

This definition allows you to work out power at a specific point in time if you know the force and the speed. This is often referred to as ‘instantaneous power’ and can be used to work out power when either the force or the velocity varies over time.

Key point 1.9

WORKED EXAMPLE 1.20

Chris is cycling at a constant speed of

on a horizontal road with a power output of

.

Calculate the total resistance to Chris and his bicycle in newtons.

From , as Chris is travelling at constant speed the resultant force is zero.

WORKED EXAMPLE 1.21

Julia is riding her motorbike along a horizontal road with constant speed , at an engine power of . Julia decides to overtake and increases to full power, . Assuming the resistance to motion is unchanged, calculate Julia’s acceleration. Julia and her motorbike have a combined mass of

.

Convert Julia’s speed to

.

Use the definition of power, but rearrange to make tractive force the subject. Calculate the tractive force of Julia and her motorbike when she is travelling at constant speed.

For the motorbike:

As Julia is cruising at constant speed the resultant force is zero. Hence calculate the resistive force.



When Julia increases the power, the tractive force increases so that it is greater than the resistive force, and she accelerates. Calculate the resultant force.

When Julia increases to full power:

resultant force

Use

Using

to calculate Julia’s acceleration.



WORKED EXAMPLE 1.22

A car of mass

is travelling along a straight horizontal road against a resistance to motion of

, where is the speed of the car and is a constant. When the engine is producing a power of , the car has speed and is accelerating at . a Find the value of . The maximum constant speed of the car on this road is

.

b Find the engine’s maximum power, giving your answer in . The car now descends a hill, which is inclined at a constant angle

below the horizontal. The car

engine is working at maximum power and the vehicle moves with constant speed

.

c Find the value of , to decimal place. R

Let be the tractive force of the car engine.

0.45m s–2 12.5m s–1

1

Resistance varies with T

2 kv N

1250g

a Rearrange the formula:

Calculate .



Use

.

.

Rearrange to find the value of .

Let the new driving force be

b

.

As the car is now moving with constant speed the resultant force is now zero.

Rearrange to find the maximum power rating of the car engine.

R

c

Let the driving force down the hill be 36.5m

kv 21 N

a°

.

s–1 T˝

1250g

As the car is moving with constant speed the resultant force is zero. Let the driving force down the hill be

Re-arrange and solve for .

EXERCISE 1E

.

EXERCISE 1E 1

A -tonne truck is able to brake from rating of the brakes.

2

A crane lifts a -tonne concrete block

3

A lift of mass more than load through

4

A car engine has a maximum driving force of average power of the engine.

5

A train engine has an average power rating of is travelling at .

6

Find the average power exerted by a climber of mass in minutes.

7

A boat is travelling at a constant speed of . The boat has mass tonnes and the engine is working at its maximum power output of . Calculate the work done when the boat is displaced .

8

Find the average power of an engine that lifts has mass .

9

A pump is used to raise water from a well that is

to rest in in

. Find the average power

. Find the average power of the crane.

can accommodate up to people, assumed to have combined mass no . Calculate the average power required by the lift motor to raise the maximum in . when travelling at

. Calculate the

. Calculate the tractive force when the train when climbing a vertical distance of

bags of flour

in hour. Each bag of flour

metres deep. Water is raised at a rate of

per second, and is ejected into a pipe at a speed of

.

a Calculate the gain of potential energy of the water per second. b Calculate the gain of kinetic energy of the water ejected per second. c Calculate the power of the pump, in watts. 10 Victoria is cycling on level ground. Victoria and her cycle have a combined mass of and she is working at a rate of . Given that Victoria is accelerating at , find the sum of the resistive forces acting on Victoria and her cycle at the instant when her speed is . 11 Stan is driving his -tonne truck on a horizontal road. Stan accelerates from to 65 , which is his maximum speed at power output. Find the maximum acceleration of the truck, assuming that total resistance is constant. 12 Val is driving her van against a constant resistance to motion of tonnes and engine power . At the instant when her speed is . Calculate .

. The van has mass , Val’s acceleration is

13 The resistance to motion of a car is , where is the speed of the car and is a constant. The power of the car’s engine is , and the car has a constant speed of along a horizontal road. Show that

.

14 A spacecraft, Athena, of mass is moving in a straight line in space, without any resistance to motion. Athena’s propulsion system is working at a constant rate of and her mass is assumed to be constant. Athena’s speed increases from to in a time seconds. a Calculate the value of . b Calculate Athena’s acceleration when her speed is

Checklist of learning and understanding

.

Gravitational potential energy: Work–energy principle: Work done by external forces change in mechanical energy If a force is acting at an angle to the direction of motion:

Mixed practice 1 1

A woman drags a suitcase at constant speed in a straight line along horizontal ground by means of a plastic tether attached to the suitcase. The tether makes an angle of with the horizontal and the tension in the tether is suitcase .

2

. Calculate the work done in moving the

A car is pulled at constant speed along a horizontal straight road by a force of inclined at to the horizontal. Given that the work done by the force is , calculate the distance moved by the car. © OCR, GCE Mathematics, Paper 4729, June 2008

3

Find the average power exerted by a rock climber of mass distance of

4

when climbing a vertical

in minutes.

A block is being pushed in a straight line along horizontal ground by a force of inclined at below the horizontal. The block moves a distance of m in with constant speed. Find: i the work done by the force, ii the power with which the force is working. © OCR, GCE Mathematics, Paper 4729/01, January 2013

5

and are two points on a line of greatest slope of a smooth inclined plane, with a vertical distance of below the level of . A particle of mass from with a speed of . Find:

is projected down the plane

i the loss in potential energy of the particle as it moves from to , ii the speed of the particle when it reaches . © OCR, GCE Mathematics, Paper 4729/01, June 2013 6

The power developed by the engine of a car as it travels at a constant speed of horizontal road is .

on a

i Calculate the resistance to the motion of the car. The car, of mass , now travels down a straight road inclined at The resistance to the motion of the car is unchanged.

to the horizontal.

ii Find the power produced by the engine of the car when the car has speed accelerating at .

and is

© OCR, GCE Mathematics, Paper 4729/01, June 2013 7

A car of mass moves along a straight horizontal road. The resistance to the motion of the car has constant magnitude and the car’s engine is working at a constant rate of . i Find the acceleration of the car at an instant when the car’s speed is

.

The car now moves up a hill inclined at to the horizontal. The car’s engine continues to work at and the magnitude of the resistance to motion remains at . ii Find the greatest steady speed at which the car can move up the hill. © OCR, GCE Mathematics, Paper 4729, June 2012 8

A stone of mass

starts from rest and is dragged

up a slope inclined at

to the

horizontal by a rope inclined at

to the slope. The tension in the rope is

resistance to the motion of the stone is

and the

. Calculate:

a the work done by the tension in the rope b the change in the potential energy of the stone c the speed of the stone after it has moved 9

up the slope.

A car of mass travels along a straight road inclined at to the horizontal. The resistance to the motion of the car is , where is the speed of the car and is a constant. The car travels at a constant speed of works at a constant rate of

up the slope and the engine of the car

.

i Calculate the value of . ii Calculate the constant speed of the car on a horizontal road. © OCR, GCE Mathematics, Paper 4729, June 2011 10 A car of mass the car is

travels along a straight horizontal road. The resistance to the motion of , where

is the speed of the car and is a constant. At the instant when

the engine produces a power of .

, the car has speed

and is accelerating at

i Find the value of . It is given that the greatest steady speed of the car on this road is

.

ii Find the greatest power that the engine can produce. © OCR, GCE Mathematics, Paper 4729/01, January 2013 11 The resistance to the motion of a car is

, where

constant. The power exerted by the car’s engine is along a horizontal road. i Show that

is the car’s speed and is a , and the car has constant speed

.

With the engine operating at a much lower power, the car descends a hill of inclination , where

. At an instant when the speed of the car is

, its acceleration is

. ii Given that the mass of the car is

, calculate the power of the engine. © OCR, GCE Mathematics, Paper 4729, January 2011

12 The maximum power produced by the engine of a small aeroplane of mass tonnes is . Air resistance opposes the motion directly and the lift force is perpendicular to the direction of motion. The magnitude of the air resistance is proportional to the square of the speed and the maximum steady speed in level flight is . i Calculate the magnitude of the air resistance when the speed is The aeroplane is climbing at a constant angle of

.

to the horizontal.

ii Find the maximum acceleration at an instant when the speed of the aeroplane is

.

© OCR, GCE Mathematics, Paper 4729, June 2010 13 The resistance to the motion of a car of mass

is

, where

is the car’s speed

and is a constant. The car ascends a hill of inclination , where exerted by the car’s engine is i Show that

. The power

and the car has a constant speed

.

.

The power exerted by the car’s engine is increased to

.

ii Calculate the maximum speed of the car while ascending the hill. The car now travels on horizontal ground and the power remains

.

iii Calculate the acceleration of the car at an instant when its speed is

.

© OCR, GCE Mathematics, Paper 4729/01, June 2008 14 A car of mass has a maximum speed of The car experiences a resistance of where constant. The maximum power of the car’s engine is i Show that

when travelling on a horizontal road. is the speed of the car and is a

.

.

ii Find the maximum possible acceleration of the car when it is travelling at horizontal road.

on a

iii The car climbs a hill, which is inclined at an angle of to the horizontal, at a constant speed of . Calculate the power of the car’s engine. © OCR, GCE Mathematics, Paper 4729/01, January 2008 15 A space shuttle of mass is moving in a straight line in space. There is no resistance to motion, and the mass of the shuttle is assumed to be constant. With its motor working at a constant rate of the shuttle’s speed increases from to in a time seconds. a Calculate the value of . b Calculate the acceleration of the shuttle at the instant when its speed is 16 A car of mass magnitude of

has maximum power of

.

. The resistive forces have constant

.

i Calculate the maximum steady speed of the car on the level. The car is moving on a hill of constant inclination to the horizontal, where

.

ii Calculate the maximum steady speed of the car when ascending the hill. iii Calculate the acceleration of the car when it is descending the hill at a speed of working at half the maximum power. © OCR, GCE Mathematics, Paper 4729, June 2009 17 A car of mass experiences a resistance of magnitude , where is a constant and is the car’s speed. The car’s engine is working at a constant rate of . At an instant when the car is travelling on a horizontal road with speed its acceleration is . At an instant when the car is ascending a hill of constant slope speed its acceleration is . i Show that

, correct to decimal places, and find .

to the horizontal with

The power is increased to

.

ii Calculate the maximum steady speed of the car on a horizontal road. © OCR, GCE Mathematics, Paper 4729, January 2009 18 A cyclist and her bicycle have a combined mass of

. The cyclist ascends a straight hill

of constant slope, starting from rest at and reaching a speed of above the level of . For the cyclist’s motion from to , find

at . The level of is

i the increase in kinetic energy, ii the increase in gravitational potential energy. During the ascent the resistance to motion is constant and has magnitude done by the cyclist in moving from to is iii Calculate the distance

. The work

.

. © OCR, GCE Mathematics, Paper 4729/01, June 2007

19 i A car of mass 800 kg is moving at a constant speed of on a straight road down a hill inclined at an angle to the horizontal. The engine of the car works at a constant rate of and there is a resistance to motion of

. Show that

.

ii The car now travels up the same hill and its engine now works at a constant rate of . The resistance to motion remains . The car starts from rest and its speed is after it has travelled a distance of . Calculate the time taken by the car to travel this distance. © OCR, GCE Mathematics, Paper 4729/01, June 2014 20 A car of mass

travels up a line of greatest slope of a straight road inclined at

to the

horizontal. The power of the car’s engine is constant and equal to and the resistance to the motion of the car is constant and equal to . The car passes through point with speed . i Find the acceleration of the car at . The car later passes through a point with speed

. The car takes

to travel from

to . ii Find the distance

. © OCR, GCE Mathematics, Paper 4729, January 2012

21 A car of mass is moving along a horizontal road against a constant resistance to motion of . At an instant when the car is travelling at its acceleration is . i Find the driving force of the car at this instant. ii Find the power at this instant. The maximum steady speed of the car on a horizontal road is

.

iii Find the maximum power of the car. The car now moves at maximum power against the same resistance up a slope of constant angle to the horizontal. The maximum steady speed up the slope is . iv Find .

© OCR, GCE Mathematics, Paper 4729, January 2010 22 A particle of mass grams moves along the -axis under the action of a propulsive force . The particle’s displacement, metres, depends on time, seconds, as follows:

Find the power of force when

.

23 A van of mass travels along a horizontal road against a constant resistive force of . The van travels with constant acceleration from rest, at time , to at time . It then travels at constant speed for before decelerating to rest over . The speed–time graph illustrates the motion. Calculate the power of the vehicle engine when: a b

.

Calculate the power of the van’s brakes when: c

. speed(m s–1)

15

O

30

150 175

time (s)

2 Dimensional analysis

In this chapter you will learn how to: understand the concept of dimensions use the language and symbols of dimensional analysis understand the connections between units and dimensions check the validity of a formula by using dimensional considerations predict formulae by using dimensional analysis.

Before you start… A Level

You should be able

Mathematics Student Book 1

to work with indices and surds.

1 Simplify: a b c d

GCSE

You should be able to rearrange formulae.

2 Make the subject of the formula:

GCSE

You should be able to solve simultaneous equations.

3 Solve the equations:

GCSE

You should be able to express direct and indirect proportion in mathematical terms.

4

GCSE

You should know common area and volume formulae.

5 For a sphere of radius expressions for:

is inversely proportional to . If what is the value of when ?

a the volume b the surface area.

when

, give in terms of

,

A Level Mathematics

You should be familiar with the

Student Book 1

standard SI units of mass , length and time

6 What are the SI units of velocity?

.

A Level Mathematics

You should know the definition of a

Student Book 2

radian.

A Level

You should know the

Mathematics Student Book 1

definitions and units of velocity and acceleration.

7 What is the angle, in radians, of a sector of a circle of radius and arc length ?

8 A particle moving in a straight line with constant velocity travels State the units.

in seconds. What is its velocity?

9 A particle moving in a straight line with constant acceleration increases its velocity from to in seconds. What is the acceleration? State the units.

A Level

You should know the

Mathematics Student Book 1

definition and units of force.

Chapter 1

You should know the definitions of kinetic energy

and

potential energy .

10 A mass of

is acted on by a constant force of

.

What is its acceleration? State the units.

11 A mass of is held at a height of metres vertically above the ground. The particle is released from rest. By equating its loss in potential energy to its gain in kinetic energy, find its speed at the instant when it hits the ground.

What is ‘dimensional analysis’? In dimensional analysis you look at the type of unit used to measure a quantity rather than the specific units. You use it as a mathematical way of checking that equations and formulae are correct, and are combining ‘like’ quantities, and also to predict and establish formulae.

Section 1: Defining and calculating dimensions The dimensions of a given quantity describe what sort of quantity you are measuring, so any distance or length, whatever its units, has the dimension of length and has the symbol . You can use square brackets to mean ‘the dimension of’ so

.

The dimension of distance is . The diameter of a pin, the radius of a circle, the length of a running track, the distance from London to Hong Kong are all distances that would be measured in different units, but which are all measurements of length or distance and have the dimension . The other common dimensions that you use in Mechanics are

for mass and for time.

The mass of a spider, the mass of an elephant, the mass of a planet might all be measured in different units – there is even a unit of mass in America called a slug – but are all measurements of mass with the dimension . Similarly time, whether measured in seconds, days or centuries, has the dimension . In some branches of science, other dimensions are used, for example, the dimensions of temperature, electric current, intensity of light and amount of matter.

Did you know? There is a connection between dimensional analysis and the greenhouse effect. The concept of dimensional analysis is often attributed to Joseph Fourier, a famous French Mathematician and Physicist. He is probably best known for Fourier series and Fourier analysis, which is widely used in Mathematics and Physics, and for his work on heat flow. Fourier is widely recognised as being the first scientist to suggest that the Earth’s atmosphere would act as an insulation layer – the idea now known as the greenhouse effect.

Key point 2.1 The dimensions of quantities in Mechanics can be expressed in terms of and for time. You use square brackets to abbreviate the phrase ‘the dimension of’, so

for mass, for length

.

Finding dimensions A scalar quantity is a quantity that only has magnitude but not direction, whereas a vector quantity is a quantity that has both magnitude and direction. Vector quantities such as velocity and displacement have the same dimension as their scalar equivalents (speed and distance). To find the dimensions of quantities that are multiplied or divided, you combine their dimensions in the same way as the quantities are combined. WORKED EXAMPLE 2.1

State the dimensions of: a velocity b acceleration c force. a If an object is moving with constant velocity in a straight line and it

State an equation for velocity.

moves a distance in time , then you have the relationship:

Write down the dimensions of the right-hand side of the equation and simplify. The dimension of distance is and of time is . You can use negative indices with dimensions. In more general terms: has the same dimension as .

b

has the same dimensions as

c

State an equation for force … and take dimensions of both sides of the equation.

WORKED EXAMPLE 2.2

Find the dimensions of: a area b kinetic energy. a

Area formulae always involve multiplying two lengths together or squaring a length.

b

State the formula for kinetic energy. To find the dimensions you need to multiply the dimension of by the square of the dimensions of . is a constant so it has no dimensions and does not affect the calculation of dimensions. Remove the brackets and simplify the result.

Dimensionless quantities If all the dimensions cancel out, then the quantity is said to be dimensionless. This is true of many quantities in Mechanics that are described as coefficients. Examples are the coefficient of restitution (which you will meet in Chapter 3, Section 3) and the coefficient of friction.

Rewind The coefficient of friction is covered in A Level Mathematics Student Book 2, Chapter 21. In this chapter, you will be given the definition when required.

The limiting value of the frictional force between two surfaces is proportional to the normal reaction force between them. This can be written , where is the coefficient of friction.

Since and are both forces:

So is dimensionless. You can leave out dimensionless quantities when you are working out the dimensions of a formula or expression, or you can put in the number to represent the dimensionless quantity.

Key point 2.2 ‘Pure’ numbers, such as and in the formula

EXERCISE 2A

, are dimensionless.

EXERCISE 2A 1

Find the dimensions of the following quantities or state if they are dimensionless: a linear acceleration b acceleration due to gravity c force d weight e momentum f

in the formula

g volume h density (mass per unit volume) i

moment of a force

j

pressure (force per unit area)

Tip Once you have established the dimensions of a quantity then the dimensions will be the same however you calculate it. Forces all have the same dimensions however they are described, for example friction, tension, thrust, reaction force. 2

What are the dimensions of sin ?

3

What are the dimensions of

4

a What are the dimensions of potential energy

? ?

b Are these the same as the dimensions of kinetic energy? 5

a What are the dimensions of work done? b How does this compare to the dimensions of kinetic or potential energy?

6

The refractive index of a material is defined as is the speed of light through the material. Find

where is the speed of light in a vacuum and .

7

What are the dimensions of

8

a Given that all forms of mechanical energy have the same dimensions, find the dimensions of mechanical energy. b

where is a force and is time?

The mechanical energy stored in an elastic string of initial length extended by a distance is where is the modulus of elasticity of the string. Find the dimensions of .

9

Newton’s law of gravitational attraction states that the force of attraction between two masses which are a distance apart is

and

where is a constant. Find the dimensions of .

10 The energy–frequency relationship for slow-moving particles is given by the formula

where

is the wavelength, is the mass of the particle, is velocity of the particle and is Planck’s constant. What are the dimensions of ?

Section 2: Units and dimensions of sums, differences and angles Sums and differences Key point 2.3 You can only add and subtract terms that have the same dimensions. If two or more quantities with the same dimensions are added or subtracted, then the resulting sum or difference will have the same dimensions. If you add two or more lengths, then the answer is also a length with dimension . If you add and subtract several forces the answer is also a force. You can add minutes to hours or you can add kilometres to miles but you cannot add metres to give any meaningful result. You can only add or subtract quantities if they have the same dimensions.

to

Only terms having the same dimensions can be added or subtracted to give a consistent formula. You can use this principle to check whether or not a formula is dimensionally consistent. This is called an error check. The sum

where and are speeds is also a speed and has the dimension of speed.

In dimensional terms:

The sum

The integral

has the same dimensions as

:

is a sum and has the same dimension as

, which is

For sums and differences, you should check that the dimensions of the terms that you are adding or subtracting are the same. Then the dimensions of the answers will also be the same.

Tip Many dimensional analysis questions look very complicated as they involve formulae, often with indices. Do not let the look of the question put you off, it’s about applying rules!

Key point 2.4 For products and quotients you multiply or divide the dimensions.

WORKED EXAMPLE 2.3

If , , , and are lengths and and are masses, check that the terms being added have the same dimensions and find the dimensions of the expression. a b

c a

The dimensions of both terms are their sum is also .

b

The dimensions of both terms in the numerator are , so the dimension of their sum is also . The dimensions of both terms in the denominator is so the dimension of their sum is also . The dimension of the quotient is .

c

The dimension of both terms of the numerator is , so the dimension of their sum is also The dimension of both terms of the denominator is so the dimension of their sum is also . You divide by giving and take the square root.

so the dimension of

WORKED EXAMPLE 2.4

Is the equation , where and are velocities, is the acceleration due to gravity and is time, dimensionally consistent? Checking the dimensions of each term:

To check an equation for consistency you need to find the dimensions of each term and show that they are all of the same dimensions.

Velocity involves dividing distance by time so

Find the dimensions of by squaring the dimensions of .

has dimensions



Similarly

The dimensions of will be the same as those of as they are both velocities. The dimension of is found by multiplying the dimensions of , (acceleration) and .

As the dimensions of the three terms are not the same the equation is not dimensionally consistent so cannot be correct.

Dimensions of angles and trigonometric functions The definition of an angle in radians is the ratio:

As both arc length and radius are lengths then the dimensions of angle are

. An angle is

therefore dimensionless. All trigonometric functions are dimensionless for the same reason – they are the ratio of two quantities with the same dimensions.

Rewind Radians were introduced in A Level Mathematics Student Book 2.

Key point 2.5 An angle has units but is dimensionless.

WORKED EXAMPLE 2.5

What are the dimensions of angular velocity, Angular velocity radians per second where is the angle in

, where is an angle in radians?

Define the quantity involved – angular velocity is rate of change of angle and has the symbol .

radians. Equate the dimensions of all terms in the equation.

Did you know? The metric system originated during the French Revolution of the 1790s in order to provide a unified system of measures that used the metre and kilogram as standard units of length and mass, respectively. SI stands for ‘Système Internationale’ d’unités, which are the units commonly used by the scientific communities of most developed nations. The main base units are metres for length, kilograms for mass and seconds for time. The system, sometimes known as MKS after the units, was the result of an initiative to standardise units started in the late 1940s, at which time the UK was using feet, pounds and seconds as standard and most of Europe were using cgs – centimetres, grams and seconds, or mixtures of centimetres and metres, grams and kilograms, and seconds.

Definitions of some SI units Some common SI units have particular names. A newton is the unit of force. newton is the force required to give a mass of kilogram an acceleration of metre per second. Force is . The joule is the unit of work and energy. joule is the work done (or energy transferred) to an object when a force of newton acts on that object in the direction of motion for a distance of metre. Work done is . The newton metre is the unit of a moment (or torque). It is the effect of a force of newton applied perpendicularly to a moment arm of metre. The moment of a force about a point is to the line of action of that force. The watt is the unit of power. watt is a rate of energy transfer or a rate of working of joule per second. The pascal is the unit of pressure. pascal is the pressure exerted by a force of newton acting on an area of square metre. Pressure is force per unit area.

Explore In this chapter, you are using , and as dimensions. These are three of the seven basic dimensions: the other four are electrical current, thermodynamic temperature, amount of substance and luminous intensity. You could use other quantities as your basic dimensions if they are independent, that is, if you cannot equate the dimensions of any one to a product of powers of the dimensions of the other two. For example force, momentum and time are not independent as Can you express quantities, such as acceleration, in terms of products of powers of density length and force ?

EXERCISE 2B In this exercise, the letters represent the following quantities:

. ,

and for velocities for acceleration , , , and for distance or displacement for angles for force for time for mass 1

If , and are measurements of length, state the dimensions of: a b c

2

If and

are masses, are distances, and and are speeds, state the dimensions of:

a b c d 3

If

4

Find the dimensions of each term in the following equations and hence determine which of the equations are dimensionally consistent.

are masses and are distances, determine the dimensions of:

a b c d Impulse e 5

in momentum ( is a time, is a length and is the acceleration due to gravity)

a What is the dimension of angular acceleration, commonly written as ? b Is the formula

6

dimensionally consistent? Give a dimensional argument for your answer.

In simple harmonic motion (SHM) the restoring force measured towards the centre of the motion is proportional to the displacement, , measured away from the centre of the motion. a Write this as an equation, in terms of , and , using as the constant of proportionality. b What are the dimensions of ?

Tip You can derive dimensions of a quantity either from its formula or from its units. Any formula for that quantity will have the same dimensions. The volume of an icosahedron will have the same dimensions as the dimensions of a cube: . You only need to know that it is a volume to state its dimensions; you do not need the specific formula.

7

Angular momentum is defined as

where is the moment of inertia

and is angular

velocity. a What are the dimensions of angular momentum? b Are these the same as the dimensions of linear momentum? c Explain why angular momentum is sometimes call ‘moment of momentum’. 8

The rotational kinetic energy of a rigid body about an axis is defined as inertia

where is the moment of

of the body about that axis and is the angular velocity.

a What are the dimensions of rotational kinetic energy? b Are these the same as the dimensions of translational kinetic energy? 9

Young’s modulus, , for a solid is defined as

. Stress is the pressure in the solid and strain is

defined as the ratio of extension to the original length. a Write a formula for in terms of , , , and , if is the force exerted on the solid, is its crosssectional area, is the original length and is the extension. b What are the dimensions of ? 10 A student writes the equation for the path of a projectile as:

a Which term in this equation is dimensionally inconsistent? b Suggest an alteration to one variable in this term that would make it dimensionally consistent.

WORK IT OUT 2.1 A particle of mass is fixed at the midpoint of an elastic string of natural length . The string is then fixed to two points and on a smooth horizontal surface, such that and . When the particle is displaced through a small distance along the perpendicular bisector of it begins to perform small oscillations. is the modulus of elasticity of the string and has the dimensions of force. Use dimensional analysis to determine which option could give the correct formula for the periodic time of these oscillations. Which solution is correct? Can you identify the errors in the incorrect solutions? Solution 1

Solution 2

Solution 3

Section 3: Finding dimensions from units and derivatives and predicting formulae Finding dimensions from units This is similar to finding the dimensions of a quantity from its formula. WORKED EXAMPLE 2.6

The pascal is a unit of pressure. Pressure is force per unit area

. The poiseuille

(very rarely used) unit of dynamic viscosity. It is equivalent to pascal seconds

is the

. What are the

dimensions of the poiseuille? To find the dimensions of dynamic viscosity you do not need to know its formula, or even what it is, as long as you know its units. You are told that it is measured in pascal seconds. A pascal is a unit of pressure so you can find its dimensions from the definition of pressure force per unit area. Simplify the dimensions of a pascal.

Then multiply by the dimension of seconds to give the required dimensions. You have seen that angles have units but not dimensions. This makes it difficult to predict units from dimensions. For example, angular velocity, , has dimension but units of radians per second. Frequency also has dimension

but has units of hertz (sometimes called cycles per second).

Key point 2.6 You can predict dimensions from units or formulae but it is not always possible to predict units from dimensions. The tension in an elastic string of initial length that has been stretched to

is given by the formula:

where is the modulus of elasticity for the string. is a physical constant, and if the string were made of a different material, then it would have a different value. has the same units and dimensions as the tension. Other examples of physical constants are surface tension and the gravitational constant.

Tip Do not assume that a quantity represented by a letter is dimensionless unless you are told specifically that it is.

Fast forward You will study the tension in an elastic string in Chapter 6.

WORKED EXAMPLE 2.7

a What are the dimensions of , the modulus of elasticity? b State the units of . a The tension (force) in an elastic

First state a formula involving .

string or spring of initial length that has been stretched to is given by the formula:

Then rearrange it to give a formula for . Find the dimensions of each term and combine. b The units are newtons as is dimensionless.

As and have the same dimensions their quotient will be dimensionless so and will have the same units, i.e. newtons, which is in keeping with the dimensions.

Finding dimensions of second derivatives You saw in an earlier example that the dimensions of acceleration are be written as

. You know that acceleration can

. How do you find the dimensions of acceleration from

?

Rewind You learned about non-uniform acceleration in A Level Mathematics Student Book 1.

Using dimensions to predict formulae Key point 2.7 You can use dimensional analysis to predict formulae by equating the dimensions of the terms of the proposed formula.

Tip Remember that numerical constants, and other dimensionless quantities such as trigonometric functions, ratios and angles, can be left out of the calculation or given the dimension . When you use dimensional analysis to construct a formula, you need to look at all possible factors that could affect the system, even if you then decide that the effects of some of them are negligible and so can be omitted. For example, if you are proposing a formula for the time of oscillation of a simple pendulum the obvious factors to include are the mass of the pendulum, the acceleration due to gravity and the length of the string. Other considerations would be air resistance, the mass of the string and the smoothness and shape of the mass at the end of the string, but you could reasonably assume that the effect of air resistance on a small, smooth object is negligible and that the mass of the string is small enough to have no real effect on the motion of the pendulum. WORKED EXAMPLE 2.8

Pressure, , is measured in newtons per metre squared. Surface tension, , is defined as force per unit length:

.

a What are the dimensions of pressure?

b What are the dimensions of surface tension? c The pressure inside an ideal soap bubble is given by the formula where is the radius of the sphere and is the surface tension. Find the values of and and hence find the formula for . a

Use the units of pressure given to write an expression for the dimensions. Newtons are units of force or mass acceleration. Metres are units of length. Simplify the indices to give a single expression for the dimensions.

b

Use the definitions of force and length to find the dimensions of surface tension in the same way.

c

Write down the formula given in the question. Write the dimensional equation. Remember that is dimensionless so can be left out.

for Equate the indices of , and .

for for and

Put these values of and back into the given formula and simplify.

WORKED EXAMPLE 2.9

A simple pendulum consists of a particle of mass suspended at the end of an inextensible string of length . The pendulum is initially hanging at rest and then it is displaced through a small angle and released to make small oscillations. Given that the formula for the periodic time, , of these oscillations is independent of for small values of , derive a formula for using a dimensional argument.

where is a constant.

You need to consider what is likely to depend on. The factors involved are mass, , length of pendulum, , and so it is sensible to propose that is proportional to a product of powers of these. Other factors, such as air resistance, are likely to be negligible. If the pendulum was in a viscous medium, such as oil, then the resistance of the medium would also have to be considered. The periodic time has dimension . The dimensions of both sides of the equation must be the same and are equal to .

is dimensionless State the dimension of each of the terms …

then combine them as the formula states and simplify. Equating indices of gives the value of .

Equating indices of

gives the value of .

Equating indices of gives the value of . The values are:   so

 

Substitute these values into the equation and state a formula for .



WORKED EXAMPLE 2.10

A particle moves in a straight line with constant acceleration . The initial velocity of the particle is . Derive a formula for the velocity, , of the particle after seconds.

where is a constant. The dimensions of dimension of

are

so the

is also

You need to consider the change in velocity after seconds. The only variables involved are and . is a constant. Again, air resistance can be regarded as negligible. The dimensions of all terms must be the same for consistency. Equate the dimensions… … and equate indices.

or constant.

EXERCISE 2C

where is a

State the equation and rearrange.

EXERCISE 2C 1

The watt is a unit of joule per second. What are the dimensions of watts?

2

The sievert is a unit of joule per kilogram. What are the dimensions of sieverts?

3

What are the dimensions of

4

A yank is defined to be the rate of change of force with time. What are the dimensions of yanks?

5

The area of a triangle can be written as

, where is an expression for angular velocity?

, where and are side lengths and is an

angle. By stating the dimensions of each of the components, i.e. area, , , and sin , and combining them, show that this formula is dimensionally consistent. 6

In a simple harmonic motion the displacement can be written as , where is the displacement and is the amplitude (greatest distance from the centre of oscillation). a Using a dimensional argument, explain why this formula is dimensionally consistent. b Is it possible to determine the dimensions of from this equation?

7

The sine formula states that

where is the radius of the circumcircle of a

triangle with angles , and , and side lengths , and , and is a constant. Find, using dimensional analysis, the value of , showing the steps in your argument clearly. 8

Decibels

are used to describe how loud a noise is. A formula for sound level in decibels is: sound

intensity in where is the sound output in watts and is the threshold-level sound output in watts. a What are the dimensions of decibels? b A speaker has a sound output of 9

times the threshold level. Express this in decibels.

a What are the dimensions of acceleration? An equation for oscillations of a damped simple harmonic motion when a system is displaced through a small distance is

where and are constants. Given that this

equation is dimensionally consistent find: b the dimensions of c the dimensions of . 10 A light inextensible string of length is fixed at one end and has a particle of mass fixed at the other end. The mass is moving at constant speed in a horizontal circle of radius and the string is fully extended. The string makes an angle with the downward vertical. Given that tan , use a dimensional argument to find a formula for tan in terms of , and , where is the acceleration due to gravity.

Section 4: Summary of dimensions and units The following exercise forms a summary of common units and dimensions.

EXERCISE 2D 1

Copy and complete the following table. Quantity

Dimension

SI unit

Time Mass Weight

newton

Length (displacement) Area Volume Velocity Acceleration Acceleration due to gravity Force

newton

Kinetic energy

joule

Gravitational potential energy Work done (

)

Moment of a force (

joule )

Power

newton metre watt

Momentum Impulse (

)

newton second

Moment of inertia Angular velocity Density Pressure

pascal

Time period (time for one complete cycle) Frequency Surface tension

hertz

Checklist of learning and understanding In Mechanics, dimensions describe a quantity in terms of three basic dimensions Mass, Length and Time. Other dimensions are used in other branches of Mathematics and Science. You use square brackets to denote ‘the dimensions of’, so

.

You can only add and subtract terms that have the same dimensions and the resulting sum or difference will also have the same dimensions. For products and quotients you multiply or divide the dimensions. A formula must be dimensionally consistent to be valid. Angles and numerical constants are dimensionless. If two quantities are equal, then they have the same dimensions. You can find the dimensions of a quantity from its definition, from an equation describing it or from its units. Quantities can have units but be dimensionless. For example, angles in radians are dimensionless. You cannot predict units from dimensions, as some dimensionless quantities have units. In dimensional calculations, you can give dimensionless quantities the dimensional value . You can use dimensional analysis to predict formulae by equating dimensions on both sides of a proposed formula.

Mixed practice 2 1

State the dimensions of: a

, where is a length and is the acceleration due to gravity

b

, where is an expression for speed.

2

Heron’s (or Hero’s) formula for the area of a triangle with sides of length , , and is where is half the perimeter. Show, with full explanation, why this formula is dimensionally consistent.

3

a Tension in a string is a force. What are the dimensions of tension? b In the following formulae, is the tension in a string, is a mass, is a velocity, is a length and is an angle. Which of the formulae is dimensionally consistent? If the formula is inconsistent, state which term is inconsistent. i ii iii

4

a What are the dimensions of potential energy? b In the cgs (centimetres, grams and seconds) system a particle has potential energy of ergs. What is this in joules?

5

In a simple harmonic motion of a mass the restoring force is proportional to the displacement. At time , the displacement and the acceleration are given by:

Find the dimensions of: a b

.

6

The radial force on a particle moving in a circle is thought to be of the form . By writing each of the components in term of its dimensions, and equating indices, form three equations to find the values of , and and hence find the formula for .

7

a What are the dimensions of force? b Newton’s law of gravitational attraction states that the force of attraction, , between two bodies of masses and is dependent on the masses, the distance between them and the constant so that it can be written as:

Write down, with reasons, the relationship between and c Given that the dimensions of are

, use a dimensional proof to find a formula for

. 8

a What are the dimensions of angular acceleration? b An equation for oscillations of a damped pendulum of length displaced by a small angle is given by

, where and are constants. Given that this equation is

dimensionally consistent, find the dimensions of . c By expressing as a product of powers of and , where is the acceleration due to gravity, find an expression for in terms of and . 9

Surface tension is defined as force per unit length. a What are the dimensions of surface tension? b When liquid forms a puddle on a clean horizontal surface, the depth of the puddle has a maximum value that can be written as:

where is a dimensionless constant, is surface tension, is the acceleration due to gravity and is the density of water. Use dimensional analysis to find a formula for . c If when find the value of .

and

of water has a mass of approximately

,

10 Tension is a force. a What are the dimensions of tension? b Frequency, , has dimension a string is of the form

. Mersenne’s law states that the fundamental frequency of , where is a dimensionless constant, is the length of

the string, is the tension in the string and is the mass per unit length of the string. Use dimensional analysis to find the values of , and and hence find the formula for . 11 a Pressure is force per unit area, measured in the unit pascal. What are the dimensions of pressure? b Dynamic viscosity is measured in pascal seconds. What are the dimensions of ? c The terminal velocity of a small spherical particle, of radius and density , falling vertically down though a medium of density and dynamic viscosity , is given by:

where is a dimensionless constant and is the acceleration due to gravity. Use dimensional analysis to find the values of and . 12 The formula for the lifting force generated on a wing of an aeroplane is of the form , where is a dimensionless constant, is the air density, is the air speed and is the surface area of the wing. Use dimensional analysis to find the values of , and and hence find the formula for . 13 Surface tension, , is defined as force per unit length. a State the dimensions of . b State the dimensions of density. c The average height, , of a liquid in a capillary tube can be written as

, where

is a dimensionless constant for the liquid, is the surface tension, is the density of the liquid, is the internal radius of the tube and is the acceleration due to gravity. Use a dimensional argument to find the values of , and and hence find the formula for . d In this question part use , and give your final answer to an appropriate degree of accuracy. of water has a mass of approximately . Water rises up a vertical capillary tube that has a diameter of . Given that and , what is the height of water in the tube, in millimetres?

3 Momentum and collisions 1

In this chapter you will learn how to: understand momentum and impulse in mathematical terms with units understand that linear momentum is conserved in a collision between objects that are free to move understand that impulse on a body is equal to the change in momentum understand Newton’s experimental law for collisions analyse and solve problems involving simple collisions in a straight line analyse and solve problems involving simple cases of connected particles.

Before you start… GCSE

GCSE

You should be able to solve simultaneous equations both for two linear equations and for one linear and one quadratic equation.

1 Solve the equations:

You should know the equations of linear motion with constant

3 A particle of mass is dropped from a height of metres above a pond.

acceleration.

A Level Mathematics Student Book 1

You should understand the definitions and units of velocity, acceleration and force.

2 Solve the equations:

What is the speed of the particle at the instant when it hits the water? 4 State the units of: a velocity b acceleration c force.

A Level Mathematics Student Book 1

You should know Newton’s second law:

5 A constant force of acts on a particle of mass . What is the acceleration of the particle?

Chapter 1

You should know the definition of kinetic and potential energy.

6 The speed of a particle of mass . What is its kinetic energy?

is

What are momentum and impulse? You use and understand the concepts of impulse and momentum instinctively in everyday life. If you hit a

ball with a tennis racquet, you know that the ball will move in the direction in which you hit it, and the harder you hit it, the faster and further it will go because it receives a greater impulse. A ball rolling down a slope gathers momentum. A hammer hitting a nail sends the nail forward in the direction of the blow and the hammer bounces back slightly in our hands. Events such as playing snooker or air hockey, applying the brakes in a vehicle, pile-driving the foundations of a building, hitting or kicking a ball, the wind blowing the sails of a boat and carrying it forward, can all be modelled using these two concepts.

Section 1: Momentum and impulse If you apply a force to a stationary object, the object will try to move in the direction of the force. If the force continues to act in the same direction and there is no resistance, the object will move more and more quickly. WORKED EXAMPLE 3.1

An object of mass is moving at

on a smooth surface in a straight line and a constant force

is applied to it in the direction of the motion for seconds. If the final velocity of the object is show that State Newton’s second law

.

The acceleration is constant, since and are both constant. Use the equations of motion in a straight line with constant acceleration to find the final velocity from the initial velocity Multiply this equation through by . Substitute for the term

.

Rearrange the equation as shown. Momentum is the product of mass and velocity and is measured in

Key point 3.1

Impulse is the product of force and time, has the symbol , and is measured in

(newton seconds).

Key point 3.2

This is called the ‘impulse–momentum principle’. Although the units for momentum and impulse appear to be different, they are in fact equivalent. Momentum, force, impulse and velocity are all vectors and can be expressed in vector format.

Tip 1

As force, impulse and momentum are all vector quantities their direction matters, so it is often helpful to draw a diagram. Remember to label a direction on the diagram as positive.

2

Always check the units and convert to kilograms, metres and seconds, if necessary.

WORKED EXAMPLE 3.2

A football of mass

is travelling along the ground at

positive direction

20m s–1

What is its momentum?

Convert the mass from to

.

State the formula. Substitute values into the stated formula.

WORKED EXAMPLE 3.3

A particle of mass

is at rest on a smooth, horizontal surface. It is hit with an impulse of

What is its velocity immediately after the impulse? State the impulse–momentum principle. Work out the change in momentum by subtracting the initial momentum from the final momentum. Equate the impulse to the change in momentum.

Modelling When you make a mathematical model you look at a simplified situation. Initially, you model all objects as point masses and do not take into account their size or what they are made of. As you learn more about the situation you can make better models and can put in more accurate data.

Focus on … You will learn more about mathematical modelling in Focus on … Modelling 1.

Tip You don’t have to draw a complicated diagram, as long as it is clear. For example, in Worked example 3.4 the boats could be replaced by dots.

WORKED EXAMPLE 3.4

A toy sailing boat of mass is blown along by a constant wind acting horizontally with a force of If the boat is initially at rest, find its velocity after seconds.

positi ve direction 0.2N

t = 0

0m s–1

t = 10

vm s–1 State the values you are given and convert to

Define any unknowns. State the formula.

.

Substitute the values and calculate. As you are asked for velocity, you must also state the direction. WORKED EXAMPLE 3.5

A football of mass is travelling in a straight line along the ground at when it is kicked along the line of motion and starts to move with a speed of Calculate the impulse given to the football by the footballer’s kick if: a the ball is now moving in the opposite direction b the ball continues to move in the same direction. a

20m s–1

before

Draw a diagram to illustrate the situation.

25m s–1 after +ve

Choose a positive direction. Check the units and change to

.

State the values you know. has a negative sign as the football is moving in the negative direction after it is kicked. To find the impulse you need to find the change in momentum. Substitute the values and calculate.

in the opposite direction to the approaching ball. b

The impulse is negative, which tells you that it is in the opposite direction to the original direction of the ball. This means the ball was moving towards the footballer, who then kicked it back along the line of motion. You need to give the direction as well as the magnitude of the impulse.

20m s–1

before

25m s–1

after

Draw a new diagram to illustrate the second situation.

+ve Check the units and change to

.

State the values you know. now has a positive sign as it is moving in the positive direction. To find the impulse you need to find the change in momentum. Substitute the values and calculate. in the same direction as the initial movement of the ball.

Tip

The impulse is positive, which tells you that it is in the same direction as the original direction of the ball. This means the ball was moving away from the footballer, who then kicked it in the same direction.

Remember that velocity and impulse are vector quantities so you must show the direction by using plus and minus signs and use these in your calculations.

EXERCISE 3A 1

Calculate the momentum of the following in before you start, if necessary. a A rocket of mass b A cat of mass

Make sure you change to standard units

and velocity and velocity

c A marble of mass

and velocity

d A car of mass

and velocity

e A rhino of mass

tonnes running at

2

A bullet of mass velocity?

3

A car is moving at a velocity of

4

What impulse is generated by an engine that exerts a force of

for

5

What impulse is generated by an engine that exerts a force of

for minutes?

6

The table shows the velocity in of a particle of mass before and after an impulse is applied to it. Calculate the impulse in each case. The arrow indicates direction and the first one is done for you.

is fired from a rifle. Its momentum is

as it leaves the rifle. What is its

Its momentum is

What is its mass in metric tonnes? ?

Positive direction  Initial velocity

Final velocity











Impulse





7

A sailing boat of mass is stationary on the sea when it is blown by a wind that exerts a constant force of What is the speed of the boat, in , after minutes?

8

A motor boat, of mass , is moving in a straight line at a speed of The drag of the water produces a constant force of in the direction opposite to the direction of motion of the boat. How long does it take for the speed of the boat to halve?

9

A gyrfalcon of mass

is flying at

with a following wind that exerts a constant force of

in the direction in which the gyrfalcon is flying. What is the gyrfalcon’s speed, in

,

later?

10 A gyrfalcon of mass is flying at into a wind that exerts a constant force of direction opposite to the flight of the gyrfalcon. What is the gyrfalcon’s speed, in ,

in the later?

11 A football of mass wall, its speed is the wall on the ball?

is kicked along the ground and hits a vertical wall at right angles. As it hits the and it bounces straight back with a speed of What is the impulse of

12 A bowling ball of mass the wall, its speed is of the wall on the ball?

is rolled along the ground and hits a vertical wall at right angles. As it hits and it bounces straight back with a speed of What is the impulse

13 A motorbike of mass slows down from newtons, of the constant braking force?

to

in

14 A motorbike of mass accelerates from to in in newtons, of the constant accelerating force produced by the engine?

 

. What is the magnitude, in

. What is the magnitude,

Section 2: Collisions and the principle of conservation of momentum In a game of snooker, if the cue ball hits a stationary red ball it exerts a force on the red ball that causes it to move along the line of the collision. From Newton’s third law, there is an equal and opposite reaction force on the cue ball. As the time for which the collision force acts is the same for both balls – i.e. the time the two balls are in contact – then the impulse on the two balls is also the same but in opposite directions.

impulse on cue ball = −I N s

impulse on red ball = +I N s positive direction

Total impulse on the two balls along the line of the collision is As the total impulse is zero, there is no change in total momentum, which means that linear momentum is conserved in a collision between two objects when they are both free to move. This is called ‘the principle of conservation of linear momentum’. We can see this by considering the following argument.

Did you know? Cannon on Spanish galleons were originally mounted on the highest part of the ship to give the best possible advantage over the enemy. It was soon realised that the impulse of the cannon’s recoil on the ship at such a height was making the ship unstable and likely to capsize, so the cannon were moved down to the lower decks.

Consider a collision between two objects, and , of mass and , moving with velocities the same straight line. If their respective velocities after the collision are and then:

and

in

Rearranging gives:

is the total momentum before the collision and

is the total momentum after the

collision.

Key point 3.3 If there are no external impulses then total momentum before collision total momentum after collision

WORKED EXAMPLE 3.6

WORKED EXAMPLE 3.6

A ball of mass is moving at when it collides with a second ball of mass moving towards it on the same straight line with a velocity of . If the second ball now moves in the opposite direction at

, find:

a the final speed of the ball of mass



b in terms of , the magnitude of the impulse on the ball of mass positive direction

.

Draw a clear diagram with the positive direction marked and values in standard units.

3m

10 m s–1 6 m s–1

2m

3m

v m s–1

2m

initial

a Let the final velocity of the mass

6 m s–1

be

Total momentum before collision total momentum after collision

Define the unknown. Use the principle of conservation of linear momentum. State the formula. Substitute the values, remembering that direction is indicated by sign. Solve for . The ball of mass moves at of its initial motion.

in the direction

Note that you were asked for speed, so you do not include the minus sign with your answer.

So the final speed is b Find the impulse on the

mass.

The impulse on the two balls is equal and opposite so it is sensible to find the impulse on the mass as you are given its velocities. This will give the correct answer even if you have made a mistake calculating the velocity of the mass . However, you can use the impulse on the ball of mass to check your answer.

The impulse on the ball of mass is equal and opposite so is in the negative

If you are asked for the magnitude, you do not give the minus sign with your answer.

direction. Its magnitude is WORKED EXAMPLE 3.7

A snooker cue ball of mass same direction at

travelling at

hits a red ball of mass

travelling in the

. If the red ball continues to move in the same direction but at

find:

a the magnitude of the impulse on the red ball b the magnitude of the impulse on the cue ball c in terms of , the speed of the cue ball after the collision. positive direction 0.16kg initial

10m s–1 0.17kg

Draw diagrams to show the situation clearly. Make sure that you label a direction as positive. 6 m s–1

The masses and

and

are written as

impulse on cue ball

impulse on red ball

collision 0.16kg

v m s–1 0.17kg

a

9 m s–1

For the red ball only, work out the impulse by finding the change in momentum.

The impulse on the red ball is

, in the

direction of the motion. b The impulse on the cue ball is equal and opposite and so it is . The magnitude of this is .

The impulse on the cue ball is equal and opposite to the impulse on the red ball. The magnitude is the modulus of this so you omit the minus sign in the answer. Only omit the minus sign if you are asked for the magnitude, otherwise leave it in as it indicates direction.

c The impulse on the cue ball Let the final speed of the cue ball be

State the values you know and define the unknown quantity. State the formula and the values.

Put impulse equal to the change in momentum. Here you do need to write the minus sign. Rearrange to find The final speed is positive showing that the cue ball continues to move in the same direction but is slower than before.

WORK IT OUT 3.1 Two particles of masses and collide. Their speeds before the collision are and , respectively, and they are moving in the same direction, which is taken as the positive direction. After the collision, their speeds are and , respectively. Solutions 1, 2 and 3 give three possible sets of values for and . Which solution is possible? Explain why the other two solutions are not possible. Solution 1

Solution 2

Solution 3

Masses that combine In some situations, two objects ‘move off together’ or ‘coalesce’ after a collision. This means that they combine into one object, rather like two raindrops running down a window that combine to form a single drop. When objects coalesce in a collision, their masses are added together after the collision. WORKED EXAMPLE 3.8

A toy truck of mass is travelling at when it hits a stationary toy truck of mass two trucks move off together, what is their speed immediately after the collision? positive direction

. If the

Draw a diagram with masses, speeds and the positive direction labelled.

2 kg

10 m s–1

5 kg

v m s–1

3 kg

0 m s–1

initial

Use the principle of conservation of linear momentum.

Total momentum before collision total momentum after collision

State the formula. Define the unknown quantity.

Let be the speed of the combined trucks.

Substitute the values and calculate.

WORKED EXAMPLE 3.9

A cat, of mass , is sitting on a toy train of mass that is free to move on a straight horizontal track. The train is at rest when the cat jumps off the train. The direction of the cat is horizontal and in the direction of the track. The velocity of the cat is . What is the velocity of the train immediately after the cat jumps off? positive direction cat train

0 m s–1 0 m s–1

cat train

Draw a diagram. 2 m s–1 v m s–1

initial

Apply the principle of conservation of linear momentum. Solve for The velocity of the train is direction to the cat’s jump.

in the opposite

Further examples: multiple collisions and using change in kinetic energy Harder examples often involve multiple collisions. Here you need to split the problem into parts and treat each collision as a separate event. As usual, you need to draw clear diagrams and set out your working clearly.

Rewind Change in kinetic energy is considered in Chapter 1.

WORKED EXAMPLE 3.10

Three particles, , and , of masses , and

and

, respectively, as shown in the diagram.



, are moving with velocities

,

positive direction m initial

2.5m s–1

2m

P

2 m s–1

Q

3m

1 m s–1

R

collides with the mass , which then goes on to collide and coalesce with particle . The combined particle moves with speed . a What is the speed of after the first collision? b What is the kinetic energy lost by in the collision with ? a

Collision 1: positive direction m initial

2.5m s–1

2m

P

m

2 m s–1

Draw the diagram for the first collision.

Q

v1 m s–1

2m

P

Always draw a diagram to show the situation. Both collisions need two sets of diagrams. Label the speeds clearly, with their directions, and mark the positive direction. The signs of the final answers will give the direction of the motion.

v2 m s–1

Let be the speed of and speed of after the collision.

be the

Q

Collision between and :

Use the principle of conservation of momentum.

Total momentum before collision total momentum after collision

State the formula.

Simplify by dividing through by . You now have one equation in two unknowns so you need more information to solve for the two velocities. Draw the diagram for the second collision.

Collision between and : positive direction 2m initial

v2 m s–1

3m

Q

1 m s–1

R

5m

1.5m s–1

Total momentum before collision total momentum after collision

Use the principle of conservation of linear momentum. Solve for

b Substituting into equation

To find the loss in kinetic energy of you need to find the velocity of after the collision with .

State the formula for kinetic energy. Find the initial

… and the initial … and subtract.

Modelling in collision questions If you are dealing with collisions between spheres, unless you are told otherwise, you assume that: the spheres are smooth the impulse during the collision acts along the line of centres none of the spheres is spinning.

Fast forward You will look at oblique collisions in Chapter 8.

EXERCISE 3B 1

Use the conservation of linear momentum to fill in the following table. and are the masses in of two particles moving in the same straight line with initial speeds and , respectively. The particles collide and their speeds after their collision are and , respectively. The speeds are given in and the signs indicate direction.

2

A ball, , of mass

travelling at

collides with a stationary ball, , of mass

. Ball is

brought to rest in the collision. What is the speed of ball immediately after the collision? 3

A ball, , of mass velocity of at a velocity of

4

moving at

collides with a ball, of mass

moving towards it at a

in the same straight line. If ball then moves in a direction opposite to its original , what is the final speed of ball ?

A cat of mass

jumps onto a stationary toy train of mass

that is free to move on a straight

horizontal track. The speed of the cat in the direction of the track immediately before it lands on the truck is . What is the speed of the cat and the train, in the direction of the track, immediately after it lands? 5

A croquet ball of mass impulse of

is at rest when a croquet mallet hits it with a force, instantly producing an

.

a What is the exact speed of the ball at the instant it leaves the croquet mallet? b The ball immediately hits another croquet ball of mass that is at rest. If the second ball moves off at , what is the speed of the first ball immediately after the collision? 6

A particle, , of mass

is moving with speed

towards particle , of mass

, which is

moving towards in the same straight line, at a speed of . The particles coalesce. What is the magnitude and direction of the speed of the combined particles? 7

Particle , of mass

is moving with speed

towards particle . Particle is moving towards

particle , in the same straight line, with speed . The two collide and after the collision the two particles move in opposite directions, each with speed . What is the mass of particle ?

8

A truck of mass

is travelling along a straight track at a speed of

when it collides with a

second, stationary, truck of mass , which then starts to move with speed speed of the first truck immediately after the impact? 9

A skateboard of mass

is moving at

when it hits a ball of mass

. What is the

which is at rest.

Immediately after the collision the skateboard moves at in the same straight line and in the same direction. What is the speed of the ball immediately after the collision? 10 A football of mass moving at it, in the same straight line, at

hits a second football of mass moving directly towards . After the collision the two footballs both have a speed of

but in opposite directions. Find the value of What assumption have you made about the two footballs? 11 A particle of mass

is moving along a straight line with speed

It collides with another

particle, , of mass moving towards on the same straight line at a speed of collision the particles coalesce and move at a speed of . a Find expressions for the two possible values of , in terms of . b If in each case the magnitude of is



, find the two possible values of .

. After the

Section 3: Restitution, kinetic energy and impulsive tension In this section you will consider examples of three different types of collisions: a perfectly elastic collision, a perfectly inelastic collision and one that is neither perfectly elastic nor perfectly inelastic. In a perfectly elastic collision, no kinetic energy is lost from the system. The kinetic energy of the individual spheres may change but the total of their kinetic energies will remain the same. Consider the example shown in the diagram.

positive direction 2kg

10m s–1

2kg

5.2m s–1 3kg

3kg

6 m s–1

initial 9.2m s–1

, so momentum is conserved.

so there is no loss in kinetic energy. In a perfectly inelastic collision, the spheres coalesce.

positive direction 2kg

10m s–1

3kg

6 m s–1

initial 5kg

7.6m s–1

Newton discovered that the outcome of a collision depends on the material that the objects are made of. The next example is of a collision that is neither perfectly elastic nor perfectly inelastic.

positive direction 2kg

10m s–1

3kg

6 m s–1

2kg

7 m s–1

3kg

8 m s–1

initial

Again momentum is conserved but kinetic energy is lost. The loss in kinetic energy is slightly lower than if the particles had coalesced. Newton discovered, by carrying out a series of experiments, that there is a constant ratio between the speed of approach and speed of separation of the spheres before and after the collision that is independent of the masses of the two spheres but depends on the materials from which the spheres are made. The constant is called the coefficient of restitution.

Key point 3.4 Newton’s experimental law of collisions states that:   or  

  

where is a constant called the coefficient of restitution. The value of varies depending on the material properties of the objects involved in the collision. For a perfectly elastic collision

and there is no loss of total kinetic energy.

If the spheres coalesce then , the collision is inelastic, and there is maximum loss of kinetic energy. This is sometimes called a perfectly inelastic collision, but for this course the phrase inelastic collision will be used. In other collisions, for which

, the total kinetic energy of the system decreases.

Applying Newton’s experimental law Consider a collision between two particles of masses velocities and .

positive direction m1

u1

m2

u2

m1

v1

m2

v2

initial

and

with initial velocities

and

and final

Did you know? Some collisions are called 'super elastic' because they cause a gain in kinetic energy in the system, usually due to another mechanism that causes potential energy to be released as kinetic energy. The study of these super elastic collisions is outside the scope of this course but information can be found on the internet. Checking that kinetic energy is not gained during collisions can be a useful error check. Applying Newton’s experimental law (NEL) gives:

Unless

the spheres will not separate. Similarly, unless

the spheres will not collide.

Tip When using Newton’s experimental law be careful to get the velocities in the correct order with the correct signs. Use a clearly labelled diagram showing directions clearly.

WORKED EXAMPLE 3.11

Two spheres of masses

and

are moving towards each other with speeds

respectively. After the collision, they move away from each other with speeds respectively.

and and

, ,

a What is the value of the coefficient of restitution between the two spheres? b In terms of , what is the kinetic energy loss for the particle of mass positive direction 3m

4

5

2m

?

Draw a clear diagram. There is no need to put units on the diagram as long as you have checked that the units are consistent. For example, use either grams and kilograms but not a mixture.

initial 2.5

3m

2m

3.5

a Newton’s experimental law:

State Newton’s experimental law.

State the values. Remember that signs indicate direction. Do not leave this step out as it helps to prevent errors with signs. Remember that the final velocities go on the top line.

Find the initial

b

… and the initial … and subtract.

Tip

Always check that is positive and between and . If it isn’t, check that you have substituted correctly and have the fraction the correct way up.

WORKED EXAMPLE 3.12

Two spheres, and , of masses and respectively, are moving towards each other with velocities and respectively. They collide and the kinetic energy lost in the collision is . a What are the speeds and directions of and immediately after the collision? b What does this tell you about the coefficient of restitution between and ? positive direction 0.4 initial

Draw a clear diagram and choose a direction to be positive. 6 m s–1 3 m s–1

A

0.4

0.5 B

v1 m s–1

0.5

v2 m s–1

a Total momentum before collision total momentum after collision

Use the principle of conservation of momentum. State the formula.

Multiply through by

Find the

to simplify the numbers.

before the collision.

Find the after the collision by subtracting the amount of lost from the initial . Write an expression for the collision and equate this to By solving the system of two equations, we find that and . The speeds of and are both in the direction of A’s initial motion. b As the velocities have the same value and are in the same direction this means that the particles coalesce (become one) so .

after the .

You now have two equations, one linear one quadratic , for the two unknown velocities. You need to solve these simultaneously.

and

You now need to state the value of , the coefficient of restitution, and comment on your answer.

WORKED EXAMPLE 3.13

Two spheres of mass

and

are moving in the same direction in the same straight line with

velocities and , respectively. The coefficient of restitution between the spheres is What are the velocities of the spheres immediately after the impact? Draw a clear diagram. Show the positive direction. Convert units as necessary.

.

positive direction 0.3

5 m s–1

0.5

4 m s–1

0.3

v1

0.5

v2

initial

State the values you are given. Total momentum before collision total momentum after collision

Use the principle of conservation of momentum.

Newton’s experimental law:

State Newton’s experimental law.

Substitute into the equation and simplify.

State the known values with their signs.

Substitute in the values and simplify. Substitute

into

: Multiply through by solve.

to make the numbers easier and

both in the same direction as the initial motion. Check back in

:

You should check your answers by substituting back into the unsimplified conservation of momentum equation .

Collisions between a moving object and a fixed object If a body hits a fixed wall, momentum is not conserved because the wall cannot usually move. However, Newton’s experimental law still holds, so you can calculate velocities for impacts involving a fixed object using the coefficient of restitution.

positive direction

initial u

m wall

v

m

Consider a particle of mass moving with velocity on a smooth horizontal plane as it collides with a fixed wall at right angles and bounces back off the wall with a velocity of . The coefficient of

restitution between the wall and the particle is .

Explore

Newton’s cradle is a set of spheres suspended from a frame. You can find videos on the internet showing it in action. Can you explain the behaviour of Newton’s cradle when one sphere is set in motion? What happens if two (or more) of the spheres are displaced at the same time? Newton’s experimental law states that:

(as the wall does not move),

and

, so the equation reduces to:

 or  where and are the velocities of the particle before and after the collision with the wall. The impulse of the wall on the particle is equal to change in momentum of the particle:

Key point 3.5 If an object, , moving with velocity , collides at right angles with a fixed object, then rebounds with velocity object.

where is the coefficient of restitution between and the fixed

WORKED EXAMPLE 3.14

A football of mass

moving with velocity

hits a fixed wall at right angles. If the

coefficient of restitution between the football and the wall is

, find:

a the speed at which the ball rebounds from the wall b the magnitude and direction of the impulse of the ball on the wall c the loss in kinetic energy of the football. Always draw a diagram so you don’t get confused with signs. Always show the positive direction with an arrow.

positive direction

before 25 m s–1 wall v1 after

a Newton’s experimental law:

You cannot use conservation of momentum as the wall is not free to move but you can use Newton’s experimental law. Substitute in the numbers.

So the ball rebounds with speed . b

The impulse of the ball on the wall is equal and opposite to the impulse of the wall on the ball. To find impulse you need force time or change in momentum. As the wall does not move, you can only find the change in momentum of the football.

The impulse of the wall on the ball is:

Find the impulse of the wall on the ball.

Therefore the impulse of the ball on the wall is towards the wall. c Initial

The impulse of the ball on the wall is equal and opposite to this.

To find the loss in kinetic energy you subtract the final from the initial the ball.

Final Loss in

WORKED EXAMPLE 3.15

Small spheres , of mass

, and , of mass

are lying in a smooth horizontal groove, which is a

straight line ending in a vertical wall that is at right angles to the groove. is projected with velocity towards , which is initially at rest. The coefficient of restitution between and is then hits the wall and rebounds. The coefficient of restitution between and the wall is a What is the velocity of as it rebounds off the wall? b State, with reasons, whether or not there will be any further collisions between and . positive direction 3m initial

u1 = u m s–1

P 3m

2m

u2 = 0m s–1

Q v1

2m

v2

Draw a clear diagram of the first collision only. Do not try to cram both collisions into one diagram. It can help to label , , and on the diagram to get Newton’s law the correct way round.

a Total momentum before 1st collision total momentum after 1st collision

Use the principle of conservation of momentum. State the equation. Substitute in the values given and divide through by as every term contains . There is no need to define the unknowns if they are clearly shown on the diagram. State the principle and the equation.

Newton’s experimental law:

Substitute in the values and simplify.

Substitute

into

Substitute the expression for into

:

.

Find . positive direction

Now deal with the collision with the wall. Draw a new diagram.

initial 0.84u

2m Q

v3

wall

2m

Momentum is not conserved as the wall is fixed so you use Newton’s experimental law.

Newton’s experimental law:

State the equation. Put in the values and solve for include with your answer. The final velocity of is from the wall. b From

away

,

Don’t forget to

Explain that the minus sign indicates direction and state that direction. To find out if there are any further collisions you need to know the velocity of after the first collision and then compare the final velocities of and .

The final velocity of is towards the wall. As is moving towards the wall and is moving away from the wall they will be moving towards each other so there will be at least one further collision.

The direction of and is important as well as the speed. Here the two particles are moving towards each other so they will collide again whatever their speeds.

WORKED EXAMPLE 3.16

A ball of mass

is dropped from rest at a height of

onto a smooth horizontal surface. It

instantly bounces back vertically and reaches a height of

.

a What is the impulse of the ball on the ground at the instant of the collision? Take as leave your answer in surd form.

and

b Taking as

, calculate the coefficient of restitution between the ball and the ground from the

information given, leaving your answer in surd form. positive direction

2m

v

1.5m

To find impulse you either need to know force and time or change in momentum. Momentum depends on velocity so you need to find the velocity at which the ball hits the ground and then the velocity at which the ball leaves the ground. To find the velocities you can either use energy equations or the equations for motion with constant acceleration.

u

a As the ball falls:

First you need to find the velocity, , at which the ball hits the ground. This example uses the work–energy principle.

By equating gain in kinetic energy to loss in potential energy you can find the value of . is negative as it is in the downward direction. You can use energy equations again to find the velocity at which the ball leaves the ground.

As the ball bounces up:

By equating the loss in the value of . Impulse on the ball is vertically up The impulse of the ball on the ground is therefore

to the gain in

, you can find

Knowing the values of and , you can use the relationship between impulse and momentum to find the impulse of the ground on the ball. This is equal and opposite to the impulse of the ground on the ball.

vertically down. b

State the formula. Substitute the values and simplify.

Range of values of The value of determines the final velocities of the colliding bodies and enables you to calculate whether further collisions will occur.

Rewind You learned about gravitational potential energy, mechanical energy and the principle of conservation of mechanical energy in Chapter 1.

Tip If two objects moving in the same straight line are going to collide, then there are two possibilities: either they are moving towards each other from opposite directions, or one is

following the other and the follower is travelling faster than the object it is following.

WORKED EXAMPLE 3.17

A sphere, , of mass

, moving at a velocity of

, hits a sphere, , of mass

at rest on a smooth horizontal surface. The coefficient of restitution between and is

, which is . then

hits a smooth vertical wall at right angles and rebounds. The coefficient of restitution between and the wall is . Find the range of values of for which there is a further collision between and . positive direction 2m initial

Draw a diagram for the first collision with the positive direction labelled. 6 m s–1

4m

P

2m

0 m s–1

Q

v1 m s–1

v2 m s–1

4m

Use the principle of conservation of linear momentum. Simplify the equation.

 

Use Newton’s experimental law.

Simplify the equation. Solve and simultaneously and check your answers in equation .

and The speed of after the collision with the wall is following . The speed of is away from the wall and the speed of is away from the wall. If collides with then:

Use Newton’s experimental law for the collision between and the wall. Now consider what is actually happening. If there is to be another collision then must be moving in the same direction as and it must be moving faster than .

Impulses transmitted through strings If two particles, and , on a smooth horizontal surface, are joined by a light inextensible string that is initially not under tension, and is projected horizontally, the string will become taut. There will be an instantaneous impulse transmitted through the string that will pull into movement. When this situation is modelled mathematically an assumption is made that starts to move in the direction of the string and that the component of the velocity of along the string equals the velocity of . The impulse is equal to the magnitude of the change in momentum of each of the particles along the string. WORKED EXAMPLE 3.18

Two particles, , of mass

, and , of mass

, are initially at rest on a smooth horizontal plane.

and are joined by a light, non-elastic string that is initially slack. is projected horizontally, away from , with a speed of . After the string becomes taut, and move at the same

velocity. Calculate: a the speed of and immediately after the string becomes taut b the impulse on and .

+ve 0

Draw a clear diagram and show the direction you have chosen to be positive.

12m s–1

2kg

4kg

A

B

2kg

4kg v

v

a Conservation of linear

As soon as the string becomes taut, you can apply the principle of conservation of momentum to the whole system to find the common velocity of and .

momentum:

b

Apply the impulse–momentum principle for just one of the particles. The magnitude of the impulse will be equal in magnitude but in opposite directions for and . It is easier to use for the impulse calculation. You can check your answer by also calculating the change in momentum of .

EXERCISE 3C 1

A small sphere, of mass , moving on a smooth horizontal plane with speed hits a vertical wall at right angles and rebounds. The coefficient of restitution between the sphere and the wall is . Find: a the speed of the sphere after the collision b the magnitude of the impulse of the wall on the sphere c the loss in kinetic energy of the sphere if: i and

2

ii

and

iii

and

iv

and

v

and

.

A mass, , moving at same straight line with speed

collides with a second mass of , , moving towards it on the . The coefficient of restitution between the two masses is . Find

the speeds and directions of and after the collision. 3

A particle , of mass it at a speed of with a speed of

, moving at

, collides with a particle , of mass

, moving towards

on the same straight line. After the collision, moves in the opposite direction . Find:

a the speed and direction of after the collision b the coefficient of restitution between and . 4

A small, smooth sphere of mass

, at rest on a smooth horizontal floor, is hit with a blow of

impulse

and immediately hits a vertical wall at right angles. If it rebounds with velocity , find:

a the coefficient of restitution between the wall and the sphere b the magnitude of the impulse of the wall on the sphere c the loss in kinetic energy of the sphere. 5

A small, smooth sphere of mass

, at rest on a smooth horizontal floor, is hit with a blow of impulse

and immediately hits a vertical wall at right angles. If it rebounds with velocity expressions, in terms of , and , for:

, find

a the coefficient of restitution between the wall and the sphere b the magnitude of the impulse of the wall on the sphere c the loss in kinetic energy of the sphere. 6

A small smooth ball-bearing, , of mass and moving with velocity , collides with another small, smooth ball-bearing, , of mass , moving in the same direction with velocity . After the collision, and are moving in the same direction, with velocity and with velocity . a Find the value of b Find the value of , the coefficient of restitution between and . c What is the total loss of kinetic energy in the collision?

7

A particle, , of mass grams, is moving at when it hits a vertical wall, at right angles to the wall, and rebounds. The coefficient of restitution between the wall and the particle is . The particle then hits another smooth particle, , of mass grams, which is initially at rest. After the collision, is at rest. What is the coefficient of restitution between and ?

8

A small sphere, , of mass

, moving in a straight line with velocity

, collides with another

small sphere, , of mass , which is moving directly towards , along the same straight line, with velocity . The coefficient of restitution between the spheres is . a Find the magnitude and direction of the velocities of the two spheres immediately after the collision. b What is the magnitude of the impulse on ? 9

A particle of mass is dropped from a height of onto a smooth horizontal surface and bounces back vertically up to a height . The coefficient of restitution between the ball and the surface is . Taking as

, find the value of .

10 Two small spheres, , of mass



and , of mass

along a smooth horizontal surface, with velocities restitution between the two spheres is .

, are moving directly towards each other



and



, respectively. The coefficient of

a Find the magnitude and direction of the velocities of the two spheres immediately after the collision. b What is the total loss of kinetic energy in the collision? 11 Two particles, and , each of mass , are moving on a smooth horizontal surface in the same direction, in the same straight line, with speeds and , respectively, when they collide. If the coefficient of restitution between the two particles is particles immediately after the collision.

, find the speed and direction of the two

12 A ball of mass is dropped from a height metres onto a smooth horizontal surface and bounces back vertically up to a height metres. The coefficient of restitution between the ball and the surface is . Find the ratio of to .

13 Two particles, and , each of mass , are moving on a smooth horizontal surface in the same direction, in the same straight line, with speeds and , respectively, when they collide. At the instant of collision, each particle receives an impulse of . a Find the magnitude and direction of the velocities of the particles immediately after the collision. b Calculate the coefficient of restitution between the two particles. 14 Two smooth spheres, and , each of mass grams, move towards each other along the same straight horizontal line and collide when they are moving with speeds and , respectively. Immediately after the collision moves with velocity

away from .

a What is the velocity of immediately after the collision? b What is the coefficient of restitution between the two spheres? c Find the magnitude of the impulse exerted on in the collision. 15 A small ball of mass is dropped from rest at a height of onto a smooth horizontal surface. It instantly bounces vertically up and reaches a height of . What is the impulse of the ground on the ball at the instant of the collision? Take as . 16 A ball bearing, , of mass is thrown vertically down with a speed of from a height of . It bounces back and just reaches its original height. Find the coefficient of restitution between and the ground. Take as . 17 Two particles , of mass



, and , of mass



, are at rest on a smooth horizontal plane. is hit

with a blow of impulse in the direction . collides with , which then hits a smooth vertical wall at right angles. The coefficient of restitution between and is and the coefficient of restitution between and the wall is . Find the range of values of for which there is at least one more collision between and . Give your answer as a fraction. 18 Three particles, of mass

, of mass



and of mass



, are at rest in the same

horizontal line on a smooth horizontal surface. is projected along the plane towards at a velocity of and the coefficient of restitution between and is . then collides with . The coefficient of restitution between and is . Find the range of values of for which there is a further collision between and . Give your answer as a fraction. 19 Two masses, , of , and , of , are initially at rest on a smooth horizontal plane. and are joined by a light, non-elastic string that is initially slack. is projected horizontally, away from , with a speed of

. After the string becomes taut, and move at the same velocity. Find:

a the speed of and immediately after the string becomes taut b the magnitude of the impulse on and .

Checklist of learning and understanding

The units of impulse

and momentum

are equivalent.

In general terms, if two objects of mass and are moving with velocities and in a straight line and they collide, then, if their velocities after the collision are and , the total momentum remains constant.

This is called the principle of conservation of linear momentum. Newton’s experimental law of collisions states that:   or   where is a constant called the coefficient of restitution and . The value of varies depending on the material properties of the objects involved in the collision. For a perfectly elastic collision,

and there is no loss of total kinetic energy.

If the colliding objects coalesce, then , and there is loss of kinetic energy. This is sometimes called a perfectly inelastic collision. In all other collisions the total kinetic energy of the system decreases and

.

If an object, , moving with velocity collides with a fixed object, which is at right angles to the plane of movement of , then the object rebounds with velocity coefficient of restitution between and the fixed object.



where is the

Mixed practice 3 1

Two particles and are projected directly towards each other on a smooth horizontal surface. has mass and initial speed , and has mass and initial speed . After a collision between and , the speed of is motion is reversed. Calculate:

and the direction of its

a the change in the momentum of b the speed of after the collision. 2

A particle of mass

is travelling with speed

on a smooth horizontal plane

towards a stationary particle of mass (see diagram). The particles collide, and immediately after the collision has speed and has speed . 10m s–1 m kg

0.8kg P

Q

a Given that both particles are moving in the same direction after the collision, calculate . b Given instead that the particles are moving in opposite directions after the collision, calculate . 3

A roller skater of mass

is moving in a straight line with speed

when she collides

with a roller skater of mass moving in the opposite direction along the same straight line with speed . After the collision the roller skaters move together with a common speed in the same straight line. Calculate their common speed, and state their direction of motion. 4

A particle of mass is moving at when it collides with a particle , of mass , moving in the same direction, in the same straight line at . The two particles coalesce to form a particle moving in the same straight line. What is the velocity of ? 9 m s–1

5 P

mkg

2 m s–1 Q

2.75m s–1

0.8kg R

0.4kg

Three particles , and , are travelling in the same direction in the same straight line on a smooth horizontal surface. has mass and speed , has mass and speed and has mass and speed (see diagram). i

A collision occurs between and , after which and move in opposite directions, each with speed . Calculate a the value of , b the change in the momentum of .

ii

When collides with the two particles coalesce. Find their subsequent common speed. © OCR, AS GCE Mathematics, Paper 4728, January 2010

6

Particles and , of masses and respectively, are moving in the same direction along the same straight line on a smooth horizontal surface. is moving with speed and is moving with speed immediately before they collide. In the collision, the speed of is reduced by and its direction of motion is unchanged. i

Calculate the speed of immediately after the collision.

ii

Find the distance

at the instant seconds after the collision. © OCR, AS GCE Mathematics, Paper 4728, January 2012

7

Two particles and are moving in opposite directions in the same straight line on a smooth horizontal surface when they collide. has mass and speed . has mass and speed

. Immediately after the collision, the speed of is

.

i

Given that and are moving in the same direction after the collision, find the speed of .

ii

Given instead that and are moving in opposite directions after the collision, find the distance between them

after the collision. © OCR, AS GCE Mathematics, Paper 4728, June 2010

6 m s–1

8

mkg

0.5kg Q

P Fig. 1

i

A particle of mass is projected with speed on a smooth horizontal surface towards a stationary particle of mass (see Fig.1). After the particles collide, has speed

in its original direction of motion, and has speed

that 4 m s–1

more than . Show

. 2 m s–1

mkg

0.5kg P

Q Fig. 2 ii

and are now projected towards each other with speeds and respectively (see Fig.2). Immediately after the collision the speed of is direction of motion unchanged and has speed relationship between and in the form

, with its

more than . Find another , where and are constants.

iii By solving these two simultaneous equations show that

, and hence find

© OCR, AS GCE Mathematics, Paper 4728, June 2009 9

A railway wagon of mass and moving with speed which has mass and is moving towards with speed

collides with railway wagon . Immediately after the

collision the speeds of and are equal. i

Given that the two wagons are moving in the same direction after the collision, find their common speed. State which wagon has changed its direction of motion.

ii

Given instead that and are moving with equal speeds in opposite directions after the collision, calculate a the speed of the wagons, b the change in the momentum of as a result of the collision. © OCR, AS GCE Mathematics, Paper 4728/01, June 2008

10 Two uniform spheres, and , have the same radius. The mass of is and the mass of . The spheres and are travelling in the same direction in a straight line on a smooth horizontal surface, with speed , and with speed , where . collides directly with and the impulse between them has magnitude . Immediately after the collision, the speed of is i

.

Calculate subsequently collides directly with a stationary sphere of mass

as and . The coefficient of restitution between and is

.

and the same radius

ii

Determine whether there will be a further collision between and . © OCR, GCE Mathematics, Paper 4729, June 2009

11 A small sphere of mass

is dropped from rest at a height of

above horizontal ground.

It falls vertically, hits the ground and rebounds vertically upwards, coming to instantaneous rest at a height of above the ground. i

Calculate the magnitude of the impulse which the ground exerts on the sphere.

ii

Calculate the coefficient of restitution between the sphere and the ground. © OCR, AS GCE Mathematics, Paper 4729, January 2010

2 m s–1

12

0.18kg

3 m s–1 m kg

Two particles of masses

and

move on a smooth horizontal plane. They are moving

towards each other in the same straight line when they collide. Immediately before the impact the speeds of the particles are and respectively (see diagram). i

Given that the particles are brought to rest by the impact, find .

ii

Given instead that the particles move with equal speeds of after the impact, find a the value of , assuming that the particles move in opposite directions after the impact, b the two possible values of , assuming that the particles coalesce. © OCR, AS GCE Mathematics, Paper 4728/01, June 2007

13 Two spheres of the same radius with masses and are moving directly towards each other on a smooth horizontal plane with speeds and respectively. The spheres collide and the kinetic energy lost is . Calculate the speed and direction of motion of each sphere after the collision. © OCR, GCE Mathematics, Paper 4729, January 2010 14 A particle of mass

is moving with speed on a smooth horizontal surface when it collides

with a stationary particle of mass . After the collision the speed of is , the speed of is and the particles move in the same direction. i

Find in terms of .

ii

Show that the coefficient of restitution between and is .

subsequently hits a vertical wall which is perpendicular to the direction of motion. As a result of the impact, loses of its kinetic energy. iii Show that the speed of after hitting the wall is iv

.

then hits . Calculate the speeds of and , in terms of , after this collision and state their directions of motion. © OCR, GCE Mathematics, Paper 4729, June 2010



4 Circular motion 1

In this chapter you will learn how to: model motion of a particle moving in a horizontal circular path at a constant speed link linear speed and angular speed of a particular moving in a horizontal circular path find the acceleration and forces acting on a particle moving in a horizontal circular path solve problems relating to motion in a horizontal circular path.

Before you start… GCSE

You should know how to calculate arc length for a given proportion of a circle.

1 A circle has radius Calculate the length of the arc of a sector if the angle subtended at the centre is

GCSE

You should know how to write column vectors and what they mean.

2 Find the vector that translates

You should be able to work with trigonometric ratios.

3 A point lies on a circle of radius at the origin. Find the angle made with the positive -axis of the straight line that goes through the origin and the point

A Level Mathematics Student Book 1

You should be able to work with simple rates of change related to speed, distance and time using calculus.

4 An object is accelerating with a constant acceleration of . Find an expression for its velocity and displacement given that the initial velocity is and the initial displacement is

A Level Mathematics Student Book 2

You should be able to work in radians as an angular measure.

5 How many radians are the same as

A Level Mathematics Student Book 1 GCSE A Level Mathematics Student Book 1

What is different about motion in a circle? In your work on kinematics so far you have considered the velocity of a particle as a vector quantity with a magnitude (speed) and direction. When a particle moves in a circular path the direction of the velocity is constantly changing. You can consider a new way of measuring how the particle is moving over time. To do this you look at how the angle changes with respect to time.

Fast forward Students studying only to AS Level should also see Chapter 9, Section 1, about motion in a vertical circle.

Section 1: Linear speed vs angular speed WORKED EXAMPLE 4.1

A particle is moving at a constant rate anticlockwise along a circle centre of radius particle takes to make one revolution of the circle.

. The

a What angle (in radians) does the particle move through in b What is the arc length traced out by the particle every second? a

P

It is a good idea to draw a diagram first.

5cm O

Write down the angle in a full turn and divide to find the angle turned through in

b

Start with the formula for arc length and then substitute the angle value found in part a

Rewind Kinematics is covered in Pure Core Student Book 1.

Rewind Arc length is covered in A Level Mathematics Student Book 2. Let be the centre of a circle of radius and let be a fixed point on the circumference. At time the particle is at an angle of measured in radians anticlockwise from the radius If the particle is travelling at a constant angular speed around the circle then the rate of change of the angle with respect to time is a constant and is denoted by , i.e.

, where is measured in radians per second.

s P

A θ

v

r

O ω

As it is equal to the arc length, the distance the particle has travelled along the circumference is given by . Given that linear speed (or tangential speed) is a change in distance with respect to change in time you can relate this to the angular speed by:

You can remove from the differentiation since it is a constant.

Key point 4.1 For a particle moving in a circular path of radius , centre and with constant angular speed :

Tip Sometimes is denoted by , which is

. This gives the formula for linear speed as

.

WORKED EXAMPLE 4.2

Two marbles are moving in two clockwise circles both centred at the origin . One circle has radius and the other has radius Calculate:

Both marbles have a constant angular speed of

a the linear speed for each marble b the time taken for each marble to complete one full circle. a

It is a good idea to draw a diagram first. 2cm O

5cm

You know that

b

You know that angular speed is the amount of turn with respect to time. Both particles have the same angular speed and, consequently, will take the same time to complete one full circle.

Consider the angle in a full turn.

WORKED EXAMPLE 4.3

A particle moves in a circular orbit of radius minute.

at a constant frequency of

a How many revolutions does the particle complete in b What is the angular speed of the particle in radians per second? c What is the linear speed of the particle?

revolutions per

a

You know that through a smaller angle in

b

Angular speed is the angle turned through in so consider what proportion of a full turn has taken place in from part a.

c

You know that

. The particle will travel than in

Tip It is important to make sure that you are using the right units for angular speed, linear speed, distance and time.

WORK IT OUT 4.1 Two particles and are moving in two clockwise circles, both centred at , of radius and , respectively. Particle moves at a linear speed of and particle moves at an angular speed of . Determine which particle has the greater angular speed. Which is the correct solution? Can you identify the errors made in the incorrect solutions? Solution 1 Angular speed: Particle A angular speed: Therefore, particle has a greater angular speed. Solution 2 Angular speed Particle angular speed: Therefore, particle has a greater angular speed. Solution 3 Angular speed: Particle angular speed: Therefore, particle has a greater angular speed.

EXERCISE 4A 1

A particle is travelling around a circular path with angular speed and linear speed . The radius of the circular orbit is . a

i If

and

find

ii If b

and

i If

and

ii If c

find find

and

find

i If the radius of the circular path is

and linear speed

ii If the radius of the circular path is d

2

and linear speed

per min, find per hour, find

i If the particle makes revolutions every second of a circular path of radius

find

ii If the particle makes

find

revolutions every minute of a circular path of radius

a A particle takes to move around in a circle at a constant linear speed of the angular speed in i

. Find:

ii the radius of the circle in . b A marble completes revolutions every i the angular speed in

with a linear speed of

. Find:

ii the radius of the circle in . c A ball takes to move round a circle at a constant linear speed of i the angular speed in

. Find:

ii the radius of the circle in . 3

A particle is travelling around a circular path of radius Calculate the angular speed.

4

A cyclist rides clockwise around a circular track of radius a Find the angular speed in

at a constant linear speed of at a linear speed of

.

.

.

b How long does it take for one circuit? 5

A spinning disc of radius

completes one revolution every

a What is the angular speed of the spinning disc in b What is the linear speed at the edge of the spinning disc? c What radius would the spinning disc need to have a linear speed of 6

Two gear wheels and , one of radius and one of radius speed of wheel is , what is the angular speed of

7

An athlete runs at

at its edge?

, are connected. If the angular

along the inside path of the track shown in the diagram.

64m 100m

a How long does it take to complete one circuit? b What is the angular speed of the athlete as he runs around the circular parts of the track? 8

Metis is a moon of Jupiter. It completes one orbit approximately every distance from Jupiter of approximately .

. The orbit has an average

a What assumptions do you need to make to be able to calculate the angular speed of Metis in its orbit? b Making these modelling assumptions, determine: i the angular speed of Metis in ii the linear speed of the moon in

.

Section 2: Acceleration in horizontal circular motion When a particle moves in a circular path there must be a resultant force that keeps the particle moving in a circle, otherwise the particle would stop turning and continue to move in a straight line.

Rewind Recall Newton’s laws of motion, which you used in A Level Mathematics Student Book 1. If there is a force acting on the particle to keep it in a circular orbit, by Newton’s second law, there must be an acceleration. The diagram shows a particle moving in a horizontal circle in the

plane.

The direction of the velocity is along the tangent to the circle. The direction of the acceleration is towards the centre of the circle. The formula for acceleration is given by where is the linear speed and the angular speed. (The proof of this result is beyond the scope of this section.)

y

v a

O

Since

θ

P r A

x

, you can write

and, since

, you can write

relationship between and any two of the three variables

. This means that you have a

and

Applying Newton’s second law, and resolving radially to a particle of mass moving in a horizontal circular path, the force that gives rise to the circular motion is proportional to the acceleration, and is in the same direction. This force is often referred to as the centripetal force.

Tip Sometimes acceleration is written as

where is

.

Focus on … In Focus on … Proof 1, you will investigate the connection between the equations for linear motion with constant acceleration in a straight line and motion in a circle involving angular equivalents of and

Key point 4.2 For a particle moving in a circular path of radius , centre and with constant angular speed , the acceleration is given by

or

, towards the centre of the circle.

This will appear in your formula book. WORKED EXAMPLE 4.4

WORKED EXAMPLE 4.4

A ball of mass is attached to a light inextensible string of length . One end of the string is fixed on a smooth horizontal table and the ball moves in a circular path with linear speed a What is the acceleration of the ball? b What is the tension in the string?

a

a

It is a good idea to draw a diagram first, labelling the direction of the acceleration.

T 150cm

Convert

to .

You know that

.

towards the centre of the circle b

The resultant force directed towards the centre of the circular motion comes from the tension in the string and using Newton’s second law.

Rewind When modelling with a rope, you can make the modelling assumptions that it is light and inextensible in order to produce a simple mathematical model of the situation, as you did in A Level Mathematics Student Book 1.

WORKED EXAMPLE 4.5

A toy car of mass

moves at an angular speed of

around a circular path of radius

.

What is the centripetal force required to keep the toy car travelling in this circular path?

a

It is a good idea to draw a diagram first.

F 20m

EXERCISE 4B 1

A particle of mass is travelling around a circular path. is angular speed, is linear speed, is the radius of the circular orbit and is the magnitude of the acceleration. a

i If

and

find

ii If b

i If

and and

ii If c

d

find

and

find

i If

and the force keeping the particle in a circular motion is

ii If

and the force keeping the particle in a circular motion is

i If

and the force keeping the particle in a circular motion is

ii If 2

find

a A ball of mass

and the force keeping the particle in a circular motion is is attached to a light inextensible string of length

, find

. One end of the

string is fixed on a smooth table and the ball moves in a circular path with linear speed i Find the acceleration due to the circular motion. ii Find the magnitude of the resultant force acting on the ball. b A car of mass

is travelling in a circular path of diameter

with a linear speed of

i Find the acceleration due to the circular motion. ii Find the magnitude of the resultant force acting on the car. c A particle of mass is attached to a light inextensible string of length that is fixed at the other end so that the particle is moving in a circular path on a smooth table at a constant rate of revolution every i Find the acceleration due to the circular motion. ii Find the magnitude of the resultant force acting on the particle. 3

A particle of mass is travelling around a circular path of radius and makes full circles every minute. Calculate the force keeping the particle in circular motion, to significant figures.

4

A car of mass travels along a horizontal road that is an arc of a circle with radius The maximum speed at which the car can travel on this circular bend without slipping outwards from the centre is . Calculate the acceleration and the maximal friction

.

force on the car towards the centre of the arc of the circle. 5

Calculate the tension required in a light inextensible string of length mass moving in a horizontal circular path if:

to keep a particle of

a the linear speed of the particle is b the angular speed is 6

.

Emily sits on a roundabout, rotating at a constant angular speed , halfway between the centre and the edge. If Emily moves to a position that is of the radius from the edge, what effect does this have on the force that Emily experiences acting towards the centre of the roundabout? Provide calculations to support your argument.

7

A car travels at a constant speed along a bend in the road that is formed by an arc of a circle of radius . a What is the greatest linear speed in at which the car can travel around the bend without moving off at a tangent, if the coefficient of friction between the tyres of the car and the surface of the road is estimated as ? b Comment on whether this seems a sensible estimate for the value of the coefficient of friction.

Rewind

You will have used the coefficient of friction in A Level Mathematics Student Book 2. 8

A marble is on a rough horizontal disc at a distance of

from its centre. When the disc is

rotating at a constant speed of the particle is on the point of moving tangentially outwards from the centre. Calculate the coefficient of friction between the marble and the disc. 9

Two particles and of mass and respectively are attached to opposite ends of a light inextensible string of length . Particle rests on a rough horizontal spinning table which has coefficient of friction and the string passes through a smooth hole in the centre of the table. Particle hangs freely below the table. Particle is moving in a circular path with constant angular speed about the centre of the table. Find the linear speed of particle , if the system is in equilibrium with particle on the point of tangential displacement out from the centre of the spinning table, and particle hanging below the table. A

B

Tip Remember that it is the resultant force acting horizontally towards the centre of the circular motion that you use for the force in 10 Road surface conditions are being assessed for a horizontal bend in a road that is formed by an arc of a circle of radius . The road surface could be made of asphalt or concrete. The coefficient of friction between car tyres and asphalt is , and between car tyres and concrete a What assumptions need to be made? b Calculate the maximum safe linear speed, in bend without slipping for: i a surface made from asphalt

, at which a car could travel around the

ii a surface made from concrete. c If the surface is wet, the coefficient of friction between the car tyres and asphalt is reduced to and that between car tyres and concrete is reduced to safe limits now?

. What are the maximum

d Given that concrete is more expensive than asphalt, which road surface would you use for the bend in the road? 11 Let the position of a particle in the diagram have coordinates constant angular speed of so that the angle measured from the line time is

Show that:

Let be the anticlockwise at

y P

O

r θ

A

x

a the position vector of is given by b the linear (tangential) velocity of is c the acceleration of is given by d the acceleration is along a radius, directed towards the centre , and of magnitude e Use the scalar product to show that the velocity vector is perpendicular to the radius.

Rewind Scalar product is covered in A Level Mathematics Student Book 1.

Explore In A Level Mathematics, you may have seen various ways to convince yourself of how the trigonometric functions are differentiated. Using the following: www.cambridge.org/links/moscmec6001, you can start with your knowledge of motion in a circle to help you to understand the derivatives of sine and cosine.

Section 3: Solving problems involving motion in a horizontal circle The examples you have seen so far in this chapter have all had a single horizontal force keeping a particle in a circular path. You are now going to consider situations where the forces are not necessarily acting horizontally but the particle is still moving in a horizontal circular path with a constant angular speed. This will require you to consider the components of a force. One example of this type of motion is the conical pendulum, shown in the diagram. The particle is attached to a light inextensible string that makes an angle of with the downward vertical, and it is moving in a horizontal circle at a constant angular speed

θ T

r

ω

O

mg

Focus on … You will learn about a simple pendulum in Focus on … Modelling 2.

WORKED EXAMPLE 4.6

A particle, , of mass is attached to the lower end of a light inextensible string with the upper end fixed at . When the particle moves in a horizontal circular path, the string traces out the curved surface of a cone and makes an angle of with the downward vertical. The centre of the circular path lies directly below the point at a distance of . a Find the tension in the string. b Find the angular speed of the particle.

B 2m

It is a good idea to draw a diagram first.

60°

T

Resolve forces vertically to form an equation involving

A 0.2g

Resolve forces horizontally and use Newton’s second law to form an equation involving the centripetal acceleration, . Use the value of acceleration with the equation for circular motion to calculate the angular speed.

a

Resolve the forces vertically. The particle is not moving in the vertical direction.

b

Resolve the forces horizontally in the direction of the acceleration.

Find the radius of the circular motion using trigonometry. You can then substitute the values for and into the equation to find

Rewind Resolving forces is covered in A Level Mathematics Student Book 2, Chapter 21.

WORKED EXAMPLE 4.7

A particle, , of mass is attached to two light rods and , as shown in the diagram, where and all lie in the same vertical plane. The lengths of and are and

metres respectively. A

0.25m B 0.07m θ C

φ P

0.24m

The particle is moving in a horizontal circle with linear speed a Calculate the tension in the rod

.

b Calculate the magnitude of the force in the rod compression.

and determine if the rod is in tension or

It is a good idea to draw a diagram first showing the forces acting on .

a

Resolve the forces at vertically. Resolve the forces horizontally in the direction of the acceleration. Use

B 0.07 C

0.25 θ 0.24

P

A 0.4

0.32 C

It is a good idea to find the trigonometric ratios for the angles in the question.

φ 0.24

P

When the values are substituted you have a pair of simultaneous equations to solve. You can find the tension b

by eliminating

You can substitute the value you found for the equations to find Magnitude of the force in rod

. into one of

is

The force is directed towards and rod is therefore in compression.

You initially assumed that was acting towards . As the value of is negative it means that the direction must have been incorrect and the rod is in compression.

Rewind Recall from A Level Mathematics Student Book 1 that a rod can be under tension or compression.

WORKED EXAMPLE 4.8

The fixed points

and are in a vertical line with above

and above . A particle of

mass is joined to , and to a particle of mass , by three light rods where the length of rod is and the length of rod is . Particle moves in a horizontal circle with centre . Particle moves in a horizontal circle with centre , at the same constant angular speed as , in such a way that and are coplanar. The rod makes an angle of with the downward vertical, rod diagram).

makes an angle of

with the downward vertical and rod

is horizontal (see

a Calculate the angular speed . b Find the tension in the rod c Find the force in the rod

A

30°

B

2m

Q 45°

C

1m P

and determine if the rod is in tension or compression.

a

It is a good idea to draw a diagram first showing the forces acting on .

Q 45°

T1 P 2g

Resolve the forces horizontally in the direction of the acceleration to get an equation that involves the tension in the rod You can find the tension in the rod forces vertically at .

by resolving the

You need to find the radius of the circular motion for . This will involve finding the horizontal distance , which is made up of two parts. You can then substitute the expression for and the radius of the circular motion into a rearranged equation to find

b A

It is a good idea to draw a diagram first showing the forces acting on

30°

T2

T3

Q 45° T1

3g Resolving the forces at vertically allows you to find the tension in the rod c

Resolve the forces horizontally in the direction of the acceleration and use the values you found for and to find , the tension in

The rod is in tension.

Since the value of the tension is positive, the rod is in tension.

Banked tracks Another example of horizontal circular motion is motion on a banked track. The two figures show an inward banked track, also known as camber.

RN

RN cosθ θ

C

C

r

RN sin θ

r θ

θ

mg

You can consider how cars move around bends that are banked at an angle to the horizontal, as shown in Worked examples 4.9 and 4.10.

Did you know? The camber of a road or a cant of a railway track allows the vehicles to turn safely through a curve at higher speeds compared with a level surface.

WORKED EXAMPLE 4.9

A remote-control toy car travels around a bend of radius angle of to the horizontal. If the car is travelling at car and the track, find the angle . R

on a track which is banked at an and there is no friction between the

It is a good idea to draw a diagram first. Remember the reaction force is always perpendicular to the road or track. mg

θ 14m a

Resolve the forces horizontally to form an equation involving and Resolve the forces vertically to form an equation involving , and . Hence

Divide the first equation by the second to eliminate the unknowns and

Use

to write an expression for . Make sure that all units

are consistent. Substitute and solve for

WORKED EXAMPLE 4.10

A racing car travels at a constant speed around a bend in a road of radius banked at an angle to the horizontal with sin tyres of the car and the road is

. The road is

. If the coefficient of friction between the

find:

a the maximum linear speed at which the car can be driven around the bend b the minimum linear speed at which the car can be driven around the bend. It is a good idea to draw a diagram first. Remember the reaction

R

a

F

force

mg

is always perpendicular to the road.

In this case, the greatest speed is the speed that is possible before slipping outwards from the centre, so friction acts towards the centre.

θ 55m

The maximum frictional force is proportional to the normal reaction force. Resolve the forces vertically to form an equation involving , and

You are told that

5

and

4

. Recall how to find exact values for

using right-angled triangles.

θ 3

Resolve the forces horizontally to form an equation involving and

Using R

b

F

.

In this case, the least speed is the speed that is possible before slipping inwards towards the centre, so friction acts away from the centre.

mg θ 55m

The maximum frictional force is proportional to the normal reaction force. Resolve the forces vertically to form an equation involving and .

Resolve the forces horizontally to form an equation involving and .

Tip If a particle is moving in a circular orbit on a rough inclined plane, then the direction of friction depends on the speed at which the particle is moving. A slowly moving particle will be on the point of slipping down the inclined plane and so friction will act up the inclined plane.

EXERCISE 4C Unless otherwise instructed, when a numerical value for the acceleration due to gravity is needed, use 1

. A particle of mass

is moving in a horizontal circle at a constant angular speed

attached to a light inextensible string of length downward vertical. a

i If

and

ii If b

, find the tension in the string.

and

i If

d

, find the tension in the string.

and

i If

and

ii If

and

i If

and

ii If 2

, find the tension in the string.

and

ii If c

that makes an angle of with the

and

, find the tension in the string. find find find the angle find the angle

Here are incomplete force diagrams for some simple conical pendulums. Each consists of a light, inextensible string of length suspended from a point and a bob of mass at point . The bob moves in a horizontal circle centred vertically below and the string forms an angle with the downward vertical. moves with a constant angular speed and the tension in the string

is

a Find the angle . A

θ

25m T

B

5g ω = 4 rad s−1

b Find the angular speed .

A

30° 1m T

B

0.2g ω unknown

c Find the tension in the string. A 13m 12m

T

B

10g ω unknown

d Find the mass of the bob.

A

25° 1.2m T = 56 N

B

mg ω = 16 rad s−1 3

A particle is moving in a horizontal circle at a constant angular speed of , attached to a light inextensible string of length . Find the angle the string makes with the downward vertical to significant figures.

Explore Watch the video at www.cambridge.org/links/moscmec6002 and use the ideas introduced and explored in this chapter to create a mathematical model of this situation. 4

A particle travels around a bend of radius horizontal, where sin

. Show that, if the surface is smooth, the particle will not slip if it

travels with a linear speed of 5

A smooth bead of mass

, with the surface banked at an angle to the

. is threaded onto a light inextensible string. The two ends of the

string are attached to fixed points and , where is vertically below . The string is taut and the bead rotates about the axis horizontal circular path of radius calculate:

. The bead moves with a constant angular speed in a . Given that angle

is

and angle

is

,

a the tension in the string b the angular speed.

Y

30°

0.8m

B

60° X 6

A bend in a road is in the form of a horizontal circular arc of radius banked outward at an angle , where force of

, with the road surface

to the horizontal. Show that, if there is a frictional

acting up the slope, a car of mass

is moving at a linear speed of

to significant figures. 10000N

a m s−2 θ

mg

7

In the diagram the fixed points and are in a vertical line, with above at a distance of . particle of mass is attached to two light inextensible strings, so that is and is . The particle rotates at a constant angular speed of with both strings

and

taut.

a Find the tension in

in terms of

and

b Find the tension in

in terms of

and

c Calculate a lower bound for the angular speed given that A

θ

remains taut.

1.44m

1.80m P 1.08m B 8

Fixed points mass

and are in a vertical line with above and above

is joined to , to and to a particle of mass

particle of

, by three light rods where the

length of rod is and the length of rod is . Particle moves in a horizontal circle with centre Particle moves in a horizontal circle with centre at the same constant

angular speed as , in such a way that make an angle of

and are coplanar. Rods

with the downward vertical and rod

and

both

is horizontal (see diagram).

a Calculate the angular speed . b Find the tension in the rod c Find the magnitude of the force in the rod compression. A

and state whether the rod is in tension or

30° 2 m B C 9

Q 1m 30°

P

A hemispherical bowl of radius is fixed with its rim horizontal. bead of mass

is

moving in a horizontal circle around the smooth inside surface of the bowl. The centre of the circle is

below the centre of the sphere of which the bowl forms a part.

a Find the magnitude of the reaction force between the bowl and the bead in terms of and b Find the linear speed of the bead in terms of and 10 The fixed points and

under , lie in the same vertical plane. A particle of mass

joined to the point by a light rod of length

is

and to the point by a light rod that is

horizontal. The particle moves in a horizontal circle with centre at a constant angular speed . a If the angle the rod the rod .

makes with the downward vertical is

b Calculate a lower bound for which would mean that the rod c If the angular speed of is the rod

, calculate the tension in is in tension.

, calculate the magnitude and direction of the force in

.

11 A conical pendulum consists of a light inextensible string of length , with the string fixed at and a small ball of mass at The ball moves in a horizontal circle, with centre vertically below the point

with constant linear speed

. Find the tension in the string

and the radius of the circle. 12 A car moves in a horizontal circular path of radius

, banked at

to the horizontal. The

coefficient of friction between the car tyres and the track is Find the maximum and minimum speeds at which the car can be driven around the circular path. 13 A car has a linear speed of

and is on the verge of slipping when driven in a horizontal

circular path of radius that is banked to the horizontal at angle Show that the coefficient of friction is

Checklist of learning and understanding For a particle moving in a horizontal circular path of radius : linear speed is given by

and at a constant angular speed

acceleration is given by motion.

,

or

and is directed towards the centre of the circular

Mixed practice 4 Unless otherwise instructed, when a numerical value for the acceleration due to gravity is needed, use 1

A particle of mass

moves in a circle of radius

with constant speed

a Find the angular speed of the particle. b Find time it takes for one revolution. c Find the force acting on the particle to keep it in a circular path. 2

A particle of mass moves with a constant linear speed of in a horizontal circle of radius on a rough table with coefficient of friction The particle is travelling at the maximum speed that will keep it in a circular path. a Draw a diagram to display the forces acting on the particle and its acceleration. b If the radius of the horizontal circle is

3

, find the linear speed.

A particle of mass is attached to a light inextensible string of length . The string is fixed on a smooth horizontal table and the ball moves in a circular path with constant angular speed a What is the acceleration of the ball? b What is the tension in the string?

4

A car of mass moves at a linear speed of around a circular path of radius . What is the frictional force required to keep the car travelling in this circular path?

5

A light rod , of length , has one particle of mass attached at and a second particle of mass attached at The rod is held fixed at a point and is free to rotate in a horizontal circle with a constant angular speed about the point Given that the tensions in parts and

6

of the rod are equal, show that the length

is

Two particles, and , are connected by a light inextensible string that passes through a smooth hole in a smooth horizontal table. Particle , of mass , moves on the table with constant angular speed in a circle of radius around the hole. Particle , of mass hangs vertically in equilibrium under the table, as shown in the diagram. Find the angular speed of .

,

P

Q

7

A small smooth ring of mass is threaded onto a light inextensible string of length . The two ends of the string are fixed at the points and , where is vertically below at a distance of . The ring is moving with constant linear speed in a horizontal circle with centre and radius . a Find the tension in the string in terms of and

b Find the linear speed in terms of and . c How did you use the assumption that the ring was smooth? 8

A car of mass

moves around a bend that is banked at a constant angle of

to the

horizontal. The car is modelled as a particle moving in a horizontal circle of radius constant angular speed. Calculate the linear speed of the car in

at a

if:

a there is no sideways frictional force on the car b the coefficient of friction between the tyres and the surface is

and the car is on the

point of i slipping down the slope ii slipping up the slope.

5m

O

9

3m

ω

P 0.2kg

4m

0.1kg Fig. 1 A particle of mass is moving on the smooth inner surface of a fixed hollow hemisphere that has centre and radius . moves with constant angular speed in a horizontal circle at a vertical distance of below the level of (see Fig. 1). i Calculate the magnitude of the force exerted by the hemisphere on . ii Calculate . A light inextensible string is now attached to . The string passes through a small smooth hole at the lowest point of the hemisphere and a particle of mass hangs in equilibrium at the end of the string. moves in the same horizontal circle as before (see Fig. 2).

O

5m

3m P

0.1kg Fig. 2 iii Calculate the new angular speed of . © OCR, GCE Mathematics, Paper 4729, January 2010 10 A particle of mass is attached to points and on a fixed vertical axis by two light inextensible strings of equal length. Both strings are taut and each is inclined at to the vertical (see diagram). The particle moves with constant speed in a horizontal circle of

radius i Calculate the tensions in the two strings. The particle now moves with constant angular speed of becoming slack.

and the string

is on the point

ii Calculate . A 60° P

0.4m

0.5kg

60° B

© OCR, GCE Mathematics, Paper 4729/01, June 2008 11 One end of a light inextensible string of length is attached to the vertex of a smooth cone of semi-vertical angle . The cone is fixed to the ground with its axis vertical. The other end of the string is attached to a particle of mass which rotates in a horizontal circle in contact with the outer surface of the cone. The angular speed of the particle is (see diagram). The tension in the string is and the contact force between the cone and the particle is i By resolving horizontally and vertically, find two equations involving and and hence show that

.

, calculate the greatest value of for which the particle ii When the string has length remains in contact with the cone.

45° l ω

m

© OCR, GCE Mathematics, Paper 4729, June 2010 12 A particle , of mass , is attached to fixed points and by light inextensible strings, each of length . and are apart with vertically above . The particle moves in a horizontal circle with centre at the midpoint of i Find the tension in each string when the angular speed of is

.

ii Find the least possible speed of . © OCR, GCE Mathematics, Paper 4729, June 2012 13 A circular cone is fixed so that the apex of the cone is sitting on a horizontal surface and the axis of the cone is perpendicular to the horizontal surface. The angle the cone makes with the horizontal surface is . A particle , of mass , moves on the inner surface of the cone. The particle is joined to by a light inextensible string , of length . The particle moves in a horizontal circle with constant linear speed and the string is taut. The inside of the cone is smooth.

a Show that the reaction force between the particle and the inner surface of the cone can be written in the form b Find the tension in the string, in terms of

and

.

c Show that the motion of the particle is only possible when

.

14 A vertical hollow cylinder of radius is rotating about its axis. A particle is in contact with the rough inner surface of the cylinder. The cylinder and rotate with the same constant angular speed. The coefficient of friction between and the cylinder is . i Given that the angular speed of the cylinder is rad downwards, find the value of .

and is on the point of moving

A 0.5m

P 0.4m

The particle is now attached to one end of a light inextensible string of length

. The other

end is fixed to a point on the axis of the cylinder (see diagram). ii Find the angular speed for which the contact force between and the cylinder becomes zero. © OCR, GCE Mathematics, Paper 4729/01, June 2013

Fast forward Students studying only to AS Level should also see Chapter 9, Section 1, about motion in a vertical circle.

5 Centres of mass 1 This chapter is for A Level students only. In this chapter you will learn how to: find the centre of mass of arrangements of particles, uniform rods and symmetrical uniform laminas find centres of mass of two- and three-dimensional objects of standard shape find centres of mass of composite bodies, including bent wires.

Before you start… GCSE

You should be able to add and subtract vectors and to multiply a vector by a scalar.

1

Where does mass act? The particle model you have used in earlier chapters assumes that mass is all located at a single point with no volume. This approximation works well for small objects. More complex objects may consist of two or more particles located in different places, or a combination of one-, two- or three-dimensional objects. For example, mass may be spread along a rod, throughout a two-dimensional shape such as a circular disc, or throughout a solid object such as a cube. For many purposes, such as when modelling linear motion, a complex shape or rigid body can be modelled as though its mass is located at a single point, called the centre of mass. In this chapter, you will learn how to find the centre of mass of a range of different objects.

Fast forward In Chapter 10 you will see that the location of the centre of mass of a complex object determines how it responds to forces that are applied to it, including its own weight. You will use your knowledge of the centre of mass, together with your knowledge of moments, to work out angles when objects are suspended in space. You will also work out whether objects placed on an inclined surface will rest in equilibrium, or topple over or slide.

Section 1: Centre of mass of a system of point masses Centre of mass of two particles The centre of mass of two identical particles lies at the midpoint of a straight line drawn between them. centre of mass is at x1 + x2 2 O

m

m

x1

x2

If the masses of the particles are different, then the centre of mass does not lie at the midpoint. It is closer to the larger mass. You find the position of the centre of mass, , by calculating a weighted average:

O

m1 x1

m2

centre of mass is at x =

m1 x1 + m2 x2 m1 + m2

x2

For example, if

,

,

and

, then

The centre of mass divides the straight line joining the particles in the ratio

.

Several particles arranged in a straight line O

m1 x1

x2

m2

mn

xn

You now extend the formula for two masses to masses:

If you write

instead of

, the formula may alternatively be written:

Key point 5.1 A combination of particles having masses arranged in a straight line, at positions , can be modelled as a single object of mass , with position where: and:

Rewind You worked with moments in A Level Mathematics Student Book 2. The formula given in Key point 5.1 equates the sum of the moments of mass of particles with the moment of mass of a combined particle acting at the centre of mass. WORKED EXAMPLE 5.1

WORKED EXAMPLE 5.1

Three point masses are attached to a light bar of length . These have mass , and and are attached at the bar at , the midpoint of the bar, and , respectively. Find the distance of the centre of mass from . 0.25kg 0.15kg A 0.25m 0.5m

0.35kg B

Draw a diagram with masses and lengths from in standard units. The modelling assumption that the bar is ‘light’ means that you do not need to include its mass. Use the formula for an arrangement of particles. is the total mass. The distance of the

In Worked example 5.1 the centre of mass is moment of mass as a single particle of mass

A

0.75kg 0.317m

mass from is zero.

from . The system of three particles has the same placed

from .

B

WORKED EXAMPLE 5.2

A light rod

of length

has three masses attached to it. A

mass is attached

from . A

mass is attached from , and an unknown mass, , is attached at end . Find the value of given that the centre of mass of the system is from point . Start by calculating the total mass. Use the formula for point masses arranged in a straight line.

Particles arranged in a plane If particles are arranged in a plane, you can find the position of the centre of mass separately for and . Vectors give you a nice way of combining these calculations.

Key point 5.2 If you have particles with masses can model these as a single mass

, at position vectors , with position vector

, you

.

WORKED EXAMPLE 5.3

Three particles are arranged in a plane. Particle has mass and is placed at has mass and is placed at . Particle has mass and is placed at . Find the - and -coordinates of the centre of mass.

. Particle

Start by calculating the total mass. Use the formula for point masses in a plane. Multiply the position vector of each point mass by the mass placed there:

Divide the result by the total mass.

EXERCISE 5A 1

A light rod of length has a mass of placed Find the distance of the centre of mass from .

2

A light rod of length

has a mass of

from and a mass of

placed at one end . A mass

placed at .

is placed

from

the other end , and the centre of mass lies in the middle of the rod. Find . 3

A light rod has masses and mass. Find the length of the rod.

4

Masses , and are placed , and centre of mass lies from . Find the value of .

5

Three point masses have position vectors in the

placed at each end, and the centre of mass lies

from the

from one end, , of a light rod. The

plane as shown in the diagram. Find the centre

of mass of the three masses combined.

y

( 66) ( 23)

( 62)

2kg

1.5kg

1.2kg x

O 6

Four point masses have position vectors in the mass of the four masses combined.

y

( 16)

( 67) 2m kg

( 31) O 7

( 64) m kg

plane as shown in the diagram. Find the centre of

m kg

3m kg

x

Four point masses have position vectors in the mass of the four masses combined.

plane as shown in the diagram. Find the centre of

y m kg 2 7

( )

2m kg 7 6

( )

m kg 4 4

( )

m kg 2

( 41)

O 8

Three point masses,

x ,

and

respectively, from an origin at

are placed in a plane at

,

, and

,

. Find the position vector of the centre of mass relative to

. 9

Three point masses,

,

and

, respectively, from an origin at and .

, are placed in a plane at . The centre of mass is at

, at

, and at . Find the values of

Section 2: Centres of mass of standard shapes Centre of mass of uniform rod An inflexible body having its mass spread along a straight line is called a rod. Its shape is defined by its length; its cross-sectional area is zero. A uniform rod has constant mass per unit length ( in standard units).

Rewind You have learned to work with uniform rods in A Level Mathematics Student Book 2.

Key point 5.3 The centre of mass of a uniform rod lies at its midpoint.

Centre of mass of a uniform lamina of standard shape A lamina is a two-dimensional object. An important modelling assumption used in calculations is that the lamina has zero thickness. A uniform lamina has constant mass per unit area ( in standard units). A compact disc is close in shape to what is meant by a lamina, as its cross-sectional area is much greater than its thickness.

Key point 5.4 The centre of mass of a symmetrical uniform lamina lies on any axis of symmetry. If there is more than one axis of symmetry, then the centre of mass lies at the intersection of these. The diagrams show the locations of the centre of mass of a uniform rectangular lamina and a uniform circular lamina, in relation to their axes of symmetry; (A circular lamina has an infinite number of lines of symmetry; only two are shown here.)

centre of mass WORKED EXAMPLE 5.4

Calculate the coordinates of the centre of mass of the uniform rectangular lamina with vertices at , ( ), and . Draw a sketch.

(5, 9)

The centre of mass lies at the intersection of the lines of symmetry shown. (9, 7)

(2, 3)

(6, 1)

Find the average of the -coordinates and the average of the -

coordinates.

The centre of mass of a uniform triangular lamina lies at the intersections of the medians. A median of a triangle joins a vertex to the midpoint of the opposite side. In the case of a lamina in the shape of an equilateral triangle, the medians are all axes of symmetry.

centre of mass

In any triangle, all three medians intersect at the same place, even when they are not axes of symmetry. The intersection of the medians divides each median in the ratio (moving from side to apex). This intersection is the location of the centre of mass of a uniform triangular lamina.

2c b 2a

a c

2b

centre of mass

Key point 5.5 The medians of a triangle intersect at

, where

,

,

are the vertices of the triangle.

Focus on … You will prove the formula for the centre of mass of a uniform triangular lamina in Focus on … Proof 2.

WORKED EXAMPLE 5.5

Find the distance of the centre of mass of the uniform triangular lamina from C 6cm

A

6cm

B

It is an isosceles triangle with angles Drop a perpendicular from meeting

,

,

at .

.

C

6cm

A

(6 sin 45°)cm G

45°

B

X

Calculate the height of the triangle, The required distance is

.

The centre of mass, , divides the median .

.

in the ratio

WORKED EXAMPLE 5.6

Calculate the coordinates of the centre of mass vertices at

,

and

of the uniform triangular lamina having

. Use the coordinates of the vertices.

WORKED EXAMPLE 5.7

Calculate the coordinates of the centre of mass

of this uniform triangular lamina.

y

8cm

5cm

θ

x

6cm Vertices are at:

,

and

Use the coordinates of the vertices.

To find : Use the cosine rule to find . Use the value of you found. Give your values here to at least significant figures so that you can give the final coordinates to significant figures.

You can find the centre of mass of a lamina in the shape of a sector of a circle. This includes half and quarter discs. You are given the formula in the formula book.

Key point 5.6 The centre of mass of a sector of a circular disc, having radius and angle centre of the circle is

radians at the

from the centre of the sector, on the axis of symmetry.

This will appear in your formula book.

Rewind Radian measure is introduced in A Level Mathematics Student Book 2, Chapter 7.

Tip The angle that appears in the formula is , which is half the angle at the centre of the sector. Make sure you know that the angle at the centre of the sector is , so you remember to halve the angle at the centre of the sector before using the formula.

WORKED EXAMPLE 5.8

Find the centre of mass of a uniform quarter disc of radius the centre of the quadrant and is the centre of mass.

. Find the distance

, where is at

O G 5cm

The angle at the centre of the sector is but the formula requires that the angle at the centre is

.

Use the standard formula: Work in radians.

Centre of mass of a uniform wire A wire is a one-dimensional but flexible solid object. The centre of mass of a uniform straight wire lies at its midpoint. A wire can be bent into several straight sections. You can combine the sections as though there are point masses at the centre of each section. Wires and rods can be combined together into a framework.

Key point 5.7 The centre of mass of a uniform wire bent to form an arc of a circle, having radius and angle radians at the centre of the circle, is

from the centre of the sector.

This will appear in your formula book.

Tip As with a sector of a circle, is the angle at the centre, measured in radians, so you must halve this angle before making use of the formula.

WORKED EXAMPLE 5.9

A length of uniform wire is bent to form an arc of a circle of radius . Find the angle, , made by the arc at the centre of the circle, and the distance of the centre of mass from , the centre of the circle. Draw a diagram.

8cm

O

θ

G

arc 5cm

8cm Use the formula for arc length to find .

Work in radians and use the formula.

EXERCISE 5B 1

A uniform rectangular lamina has vertices at centre of mass of the lamina.

2

A uniform square lamina has three of its vertices at the centre of mass.

3

A uniform lamina is in the shape of an equilateral triangle of side . It is placed with one vertex at and one edge along the -axis. Determine the coordinates of the centre of mass.

4

Find the centre of mass of a uniform triangular lamina with vertices as follows: a

,

b

,

and and

. Find the coordinates of the . Determine the coordinates of

and

,

and

d

,

e

,

and ,

c

,

,

and and

5

A uniform semicircular lamina has radius centre of mass.

6

Find the coordinates of the centre of mass of the following uniform triangular laminas by first finding the coordinates of the vertices. a

y

4cm

6cm

x

. Find the distance from the centre of the circle to the

b y

5cm

5cm

x

6cm c y

4cm

3.5cm

1.5cm

x

d y

9cm

8cm

7

6cm x

A length of uniform wire is bent to form an arc of a circle. The radius of the corresponding circle is , and the arc makes an angle of  at its centre. Find the distance of the centre of mass from the centre of the circle.

8

A length of uniform wire is bent to form an arc of a circle. The arc is of length

, and makes an

angle at the centre of the corresponding circle of  . Find the distance of the centre of mass from the centre of the circle. 9

The centre of mass of a length of uniform wire bent to form an arc of a circle is

from the centre of

the circle, where is the radius of the circle. Find numerically the angle made by the arc at the centre of the circle, giving your answer in radians to decimal places.

Section 3: Centres of mass of composite bodies Key point 5.8 A composite body is one made from a combination of shapes. For a composite body:

Rewind You met this formula in Section 1, Key point 5.2. You have already used a similar formula to find an equivalent centre of mass of a system of particles. The same approach can be applied to calculating the centre of mass of a composite body made from any combination of the shapes you have worked with so far. This time you work from the centres of mass of the component parts

.

WORKED EXAMPLE 5.10

A uniform rod , of mass and length , has three masses attached to it. A mass is attached from and a mass is attached from . A mass is attached at point . The centre of mass of the system is from . Find the length . As the rod is uniform its centre of mass is at its midpoint.

For the uniform rod 0.5m A

0.5m C

3kg

2kg 4kg 5kg 4m

B

Measure distances from and use:

WORKED EXAMPLE 5.11

A composite body is made from a uniform rectangular lamina of mass with side lengths and placed with one vertex at and one of its longer sides along the -axis. Point masses, , and , are added at , and , respectively. Find the centre of mass of the composite body. For the uniform rectangular lamina

As the rectangular lamina is uniform its centre of mass is at its geometric centre. Find the total mass.

y

m kg

40cm

2m kg 2m kg 3m kg x

25cm

Use:

WORKED EXAMPLE 5.12

A composite body consists of a uniform rectangular lamina with dimensions by and a uniform circular lamina with diameter joined on. A diameter of the circular lamina coincides with edge of the rectangle as shown. The mass density per unit area is the same for the rectangular and circular laminas. Find the distance of the centre of mass of the composite lamina from and . D

C

20cm A

B

30cm

y D (0, 20)

C (30, 20) E (30, 10)

A (0, 0)

In this question you are not given the coordinates of the vertices. You can introduce the coordinate system with the origin at and the axes along and . marks the position of the centre of the circle and the horizontal dashed line shows the axis of symmetry.

x B (30, 0)

For the composite body,

The centre of mass lies on the axis of symmetry at

.

Use the composite body formula. As the composite lamina is uniform, mass is directly proportional to area so you can work with areas.

The centre of mass of the composite body is from and from .

Check that the -coordinate of the centre of mass is sensible.

When part of a larger shape has been cut out, you can use the usual formula for a composite lamina. This time

is the mass of the original lamina before removal of the part of mass

and

.

WORKED EXAMPLE 5.13

The rectangular uniform lamina has had a square cut out. Find the distance of the centre of mass of the composite lamina from and . D

C

Let

be the -axis and

Let be

4cm

be the -axis.

.

2cm

5cm

2cm A

B

7cm

The mass of the lamina and the square cut out are directly proportional to area, so you can work with areas. Check with the diagram to make sure that the centre of mass looks sensible.

A wire can be bent into several straight sections. You combine the sections as though there are point masses at the centre of each section. WORKED EXAMPLE 5.14

A uniform wire of length is bent to form three sides, , and of a rectangle as shown. Find the distance of the centre of mass from the straight line passing through . D

45cm

Let

C

be the -axis and

Let be

be the -axis.

.

30cm

A

45cm

B

The centres of mass of from .

and

are

As the sections of wire are uniform, the centres of mass are at the midpoints. As the wire is uniform, its mass is directly proportional to length.

The centre of mass is line passing through

from the straight .

The centre of mass of the wire does not lie on the wire itself.

WORKED EXAMPLE 5.15

A uniform wire is bent into a framework consisting of a semicircular arc of radius with the diameter joining . Find the distance of the centre of mass from . Let

together

be the -axis.

Let the -axis pass through the midpoint of

.

3cm

A

B

As the wire is uniform mass is proportional to length. Use

, where

Use the formula for the centre of mass of a uniform arc of wire.

and

The centre of mass of the arc is from Combine the straight edge and the arc.

The centre of mass of the framework is from .

Centres of mass of standard three-dimensional figures You need to know how to use formulae for centres of mass of a solid hemisphere, hemispherical shell, solid right cone or pyramid and conical shell. The formulae are all given in the formula book. Solid hemisphere, radius

from centre

Hemispherical shell, radius

from centre

Solid cone or pyramid of height

above the base on the line from centre of base to vertex

Conical shell of height

above the base on the line from centre of base to vertex

Did you know? You may know about beds that can be folded up by hand into a wall cupboard when not in use. The design of the folding mechanism means that little lifting force is needed. This is remarkable when you consider the significant weight of the bed, including the mattress and bedding. How is it possible to achieve this? The answer lies in ‘counter-weighting’, which enables the bed to be fairly well balanced in all positions. Counter-weighting relies on an understanding of how mass is spread over the object to be lifted and a means of counteracting its weight in all positions. To start with, you would need to know the location of the centre of mass of the object to be lifted.

Fast forward In Chapter 10 you will learn how to find the centre of mass of a solid hemisphere and a solid cone by integration.

WORKED EXAMPLE 5.16

A uniform conical shell of perpendicular height and radius is joined to a uniform disc of radius . The mass per unit area of the shell and the circular base are the same. Find the distance of the centre of mass of the composite shell from the base.

Let the centre of the circular base be

.

Let the main axis of the cone be the -axis.

24cm

10 cm

Curved surface area of cone As the shell and base are uniform and have the same mass per unit area, mass is proportional to surface area. The centre of mass of a conical shell is

above the base

on the line from centre of the base to vertex. Cancel and simplify:

Note: The centre of mass lies along the central line of symmetry for all of these standard solids. WORKED EXAMPLE 5.17

A solid hemisphere of radius is joined to a solid cone of radius and height . Both solids are uniform with the same mass per unit volume. The base of the cone coincides with the base of the hemisphere. Show that the centre of mass is

from the vertex of the cone. Draw a diagram. Let the vertex of the cone be

.

Let the main axis of the cone be the -axis.

r

3r

As the solids are uniform, mass is proportional to volume.

For the cone,

The centre of mass of a uniform solid cone is from the centre of its base.

For the hemisphere,

The centre of mass of a uniform hemisphere is from the centre.

WORK IT OUT 5.1

, as required.

Cancel

and simplify.

A brooch is modelled as a lamina in the shape of a sector of a circle, together with an arc of wire on the curved edge of the sector, as shown. The angle at the centre of the sector is radians, the mass of the sector is

and the mass of the arc of wire is

.

O r cm

π 3

200g

300g

Form an equation that could be used to find the distance of the centre of mass of the brooch, , from . Which solution is correct? Can you identify the errors made in the incorrect solutions? Solution 1

Solution 2

Solution 3

Did you know? You can see the Moon moving across the night sky as it revolves around the Earth, but the Earth and Moon together revolve around the Sun. In order to calculate the orbit of the Earth around the Sun, astronomers sometimes calculate the position of centre of mass not just of the Earth, but of the Earth and Moon taken together. The centre of mass of the Earth and Moon taken together actually lies within the Earth, about below the surface. This centre of mass moves as the Moon rotates. This moving centre of mass causes deviation in the path of the Earth from a smooth curve around the Sun. The Earth wobbles in its orbit.

EXERCISE 5C 1

Find the centres of mass of the uniform laminas in the following diagrams.

a

y

O

b

y

O

c

x

y

O

f

x

y

O

e

x

y

O

d

x

x

y

O

x

g

y

x

O

h

y

x

O

2

A composite body is made from a uniform rod of mass masses, , and attached to it. has mass and . has mass from .

3

and

is

and length with three point is . has mass and is

. Find the distance of the centre of mass of the composite body

A composite body is made from a rectangular lamina

of mass

with side lengths

and

together with point masses, each, added at and , as shown. Find the distance of the centre of mass of the composite body from and . D

C 8cm

20cm

A

4

12cm

Y

5cm

10cm B

30cm

When a point mass of is added to a uniform rod of mass and length metres, it moves the position of the centre of mass. Find where the additional mass must be added to move the centre of mass to

5

X

metres from .

A composite body is made from a rod

of length

and mass

that has a disc of mass

attached to end , with its centre placed at the end of the rod. A point mass of of the rod. Find the value of if the centre of mass is to be 6

from .

A length of uniform wire is bent to form three sides of a rectangle , and . is the base of the rectangle and the rectangle is open at . Find the distance of the centre of mass from

7

is attached to end

.

Three uniform rods are joined together to make a right-angled triangular framework. Edge

is of

length , edge is of length and edge is of length . The rods have equal mass density per unit length. Calculate the distance of the centre of mass from and . 8

A uniform triangular lamina of mass

has vertices at

attached to the lamina at and a of mass of the composite body.

mass is attached at

,

and

. A

mass is

. Find the coordinates of the centre

9

A uniform composite lamina consists of a rectangle

and a semicircular lamina.

is of length

and is of length . The rectangular lamina and the semicircular lamina have the same mass per unit area. The semicircular lamina has diameter and is joined on so its diameter coincides with

(see diagram). Find the distance of the centre of mass from

D

.

C

20cm

A

B

30cm

10 A shop sign consists of a uniform horizontal rod

together with a lamina in the shape of a

trapezium, as shown in the diagram. The rod is in length and has a mass of . The lamina has a mass of and hangs with vertical. Find the distance of the centre of mass of the shop sign from edges

and

.

1.5m

A

B Dave’s Mountain 0.6m Bikes

0.3m

0.9m

C

11 A garden ornament is made from a solid cylinder with a solid hemisphere placed on top. The radius of the cylinder is

, its height is

, and it mass is

. The hemisphere has radius

and mass

. The main axis of the cylinder passes through the centre of the hemisphere. The ornament is placed on level ground with the hemispherical part uppermost. Find the height of the centre of mass of the ornament above the ground.

12 A solid uniform frustum of a cone has been made from a solid cone of base radius

and height

. A cone of height has been removed from the vertex end, as shown in the diagram. Calculate the distance of the centre of mass of the frustum from the base.

30cm 50cm

frustum 15cm

13 A hat is modelled as a conical shell together with a brim, both made of the same uniform fabric. The conical shell has a radius of and a slant height of . The brim is an annulus (a plane figure made by cutting out a concentric disc from a larger disc) with inner radius and outer radius

. The diagram shows the shape of the hat. Find the vertical distance of the centre of

mass of the hat above its brim.

14 A uniform solid is made from a solid hemisphere and a solid cone. The hemisphere and cone have the same base radius and the centre of the circular plane face of the hemisphere coincides with the centre of the circular base of the cone. The centre of mass of the composite solid lies in the plane of the join. Show that the height of the cone is given by

.

Checklist of learning and understanding A combination of point masses can be modelled as a single mass

etc. arranged on a straight line at , with position , where

,

. A combination of point masses

etc. arranged in a plane at positions

can be modelled as a single mass

, with position

vector:

The centre of mass of a uniform rod lies at its midpoint. The centre of mass of a symmetrical uniform lamina lies on any axis of symmetry. If there is more than one axis of symmetry the centre of mass lies at the intersection of these. The centre of mass of a uniform triangular lamina lies at the intersection of its medians. The medians intersect at

where

,

and

are the

vertices of the triangle. The centre of mass of a uniform sector of a circle, having radius and angle centre of the circle, is

radians at the

from the centre of the sector.

The centre of mass of a uniform wire bent to form an arc of a circle, having radius and angle radians at the centre of the circle, is

from the centre of the circle.

The rule for combination of point masses may be extended to composite bodies comprising point masses, wires, and laminas.

, where

The centre of mass of a solid hemisphere radius lies

from the centre.

The centre of mass of a hemispherical shell radius lies

from the centre.

The centre of mass of a solid right cone or pyramid of height lies

above the base on the

line from the centre of the base to the vertex. The centre of mass of a conical shell of height lies centre of the base to the vertex.

above the base on the line from the

Mixed practice 5 1

A uniform rod , of mass and length , has three masses attached to it. A mass is attached at the end and mass is attached at the end . A mass is attached at a point on the rod. Find the distance

2

if the centre of mass of the system is

Three particles are attached to a light rectangular lamina as the -axis, as shown.

from point . . Take

as the -axis and

y

B

C P Q R O

A

x

Particle has mass

and is attached at

Particle has mass

and is attached at

Particle has mass

and is attached at

. . .

Find the coordinates of the centre of mass of the system. 3

Four tools are attached to a board. The board is to be modelled as a uniform lamina and the four tools as four particles. The diagram shows the lamina, the four particles , , and , and the - and -axes. The board has mass

and its centre of mass is at the point

Particle has mass

and is at the point

Particle has mass

and is at the point

.

Particle has mass

and is at the point

.

Particle has mass

and is at the point

.

.

.

Find the coordinates of the centre of mass of the system of board and tools.

y

B

A

C

D O

A

x

4

A uniform lamina an isosceles triangle

consists of a rectangular lamina and a lamina in the shape of joined together along , as shown. ,

and from

. Find the distance of the centre of mass of the lamina

.

D 5cm E

C

4cm A 5

B

6cm

A uniform square lamina of side has a half disc, with the distance of the centre of mass of the remaining lamina from

D

8cm

as diameter, cut out. Find .

C

8cm

A 6

B

A lamina is made from a uniform rectangular lamina, with side lengths together with a uniform lamina in the shape of a quarter disc of radius

and , , as shown.

D 9cm E

C

18cm

A

9cm

B

Find the coordinates of the centre of mass of the lamina, taking as as . 7

A composite body is made from a uniform rod, of length

and mass

,

as

and

, together with a

semicircular arc of wire of diameter , its ends fixed to the ends of the rod, of mass Find the distance of the centre of mass from the rod, giving your answer to significant figures.

.

wire

0.5m

8

A composite body is made from two uniform rods, of length , joined together at right angles. Rod has mass and rod has mass . A length of wire of mass , is bent to form a quarter circle of radius and is joined to the rods at and . Find the distance of the centre of mass from , giving your answer to significant figures.

Y wire 1.2m

O 9

X

1.2m

Two uniform right-angled triangular laminas are joined together to form one shape , as shown. , and . Find the distance of the centre of mass of the combined shape from and , giving your answers to significant figures. D 6cm A

12cm

B

9cm

C

10 A child’s toy is a uniform solid consisting of a hemisphere of radius

joined to a cone of

base radius . The curved surface of the cone makes an angle of with its base. The two shapes are joined at the plane faces with their circumferences coinciding (see diagram). The distance of the centre of mass of the toy above the common circular plane face is . Calculate the value of .

65°

5cm

[The volume of a sphere is

and the volume of a cone is

.]

11 A uniform conical shell has mass , height and base diameter . A uniform hollow cylinder has mass , length and diameter . The conical shell is attached to the cylinder, with the circumference of its base coinciding with one end of the cylinder (see diagram). Calculate the distance of the centre of mass of the combined object from the vertex

of the conical shell. 0.6 m

1.4m

1.6m

0.5kg

0.75kg

FOCUS ON … PROOF 1

Is there a connection between the equations for linear motion with constant acceleration in a straight line and for motion in a circle involving angular equivalents of equations for constant angular acceleration?

and ? Can you prove an equivalent set of

The linear equations, sometimes known as the SUVAT equations, are: where is the initial velocity, is the final velocity, is the acceleration, is the displacement and is the time taken. You obtain equation (1) by applying calculus to basic definitions.

Angular displacement is usually represented by the symbol . Angular velocity is the rate of change of angle and is denoted by the symbol or ; angular acceleration, which is the rate of change of angular velocity, is denoted by or (to distinguish it from linear acceleration ). Again, you use the basic definition to obtain the equation. You use for the initial angular velocity and for the final angular velocity after the constant acceleration has been acting for time .

Questions 1

Show, by integration, that the equivalent of equation (3) for motion in a circle with constant acceleration is:

2

Find the angular equivalents of equations

3

A particle is moving with constant acceleration of What is its angular velocity after

,

and

. . Initially its angular velocity is

.

?

Although there is a distinction between a velocity vector and its magnitude, which is called speed, there is not an appropriate word to make the same distinction between an acceleration vector and its magnitude. You call them both acceleration and allow the context to tell you whether you are referring to a scalar or a vector quantity. When you are dealing with Cartesian coordinates, the directions with which you relate the acceleration are obvious . When you refer to a particle moving in a circle with constant acceleration, you are referring to a scalar quantity because the direction of the acceleration is changing so cannot be a constant vector relative to Cartesian coordinates.

FOCUS ON … PROBLEM SOLVING 1

It is very easy to think of Mechanics in separate blocks, rather like the chapters in this book, but sometimes you have to use more than one principle to solve a problem in Mechanics. The best way to solve a Mechanics problem is to split the problem up into a series of logical steps. In this Focus on … section, you are going to look at problems involving more than one principle. Consider this problem: A small, smooth sphere of mass

is free to move in a smooth vertical groove, which is in the shape of a

circle of radius . At the sphere is at rest at a point at the lowest point of the groove when it is hit with a blow of impulse in a direction that is a tangent to the circular groove at . What is the speed of the sphere when it reaches the highest point of the groove? Take as .

Visualising You need to have a clear picture in your mind of what is happening not the maths but the actual physical situation. Without this picture you cannot represent what is happening on a diagram. You need to put as much relevant information as you can on the diagram so that it makes sense to you. You may need more than one diagram. Step 4

Step 2 Step 3

Draw a diagram change units if necessary

Step 1 Interpret the situation

Define any unknowns

Use impulse = change in momentum to find the initial velocity

Step 5 State and use the principle of conservation of energy to find the velocity at the top of the groove

Step 6 State the answer clearly to the required degree of accuracy

Step 1 Consider the connections to find the principles needed. You need to work out the connection between impulse, height and velocity. Impulse produces momentum and velocity so you need to use the impulse momentum principle to find initial velocity. As the sphere rises, it slows down and the principle of conservation of energy will give us the relationship between velocity and height.

Step 2 A′

v

1m

I

A

2m

u

Step 3 Let be the initial velocity of the particle. Let be the velocity of the particle at the highest point.

Step 4

Step 5

Step 6 The speed of the sphere at the top of the groove

is

(3 d.p.)

So look at the problem and make a plan. Remember that you may need to use more than one principle to solve a Mechanics problem.

Questions 1

A particle of mass with an impulse of

is at rest on the edge of a smooth horizontal table which is high. It is hit in a horizontal direction away from the table and at right angles to the edge of

the table. How far from the edge of the table does the particle land? 2

A particle of mass is at rest at a point in a smooth, horizontal groove when it is hit with a blow of impulse . The groove is in the shape of a circle of radius . In how many seconds does the particle return to ?

3

A particle of mass is dropped from rest and falls a vertical distance of to the horizontal ground. The constant air resistance acting on the particle during the fall is . When it hits the ground, it rebounds with velocity . Given that the coefficient of restitution between the particle and the ground is , find the magnitude of . Use and give your answer to an appropriate degree of accuracy.

FOCUS ON … MODELLING 1

A fairground game of ‘test your strength’ involves a competitor using a hammer to hit one side of a platform, which is balanced on a pivot. As the force of the hammer sends one side of the platform down, the other side rises up and sends a ball up a vertical tube towards a bell. If the ball hits the bell, then you win a prize. In this Focus on … section we are going to look at how we can model this situation and how we can then improve the model to make it more realistic.

Model 1 vertical tube

h hammer 22N s

pivot

A

u

2kg

22N s

In this model, the ball is assumed to be a particle, i.e. its mass acts at one point. The impulse generated by the hammer is instantaneous and is . All of the impulse is transmitted via the pivot to a ball which is of mass . The vertical tube is smooth and the bell is above the ground. Does the ball reach the bell? Let be the initial velocity of the particle. We need to use the impulse–momentum principle to find and then energy equations to find .

The final velocity of the particle is

:

Loss in

The ball hits the bell. How do we make a better model? How realistic is Model 1? What assumptions have we made? Which physical details have we left out? We have assumed that: the contact of the hammer is instant it takes no time for the platform to reach the ground the vertical tube is smooth the initial direction of the ball would be vertical if it were free to move. We have left out: the length of the platform the height of the ball above the ground when it leaves the platform

the radius of the ball. Let’s do the calculation again with the following changes.

Model 2 The platform is of length

and is supported at its centre by a pivot of height

transmitted to the ball is

. The impulse

and is at right angles to the platform. The ball is at the end of the platform

at the base of the tube. The radius of the ball is

.

u A

Questions 1

Find the component of the initial velocity of in the vertical direction.

2

Work out the total height it now needs to rise through to reach the bell.

3

Does the ball still reach the bell?

Model 3 To improve the model further, we need to look at the possibility of friction between the ball and the tube that it moves through. Frictional forces exist between most surfaces in contact and should be included to make this model more realistic.

Question 4

A constant frictional force of the ball now rises.

acts on the ball whilst it is in the tube. Calculate the height to which

CROSS-TOPIC REVIEW EXERCISE 1 1

The terminal velocity of a falling object is reached when the upward force on the object (the drag) is equal to the weight of the object. The formula for the drag force is of the form:

where is a dimensionless constant, is the density of the atmosphere, is the terminal velocity and is the exposed area of the falling object. Use dimensional analysis to find the values of 2

and and hence find a formula for the drag force.

A small toy boat of mass is sailing in a straight line with initial velocity on by a constant wind with force for . a What is the total impulse on the boat in these

. It is acted

?

b Hence, what is its final velocity? 3

A group of children are playing on a playground roundabout that is spinning at a constant rate of revolutions per minute. A small parcel of mass is at rest on the roundabout at a distance of from the centre. If the parcel is at rest, what is the frictonal force acting on the parcel? You can model the roundabout as a horizontal disc and the parcel as a point mass.

4

A particle, , of mass is attached to one end of a light inextensible string of length The other end of the string is attached to a fixed point , vertically above .

.

is hit with a blow of impulse in a horizontal direction. Find the angle that the string makes with the vertical when the particle first comes to rest. 5

A car of mass is going up a hill at an angle to the horizontal where is travelling against a constant frictional force of with a constant speed of is the power output of the car’s engine as it goes up the hill? (Take )

6

Prove by induction that

7

Two smooth spheres, and , both of mass , are at rest on a smooth horizontal table that is high. The line joining and is at right angles to the edge of the table. is from the edge of the table and is from the edge of the table. is projected towards at . is brought to rest by the collision and subsequently moves to the edge of table and falls off it. What is the time from the collision to the moment that hits the ground? (Take )

8

Two particles, of mass and of mass , are on a smooth horizontal plane. is moving at and at . The particles collide and coalesce to form a new particle , which then hits a vertical wall at right angles. The coefficient of restitution between and the wall is . What is the velocity of as it leaves the wall?

9

A uniform circular lamina , with equation

, has the circle

. The car . What

removed from it. What are the coordinates of the centre of mass of the new shape? 10 Brinell’s test evaluates the hardness of a material using a Brinell hardness number

. The

test involves forcing a steel or tungsten carbide sphere into the material being tested and measuring the diameter of the indentation. The formula is:

where is the applied force, is the diameter of the sphere and is the diameter of the indentation. a What are the dimensions of force? b What are the dimensions of BHN? 11 Particles and , both of mass

, are joined by a light inextensible string of length

. is

moving in a circle on a smooth horizontal surface at a constant speed of . The string goes through a small, smooth hole in the surface and hangs at rest at a distance below it. Find the value of . (Take

) A

x

B

12 A conical pendulum consists of a particle , of mass , attached by a light inextensible string, of length , to a point The particle moves in a horizontal circle at a constant speed of so that the string describes a cone. What is the volume of the cone?

B

1.5m

A

13 A square lamina with vertices at coordinates , , and has a smaller square lamina, made of material of double the mass per unit area of the first lamina, added to it. The coordinates of the vertices of the second square are , , and What are the coordinates of the centre of mass of the combined shape?



6 Work, energy and power 2 This chapter is for A Level students only. In this chapter you will learn how to: calculate the work done by a variable force -axis

when displacement is along the

understand and use Hooke’s law for elastic strings and springs calculate the work done extending an elastic string calculate the work done extending or compressing an elastic spring include elastic potential energy in problems involving conservation of energy use vectors to calculate work done, kinetic energy and power.

Before you start… Chapter 1

You should know that work done is the product of force and displacement in the direction of motion and that work done is measured in joules.

1 Calculate the work done by gravity when a stone of mass falls vertically .

Chapter 1

You should know that kinetic energy is joules.

2 Calculate the change in kinetic energy when a boy of mass increases his running speed from to .

Chapter 1

You should know that the work done by a force acting on an object causes a change in its kinetic energy. This is the work–energy principle.

3 Find the horizontal resistive force that causes an ice hockey puck of mass to reduce speed from to over .

A Level Mathematics Student Book 2

You should be able to integrate a function between limits, and , to find a quantity:

4 Find

defined as

and is measured in

.

Extending your knowledge of work and energy The weight of an object is approximately constant close to the Earth’s surface and the change in gravitational potential energy caused by a change in height can be accurately modelled as the work done

by or against constant weight. When a rocket is fired into space its weight reduces as its distance from the centre of the Earth increases. Weight is an example of a force that varies with distance.

Rewind You learned about the work done by a constant force and the work–energy principle in Chapter 1. You also learned about conservation of mechanical energy. These ideas are crucial to understanding how energy is converted to useful work. Other forces can vary. In this chapter, you are going to learn how to work with forces that vary with distance. A particularly important example is the force in an elastic spring or string, which increases with extension (Hooke’s law). As a spring is extended, you do work against an increasing tension. The work is stored as elastic potential energy. Springs have many practical uses. In mountain bikes springs called shock absorbers are used to help smooth out the effects of unevenness in the road surface. When your bike hits a bump, the spring compresses. The compression absorbs some of the energy that would otherwise pass straight through to the rider. There is elastic potential energy in the spring. If you hit a bump too hard you can break your shock absorber – you have gone beyond its elastic limit. Force, displacement and velocity can all be expressed as vector quantities. Work done, the product of force and distance, can be defined more precisely as the scalar product of vector force and vector displacement. Power can likewise be determined from the scalar product of vector velocity and vector force.

Fast forward You will learn about Hooke’s law in Section 2. You will learn to work with the scalar product in Section 4.

Section 1: Work done by a variable force You know that for a constant driving force acting in the direction of motion:

If the force is always parallel to the motion but the magnitude of the force is changing you need to use integration to find the work done.

Key point 6.1 If an object is moved in a straight line from a position variable force

to a position

by the action of a

that depends on displacement, , work done is defined as:

WORKED EXAMPLE 6.1

Find the work done by a force

that displaces an object from

to

.



Let

and

.

Integrate and substitute the limits of integration.

WORKED EXAMPLE 6.2

A car of mass

moves from rest at on a horizontal surface. The driving force is constant at

and resistance to motion is modelled as

The car moves

to

a Find the work done by the driving force and the work done against resistance as the car travels from to b Find the speed of the car at . a The work done by the driving force

As the driving force is constant, you can use the definition of the work done by a constant force. Use integration to find the work done against the variable resistive force. Let

and

.

Integrate and substitute the limits of integration.

b



Use the work–energy principle. Car moves from rest

.

Rearrange to find the car’s final speed.

WORKED EXAMPLE 6.3

WORKED EXAMPLE 6.3

An object is moving in a horizontal straight line against a resistive force that is directly proportional to its distance from its starting point, . If the work done against resistance as the object travels from the origin,

, to a point

away,

, is

:

a find the magnitude of b state the units of . a

Use the definition of work done by a force that depends on displacement.



Integrate between the given limits to find an expression for work done in terms of .

Use the given value for work done to calculate . b Since is a force have units

EXERCISE 6A

.

, must

EXERCISE 6A For questions to , calculate the work done by force . Use the formula:

moving an object along the -axis from

to

work done

1 2 3 4 5

The work done by a force

moving an object from

to

is

6

The work done by a force

moving an object from

to

is ln Find an exact value for

7

A particle moves along the -axis under the action of a force in from

8

. Find the value of

newtons, where is measured

Find the work done by the force moving the particle from

to

.

A vehicle of mass moves along a horizontal road with driving force . It starts from rest at and experiences a resistance to motion of . Find the speed of the vehicle when it reaches .

9

A particle of mass moves along a horizontal axis with driving force The particle starts from rest at and experiences a resistance to motion of Find the speed of the particle when it reaches

.

10 A piston of mass moves in a straight line inside a cylinder against a resistance of where is measured in metres. When the speed of the piston is , and when the speed of the piston is

,

. Find the value of

11 A truck of mass tonnes experiences a resistance to motion of travelled after the brakes are applied. The truck is travelling at brakes with a constant braking force of

, where is the distance when the driver applies the

. Find how far the truck travels before coming to rest.

12 A vehicle of mass starts from rest at and moves in a straight line parallel to the -axis. There is a driving force of and a resistive force of . Derive an expression to represent work done after the vehicle has travelled . Numerically find the value of when the vehicle is travelling at

Section 2: Hooke’s law, work done against elasticity and elastic potential energy Robert Hooke was an English experimental scientist, born in 1635. He found that the extension caused when stretching an elastic string obeyed a simple rule. The same rule applied to the extension or compression (reduction in length) of an elastic spring The rule is known as Hooke’s law. Hooke measured the force required and found that, providing the object was not deformed past its elastic limit: when stretching: when compressing:

.

Elastic strings may be extended but not compressed. Elastic springs may be extended and compressed. Extension or compression must occur within the elastic limit of the string or spring. When an elastic string is stretched beyond its elastic limit it does not return to its original length when the force is removed. As you stretch or compress an elastic object, you do work against elasticity and this work is stored as elastic potential energy:

You can use this extended definition of mechanical energy to solve problems about the motion of objects attached to elastic strings or springs.

Rewind In Chapter 1 you learned that when gravity is the only force acting on an object: total mechanical energy kinetic energy gravitational potential energy

Hooke’s law for elastic strings and springs When an elastic string or spring is extended from its natural length, there is a tension in the string or spring in the opposite direction to the extending force.

string extended string fixed here

tension

force causing extension

natural length, l When an elastic spring is compressed from its natural length, there is a thrust in the spring in the opposite direction to the compressing force.

natural length, l spring compressed spring fixed here

thrust force causing compression

You can relate the extension or compression of an elastic spring, to the tension or thrust in the spring, , by the formula:

where is the natural length of the spring. The modulus of elasticity, , is the force required to double the length of the spring, assuming that such an extension does not cause the spring to exceed its elastic limit. If a spring does exceed its elastic limit, then Hooke’s law no longer applies. The same formula applies to the extension of an elastic string. The modulus of elasticity, , is measured in .

Did you know? In A Level Physics you may see an alternative formula for Hooke’s law,

, with being the

stiffness of the object being compressed or extended.

Key point 6.2 Hooke’s law for an elastic string or spring is: A low value for means that the string or spring is quite flexible and easy to extend or compress.  A high value for means the string or spring is quite stiff and difficult to extend or compress. An important assumption is that the elastic string or spring is ‘light’, that is, its mass can be ignored. If the string were not ‘light’ then the tension or thrust through the string or spring could vary along its length. WORKED EXAMPLE 6.4

A light elastic spring, which has modulus of elasticity and natural length , has one end attached at a fixed point A horizontal force, of magnitude , is applied to the spring causing a compression. The spring rests in equilibrium. Find the distance that the spring is compressed from its natural length.

natural length 1.8m A

40N Since the spring is in equilibrium there is zero resultant force. Rearrange the formula for Hooke’s law to make extension the subject. The compression of the spring is less than its natural length. This is a simple check of validity.

The compression of the spring is WORKED EXAMPLE 6.5

A light elastic string is attached to a fixed point and hangs vertically, in equilibrium, with an object of mass attached to its lower end. The string has natural length and the object is resting below the point of suspension. Find the modulus of elasticity of the string. Draw a diagram showing the forces acting on the particle.

T

1.3 m

0.4 g

Convert grams to kilograms. Since the string is in equilibrium there is zero resultant force.



Rearrange the formula for Hooke’s law to make the subject.



WORKED EXAMPLE 6.6

An object of mass is attached to the end of a light elastic string of length . The other end of the string is fixed to point . The object is held at and released. The modulus of elasticity of the string is . Find the extension of the string when the object reaches its maximum speed, and hence its distance below at this time. Using To start with the object falls under gravity for tension in the string.

. There is no

As the string extends beyond its natural length the increasing tension in the string reduces the acceleration. When

Speed reaches its maximum value when

.

Use Hooke’s law for . Rearrange to find the extension. So the object is below when it reaches maximum velocity.

Find the total distance fallen.

WORKED EXAMPLE 6.7

An object of mass is attached to the lower ends of two light elastic strings. One string is of natural length with modulus of elasticity The other string is of natural length with modulus of elasticity . The free ends of the strings are attached to a point and hangs in equilibrium vertically below Find the distance .

A natural length 0.4m

natural length 0.5m

T1

T2

P 10g Let the tensions in the strings be and The mass hangs in equilibrium so there is zero resultant force on . Let distance

be metres.

Use

for the tensions in the strings. in each case.

Substitute your expressions for tension. Rearrange to find the value of .

Work done extending an elastic string The force required to extend an elastic string or spring varies with the extension of the string. This means that you cannot find the work done by multiplying force and distance. You need to integrate to find the work done against a variable force.

Key point 6.3 Work done extending an elastic string from extension

to

is given by:

The same formula is used to calculate the work done compressing an elastic spring from compression compression .

to

PROOF 1

An elastic string has modulus of elasticity and natural length . Prove that the work done extending from extension

to extension

is

.

The string is extended from from to .

to

. Integrate your expression for

Take the constants out of the integration. Substitute the limits.

Tip When calculating the work done against elasticity, make sure you take the difference of the

squares of the extensions, not the square of the difference:

WORKED EXAMPLE 6.8

Find the work done when a light elastic string of natural length is stretched from a length of to .

and modulus of elasticity

Work out the starting and ending extensions. Use the formula for work done against elasticity. Substitute in the values.

Elastic potential energy The work done extending a light elastic string or spring is stored as elastic potential energy. When the string or spring is released, it will contract towards its natural length and elastic potential energy is converted to kinetic energy. Similarly, the work done compressing a spring is stored as elastic potential energy. When the spring is released it will expand towards its natural length, converting elastic potential energy to kinetic energy.

Key point 6.4 Elastic potential energy (EPE) is the energy stored in a string or spring extended by , or in a spring compressed by .

Key point 6.5 Using the principle of conservation of energy, when an object is acted on only by its weight and the force in an elastic string or spring:

where energy.

is gravitational potential energy,

is elastic potential energy and

is kinetic

Rewind In Chapter 1 you learned that work done against gravity is stored as gravitational potential energy and that when the only force acting on an object is its weight, the sum of kinetic and gravitational potential energy is conserved.

WORKED EXAMPLE 6.9

A light elastic spring with natural length rests on a smooth horizontal table. One end is attached to a fixed point A and a mass is attached at the other end , held from . The modulus of elasticity of the spring is . a Find the elastic potential energy in the spring. The spring is released and moves horizontally away from . b Find how fast the mass is travelling when the spring reaches its natural length.

a



Convert centimetres to metres.

b

When the spring reaches its natural length, all its EPE has been converted to . and

.

WORKED EXAMPLE 6.10

One end of a light elastic string of natural length and modulus of elasticity is attached to a fixed point . A particle of mass is attached to the other end of the string. is released from rest at and falls vertically. Assuming there is no air resistance, find: a the extension of the string when is at its lowest position b the acceleration of at its lowest position. a

Use the principle of conservation of energy:

Let the lowest point be the zero level for

.

Then , where is the height fallen. is stationary at and at its lowest position:

At the string is not extended, so

.

Take to represent extension, as usual.

Substitute in the given values and rearrange the quadratic equation.

Solve the quadratic, taking the positive solution for . This is the maximum extension of the string.

b

At the maximum extension of the string, take upwards as positive, and use Use Hooke’s law When

WORKED EXAMPLE 6.11

for the tension.

Use the value for the maximum extension found in part .

WORKED EXAMPLE 6.11

A light elastic spring, of natural length , has one end fixed to a horizontal surface with the other end vertically above. A sphere of mass rests on the top of the spring, which is in equilibrium,

above the surface.

0.8kg

12cm

a Show that modulus of elasticity of the spring is The spring is compressed a further

.

.

b Find the maximum speed of the sphere in the subsequent vertical motion. a

The sphere rests in equilibrium so the resultant force is zero.  

Use Hooke’s law

.

   

, as required

b

Use the principle of conservation of mechanical energy for the upward movement. Let be the final compression of the spring:

Let the be zero

at the start of the upward movement . , where is the distance moved:

starting compression − final compression   Simplify, substituting

.

Rearrange to express in terms of a quadratic in . Complete the square to find the stationary point for . The maximum speed arises at the equilibrium compression of the spring.

Focus on … You will use an alternative method of finding the maximum speed in Focus on … Problem solving 2.

WORK IT OUT 6.1 A light elastic string has one end attached to a fixed point and the free end attached to a particle of mass

. Particle is released from rest at and falls a distance

metres,

where is the natural length of the string and is its extension. David wants to work out an expression for ’s kinetic energy. Which of the following energy equations should he use? Which solution is correct? Can you identify the errors in the incorrect solutions? Solution 1

Solution 2

Solution 3

EXERCISE 6B 1

Calculate the tension in a light elastic string when it is extended from a natural length of . The modulus of elasticity is

by

2

Calculate the thrust in a light elastic spring that is compressed from its natural length of . The modulus of elasticity is .

to

3

A light elastic string of natural length

is extended by

. The tension in the string is

. Find

the modulus of elasticity. 4

A light elastic spring is compressed from its natural length of . The thrust in the spring is the modulus of elasticity is . Find the compression of the spring.

5

A light elastic string is attached to a fixed point and hangs vertically, in equilibrium, with an object of mass attached to its lower end. The string has natural length and the modulus of elasticity is

6

and

. Find the extension of the string.

A light elastic string is attached to a fixed point and hangs vertically, in equilibrium, with an object of mass attached to its lower end. The string has natural length and the object is resting below the point of suspension. Find the modulus of elasticity of the string.

7

A light elastic spring of natural length

has one end attached to a fixed point on a smooth

horizontal surface. A horizontal force is applied to the other end of the spring causing a compression of . The modulus of elasticity of the spring is . Find the magnitude of the force causing the compression. 8

A light elastic spring of natural length has one end attached to a fixed point on a smooth horizontal surface. The spring is extended by a horizontal force of magnitude . The modulus of elasticity is

9

. Find the extension of the spring.

Find the elastic potential energy stored in a light elastic spring of natural length elasticity

when it is compressed to a length of

and modulus of

.

10 Find the increase of elastic potential energy when a light elastic string of natural length modulus of elasticity is extended from to .

and

is extended from to . The work done against 11 A light elastic string of natural length elasticity extending the string is . Find the modulus of elasticity of the string. and natural length is compressed from its 12 A light elastic spring having modulus of elasticity natural length. The elastic potential energy stored within the spring is . Find the length of the compressed spring. is attached to one end of a light elastic string of natural length with 13 An object of mass its other end attached to a fixed point, . The modulus of elasticity of the string is . is dropped from . Find the extension of the string when the object reaches its maximum velocity.

and modulus of elasticity is attached to a 14 One end of a light elastic string of natural length fixed point A particle of mass is attached to the other end of the string. is released from rest at and falls vertically. Assuming there is no air resistance, find: a the extension of the string when is at its lowest position b the acceleration of at its lowest position, stating the direction. 15 A light elastic string has natural length , modulus of elasticity and extension Show that the work done extending from to can be expressed as the change in string tension multiplied by the mean extension of the string. is attached to the free end of an elastic string of natural length . The 16 A particle of mass other end is attached to a fixed point, The particle is held below and released. Given the modulus of elasticity of the string is instantaneous rest.

, find how far the particle is from when it comes to

is attached to the lower ends of two parallel light elastic strings. One string is 17 An object of mass of natural length with modulus of elasticity The other string is of natural length with modulus of elasticity . The free ends of the strings are attached to a point , and hangs in equilibrium vertically below . Find the distance

.

rests on a smooth horizontal table. One end is attached 18  A light elastic spring with natural length to a fixed point and a mass is attached at the other end held from The modulus of elasticity of the spring is a Find the elastic potential energy in the spring. The spring is released and moves horizontally away from b Find how fast the mass is travelling when it is

from .

Section 3: Problem solving involving work, energy and power More complex problems may combine the work–energy principle and the principle of conservation of energy. You may be working with any of the propulsive or resistive forces you have met in Chapter 1, Sections 1 and 2. WORKED EXAMPLE 6.12

An object, , of mass

, is attached to the ends of two light elastic strings with the same natural

length, , but different modulus of elasticity. One of the strings is attached to a point and the other is attached to point on the same horizontal level as , such that the distance is . hangs in equilibrium. The distance is and is . Calculate the modulus of elasticity of each of the strings.

2m

A

α

β

TA

TB

1.5m

B

Draw a clearly labelled diagram.

0.9m

P

2g Use the cosine rule to find angles significant figures at this stage.

, , and

, . Keep

Resolve horizontally and get an expression for .

in terms of

Resolve vertically and then substitute for

Rearrange to make But

Now calculate

the subject and calculate

.

.

Use Hooke’s law. Use Hooke’s law for each string to find each modulus of elasticity.

WORKED EXAMPLE 6.13

A light elastic string, of natural length , has one end fixed to point on a rough plane inclined at to the horizontal. The string has modulus of elasticity . A particle of mass is attached to the free end of the string. is released from rest at and descends the plane to , where it comes to rest. Given that the coefficient of friction between and the plane is : a find the distance

b determine whether the particle remains stationary at or starts to travel back up the plane.

R

T F

P

30°

Draw a diagram showing the forces acting on as it slides down the plane from .

2.1 g

a Work done by gravity – work done against tension – work done against friction increase in kinetic energy

Use the work–energy principle: net work done change in kinetic energy

Work done by gravity – work done against tension – work done against friction

The particle is at rest at and at so the increase in kinetic energy is zero.

Work done by gravity

The distance travelled down the slope by is the natural length, , plus the extension in the string . Let be the frictional force.

While is moving,

µ

µ , where

Substitute for in the work–energy equation. Simplify to get a quadratic equation and take the positive value for the extension.

b At

Find the tension in the string when it is at its maximum extension.

Calculate the total force parallel to the plane without friction. Using the expression for from part a. Consider the resultant force up the plane.

the particle will start to travel back up the plane. WORKED EXAMPLE 6.14

One end of a light elastic string, of natural length and modulus of elasticity , is attached to a fixed point on a smooth plane inclined at an angle to the horizontal, where



. A particle , of mass

, is attached to the other end of the string. is released from rest at

and travels down the plane without reaching the bottom. Find the maximum speed of as it travels down the plane.

R

Draw a diagram showing the forces acting on as it slides down the plane from .

T P

α

mg

Let the extension of the elastic string be . Component of weight of parallel to the plane is At maximum speed: has maximum speed when acceleration is zero because there is no resultant force parallel to the plane. Use Hooke’s law.

and

is the extension of the string.

has travelled

from when it

Distance from is natural length plus extension.

reaches its maximum speed. is constant. Let speed)

(when is travelling at its maximum ( moves from rest)

Substitute and rearrange to find .

WORKED EXAMPLE 6.15

a A car of mass moves along a straight horizontal road. The resistance to motion is ; the engine is working at and the car is moving with constant speed. Find the constant speed of the car in . b The same car now moves up a hill inclined at to the horizontal. The car’s engine continues working at and the resistance to motion is unchanged. Find the new, constant, speed of the car up the hill, in . R

a

Draw a clearly labelled diagram. is the constant tractive force. T

580N

1150g

 

Convert kilowatts to watts.

Rearrange the formula to find the speed. Convert b When travelling uphill:

to

.

Draw a clearly labelled diagram.

R

Use

to represent new tractive force.

T'

580N 4°

1150g

The component of the car’s weight acting parallel to the road surface is , as the road surface is inclined at to the horizontal.

Rearrange to find the tractive force.

.

Rearrange to find speed.



Convert

to

.

WORKED EXAMPLE 6.16

A car and driver of combined mass

accelerate from rest up a road inclined at

to the

horizontal, with average resistance to motion of . The car engine is working at a constant rate of . The car reaches a speed of after it has travelled . Calculate the time taken. To find the time taken when the vehicle engine is operating at constant power you can find the total work done and use the definition:

Work done by engine – work done against resistance

Mechanical energy is increased because the car is accelerating up a hill. This increase in mechanical energy comes from the car engine. But some of the work done by the engine is expended on resistance to motion.

Work done by engine = work done against resistance

Calculate the total work done by the car’s engine. (car moves from rest) Let

Work done against resistance

(at the start of the travel)

Convert kilowatts to watts.

s

Divide the work done by the engine’s constant power output.

EXERCISE 6C 1

A block of mass is being pushed in a straight line along horizontal ground by a force of inclined at above the horizontal. The block moves a distance of in . Find: a the work done by the force b the power with which the force is working.

2

Darius is pulling a wheeled suitcase of mass up a plane inclined at to the horizontal. The strap he is holding is taut, with tension , and angled at to the horizontal. The resistance to motion is a Calculate the increase in the suitcase’s kinetic energy as it moves

up the slope.

Darius trips slightly and releases the strap. The suitcase comes to rest before rolling back down the slope against the same resistance to motion. b Find the speed of the suitcase after it has travelled

down the slope from rest.

3

A block of mass is projected up an inclined plane at and comes to rest after travelling up the plane. Given that the resistance to motion up the plane is constant at , find the inclination of the plane, to the nearest degree.

4

A parcel of mass . Find:

is projected up a smooth plane inclined at

a the speed of the parcel after it has travelled

to the horizontal with a speed

up the plane

b how far the parcel travels up the plane before it stops moving. 5

A car and driver of combined mass accelerate from rest down a road inclined at to the horizontal, with average resistance to motion of . The car engine is working at a constant rate of . The car reaches a speed of

after it has travelled

; calculate the time taken.

6

A package of mass is projected up a rough plane inclined at projection is and the resistance to motion is constant at package when it returns to its starting point.

to the horizontal. The speed of . Calculate the speed of the

7

An object of mass is attached to the ends of two light elastic strings having the same modulus of elasticity. One of the strings has natural length and the other has a natural length of . The longer string is attached at and the shorter string is attached at on the same horizontal level. The object hangs below a point on the same level as and , from and from . Find the modulus of elasticity of the strings.

8

A light elastic string , of natural length , is fixed at point on a rough plane inclined at to the horizontal. The string has modulus of elasticity . A particle of mass is attached to end and the particle is released from rest to descend the plane from to The particle descends m from a Show that the coefficient of friction between the particle and the inclined plane is

.

b Find the acceleration of the particle at 9

A car of mass

moves along a straight horizontal road. The resistance to motion is constant,

, and the car’s engine is working at a constant rate of a Find the acceleration of the car when the car’s speed is The road now ascends a constant slope inclined at

.

to the horizontal. The car’s engine continues

working at

and the resistance to motion remains

b Find the greatest steady speed of the car as it ascends the hill. is ascending a hill inclined at to the horizontal. The power exerted by the 10 A car of mass engine is and the car has a constant speed of It is assumed that resistance to motion is where is the car’s speed and is a constant value. a Show that The power of the engine is now increased to

.

b Calculate the maximum speed of the car while ascending the hill.

Section 4: Using vectors to calculate work done, kinetic energy and power Force, displacement and velocity are all vector quantities. This means that you can use your ability to manipulate vectors to solve problems connected with work and energy.

Rewind In Chapter 1 you learned about work, kinetic energy and the work–energy principle.

Using the scalar product to calculate work done by a constant force If a constant force is directed at an angle to the direction of motion then:

fo r

ce

θ

fo r

ce

θ

starting position distance

final position

You can use the scalar product to find the work done by a force.

Tip The work done by a force is a scalar quantity.

Key point 6.6 The work done by a constant force that causes a displacement is given by the formula: This is the scalar product of the force and displacement vectors.

Rewind Remember the definition of the scalar product that you learned in Pure Core Student Book 1: , where is the angle between vectors and .

WORKED EXAMPLE 6.17

A force is acting on an object that moves from , with position vector position vector . Find the work done by the force.

, to with

Displacement final position – starting position Calculate the scalar product of force and displacement.

Work done is a scalar quantity.

WORKED EXAMPLE 6.18

An object, of mass , is at rest at when a constant force . has position vector and has position vector

causes the object to move to . Given that no other forces

act on the object, use the work-energy principle to find the speed of the object at .

Calculate the work done by the force.

The work–energy principle states that the work done by a propulsive force is equivalent to the increase in kinetic energy.

Tip Work done on an object by a propulsive force is positive in sign. Work done on an object by a resistive force is positive in sign.

WORKED EXAMPLE 6.19

A force is acting on an object that moves from , with position vector , to with position vector . Find the work done by the force and state whether the force is propulsive or resistive. Displacement = final position − starting position Calculate the scalar product of force and displacement. Work done is negative in sign so the applied force is resistive.

Using the scalar product to calculate kinetic energy You have previously learned how to calculate the kinetic energy of a moving body in terms of its speed. If instead of working with speed (a scalar quantity) you are working with a velocity (a vector quantity), you can calculate by using the scalar product. where is the velocity vector. This means that a velocity vector, rather than speed, can be used to calculate the kinetic energy of a moving object.

Rewind In Chapter 1, Section 2, kinetic energy is defined as , where

is mass and is speed.

If mass is measured in

and speed in

then kinetic energy is measured in joules

Key point 6.7 The kinetic energy of an object of mass where

moving with velocity

is the scalar product of velocity with itself.

is defined as

.

Tip Remember that we use bold letters to represent vectors and italic letters to represent their magnitudes; so is the magnitude of

WORKED EXAMPLE 6.20

A rocket of mass tonnes is moving with velocity rocket, giving your answer in .

. Find the kinetic energy of the

Find using the scalar product.

tonnes is

.

Equations of motion can also be written in vector form, using scalar products.

Rewind In A Level Mathematics Student Book 1, Chapter 20, you learned to use the formula for motion in a straight line with constant acceleration: where and represent starting and final velocity, is constant acceleration and is displacement.

Key point 6.8 If an object is moving with constant acceleration, , then: and represent final velocity, starting velocity and displacement, respectively, in vector form. The formula can also be written:

WORKED EXAMPLE 6.21

A small object of mass accelerates across a horizontal surface due to the action of a force that is acting at an angle to the resulting displacement. The driving force is , and the displacement is . Given that the starting speed is , find the final speed of the object. Use

to find the acceleration vector.

Substitute in the formula

Using the scalar product to calculate the power of a driving force

If driving force and velocity are both given in vector form, then power is the scalar product of the driving force and velocity vectors.

Rewind In Chapter 1 you learned that power can be expressed as driving force speed, given that the driving force is acting in the direction of motion.

Key point 6.9 The power of an engine producing a driving force

on a vehicle moving with velocity

can be calculated from the formula:

Tip The power of a propulsive force acting on an object is positive in sign. The power of a resistive force acting on an object is negative in sign.

WORKED EXAMPLE 6.22

A vehicle is moving under the action of a driving force in a horizontal plane with velocity . Given that no other forces are acting on the vehicle, find the power of the vehicle engine. Use the formula for power using the scalar product

.

Evaluate the scalar product to find the power of the vehicle engine.

WORKED EXAMPLE 6.23

A vehicle of mass

a Find the power, in

is moving in a horizontal plane, at time seconds,

, of the vehicle engine when

b Find the work done, in a

, with velocity

.

, by the vehicle engine between

and

seconds.

Find the velocity of the vehicle when

.

When Differentiate to find the acceleration vector. Use to find the tractive force of the vehicle engine when

Evaluate the scalar product to find the power of the vehicle engine.

b Net work done by propulsive

The driving force is variable but the total work done can be calculated using the work–energy principle. Calculate the starting and final velocities.

EXERCISE 6D

EXERCISE 6D 1

a Calculate the work done when: causes a displacement of i a force ii a force

causes a displacement of

causes an object to move from , with position vector iii a force position vector . b The work done by a force c The work done by a force 2

causing a displacement

is

causing a displacement

is

, to with . Calculate .

. Calculate .

a Calculate the kinetic energy when: is moving with a velocity i an object of mass ii an object of mass

is moving with velocity

.

b Use work and energy to calculate the final speed when: and starting velocity i a particle having mass

is acted on by a force of

causing a displacement of and starting velocity ii a particle having mass causing a displacement of . c An object of mass is moving with speed causing a displacement of 3

is acted on by a force of

after being acted upon by a force of . Find the starting speed of the object.

a Calculate the power when: i an object acted on by a driving force of

is travelling with velocity

ii an object acted on by a driving force of

is travelling with velocity

b A particle acted on by a driving force of

is moving with velocity

Find the value of , given the power is c A particle acted on by a driving force of Find possible values of , given the power is 4

An object is acted on by a force done is

5

is moving with velocity

and is moving parallel to the vector

. Given the work

, find the displacement vector.

A particle of mass

is displaced

while moving with acceleration

. Find the

change in kinetic energy due to action of the force. 6

A particle is displaced while moving with acceleration velocity is . Find the final speed of the particle.

7

A vehicle engine is providing a driving force of

. The starting

as the vehicle travels with velocity

. Find the power of the vehicle engine at this time. 8

A vehicle engine is providing a driving force of with engine power

9

A particle of mass

as the vehicle travels parallel to

. Find the velocity of the vehicle. is moving with velocity

.

a Find the force acting on the particle when

.

b Find the power exerted by the force when

.

c Find the work done by the force between

and

10 A particle of mass

at rest at

a Find the velocity of the particle when

is acted on by a force .

b Find the power exerted by the force when c Find the work done by the force between

.

. and

.

Checklist of learning and understanding For a variable force

that depends on displacement, , work done is defined as:

Hooke’s law for an elastic string or spring is:

where is the modulus of elasticity

Work done against elasticity is: Elastic potential energy is: which is the scalar product of the force and displacement vectors joules, where

is the scalar product of velocity with itself

When an object moves with constant acceleration: , the scalar product of vector force and vector velocity

Mixed practice 6 1

An object, moves under the action of a force, position vector , to , having position vector

is displaced from , having . Calculate the change in kinetic

energy of as it moves from to . 2

A body, is

, of mass , where

moves under the action of a force . At time

, the velocity of

a Find in terms of . b Calculate the rate at which the force is working when c By considering the change in kinetic energy of during the time interval . 3

A body

of mass

. When

,

.

, calculate the work done by the force

, moves under the action of a force

, at time

where

velocity is

a Determine the acceleration of b Determine the velocity of when

.

c Calculate the rate at which the force is working when place. 4

giving your answer to decimal

A small object of weight is attached to one end of each of two parallel light elastic strings. One string is of natural length and has modulus of elasticity ; the other string is of natural length and has modulus of elasticity . The upper ends of both strings are attached to a horizontal ceiling and hangs in equilibrium at a distance m below the ceiling (see diagram). Find

dm

W © OCR, GCE Mathematics, Paper 4730/01, June 2013 5

A particle of mass

moves in the

plane with acceleration

a Find the force acting on the particle when significant figures. When

the particle has velocity

b Find the power of force when 6

.

, giving your answer to

. , giving your answer to significant figures.

A particle of mass is attached to one end of a light elastic string of natural length and modulus of elasticity . The other end of the string is attached to a fixed point on a smooth horizontal surface. is held at rest at a point on the surface from . The particle is then released. Find i the initial acceleration of

ii the speed of at the instant the string becomes slack. © OCR, GCE Mathematics, Paper 4730/01, June 2008 7

One end of a light elastic string, of natural length

and modulus of elasticity

attached to a fixed point . A particle of weight

is attached to the other end of the

string. is released from rest at a point

vertically below

, is

Subsequently just reaches

i Find ii Find the magnitude and direction of the acceleration of when it has travelled its point of release.

from

© OCR, GCE Mathematics, Paper 4730/01, June 2014 8

One end of a light elastic string, of natural length

and modulus of elasticity

, is

attached to a fixed point . A particle of mass is attached to the other end of the string. is released from rest at and falls vertically. Assuming there is no air resistance, find i the extension of the string when is at its lowest position, ii the acceleration of at its lowest position. © OCR, GCE Mathematics, Paper 4730, January 2012 9

One end of a light elastic string, of natural length and modulus of elasticity is attached to a fixed point on a smooth plane inclined at an angle to the horizontal, where . A particle of mass

is attached to the other end of the string. is released

from rest at and moves down the plane without reaching the bottom. Find i the maximum speed of in the subsequent motion, ii the distance of from when it is at its lowest point. © OCR, GCE Mathematics, Paper 4730, June 2012 10 A particle , of mass

, is in equilibrium suspended from a fixed point by a light elastic

string of natural length 3 m and modulus of elasticity . Another particle , of mass is released from rest at and falls freely until it reaches and becomes attached to it.

,

i Show that the speed of the combined particles, immediately after becomes attached to is The combined particles fall a further distance ii Find a quadratic equation satisfied by

before coming to instantaneous rest.

and show that it simplifies to

.

© OCR, GCE Mathematics, Paper 4730/01, January 2013 11 A light elastic string of natural length

has modulus of elasticity

One end of the

string is attached to a fixed point and the other end is attached to a particle of weight . The particle is released from rest at the point which is vertically below It comes instantaneously to rest at , which is vertically above . i Verify that the distance

is

.

ii Find the maximum speed of during its upward motion from to . © OCR, GCE Mathematics, Paper 4730, January 2010 12 A vehicle of mass

is supplied with a constant power of

to move it along a straight

horizontal road against a variable force that may be approximated as

where

represents the distance from a fixed point on the road and Initially the vehicle is at from travelling with speed . After , the vehicle is at , from , travelling with speed . Find the value of . 13 A vehicle of mass moves along a horizontal road with driving force . It starts from rest and experiences a resistance to motion of . Find the speed of the vehicle after it has travelled . 14 A truck of mass tonnes experiences a resistance to motion of , where is the distance travelled after the brakes are applied. The truck is travelling at when the driver applies the brakes with a constant braking force of . Find how far the truck travels before coming to rest.

A

15

P

θ A particle , of mass

, is in equilibrium suspended from the top of a smooth slope

inclined at an angle to the horizontal, where sin

, by an elastic rope of natural

length and modulus of elasticity (see diagram). Another particle , of mass , is released from rest at and slides freely downwards until it reaches and becomes attached to it. i Find the value of , where is the speed of immediately before it becomes attached to , and show that the speed of the combined particles, immediately after becomes attached to , is

.

The combined particles slide downwards for a distance of

, before coming instantaneously

to rest at . ii Show that

. © OCR, GCE Mathematics, Paper 4730, January 2011

16 A bungee jumper of weight is joined to a fixed point by a light elastic rope of natural length and modulus of elasticity . The jumper starts from rest at and falls vertically. The jumper is modelled as a particle and air resistance is ignored. i Given that the jumper just reaches a point

below , find the value of

ii Find the maximum speed reached by the jumper. iii Find the maximum value of the deceleration of the jumper during the downward motion. © OCR, GCE Mathematics, Paper 4730, June 2010 17 A particle of mass

is attached to one end of a light elastic string of natural length

and

modulus of elasticity . The other end of the string is attached to a fixed point . The particle is held at rest at and then released. When the extension of the string is , the particle is moving with speed . i By considering energy show that

.

ii Hence find a the maximum extension of the string, b the maximum speed of the particle, c the maximum magnitude of the acceleration of the particle. © OCR, GCE Mathematics, Paper 4730, January 2009 18 A space vehicle of constant mass

tonnes is fired vertically upwards from sea level on the

Earth's surface. The weight of the vehicle varies but can be modelled approximately as



where represents its mass in kilograms, represents its distance in metres from the centre of the Earth, and is a constant. a Explain why

, where is the radius of the Earth in metres.

b Given that the radius of the Earth is approximately for

, find an approximate value

c Find an approximation for the work done against gravity by the rocket in propelling the space vehicle as it climbs from sea level to a height of above sea level. 19 A particle of mass

has the following displacement vector relative to an origin given

by: a Derive an expression for , the force acting on particle . b Find an expression for the magnitude of . c Use the scalar product

to find the kinetic energy of particle .

d Show that the power of force is zero for all values of .

7 Linear motion under variable force This chapter is for A Level students only. In this chapter you will learn how to: solve equations of motion of a particle when the velocity is given as a function of displacement solve equations of motion of a particle when the acceleration is a function of velocity or displacement use connected rates of change to solve linear motion problems set up and solve problems which can be modelled as linear motion of a particle acting under a variable force.

Before you start… A Level Mathematics Student Book 1

You should be able to calculate the force acting on a particle moving in a straight line.

1 A particle of mass is moving in a straight line along a smooth plane with constant acceleration . Calculate the force acting on the particle.

A Level Mathematics Student Book 2

You should know the relationship between acceleration , velocity and displacement for variable acceleration as a

2 A particle is moving along the -axis in the positive -direction. At time the particle is metres from the origin and its velocity is given by for . When , is from the origin.

function of time.

a Find the magnitude and direction of the acceleration when . b Find the displacement of when

A Level Mathematics Student Book 2

You should be able to separate variables to solve differential equations.

3 Solve the differential equation

Chapter 3

You should be able to use the impulse and conservation of linear momentum formulae for two particles colliding.

4 A particle of mass is moving with a constant speed of when it hits a stationary particle of mass . After impact both particles move in the same direction with speed .

.

a Find speed of each particle after impact. b Find the magnitude of the impulse given to the stationary particle.

Pure Core Student

You should be able to use an

Book 2

integrating factor to solve linear differential equations.

5 Find the particular solution, given that when , for

Why do we need to study variable force? In A Level Mathematics you extended your work on linear motion of particles to include variable acceleration, where acceleration, velocity and displacement all vary with time. This enabled you to solve problems using connections in the rates of change, as shown here.

I N T ∫ v dt E G R A ∫ a dt T E

displacement (x) velocity (v) acceleration (a)

dx dt dv dt

D I F F E R E N T I A T E

By Newton’s second law of motion you know that , and consequently, if acceleration varies with time, then force also varies with time. This enables you to work with force as a function of time in linear motion. Acceleration, velocity, displacement and force are not always defined as a function of time. For example, the force of gravity acting on a rocket as it leaves the Earth decreases with displacement, the force between two charged particles varies with their distance apart, and the force in a stretched string or compressed spring depends on the displacement. This means that you will need to revisit a mathematical model given for a particular situation and apply your knowledge of connected rates of change for problems that can be modelled as linear motion under variable force. Section 1 extends the work on variable acceleration that you have completed at A Level. Section 2 uses all the processes developed in Section 1 and brings in Newton’s second law.

Section 1: Working with acceleration, velocity and displacement When acceleration, , velocity, , and displacement, the following formulae to solve problems:

, are given as functions of time you can use

However, sometimes acceleration and velocity are functions of displacement, for example when using Hooke’s law. This means that these relationships cannot be applied in the same way.

Rewind You worked with Hooke’s law in Chapter 6.

Velocity as a function of displacement When velocity is given as a function of displacement, , you can find the associated acceleration function by differentiating with respect to . You can use the chain rule to derive a relationship for :

and given that

,

.

Rewind Related rates of change are dealt with in A Level Mathematics Student Book 2.

Key point 7.1 When velocity is given as a function of displacement, you can use the relationship:

to find an expression for acceleration.

WORKED EXAMPLE 7.1

A particle is travelling along the -axis in the positive -direction with velocity that measures the displacement from the origin, find:

. Given

a the velocity when b the magnitude and direction of the acceleration when c the displacement when a

b

Velocity is given in terms of displacement, so you can use the formula in Key point 7.1 to find . To use the formula you need to find

.

When Hence the magnitude of is Acceleration is in the negative direction. c

Given that velocity is a function of displacement

and that

, we can use these to find a

relationship between time and displacement, via velocity:

Separating variables gives Hence

.

.

Rewind Separation of variables is dealt with in A Level Mathematics Student Book 2.

Key point 7.2 If velocity is a function of displacement then:

WORKED EXAMPLE 7.2

A particle is moving along the -axis in the positive -direction. Initially the particle is from the origin. The velocity of is given by where is the displacement in metres from the origin. a Find the time at which is

from the origin.

b Find as a function of The velocity is given as a function of displacement and we wish to find a time so we can use the formula given in Key point 7.2.

a

Integrals of this form require the use of partial fractions.

Equating the numerators:

Substitute in the boundary condition to find the value of the constant. Since at , we will choose the modulus in the logarithm functions to be and ln . However, this does mean our solution is only valid for .

When Hence:

We can use the laws of logarithms to help simplify the right-hand side.

We can use inverse function for natural logarithm to rearrange to make the subject.

b

This solution is only valid for

. As

increases (remembering that we are talking about motion) the velocity tends to infinity as tends to

.

Tip When working with the logarithm functions as the result of an integral, we need to correctly identify the branch of the result we need. This then gives conditions on when the solution we have found is valid.

Acceleration as a function of displacement When acceleration is given as a function of displacement, 7.1,

, you can use the relationship from Key point

, to find velocity as a function of displacement.

Starting with

Hence

you can separate variables to give:

.

You can then rearrange to make the subject.

Key point 7.3 If acceleration is given as a function of displacement then:

WORKED EXAMPLE 7.3

WORKED EXAMPLE 7.3

The acceleration of a particle moving in a straight line is given by travelling along the -axis in the positive -direction and initially

and

. The particle is .

a Find an expression for as a function of . b Find an expression for as a function of The acceleration is given as a function of displacement and you want to find the velocity, so you can use the formula given in Key point 7.3.

a

In order to find an expression for as a function of you need to substitute in the boundary condition to find the value of .

In order to rearrange to make the subject you need to square root the function in . However, you can see it is a perfect square.

Only the positive solution is required because the particle is moving in the positive -direction. b

Recall that

.

In order to perform this integration, you need to separate variables.

In order to find an expression for as a function of you need to substitute in the boundary condition to find the value of .

Acceleration as a function of velocity When acceleration is given as a function of velocity,

, you can use

to find in terms of by

separating variables:

Key point 7.4 If acceleration is given as a function of velocity then:

WORKED EXAMPLE 7.4

A particle starts moving from travelling at

along the -axis in the positive -direction. At

. Given that its acceleration is

it is

with a direction towards the origin and

: a find the initial velocity of the particle b express its displacement as a function of time. The acceleration is given as a function of velocity and you wish to

find the initial velocity. So, you can relate the acceleration to time by using the formula given in key point 7.4.

a

You are told that the direction of acceleration is towards the origin so

.

Substitute in the boundary condition to evaluate .

Substitute in

and solve to find the initial velocity.

You want to find displacement as a function of time, so you can use

b

to relate the function you have found to displacement.

You can integrate with respect to to find an expression for .

To find , substitute the boundary conditions,

and

.

Substitute for and simplify to express displacement as a function of time.

An alternative approach when acceleration is given as a function of velocity is to use

from Key

point 7.1. After separating variables:

You can then find in terms of .

Key point 7.5 If acceleration is given as a function of velocity then:

WORKED EXAMPLE 7.5

A particle is travels along the -axis in the positive -direction after starting at the origin at rest.

Its acceleration is given by a Find the displacement from the -axis when is travelling at

.

b Find the velocity of as tends to infinity. The acceleration is given as a function of velocity and you wish to find the displacement. You can use the formula given in Key point 7.5.

a

Recognise that this integral is of the form

and hence

find by inspection. Substitute in the boundary conditions to find the unknown value . Hence Substitute

to calculate displacement from the origin.

Rearrange to make the subject.

b

Only the positive solution is required.

v –2x v = 2 1 – e

2

x

O As tends to infinity, the velocity tends to This is because as

. ,

Hence

EXERCISE 7A all involve a particle moving along the -axis in the positive -direction. Assume that all units are units. 1

Given a b

, find a general expression for in terms of when:

2

Given

, find a general expression for in terms of when:

a b 3

Given

find a general expression for in terms of when:

a b 4

Given

, find a general expression for in terms of when:

a b 5

Given

, find a general expression for in terms of when:

a b 6

Given

, what is when

7

Given

and when

8

Given

9

Given

what is when

, and when

, then

, and when

10 Given

and when

11 Given

and

, what is the velocity when , what is when find when

when

.

, find when

.

12 A particle is moving along the -axis in the positive -direction. The particle is initially at the origin and at time

the particle is

direction with

from the origin. The particle is accelerating in the positive -

.

a State, giving a reason for your answer, the value of at which the maximum speed of occurs. b Given that the maximum speed of is

, find an expression for in terms of .

13 A particle is moving along the -axis in the positive -direction. After through the origin and its velocity is given by

the particle passes

.

a Find the distance and direction of the particle from the origin when the particle is accelerating at . b Find the time taken for the particle to travel from the origin to

in the positive -direction.

14 A particle is moving along the -axis in the positive -direction. At at

. Acceleration is given by

, the particle is travelling

. Find the displacement of when it is travelling at

.

Tip Recall that expressions involving products can often be integrated using integration by parts. 15 A particle is travelling in a straight line away from a fixed point . At time the particle is from and travelling with velocity given by . After is from the fixed point . a State the magnitude and direction of the acceleration for when it is

away from .

b Express as a function of . c State a bound on the distance travels from , giving reasons for your answer.

Tip

Recall that integrals with a factorisable polynomial denominator require splitting into partial fractions in order to integrate. 16 A particle moves right and left along the -axis periodically with maximum displacement from the origin. Its velocity changes based on its displacement. Suggest a function that could model this movement and hence find general expressions for its acceleration with respect to and velocity with respect to time.

Explore Underground Mathematics has reproduced some examples of exam questions from the Cambridge Assessment Group Archives. Have a go at this question from 1966: www.cambridge.org/links/moscmec6003

Section 2: Variable force In Section 1 you worked with variable acceleration to solve kinematic problems. This was an extension to the work you completed in A Level Mathematics on variable acceleration. In this section, you will use the relationships found in Section 1 to solve problems involving forces using Newton’s second law. Since the mass of an object will be assumed constant in this section, the theory of variable acceleration extends naturally to an object moving under a variable force.

Force as a function of time For a particle of mass travelling in a horizontal line acting under a force, , you can use Newton’s second law and knowledge of connected rates of change to form a differential equation. and

Separating variables gives Hence

Fast forward Integrating a force over a time period gives rise to a change in momentum, which is equal to an impulse. You will study this in Chapter 8.

Key point 7.6 If force is given as a function of time, using Newton’s second law and

you can find the

velocity of the particle using:

WORKED EXAMPLE 7.6

A particle of mass starts at the origin and travels along the -axis in the positive -direction with initial speed . At time the particle is from the origin, and for a single force acts against the direction of motion magnitude . a Find the velocity of when

.

b Find the displacement of when a

You have the force given as a function of time, so you can use Key point 7.6, noting that the force is acting in the opposite direction.

Substitute in the boundary condition to evaluate .

You now have the equation for as a function of

Substitute

.

You want to find the displacement and your function from part a relates

b

velocity to time. Use

to relate displacement to time.

Integrate.

Substitute in the boundary condition to evaluate .

You now have the equation for as a function of Substitute

.

Force as a function of displacement Using

and

you get

.

Separating variables gives

.

Key point 7.7

To find velocity as a function of displacement, rearrange to make the subject.

Rewind This links the change in kinetic energy to the work done by or against a variable force, as seen in Chapter 6.

WORKED EXAMPLE 7.7

A particle of mass is initially at rest at the origin and then moves along the -axis in the positive -direction under the action of a single force. When is the force is given by . a Show that

.

b Calculate the maximum speed reached by a

Make sure units are the correct standard units. Since you are given the force as a function of displacement you can use Key point 7.7.

Integrate and substitute in the boundary condition to evaluate .

You now need to rearrange to get the format given in the

question. b

Maximum speed occurs when is maximised. This happens when .

Force as a function of velocity When force is given as a function of velocity, there are two alternative differential equations you can form, depending on how the acceleration is written. Using

,

Separating variables gives Hence Using

,

Separating variables gives

.

Hence The choice of form for acceleration depends on the demands of the problem being solved.

Key point 7.8 If force is given as a function of displacement, using Newton’s second law either can be used to find

or

or

: and

WORKED EXAMPLE 7.8

A particle of mass

is travelling along the -axis in the positive -direction. Initially the particle

is at rest at the origin, until acted upon by a single force in the positive -direction, where . a Find in terms of and calculate the velocity when b Calculate the displacement of when

.

. Given that you have force as a function of velocity and you require velocity in terms of time, you can use the first formula from Key point 7.8.

a

Recall that integrals of this form require partial fractions. When When

, hence , hence

When

Substitute in the boundary condition to evaluate .

Hence Rearrange to make the subject.

Given that you have force as a function of velocity and you require velocity in terms of displacement, you can use the second formula from Key point 7.8.

b

You should recognise that this integral is of the form and hence find by inspection the integral. When

Substitute in the boundary condition to evaluate .

Substitute in to calculate

Note: it is also possible to use your answer to part a and

to find . However, this would give , which is a far more involved function to

integrate.

Vertical motion If a particle is moving in a vertical straight line then its weight must also be taken into account when using Newton’s second law. WORKED EXAMPLE 7.9

A ball of mass is released from rest and falls vertically downwards. The ball experiences a single resistive force against its motion that is proportional to the speed of the object. a Find an expression of the velocity at time for the ball. b What is the long-term behaviour of the velocity of the ball?

c Find the distance the object falls after a

.

If the resistive force is proportional to the speed, you need a constant of proportionality to make this relationship into an equation. where is a constant of proportionality.

R = kvN

a

mg N Use Newton’s second law with the resistive force.

the weight of the particle and

You require velocity as a function of so you need to use

.

Manipulate the differential equation into a form that you can integrate.

Since , and are all constant, you can solve this differential equation by separating the variables, but notice that it is also in

The integrating factor is

the form You now have

. This type of differential equation can

be solved using an integrating factor. Once you have found the integrating factor you can then find a general solution for the differential equation.

where is a constant of integration. At

Use the initial conditions in the question to find the constant of integration.

so:

and hence

After some manipulation, you have found as a function of .

Therefore

where is a constant of proportionality. b As

, you have

.

c To find distance fallen as a function of time :

In the long term you need to look at what happens to .

as

To find the distance as a function of time you need to use the relationship of .

and integrate to derive a function of in terms

where is a constant of integration. At

so:

Once integrated, you again have to find the constant of integration using the initial conditions.

Finally, you can write the distance the ball has fallen as a function of .

Therefore

Rewind The integrating factor method for solving differential equations is dealt with in A Level Mathematics Student Book 2.

Explore In Worked example 7.9, you looked at a model of an object under a resistive force proportional to its speed. Explore a model of a falling object where the resistive forces are proportional to . You might like to experiment by dropping from a height various objects that have different crosssectional areas from their plan view. Can you find out the same information about the falling object as you did in Worked example 7.9?

WORKED EXAMPLE 7.10

A parachutist falls from a plane and moves in a vertical straight line towards the ground. The parachutist is acted upon by a single resistive force of magnitude

where is the mass,

is the speed and is the time since falling of the parachutist. Calculate the velocity of the parachutist after . Give your answer to significant figures. The resistive force is equal to a function involving mass, speed and time.

5mv R = N t + 1

a

mg N The weight of the particle and the resistive force act vertically. Use Newton’s second law to form the equation of motion. You require velocity as a function of time so need

.

You can manipulate the differential equation into a form that

you can integrate.

The integrating factor is:

You have formed a differential equation of the form . This type of differential equation requires an integrating factor to solve it.

You now have

Once you have found the integrating factor, you can then find a general solution for the differential equation.

where is a constant of integration. At

You can use the initial conditions in the question to find the constant of integration.

, so

Therefore

At

You can then substitute into the equation you have found to calculate the velocity to significant figures.

:

Power under a variable force Rewind Recall that power is measured in watts and defined as

. You used this in Chapter 1.

WORKED EXAMPLE 7.11

A car of mass

is travelling in a straight horizontal line. It has a power output of

resistance to motion magnitude . When the car passes point it is travelling at Calculate the speed of the car when it has travelled a further . Convert

and .

.

Recall the formula for power from Chapter 1. Use this to find an expression for the driving force . Hence Draw and label a diagram to show all forces acting on the car.

RN

a

120 F = N v

R = 0.15v2 N

mg

Use Newton’s second law to form the equation of motion. Use

to form a differential equation.

Recognise that this integral is of the form and hence find by inspection.

Substitute in the boundary conditions and evaluate .

When

When

EXERCISE 7B

:

Substitute

and solve to find .

EXERCISE 7B 1

A particle of mass

moves along the -axis in the positive -direction. A single horizontal force

acts on the particle away from the origin such that rest at the origin and that 2

, find velocity as a function of time.

A particle of mass moves horizontally from a fixed point on a smooth horizontal plane. At time , is metres and a single force acts on the particle in the direction of motion. Given that when the particle is as a function of velocity.

3

. Given that the particle starts at

A particle of mass

from the origin the speed of the particle is

, find its displacement

is travelling along the -axis in the positive -direction. At time

the

particle is metres away from the origin, from where it starts with velocity . The particle moves under a single force against the direction of motion. Find the speed and direction of the particle when 4

.

A particle of mass

is travelling in a horizontal line across a smooth horizontal plane away from

a fixed point . At time starts

the particle is

from with velocity

from with velocity

direction of motion, find the speed of when 5

A particle of mass single force and after

. Given the particle

and moves under the act of a single force

in the

.

moves along the -axis in the positive -direction under the motion of a

against the direction of motion. Given the particle starts at rest at the origin the particle has a displacement of from the origin with velocity , what

is the speed and direction of when 6

A particle of mass is travelling along the -axis in the negative -direction. It has velocity at a displacement of from the origin when it is acted upon by a single force in the positive -direction. Find the displacement and direction of from the origin when

7

A particle of mass

.

is travelling along the -axis in the positive -direction. Initially, the particle is

at the origin with speed direction such that at time

and moves under the act of a single force in the positive it is from the origin. Find the displacement of as a

function of velocity. 8

A particle of mass , the particle is direction of motion,

travels in a horizontal straight line away from a fixed point . At time from . The particle moves under the act of a single force in the . Given that the particle starts at with velocity

show that

. 9

A single force



acts on a particle of mass

in the direction of motion. At time

is from a fixed point after initially starting at with velocity velocity as a function of displacement. 10 A particle of mass

. Find

travels along the -axis in the positive -direction under the action of a force

in the direction of motion and with resistance to motion magnitude . The particle starts at the origin with velocity and after time the particle is from the origin. Find the velocity of as a function of time. 11 A stone of mass . After

is released from rest and falls freely under gravity with resistance to motion the stone has fallen

and has velocity

a Find the stone’s velocity as a function of time. b Find its displacement as a function of time. The stone falls

until it hits the ground.

c How long is the stone falling before it hits the ground? d What is the velocity of the stone when it hits the ground?

.

Tip If it is not possible to separate the variables, try the integrating factor method for solving differential equations that is given in Pure Core Student Book 2.

Checklist of learning and understanding For a particle moving in a straight line with variable acceleration: if velocity is a function of displacement, then if velocity is a function of displacement, then if acceleration is a function of displacement, then if acceleration is a function of velocity, then if acceleration is given as a function of velocity, then

.

For a particle moving in a straight line under a variable force: if force is given as a function of time, then from Newton’s second law you can find the velocity of the particle using

.

if force is given as a function of displacement, then from Newton’s second law you can find the velocity of the particle using

and rearranging to make the subject

if force is given as a function of velocity, then from Newton’s second law you can use either or Then

to find

or

.

and

if the linear motion of the particle is vertical, then its weight has to be taken into account when using Newton’s second law to form the equation of motion.

Mixed practice 7 1

A particle is moving in a straight, horizontal line such that at time it is away from a fixed point . Velocity is given by and the particle starts from the point . a Find an expression for the acceleration of in terms of displacement. b Hence calculate the acceleration of when it is

from .

c Find an expression for the displacement of in terms of time. d Hence calculate the displacement of from after 2

A particle is moving along the positive -axis in the positive -direction with acceleration given by . At time the particle is from the origin and when passed through the origin it had speed

3

.

. Show that

.

A particle is travelling in a horizontal straight line away from a fixed point . At time it is from , moving with velocity . Acceleration is given by and the particle started at the origin with velocity . Calculate the displacement and direction of acceleration of from the origin when it is instantaneously at rest.

4

A particle is travelling along the -axis in the positive -direction. At time particle is origin at time

from the origin with velocity . Given that

the

. The particle passes through the find:

a an expression for time in terms of displacement b an expression for displacement in terms of time and hence the displacement of when

c an expression for the acceleration of d the magnitude and direction of the acceleration when 5

A particle of mass is projected horizontally with speed from a fixed point on a smooth horizontal surface. moves in a straight line and, at time after projection, has speed and is from . The only force acting on has magnitude and is directed towards . i

Show that

ii

Hence show that

iii Find , given that

. . when

. © OCR, GCE Mathematics, Paper 4730/01, June 2007

6

is a fixed point on a horizontal plane. A particle of mass is released from rest at and moves in a straight line on the plane. At time after release the only horizontal force acting on has magnitude:

The force acts in the direction of

motion.

velocity at time is

i

Find an expression for in terms of , valid for greater when than it is when .

ii

Sketch the

graph for

.

, and hence show that is three times

. © OCR, GCE Mathematics, Paper 4730, June 2010

7

A duck of mass is travelling with horizontal speed when it lands on a lake. The duck is brought to rest by the action of resistive forces, acting in the direction opposite to the duck’s motion and having total magnitude

, where

is the speed of the duck.

Show that the duck comes to rest after travelling approximately initial contact with the surface of the lake.

from the point of its

© OCR, GCE Mathematics, Paper 4730, June 2006 8

A stone of mass

falls freely under gravity, from rest, until it has travelled a distance of

. The stone then continues to fall in a medium which exerts an upward resisting force of , where is the speed of the stone after the instant that it enters the resisting medium. i

Show by integration that

.

ii

Find how far the stone travels during the first

in the medium.

© OCR, GCE Mathematics, Paper 4730, January 2009 9

A motor-cycle, whose mass including the rider is , is decelerating on a horizontal straight road. The motor-cycle passes a point with speed and when it has travelled a distance of beyond its speed is . The engine develops a constant power of and resistances are modelled by a force of

opposing the motion.

i

Show that

.

ii

Find the speed of the motor-cycle when it has travelled

beyond .

© OCR, GCE Mathematics, Paper 4730, June 2009 10 A particle of mass is released from rest at a fixed point and falls vertically. The particle is subject to an upward resisting force of magnitude of where is the velocity of the particle when it has fallen a distance of i

from .

Write down a differential equation for the motion of the particle, and show that the equation can be written as

ii

.

Hence find an expression for in terms of . © OCR, GCE Mathematics, Paper 4730/01, January 2008

8 Momentum and collisions 2 This chapter is for A level students only. In this chapter you will learn how to: find the impulse of a variable force apply the principles of impulse, conservation of momentum and Newton’s experimental law in two dimensions using vector notation calculate the result of oblique impacts.

Before you start… Chapter 3

You should know the relationship between force, impulse and momentum.

1 A force of acts on a mass for . What is the change in momentum of the mass during those ?

Chapter 3

You should know the principle of

2 A particle of mass

conservation of momentum and Newton’s experimental law.

travelling in a straight

line on a horizontal surface hits a vertical wall with speed and rebounds with speed . What is the coefficient of restitution between the wall and the particle?

A Level Mathematics Student Book 1

You should understand how to combine vectors, including drawing triangles to find the sum of two vectors.

3 Draw a sketch of a triangle to show the sum of two vectors and .

A Level Mathematics Student Book 2

You should be able to resolve a vector into two perpendicular components and be able to find the magnitude of a vector from its components.

4 The force acts at to the horizontal. What are the horizontal and vertical components of ?

A Level Mathematics Student Book 1

You should be able to integrate simple functions and apply limits.

5 Find

.

Variable forces and oblique impacts In Chapter 3 you learned about momentum and impulse involving constant forces and direct collisions. Collisions are not restricted to situations where two objects are moving in the same straight line; collisions can also occur when the objects are moving at an angle to each other. These are called oblique collisions. You see this in tennis matches when the racket head does not meet the tennis ball straight on but sends it

in a completely different direction. In this chapter, you will analyse more realistic situations such as those involving variable forces or oblique impacts.

Section 1: Variable force and vector notation For a constant force acting for time , the impulse of the force is change in momentum.

and this is equal to the

Key point 8.1 For a constant force, ,

Rewind You learned about impulse and momentum in Chapter 3. A graph of force against time for a constant force will be a horizontal line and the impulse is given by the area between the force line and the time axis. force

F

I = Ft

time

t

Not all forces are constant. The more you stretch an elastic band, the greater the force trying to pull it back. The force increases with the extension of the band. If the force is not constant, you may need to find the impulse using integration.

Key point 8.2 The impulse, , of a variable force acting for a time

is:

WORKED EXAMPLE 8.1

The force–time graph for the force acting on a mass of

is shown.

The mass, , is initially moving in a straight line with velocity the same straight line in the direction of motion. force, F (N) 6

O

4

10

time, t (s)

Calculate the speed of after: a

and the force is acting along

b a Let be the speed of after

First define the unknown quantity.

. Use The impulse in the first between and

Total impulse in the first

is the area under the graph

Use change in momentum final momentum initial momentum Use Therefore You could use this result to help you to calculate the answer for part but it is safer to start again in case you have made an error. b Let be the speed of after .

Define the unknown. The impulse in the first is the area under the graph between and , which is a trapezium of height and sides and .

Total impulse in the first is:

Use

and solve for .

Therefore WORKED EXAMPLE 8.2

A smooth sphere of mass motion of the sphere.

is acted on by a variable force

a Find the magnitude of the impulse between b When

the speed of the sphere is

and

in the direction of .

. What is the speed of the sphere after

?

To find the change in momentum you need to integrate from to

a Impulse

Integrate between

and

Substitute in the limits and calculate .

b Let the speed of the sphere after .

be

Define the unknown quantity. Write an expression for the change in momentum. Equate to the impulse and work out .

Therefore

WORKED EXAMPLE 8.3

A particle, , of mass

, is acted on by a variable force

, which is defined as:

Find the speed of after

if the speed is

when

.

To find the increase in momentum of you need to know the total impulse on during the . Change to Let the speed of after

Define the unknown.

be

Integrate the force between total impulse on in that time.

and

to find the

Use Solve for .

Vector notation The same principles and equations that you learned in Chapter 3 can be applied when velocity and impulse are given in vector format.

Explore Rockets are an example of a momentum–impulse problem where there is a constantly changing situation because fuel is expelled from the back of the rocket, so the mass of the rocket decreases. Look up the Tsiolkovsky rocket equation on the internet.

WORKED EXAMPLE 8.4

A toy sailing boat of mass is blown along by a constant wind, which produces a force on the boat. If the boat is initially at rest, find its velocity after .

F = i + 2j 0.8kg

u

t = 0

Draw a diagram with arrows to show the direction of motion.

0.8kg

v

t = 10 Convert to to

Let final velocity be

.

Write an expression for the final velocity vector.

The formula for impulse can be written with as vectors.

and

Substitute in the values. Equate the values of both sides of the equation and the values of both sides of the equation to form two scalar equations. You should not write and in these equations. Solve to find the components of . State the value of in vector format. You may find it easier to work with column vectors.

Alternative method:

Convert to

.

Write an expression for the final velocity vector.

Let final velocity

The formula for impulse can be written with as vectors.

and

Substitute in the values. Equate the values (top line) and the values (bottom line) of both sides of the equation. You should not write and in these equations. Solve to find the components of . State the value of the velocity vector in the format of the question. WORKED EXAMPLE 8.5

A particle , of mass mass

, is moving with velocity

, which is moving with velocity

when it collides with particle , of . If the particles coalesce during the collision:

a find the velocity of the combined particle after the collision b find the total loss in kinetic energy as a result of the collision. a Let the final velocity of the combined particle be

Use the principle of conservation of linear momentum in vector format.

Substitute in the values. It is easier to work with column vectors.

Equate the components to find .

Equate the components to find .

Give the answer in the original format. b

You need to be able to find , which is a scalar

quantity. Find the square of the magnitude of the velocity of (i.e. the square of the speed of ) … … and the square of the magnitude of the velocity of (i.e. the square of the speed of ). The initial

is the sum

Find the square of the magnitude of the velocity of the combined particle …

… and use this to find the final combined particle. Loss in

initial

final

of the .

EXERCISE 8A 1

Find the impulse generated by the forces shown by the solid lines between a force(N)

and

.

6

O

5

10

time(s)

b force(N) 6

O

10

time(s)

c force(N)

4

O

3

6

10

3

5 6

10

time(s)

d force(N) 4

O

time(s)

–3

2

A particle of mass

is moving in a straight line on a smooth horizontal plane, when it is acted on by

a force in the same straight line. If the speed of the particle is when , find the speed of the particle after for each of the forces shown in the diagrams in Question 1. 3

Fill in the gaps in this table. Mass

Initial velocity

Final velocity

Constant force

Time

4

A particle , of mass , is moving with velocity when it collides with particle , of mass , which is moving with velocity . If the particles coalesce during the collision, find the velocity of the combined particle after the collision.

5

A particle , of mass , is moving with velocity when it collides with particle , of mass , which is moving with velocity . If the velocity of after the collision is , find the velocity of after the collision.

6

A particle , of mass , moving in a straight line on a smooth horizontal plane, is acted on in the same straight line by a force for , such that . a Find the total impulse of the force on between b If the speed of is

7

when

and

.

, find the speed of when

A particle , of mass , is moving with velocity mass , which is moving with velocity

.

when it collides with particle , also of . The two particles coalesce.

a Find the velocity of the combined particle after the collision. b What is the loss in kinetic energy as a result of the collision? 8

A mass, , of , moving in a straight line on a smooth horizontal plane at force along its line of motion. is defined as:

, is acted on by a

a Find the magnitude of the total impulse on : and i between ii between

and

b Hence find the speed of when 9

At time a particle , of mass , is moving in a straight line at a constant speed of on a smooth horizontal plane when it is acted on by a force in the same straight line. a Find the total impulse on : i between

and

ii between

and

b What is the speed of when 10 At time , a particle , of mass , is moving in a straight line at a constant velocity of on a smooth horizontal plane when it is acted on by a force acting in the same straight line. At time

later, the particle is moving with velocity

. Find the value of .

Section 2: Oblique impacts and the impulse–momentum triangle So far, you have considered direct impacts with objects moving along their line of centres or hitting walls at right angles. You are now going to look at oblique impacts. Consider a football, of mass , kicked along the ground so that it hits a wall at an angle of magnitude . If the wall is smooth and has a coefficient of restitution equal to the ball will bounce back at the same angle at which it hit the wall, as shown in the diagram.

θ

u

θ

v

I

Rewind You looked at coefficient of restitution in Chapter 3. The impulse that the smooth wall exerts on the ball will be perpendicular to the wall. The vector diagram shows what happens when . The component of the velocity parallel to the wall, , is unchanged by the impact. The velocity perpendicular to the wall, , has the same magnitude after the collision, but is in the opposite direction, . The velocity of approach and the velocity of rebound therefore have the same magnitude of , but are in different directions. before impact a θ

b

after impact a θ

u

v

- b e = 1

If the coefficient of restitution is not equal to but the wall is still smooth: the impulse is again perpendicular to the wall so has no effect on the component of the velocity parallel to the wall the impulse changes the magnitude of the component of the velocity perpendicular to the wall. before impact a θ

b u

after impact a φ v

–eb e < 1

The velocity, , perpendicular to the wall is changed by the impact with the wall to The magnitude of the angle that the direction of motion of the football makes with the perpendicular to the wall will change.

Rewind You learned about collisions at right angles to walls in Chapter 3.

Before the impact the football is travelling towards the wall with speed perpendicular with the wall where

at an angle to the

.

After the collision the football is travelling away from the wall with speed the perpendicular with the wall where

at an angle to

.

If we compare these two we can see that

.

The impulse acting on the football is perpendicular to the wall and is equal to the change in momentum in that direction. In this case, taking the direction away from the wall as positive:

We can apply this principle to all impacts of spheres with smooth surfaces.

Key point 8.3 When an object moving at velocity collides at an angle with a smooth, flat surface and rebounds: The impulse acts at right angles to the surface and is equal to the change in momentum in that direction. The component of the velocity parallel to the surface remains unchanged. The component of the velocity perpendicular to the surface is multiplied by coefficient of restitution between the object and the surface.

, where is the

before impact after impact u cos θ u cos θ θ

u sin θ

θ

- eu sin θ

v

u

WORKED EXAMPLE 8.6

An exercise ball, , of mass , is moving in a straight line at on a smooth horizontal surface when it collides with a smooth wall. The line of motion of the ball makes an angle of with the wall. If the coefficient of restitution between the wall and the ball is , find the speed and direction of the ball when it rebounds from the wall. Give your speed correct to decimal places and the angle correct to decimal place.

before impact a 40° b

after impact a φ v

5m s- 1

Draw a clear diagram showing the component of velocity parallel to the wall unchanged and the component of velocity perpendicular to the wall changed to . You can write the components in as and or and

- eb

Resolve out .

to find the values of and and work

Use Pythagoras’ theorem to find .

Use trigonometry to find the angle.

WORKED EXAMPLE 8.7

A smooth vertical wall is parallel to the direction. A smooth sphere of mass moving with velocity on a smooth horizontal plane collides with the wall and rebounds with velocity . If the coefficient of restitution between the wall and the sphere is , find the values of and As the velocity is given as a vector, you know the components of the velocity so there is no need to resolve. There is no impulse parallel to the wall so there is no change of speed parallel to the wall. Perpendicular to the wall, use the relationship Note that here you include the minus sign in since you are modelling the motion in the positive direction. A negative answer tells you the sphere is now moving in the negative direction. You were asked for the values of and so you do not need to write the answer as a vector.

Finding the impulse In Worked example 8.6, the impulse on the

a

ball was acting perpendicular to the wall, as in the diagram.

a φ

40° b

v

5m s- 1 I

- eb

e = 0.6

Taking the direction away from the wall as the positive direction, the perpendicular component of impulse is the change in momentum perpendicular to the wall, which equals

There is no component of impulse parallel to the wall. An alternative approach is to use vectors to find from an impulse–momentum vector triangle. The vector triangle needs to show the final momentum as the resultant of and . In vectors or or as shown in the diagram. The easiest way to get the sides in the correct order is to draw the initial and final momentum from the same point.

I mu mv

The double arrows on the expect.

side of the triangle indicate that it is the resultant of

In Worked example 8.6, the mass of the ball was and . Therefore look like this:

and

and

was

and , as you would

and you found that

. Using these values, the triangle would

40° 10 kg m s- 1

I

26.7° 8.58 kg m s- 1

and

.

Tip In a problem involving a simple impact with a wall, when the impulse acts at right angles to the wall, then the method of splitting into components works best but in some problems drawing the impulse–momentum triangle can save pages of calculation. These are usually problems with a single moving object being hit so that it is deflected through a given angle, or collisions with a rough wall when you are specifically told that the impulse is not perpendicular to the wall.

WORKED EXAMPLE 8.8

A small smooth sphere of mass is moving in a horizontal plane at to a smooth vertical wall when it collides with the wall. The speed of the sphere immediately before the collision is . The coefficient of restitution between the sphere and the wall is

60° 10m s- 1

θ I

v

a Find the magnitude and direction of the sphere’s velocity immediately after the collision. b Find the impulse of the wall on the sphere. This is an oblique collision between a sphere moving horizontally and a vertical wall so can you solve it by resolving parallel and perpendicular to the wall. Draw clear diagrams with arrows showing the direction and values of the velocities before and after the collision, split

10 cos 60

a

60°

v cosθ

into the components of their speeds parallel and perpendicular to the wall.

θ

10m s- 1

10 sin 60 I initial

vm s- 1

v sin θ final

Let the speed of the sphere after the collision be at angle to the wall.

Define any unknown values.

The component of the initial velocity parallel to the wall remains unchanged in the collision, as there is no impulse in that direction. Find the component of the velocity perpendicular to the wall using Calculate the value of from its components. Find

b

and hence .

Convert to

.

Use impulse change in momentum

.

Decide on a direction to be positive – it is usually best to take away from the wall as positive – and substitute in the values.

WORKED EXAMPLE 8.9

A cricket ball of mass is travelling horizontally at speed of when it is hit by a cricket bat. It leaves the bat, horizontally, at a speed of travelling at to the line of its original path as shown in the diagram. What is the magnitude of the impulse of the bat on the ball?

30m s–1

40m s–1 40°

bat

Convert to Calculate

. and

.

Draw a momentum–impulse triangle, remembering to keep the directions of the two velocities and to draw and coming away from the same point. You need to find the obtuse angle in the triangle.

mv 40° 140°

I

mu

Use the cosine rule to find .

WORKED EXAMPLE 8.10

A sphere of mass is moving with velocity in the positive direction when it is hit with a blow of impulse in the positive direction. What is the speed and direction of the sphere immediately after the impulse? j

Draw the impulse–momentum triangle, remembering to multiply the velocity by the mass to get the momentum. You are given the magnitude and direction of the impulse and the magnitude and direction of Remember to show and coming away from the same point.

i mv 15

θ mu = 20

Use Pythagoras’ theorem to find

.

Divide by to find the speed.

Use trigonometry to find the angle. Immediately after the

State the final speed and direction of the sphere.

collision the sphere moves with speed at to the positive direction.

Focus on … You will use an impulse–momentum triangle to solve a problem involving conservation of energy in Focus on … Problem solving 2.

EXERCISE 8B 1

A smooth vertical wall is parallel to the direction. A sphere of mass

moving with velocity

on a smooth horizontal plane collides with the wall and rebounds with velocity .

a If the coefficient of restitution between the wall and the sphere is

, find the values of and .

b What is the magnitude of the velocity of the sphere when it rebounds from the wall? 2

A smooth vertical wall is parallel to the direction. A sphere of mass

moving with velocity

on a smooth horizontal plane collides with the wall and rebounds with velocity . a If the coefficient of restitution between the wall and the sphere is

, find the values of and .

b What is the loss of kinetic energy of the sphere as a result of its collision with the wall? 3

A particle, , of mass is moving in a straight line at on a smooth horizontal surface when it collides with a smooth vertical wall. The line of motion of the particle makes an angle of with the wall. a What is the component of the speed of the particle parallel to the wall? b What is the component of the speed of the particle perpendicular to the wall before the collision? c If the coefficient of restitution between the wall and the particle is speed perpendicular to the wall immediately after the collision.

, find the component of the

d Find the speed and direction of the particle when it rebounds from the wall. Give your speed correct to 4

places and the angle correct to

A sphere of mass

place.

is moving in a straight line at

on a smooth horizontal surface when it

collides with a smooth vertical wall. The line of motion of the sphere makes an angle of with the wall. If the coefficient of restitution between the wall and the sphere is , find the speed and direction of the sphere when it rebounds from the wall. Give your speed correct to the angle correct to place. 5

A particle, of mass

, is moving in a straight line at

places and

on a smooth horizontal surface when it

collides with a smooth vertical wall. The line of motion of the sphere makes an angle of with the wall. a What is the component of the velocity of the particle parallel to the wall, immediately before the collision? b What is the component of the velocity of the particle parallel to the wall, immediately after the collision? c What is the component of the velocity of the particle perpendicular to the wall, immediately before the collision? d If the coefficient of restitution between the wall and the sphere is , what is the component of the velocity of the particle perpendicular to the wall, immediately after the collision? e Find the magnitude of the velocity of the particle immediately after the collision. f Show that the loss in kinetic energy as a result of the collision is 6

.

A small smooth ball of mass is moving in a straight line at on a smooth horizontal surface when it collides with a smooth vertical wall. The line of motion of the ball makes an angle of with the wall. If the coefficient of restitution between the wall and the sphere is a find the speed and direction of the ball when it rebounds from the wall b find the impulse of the wall on the ball.

7

A sphere of mass is moving in the positive direction with speed when it receives a blow of impulse acting in the negative direction. Find the magnitude and direction of the velocity of the sphere immediately after the impact.

8

A sphere of mass is moving in the positive direction with speed when it receives a blow of impulse acting in the direction. Find the change in kinetic energy of the sphere as a result of the impulse.

9

A football of mass

is travelling horizontally at speed of

. The football is then kicked, and it

immediately moves horizontally, at a speed of , travelling at to the line of its original path, as shown in the diagram. What is the magnitude of the impulse of the kick?

20m s–1

30°

25m s–1

10 A snooker ball of mass angle of

hits a vertical side cushion of a snooker table at a speed of

at an

to the cushion, as shown in the diagram. The impulse of the cushion on the ball is

perpendicular to the side of the table and the coefficient of restitution between the ball and the side cushion is . a At what speed and angle to the side of the table does the ball leave the side cushion of the table? b What is the kinetic energy lost by the snooker ball as a result of the collision? side cushion 35° 5m s–1

11 A snooker ball of mass hits a vertical side cushion of a snooker table at a speed of angle of to the cushion as shown in the diagram. The impulse of the cushion on the ball is

at an

perpendicular to the side of the table and the coefficient of restitution between the ball and the side cushion is . The ball then hits the side cushion that is at right angles to the original cushion. The coefficient of restitution between the ball and this side cushion is also . At what speed and angle to the side of the table does the ball leave the second side cushion? side cushion θ side cushion

u

Section 3: Oblique collisions of two spheres and impulsive tensions in strings You are now going to consider the oblique impact of two moving spheres. When two smooth spheres collide, the force of the collision will act along the line joining the centre of the spheres – at right angles to the tangent at the point where they meet, as shown in the diagram. A

B

2 kg I

3 kg

I

line of centres

Did you know? In a standard game of snooker all the balls are the same size and weight, but in coin-operated pool and barbilliards machines the white cue ball is a different weight and size from the rest of the colours. This is so that the machine can detect if it goes down a pocket and can return it.

Key point 8.4 When two smooth spheres collide, the impulse is along the line of centres of the spheres. You can apply the principle of conservation of momentum and Newton’s experimental law to the components of velocity along the line of centres. There is no change in the components of velocity perpendicular to the line of centres as there is no impulse in that direction.

WORKED EXAMPLE 8.11

A sphere, of mass , is moving on a smooth horizontal surface with velocity when it collides with a second sphere, , of equal size and mass , moving on the same surface with velocity . The line of centres of the two spheres at the moment of collision is parallel to the direction and the coefficient of restitution between the two spheres is . Find the velocities of and immediately after the collision. Let the velocity of after the collision be and let the velocity of after the collision be i 5

Draw a clear diagram.

1 4

2 kg

I

A

3 I

3 kg

line of centres

B

b

d a

Define any unknowns. Put both and in component form so you have four unknowns.

c

Deal with the components of the velocity perpendicular to the line of centres first. These components remain unchanged. Apply the principle of conservation of linear momentum to the components along the line of centres. State the equation and substitute the values you know. Apply Newton’s experimental law. State the equation and substitute the values you know.

Solve

and

simultaneously to find and

Give the velocities of and in vector format.

Tip If you are not given the velocities in component form, then you have to resolve the velocities parallel and perpendicular to the line of centres before you start the calculation.

Rewind You learned how to resolve a force into components in A Level Mathematics Student Book 2.

WORKED EXAMPLE 8.12

Two smooth spheres, and , of masses collision, is moving at at

and

, respectively, collide. Immediately before the

at an angle above the line of centres, where

towards , at an angle below the line of centres, where

. is moving

, as shown in the

diagram. The coefficient of restitution between the two spheres is a What are the magnitudes and directions of the velocities of the two spheres immediately after the collision? b What is the magnitude of the impulse on ?

θ

A

B

4 kg

1 kg φ

4m s–1

a

line of centres

3m s–1

Calculate the components of the initial velocities for and

Put the components of the initial velocities above the relevant sphere and the final components below, so you don’t get confused. The components of the velocities perpendicular to the line of centres remain unchanged and can be put on the diagram as final velocity components immediately.

1.8

2.4 3.2

θ

2.4 A

B

4 kg

1 kg

4

a

φ

line of centres

1.8 b

3.2 positive direction

Apply the principle of conservation and momentum along the line of centres. Apply Newton’s experimental law along the line of centres.

Solve the simultaneous equations. The components of the velocity of are and so

at an angle of the line of centres

to

The components of the velocity of are

and

Find the velocity of using Pythagoras’ theorem and the direction using trigonometry.

so

Find the velocity of using Pythagoras’ theorem and the direction using trigonometry.



at an angle of the line of centres

to

b

Use impulse = change in momentum. The impulse on the spheres is along the line of centres and the impulse on is equal and opposite to the impulse on . If one is easier to work out than the other you choose that one. In this case, there is nothing to choose between them. Substitute in the numbers and state the magnitude of the impulse. Be careful with the signs.

For :

along the line of centers

Impulsive tensions in strings In the diagram, the particles and , of masses and respectively, are joined by a light, inextensible string that is straight but not under tension. Particle is initially at rest and particle is projected at an angle of to the string with speed

before string is taut

45°

B 1.5 kg 0 B

4√2 m s- 1 A 2.5 kg A

v1

after string is taut

v1

v2

The diagram shows the velocities of and at the instants before and after the string becomes taut.

Rewind In Chapter 3 you saw that when two particles are attached to the ends of a light elastic string and one is set in motion, an impulsive force was generated in the string when it becomes taut. When the string becomes taut, there will be an impulse on both and generated by the tension in the string and the velocities of and will change. The new velocity of along the string will be the same as the component of the new velocity of along the string and in the same direction. The component of the velocity of perpendicular to the string will be unchanged as the impulsive tension in the string acts along the string. You can find and by using the principle of conservation of momentum. Parallel to

:

Perpendicular to

:

starts to move along the length of the string with speed at an angle Impulsive

₁

and then starts to move with speed

to the line of the string. . This will be the same for both and .

Key point 8.5 For particles connected by a light inextensible string, you need to split the velocities into components parallel and perpendicular to the string. You then apply the principle of conservation of momentum parallel and perpendicular to the string.

WORKED EXAMPLE 8.13

A light, inextensible string has particles and , each of mass , attached to either end. The string and particles rest on a smooth table. The particle at is hit with an impulse of at an angle of to the string, which is straight but not taut and at rest. What is the velocity of after the impulse at the instant the string becomes taut? Convert to

.

Calculate the initial speed of using impulse = change in momentum. Draw a clear diagram showing the velocity of immediately before the string starts to move and the components of the velocities of and , parallel and perpendicular to the

B

before v1

after

string, immediately afterwards.

20m s–1 30° A

v1 v2

Parallel to the line of the string:

The velocity of after the impulse is along

.

EXERCISE 8C 1

A smooth sphere, , of mass , is moving on a smooth horizontal surface with velocity when it collides with a second smooth sphere, , of mass , moving on the same surface with velocity . The line of centres of the two spheres at the moment of collision is parallel to the direction and the coefficient of restitution between the two spheres is and immediately after the collision.

2

. Find the velocities of

A smooth sphere, , of mass , is moving on a smooth horizontal surface with velocity when it collides with a second smooth sphere, , of mass , moving on the same surface with velocity . The line of centres of the two spheres at the moment of collision is parallel to the direction and the coefficient of restitution between the two spheres is . Find the velocities of and immediately after the collision.

j

3j i

3i 5kg

2kg

P

Q

3

A smooth sphere, , of mass , is moving on a smooth horizontal surface with velocity when it collides with a second smooth sphere, , of mass , which is at rest. The line of centres of the two spheres, at the moment of collision, is parallel to the direction and the coefficient of restitution between the two spheres is . Find the magnitude and direction of the velocities of and immediately after the collision.

4

Two smooth spheres, and , of masses and Immediately before the collision, is moving at

, respectively, collide as shown in the diagram. at an angle above the line of centres, where

. is initially at rest. The coefficient of restitution between the two spheres is

. Find the

magnitude and direction of the velocities of and immediately after the collision. 5m s–1 θ A

5

B

A smooth sphere, , of mass , is moving at on a smooth horizontal surface when it collides with a second smooth sphere, , of mass , moving on the same surface with velocity . At the moment of collision, is moving at to the line of centres and is moving at to the line of centres towards , as shown in the diagram. If the coefficient of restitution between the two spheres is

, find the magnitude and direction of the velocities of and immediately after the collision.

6m s–1

3m s–1 40°

30° A

6

B

A smooth plastic ball, , of mass , is moving at on a smooth horizontal surface when it collides with a smooth metal ball, , of mass , moving on the same surface with velocity . At the moment of collision, is moving at to the line of centres and is moving at to the line of centres towards , as shown in the diagram. If the coefficient of restitution between the two spheres is , find the magnitude and direction of the velocities of and immediately after the collision.

30°

30° 6m s

–1

P

7

10m s–1

Q

A particle , of mass , is attached by a straight, light inextensible string to a particle of mass is initially at rest and is projected with speed so that, when the string is about to move, is moving at to the string, as in the diagram. Find: a the speed of immediately after the string becomes taut b the magnitude of the impulsive tension in the string.

5m s–1 45°

B 8

A

A particle , of mass , is attached by a light inextensible string to a stationary particle of mass . The string is initially straight but not under tension. is hit with a blow of impulse to the string and away from , as in the diagram Find: a the initial speed of as a result of the impulse b the speed of immediately after the string becomes taut.

5m s–1 45°

B 9

A

A particle , of mass , is attached by a light inextensible string to a stationary particle of mass . is projected with speed so that, when the string is about to move into tension, is moving at

to the string, as in the diagram. Find:

a the magnitude and direction of the speed immediately after the string becomes taut b the magnitude of the impulsive tension in the string c the loss in kinetic energy in the system as a result.

5m s–1 B

45°

A

Checklist of learning and understanding The impulse, , of a constant force acting for a time is

.

The impulse, , of a variable force acting for a time

is

.

The impulse of a force acting on a body is equal to the change in momentum. When an object moving at speed collides at an angle with a smooth, flat surface and rebounds: the impulse acts at right angles to the surface and is equal to the change in momentum in that direction the component of the velocity parallel to the surface remains unchanged the component of the velocity perpendicular to the surface is multiplied by the coefficient of restitution between the object and the surface.

, where is

For particles connected by a light inextensible string, you need to split the velocities into components parallel and perpendicular to the string. You then apply the principle of conservation of momentum in directions parallel and perpendicular to the string, for the situations before and after the string becomes taut.

Mixed practice 8 1

A particle of mass is moving in a straight line with speed when it is deflected through an angle by an impulse of magnitude . The impulse acts at right angles to the initial direction of motion of (see diagram). The speed of immediately after the impulse acts is . Find the value of and the value of

8m s–1 6m s–1

P

θ

I Ns

2

A particle of mass is moving horizontally with speed when it receives an impulse of magnitude , in a direction which makes an angle with the direction of motion of . Immediately after the impulse acts moves horizontally with speed . The direction of motion of is turned through an angle of by the impulse (see diagram). Find and .

I N s 3m s–1 60°

θ° P

4m s–1

© OCR, GCE Mathematics, Paper 4730, January 2010 3

A particle of mass is moving in a straight line with speed . An impulse of magnitude applied to deflects its direction of motion through an angle of , and reduces its speed to (see diagram). By considering an impulse–momentum triangle, or otherwise, i

show that

,

ii

find the angle that the impulse makes with the original direction of motion of .

2.5m s–1 6.3m s–1

P

θ 2.6N s © OCR, GCE Mathematics, Paper 4730, January 2009

4

A small ball of mass is moving with speed when it receives an impulse of magnitude . The speed of the ball immediately afterwards is . The angle between the directions of motion before and after the impulse acts is . Using an impulse–momentum

triangle or otherwise, find . 5

A ball of mass is moving with speed in a straight line when it is struck by a bat. The impulse exerted by the bat has magnitude and the ball is deflected through an angle of

(see diagram). Find

i

the direction of the impulse,

ii

the speed of the ball immediately after it is struck.

22m s–1 15N s

© OCR, GCE Mathematics, Paper 4730, January 2011 6

A particle of mass is moving on a smooth horizontal surface with speed , when it is struck by a horizontal blow in a direction perpendicular to its direction of motion. The magnitude of the impulse of the blow is . The speed of after the blow is . i

Find the value of .

Immediately before the blow is moving parallel to a smooth vertical wall. After the blow hits the wall and rebounds from the wall with speed ii

.

Find the coefficient of restitution between and the wall. © OCR, GCE Mathematics, Paper 4730, January 2012

7

Two uniform smooth spheres and , of equal radius, have masses and respectively. They are moving on a horizontal surface when they collide. Immediately before the collision the velocity of has components along the line of centres towards and perpendicular to the line of centres. is moving with speed along the line of centres towards (see diagram). The coefficient of restitution between the spheres is . i

Find, in terms of , the component of the velocity of along the line of centres immediately after the collision.

ii

Given that the speeds of and are the same immediately after the collision, and that , find .

4m s–1

A

B

6kg

3kg

8m s–1

vm s–1

© OCR, GCE Mathematics, Paper 4730/01, January 2008 8

Two uniform smooth spheres and

of equal radius, have masses

and

, respectively,

They are moving on a horizontal surface, and they collide. Immediately before the collision, is moving with speed at an angle to the line of centres, where and is moving along the line of centres with speed (see diagram). The coefficient of restitution between the spheres is Find the speed and direction of motion of each sphere after the collision. A

B

5kg

3kg

15m s–1

α 10m s–1

9

Two uniform smooth spheres and of equal radius are moving on a horizontal surface when they collide. has mass and has mass . Immediately before the collision is moving with speed along the line of centres, and is moving with speed at an angle to the line of centres, where (see diagram). Immediately after the collision is stationary. Find i

the coefficient of restitution between and ,

ii

the angle turned through by the direction of motion of as a result of the collision.

A 2.8m s–1

B θ

0.1kg

0.4kg 1m s–1

© OCR, GCE Mathematics, Paper 4730/01, June 2014 10 Two uniform smooth spheres and of equal radius are moving on a horizontal surface when they collide. has mass and has mass . Immediately before the collision is moving with speed along the line of centres, and is moving away from with speed at an acute angle to the line of centres, where (see diagram). The coefficient of restitution between the spheres is

Find

i

the velocity of immediately after the collision,

ii

the angle turned through by the direction of motion of as a result of the collision. A 3m s–1

0.1kg

B θ

0.2kg

1m s–1

© OCR, GCE Mathematics, Paper 4730/01, June 2013 11 Two uniform smooth spheres and , of equal radius and equal mass, are moving towards each other on a horizontal surface. Immediately before they collide, has speed along the line of centres and has speed at an angle of to the line of centres (see diagram). After the collision, the direction of motion of is at right angles to its original direction of

motion. Find i

the speed of after the collision,

ii

the speed and direction of motion of after the collision,

iii the coefficient of restitution between and . A

B 30°

0.3m s–1

0.6m s–1

© OCR, GCE Mathematics, Paper 4730/01, January 2013 12 Two uniform smooth spheres and , of equal radius, have masses and respectively. They are moving on a horizontal surface when they collide. Immediately before the collision, has speed and is moving towards at an angle of to the line of centres, where has speed and is moving towards along the line of centres (see diagram). As a result of the collision, ’s loss of kinetic energy is s direction of motion is reversed and s speed after the collision is . Find i

the speed of after the collision,

ii

the component of s velocity after the collision, parallel to the line of centres, stating with a reason whether its direction is changed to the left or to the right,

iii the value of , iv the coefficient of restitution between and . A

B 2kg

m kg

2m s–1

α 5m s–1

© OCR, GCE Mathematics, Paper 4730, June 2012 13 Two uniform smooth spheres and , of equal radius, have masses and respectively. They are moving in opposite directions on a horizontal surface and they collide. Immediately before the collision, each sphere has speed in a direction making an angle with the line of centres (see diagram). The coefficient of restitution between and is i

Show that the speed of is unchanged as a result of the collision.

ii

Find the direction of motion of each of the spheres after the collision.

A 2m kg α

B

u m s–1 α

m kg

u m s–1

© OCR, GCE Mathematics, Paper 4730, January 2012

14 A particle , of mass , is attached by a light inextensible string to a particle , of mass . is initially stationary and is projected with speed so that when the string is about to move into tension, is moving at to the string. Find, correct to three significant figures: i

the speed of immediately after the string becomes taut

ii

the magnitude of the impulsive tension in the string.

9 Circular motion 2

In this chapter you will learn how to: work with a particle moving in a circle with variable speed model the motion of a particle moving in a circle in a vertical plane use the principle of conservation of mechanical energy to solve problems involving a particle moving in a vertical circle. If you are following the A Level course, you will also learn how to: work with the radial and tangential components of the acceleration solve problems involving moving particles where only part of their path is a vertical circle.

Before you start… Chapter 1

You should know the principle of conservation of mechanical energy (using kinetic energy and gravitational potential energy).

Chapter 4

You should be able to model motion in a horizontal circle.

A Level Mathematics Student Book 1

You should be able to model motion of a particle under constant acceleration using the equations of motion.

A Level Mathematics Student Book 2

You should be able to label the forces acting on an object resting on an inclined plane at an angle to the horizontal.

1 A cyclist is travelling at

along a road

when he reaches an incline making an angle of with the horizontal. If the cyclist does not pedal to maintain his speed, how far along the road will he reach, assuming there is no resistance? 2 A particle moves in a circular orbit of radius at a constant angular speed of . What is the linear speed of the particle? 3 A ball is hit at an angle of to the horizontal at a speed of from a height of above the ground. Calculate the maximum height above the ground reached by the ball.

4 A box is on the point of sliding down a plane inclined at to the horizontal. Find the normal reaction between the box and the inclined plane. 2kg

30°

What is circular motion with variable speed? In Chapter 4 you considered the speed of a particle moving in a horizontal circle as constant, however the

velocity was not since the direction of the tangential vector (the velocity vector) changes as the particle moves in a horizontal circle. An important example of circular motion with variable speed is when a particle is moving in a vertical circle. You can consider what is happening to a particle when it moves around a vertical circle, from an initial speed to a final speed . v m s–1

change in vertical height

O u m s–1

Section 1: Conservation of mechanical energy Key point 9.1 You can use the principle of conservation of mechanical energy to determine the speed of a particle at any point in a vertical circular orbit.

Rewind You learned about the conservation of mechanical energy in Chapter 1.

WORKED EXAMPLE 9.1

A smooth bead of mass is threaded onto a smooth circular wire of radius in a vertical plane. The bead is projected from its lowest point with speed

and centre fixed .

O

A 12 m s–1

Find the speed of the bead when it reaches its highest point in its motion.

travels,

and the vertical distance bead , is .

Set the gravitational potential energy to equal zero at the horizontal line passing through the lowest point of the vertical circle, .

Write down any information that may be helpful from the diagram. Include the zero level for gravitational potential energy.

At the point :

Calculate the gravitational potential energy and kinetic energy of the bead at .

At the point :

Calculate the gravitational potential energy and kinetic energy of the bead at .

Since energy is conserved:

By the principle of conservation of mechanical energy, the total energy at is equal to the total energy at . You can rearrange to make

the subject.

If you want to find the forces acting on a particle as it moves in a vertical circle, you can use the principle of conservation of mechanical energy to find the speed at any point and then apply Newton’s second law .

Acceleration towards the centre of circular motion is

, so you can calculate the force towards the centre.

Focus on… In Focus on … Problem solving 1 you solve a problem involving the impulse–momentum principle as well as conservation of mechanical energy to find the speed of a particle moving in a vertical circular orbit. Acceleration towards the centre of motion is known as a radial acceleration.

v m s–1

r v2 m s–2 r O u2 m s–2 r

u m s–1

Rewind You learned about acceleration in horizontal circular motion in Chapter 4.

Key point 9.2 Once you know the speed of a particle at a particular point in a circular path, you can: find the acceleration towards the centre by using use Newton’s second law to find the force towards the centre of the circle.

Fast forward The linear speed of the particle moving in a vertical circle is changing, so there is a component of acceleration (and hence also a force) acting on the particle in the direction tangential to the circle. You will find out about this in Section 2.

WORKED EXAMPLE 9.2

A smooth bead of mass is threaded onto a smooth circular wire of radius in a vertical plane. The bead is projected from its lowest point with speed

and centre fixed .

O

A 12 m s–1

What is the magnitude of the normal reaction force of the wire on the bead when the bead is in the same horizontal line as the centre of the circle? Write down any information that may be helpful on the diagram.

O

R

2 m

C

mg

A and the vertical distance the bead travels,

,

is . Set the gravitational potential energy to equal zero at the horizontal line passing through the lowest point of the vertical circle, . Let denote the position of the bead on the circular wire and the normal reaction force of the wire

Decide on the zero level for gravitational potential energy. You first need to find the speed of the bead at the point when the bead is in the same horizontal line as the centre of the circle.

on the bead. At the point :

Calculate the gravitational potential energy and kinetic energy of the bead at .

At the point :

Calculate the gravitational potential energy and kinetic energy of the beadat the new point .

Since energy is conserved:

By the principle of conservation of mechanical energy, the total energyat is equal to the total energy at .

Using

towards the centre :

is equal to the resultant force towards the centre of the circular motion. It is the normal reaction of the wire on the bead that provides this force.

When a point in the circular motion is being considered whose radius is at an angle to the horizontal,you need to include in the component of the weight towards the centre. This is shown in Worked example 9.3. WORKED EXAMPLE 9.3

A particle of mass is attached to one end of a light inextensible string of length . The other end of the string is attached to a fixed point and is free to rotate in a vertical circle. The particle is hanging in equilibrium at its lowest point when it is projected with a horizontal speed of . a Find an expression for the tension in the string when it makes an angle of with the downward vertical through . b Find the range of values of for which the particle will perform a complete circle. a Set the gravitational potential energy to equal zero at the horizontal line passing through the lowest point of the vertical circle. Call this point .

a

O θ

uP m s–1

Write down any information that may be helpful and draw a diagram. Include the zero level for gravitational potential energy. Note that only the radial acceleration is marked on the diagram since this is all you need to solve the problem, but there will also be a tangential component.

T

θ

P

A

0.2g

Let the particle be at position in its circular orbit when it makes an angle of with the downward vertical.

Include any extra labels that might be helpful when trying to refer to position in the circular orbit.

At the lowest point, :

Calculate the gravitational potential energy and kinetic energy of the particle at .

At the point :

Calculate the gravitational potential energy and kinetic energy of the particle at , where is the vertical height above point .

Since energy is conserved:

By the principle of conservation of mechanical energy, the total energy at is equal to the total energy at . You now have an expression for the tangential speed at Use this speed in the formula for acceleration in circular motion.

Resolving the forces in the radial direction of the circular motion,

You can now resolve the forces in the same direction as the tension.

Rearrange to find an expression for tension in terms of and b For the particle to make a full circle the speed must be large enough so that the string is taut at the highest point in the circular motion. You want

when

For the particle to move in a full circle the string must have some tension in order to keep the string taut. At the highest point in the movement the angle made with the downward vertical is .

:

You can rearrange to find . This will give a lower bound for the value that the initial speed can take in order for the particle to complete a full circle.

Key point 9.3 If a particle connected to a light inextensible string moving in vertical circles is to complete full circles, the tension in the string must be greater than or just equal to zero at the highest point of the vertical circular orbit.

WORKED EXAMPLE 9.4

A particle of mass is attached to one end of a light rod of length . The other end of the rod is attached to a fixed point and is free to rotate in a vertical circle. The particle is hanging in equilibrium at its lowest point when it is projected with a horizontal speed of . Find: a an expression for the speed of the particle when the rod is at an angle of with the upward vertical through b the set of values of for which the particle will perform a complete circle. a

, length of rod is

and initial speed is

.

Write down any information that may be helpful from the diagram. Include the zero level for gravitational potential energy.

Set the gravitational potential energy to equal zero at the horizontal line passing through the lowest point of the vertical circle. Call this point .

O θ

u P m s–1

a T

P

0.4 m

A 0.1g

Note that only the radial acceleration is marked on the diagram since this is all you need to solve the problem, but there will also be a tangential component.

Let the particle be at position in its circular orbit when it makes an angle of with the upward vertical through

Write down helpful information from the question. Include any extra symbols that might be helpful when trying to refer to position in the circular orbit.

At the lowest point, :

Calculate the gravitational potential energy and kinetic energy of the particle at , where is the vertical height at (the bottom of the circle is taken to be height ).

At the point :

Calculate the gravitational potential energy and kinetic energy of the particle at , where is the vertical height above point .

Since energy is conserved:

By the principle of conservation of mechanical energy, the total energy at is equal to the total energy at .

b For the particle to make a full circle the speed must be greater than zero when the particle is at the highest point in the circular orbit.

You want

when

,

Unlike a string, a rod cannot go slack. Therefore, the speed of the particle at the highest point must be greater than zero for the particle to move in complete vertical circles. At the highest point in the movement the angle made with the downward vertical is

You can rearrange to find . This will give a lower bound for the value that the initial speed can take in order for the particle to complete a full circle.

Key point 9.4 If a particle connected to a light rod moving in vertical circles is to complete full circles, the speed of the particle must be greater than zero at the highest point of its circular orbit. The same condition is true if the particle is representing a bead threaded onto a smooth circular wire.

WORK IT OUT 9.1 A particle of mass

is attached to one end of a light inextensible string of length

. The

other end of the string is attached to a fixed point . The particle hangs in equilibrium with the string taut at position . The particle is then set in motion with a horizontal speed of so that the particle moves in a vertical circle. What conditions are necessary for the particle to complete the vertical circle? Which solution is correct? Can you identify the errors in the incorrect solutions?

.

Solution 1 The particle will complete vertical circles if the speed at the top of the circle is greater than zero. Solution 2 The particle will complete vertical circles if the initial speed

.

Solution 3 The particle will complete vertical circles if the speed of the particle is enough to keep the string taut at the top of the circle.

EXERCISE 9A Unless otherwise instructed, when a numerical value for the acceleration due to gravity is needed, use . 1

A particle of mass

is attached to one end of a light inextensible string of length

. The other

end of the string is attached to a fixed point . The particle hangs in equilibrium with the string taut at position . The particle is then set in motion with a horizontal speed of moves in a vertical circle.

O

so that the particle

r m T

P

A

mg u m s

–1

a i ii b i ii c i ii d i ii

If

and

, find the speed of the particle when angle

is

If

and

, find the speed of the particle when angle

is

If

and

If If If

and

and

of the particle is 2

, find the tension in the string when angle

is

, find the tension in the string when angle

If and , find the angle made between speed of the particle is . If

is

, find the acceleration towards centre when angle

and ,

.

, find the acceleration towards centre when angle

and ,

.

, find the angle made between

.

is

.

. is

.

and the downward vertical when the and the upward vertical when the speed

.

A smooth bead is threaded onto a smooth circular wire fixed in a vertical plane, with centre and radius . The bead is projected from the lowest point with initial speed a If

.

, find the speed of the bead when it passes through the general point where

makes

an angle of with the downward vertical. b If

, find the greatest height reached by the bead above the lowest point of the circular

wire. c If , find the speed of the bead when it passes through the point that is level with the centre .

d If

, find the speed of the bead when it passes through the highest point of the circular

wire.

O r m bead

B

3

u m s–1

A smooth bead is threaded onto a smooth circular wire fixed in a vertical plane, with centre and radius . The bead is projected from the lowest point with initial speed

. Determine if the bead

will make a full circle and, if not, find the maximum vertical height reached by the bead, when the initial speed is given by: a b c 4

A particle of mass is attached to one end of a light rod of length . The rod is free to rotate in a vertical plane about . The particle is held at rest with horizontal and then released. a Calculate the speed of the particle as it passes through the lowest point. b Find the tension in the rod at this lowest point.

5

A particle of mass

is attached to one end of a light inextensible string

of length

. The

particle is hanging in equilibrium at the lowest point when it is set in motion with a speed of . If the string remains taut during the particle’s motion, write an expression for the speed of the particle when 6

makes an angle of with the downward vertical

A particle of mass

.

is attached to one end of a light rod of length

. The other end of the rod

is attached to a fixed point and it is free to rotate about . The rod is hanging vertically with below when the particle is set in motion with a horizontal speed of . Find the minimum value of for which the particle will perform a complete circle. 7

A light inextensible string of length has a particle of mass attached at one end. The other end is attached to a fixed point and the particle describes complete vertical circles, centre . Given that the speed of the particle at the lowest point is times the speed of the particle at its highest point, find the tension in the string when the particle is at the highest point.

8

A light inelastic string of length has one end attached to a fixed point . A particle of mass is attached to the other end. The particle is held with horizontal and the string taut. a If the particle is released from rest, what is the maximum speed and where will this occur in the circular orbit? b If the particle is projected vertically downwards with speed when

makes an angle of with the horizontal.

c Given that the string will break when the tension in the string is string and the horizontal when the string breaks. 9

, find the tension in the string

A smooth hemispherical bowl, centre and of radius

, find the angle between the

, is fixed on a horizontal surface such that

the top of the bowl is parallel with the horizontal surface. A smooth marble of mass

is held in

place on the inner surface such that, for the plane containing the centre and the marble, the line makes an angle with the downward vertical. The particle is released from rest. a Calculate the speed of the particle as it passes through the lowest point.

b Calculate the normal reaction force acting on the marble when the marble passes through the lowest point. c If the model is refined to include friction, calculate the constant frictional force such that the marble comes to rest directly below the centre and nowhere else before.

Explore As you have seen, you can model motion in a vertical circle using energy. However, sometimes you just need to think about the limiting factors for an object to be able to move in a vertical circular path. This idea will help you tackle this problem from NRICH: www.cambridge.org/links/moscmec6004.

Section 2: Components of acceleration (a general model) In the previous section, you used the conservation of mechanical energy to find the speed of a particle moving in a vertical circular orbit and used this to calculate the force directed towards the centre of motion (the radial direction). You also need to be able to model the motion of the circular orbit and determine the tangential and radial components of the acceleration, not just the acceleration directed towards the centre. The formula for acceleration of a particle moving in a circular orbit in Chapter 4 was based on the fact that the angular speed of the particle was constant. If you now consider that the angular speed is given by a function of time, you can create a general model for particles moving in a circular orbit. y

P r θ O

x

If you no longer have a constant linear or angular speed, you need to consider the angle as a function of time, . So you can say that the position vector is given by:

Rewind Remember, here angular speed is not constant. Differentiating with respect to time gives the velocity vector for angular speed:

Rewind You used vectors in A Level Mathematics Student Book 2. You can see that the magnitude of the velocity is time you get the acceleration vector for angular speed:

. If you differentiate again with respect to

The two most important directions as a particle moves in a circular path are the radial direction and the tangential direction. Now that you have found the velocity and acceleration vectors, you can look for the radial and tangential components of each.

Did you know? You can prove that the velocity vector is tangential to the circular path by using the displacement vector and the scalar product. For the velocity vector, you have a vector of magnitude

moving only in the direction

, which

is the tangential direction. For the acceleration vector, there are two parts. The first part is a component of the vector in the direction

, with magnitude

and this is

the tangential component of the acceleration. The second part gives the radial component of the acceleration as

(since

is the radial

vector). If you want to find the magnitude of the acceleration vector, since the radial and tangential components are perpendicular, you can use Pythagoras’ theorem.

Rewind This model is a more sophisticated model for circular motion, when angular speed is not constant. However, if you did have constant angular speed,

is equal to a constant, then

and the vector equation for the acceleration would simplify to the case you studied in Chapter 4.

Key point 9.5 is the tangential (or transverse) component of the acceleration and is the radial component of the acceleration.

Formula book Radial acceleration is

or

Tangential acceleration is

towards the centre. .

or

Using the notation developed in A Level Mathematics Student Book 1 you can refer to as

as

and

where the dot denotes differentiation with respect to time.

WORKED EXAMPLE 9.5

An athlete is running around the circular part of a running track of radius speed uniformly from to in a second period.

. He increases his

a Find an expression for the tangential and radial parts of the athlete’s acceleration over this time period. b What is the magnitude of the acceleration seconds after the athlete starts to increase his speed? a The acceleration of the athlete will be given by

in the tangential

direction and

If a particle is moving in a circular path where speed is not constant then you need to look at the components of the acceleration in the radial and tangential directions.

in the radial

direction. The athlete’s speed changes uniformly from to over seconds.

You can calculate the athlete’s tangential component of the acceleration by using the uniform increase in speed from to in seconds. is the change in speed per change in time and this will give the tangential component of the acceleration. You can integrate

At

,

so

to get an expression for in termsof .

You can use the initial conditions to find the constant of integration. You can now substitute into

to find the radial

component of the acceleration. b At : Radial acceleration is

You can substitute the value for time into each component of the acceleration.

Tangential acceleration is Thus:

To find the magnitude of the acceleration, you calculate the magnitude of the vector.

WORKED EXAMPLE 9.6

A smooth bead of mass is threaded onto a smooth circular wire fixed in a vertical plane, centre and radius The bead is projected from its lowest point on the circular wire with speed . a Find the speed of the bead when it passes through the point marked on the diagram. b Find the radial and transverse components of the acceleration of the bead at .

O

r θ

P

mg

A

a The bead is at position in its circular orbit when it makes an angle of with the downward vertical at . Set the gravitational potential energy to equal zero at the horizontal line passing through the lowest point of the vertical circle at .

Write down any information that maybe helpful from the diagram. Include where you will be measuring your gravitational potential energy from. To find the speed of the bead, you can use the principle of conservation of mechanical energy.

At the point :

Calculate the gravitational potential energy and kinetic energy of the bead at .

At the point :

Calculate the gravitational potential energy and kinetic energy of the bead at .

Since energy is conserved:

By the principle of conservation of mechanical energy, the total energy at is equal to the total energy at . You can rearrange to find

b At the point :

You need to find the components of the acceleration of the bead in the radial and tangential directions. It is more convenient, for this part, to use rather than for the velocity at .

v

v2 r

O θ

θ

P

dv dt

mg

For the radial direction you can use your calculation from part to find Using

,

so

EXERCISE 9B

. For the tangential component, use the force acting on the bead in tangential direction, with Newton’s second law.

EXERCISE 9B Unless otherwise instructed, when a numerical value for the acceleration due to gravity is needed, use . 1

A particle is moving in a circular path of radius . a The particle’s speed uniformly increases from to in seconds. i If , , and , find an expression for the tangential and radial parts of the acceleration. ii

If , acceleration.

,

and

, find an expression for the tangential and radial parts of the

b The particle’s speed uniformly decreases from to in . i If , , and , find an expression for the tangential and radial parts of the acceleration. ii

If , acceleration.

,

and

, find an expression for the tangential and radial parts of the

c The particle’s tangential component of the acceleration is given by . i If , find an expression for the linear speed of the particle if at ii 2

If

, find an expression for the linear speed of the particle if at

A car is driven around a roundabout of radius

. ,

.

. Its speed increases uniformly from

to

in seconds. a Find expressions for the radial and tangential acceleration of the car. b Find the magnitude of the acceleration after seconds. 3

A rally car travelling at

is accelerating around a circular bend of radius

car is increasing at a rate of

. If the speed of the

, find the magnitude of the acceleration of the car after seconds.

4

A particle of mass describes complete vertical circles while attached to one end of a light inextensible string of length . The other end of the string is fixed at the point . If the speed of the particle is at the highest point in the circular orbit, find the magnitude of the tangential acceleration when the string is horizontal.

5

A smooth bead of mass

is threaded onto a smooth circular wire fixed in a vertical plane with

centre and radius . The bead is projected from its lowest point on the circular wire with a speed of . Find an expression for the magnitude of the acceleration when the bead and the centre of the circular wire are in the same horizontal line. 6

One end of a light inextensible string of length is attached to a fixed point and the other end is attached to a particle of mass . With the string taut and horizontal, the particle is projected with a velocity of , vertically downwards. The particle begins to move in a vertical circle with centre . While the string remains taut, the angular displacement of from its initial position is radians and the speed of is acceleration of .

. Find, in terms of , the radial and tangential components of the

Section 3: Problem solving situations Leaving a circular path Sometimes a particle will only follow a circular path for a short period of time. Once it is no longer moving in a circle, it will need to be modelled as a particle moving freely under gravity. WORKED EXAMPLE 9.7

A smooth solid hemisphere with radius and centre is resting on a horizontal table with its flat face in contact with the table. A particle of mass starts to slip from rest at the highest point on the hemisphere. a If the hemisphere stays in a fixed position, find an expression for the normal reaction force of the particle to the surface of the hemisphere if the angle between and is . b Find the angle between

and

when the particle leaves the surface of the hemisphere.

c Once the particle leaves the hemisphere, how could you model its subsequent movement? d Find the distance away from the centre of the hemisphere when the particle first hits the table.

A

P

θ O a

,

and

.

Set the gravitational potential energy to equal zero at the horizontal line passing through the point .

Write down any information that may be helpful from the diagram. Include where you will be measuring your gravitational potential energy from.

At the point :

Calculate the gravitational potential energy and kinetic energy of the particle at .

At the point :

Calculate the gravitational potential energy and kinetic energy of the particle at .

Since energy is conserved:

By the principle of conservation of mechanical energy, the total energy at is equal to the total energy at . Once you have calculated the speed you can use Newton’s second law and

to find the

normal reaction force by resolving perpendicularly to the surface of the hemisphere. b When the particle leaves the surface,

.

The angle when the particle leaves the surface will be the instant when the normal reaction force between the particle and the surface is zero.

So the angle between

and

is

c You can model the particle as a projectile falling freely under gravity.

When the particle is no longer moving in a circular path it is now free to move as if it were a projectile, falling freely under gravity.

d The particle leaves the surface when

You first need to calculate the horizontal distance the particle travels before it leaves the surface of the hemisphere.

. The horizontal distance between and this point is given by Once the particle leaves the surface of the hemisphere it falls under gravity. Let the positive vertical direction be in the direction to and the positive horizontal direction be away from :

When the particle leaves the surface of the hemisphere it can be modelled as a projectile. You need to set up a direction for the horizontal and vertical components of the velocity and displacement for a projectile.

Acceleration due to gravity

You can use the equations of motion for a particle moving with a constant acceleration to calculate horizontal and vertical components of velocity and displacement and the time of flight.

Using

Using the quadratic formula:

which gives

You have a quadratic equation in and so can use the quadratic formula to find the positive value for .

seconds

The horizontal distance travelled from the point where the particle leaves the surface of the hemisphere is given by :

Total horizontal distance from is given by:

Finally, you can combine the horizontal distances travelled by the particle before leaving the surface and after leaving the surface of the hemisphere.

Key point 9.6 When the normal reaction force between the particle and a surface is equal to zero, a particle loses contact with the surface.

WORKED EXAMPLE 9.8

A particle of mass is attached to one end of a light inextensible string of length metres. The other end of the string is attached to a fixed point such that the particle hangs in equilibrium

directly below at . The particle is set in motion with a horizontal speed of the string first goes slack.

. At the point ,

a Find the vertical height of above the starting position where the string first goes slack. At the point when the string first goes slack, the particle is released from the string. b Find the maximum height the particle reaches above its starting position.

O θ l

uP m s–1

T

P 2 gl m s–1

A mg

a Set the gravitational potential energy to equal zero at the horizontal line passing through the point . Let be the angle formed between and the downward vertical . Let be the tension in the string.

Write down any information that may be helpful from the diagram. Include where you will be measuring your gravitational potential energy from. Include any extra labels that might be helpful when trying to refer to position in the circular orbit.

At the point :

Calculate the gravitational potential energy and kinetic energy of the particle at .

At the point :

Calculate the gravitational potential energy and kinetic energy of the particle at .

Since energy is conserved:

By the principle of conservation of mechanical energy, the total energy at is equal to the total energy at . Once you have calculated the speed you can use Newton’s second law and

to find the

tension by resolving in the radial and tangential directions. When the string goes slack there is no tension in the string. You can rearrange to make calculate a value for .

The vertical height above where the string first becomes slack is:

the subject and

You can find the vertical height using

b When the particle is released from the string the particle can be modelled as a projectile. For the motion of the particle, take the upward vertical as the positive direction.

It is important to define a direction to the projectile motion.

.

The maximum height reached by the projectile is when the final vertical velocity equals zero. Using

:

You can find the total vertical distance travelled by the particle by combining the projectile motion with the motion in its circular path.

WORK IT OUT 9.2 A smooth solid hemisphere with radius and centre is placed in a fixed position on a horizontal plane with its flat face in contact with the horizontal plane. A particle of mass starts to move from rest from the highest point. When has turned through an angle of , if the particle is still on the surface of the hemisphere, find an expression for , the normal reaction force of the hemisphere on the particle. Which solution is correct? Can you identify the errors in the incorrect solutions? Solution 1

At :

Solution 2

At :

Solution 3

At :

EXERCISE 9C Unless otherwise instructed, when a numerical value for the acceleration due to gravity is needed, use . 1

A smooth solid hemisphere with radius

and centre is resting on a horizontal table with its

flat face in contact with the table. A particle of mass is projected from a point , the highest point of the hemisphere, parallel with the horizontal surface at an initial speed of a i ii b i

If

and

If

and

If

,

, find the speed of the particle if the angle between , find the speed of the particle if the angle between and

and and

is is

. .

, find the normal reaction force acting on the particle at the surface

of the hemisphere if the angle between ii c i ii

and

is

.

If , and , find the normal reaction force acting on the particle at the surface of the hemisphere if the angle between and is . If and , find the angle between of the hemisphere. If

and

, find the angle between

and

when the particle leaves the surface

and the horizontal when the particle leaves the

surface of the hemisphere. d i ii

If and hemisphere. If

, find the speed of the particle when it leaves the surface of the

and

find the speed of the particle when it leaves the surface of the hemisphere. u m s–1 A P

O

2

A particle of mass

is released from rest at the top of a smooth track which forms a quarter of

circle, centre of radius

, followed by a drop of

to the ground.

O

50cm

2m

a Calculate the speed of the particle as it leaves the quarter circle part of the track. b What is the total horizontal distance that the particle travels before it hits the ground? 3

A smooth piece of track

is constructed so that it is in the shape of a circular arc. The arc

has a radius of and subtends an angle of at its centre line that is parallel to a horizontal surface, which is vertically

. The points and below the point

are on a , and

all lie in the same vertical plane. A particle is released from rest at a Find the speed of the particle as it leaves the arc

.

b Find the time taken for the particle to hit the ground after it has left c Find the horizontal distance that the particle travels after it leaves the track before it hits the horizontal surface.

X

P

2m

OXY

30°

Y 5m

4

A smooth piece of track

is constructed so that it is in the shape of a circular arc. The arc

has a radius of and subtends an angle of at its centre line that is parallel to a horizontal surface, which is vertically horizontal surface. at

. The points and are on a below point parallel to a

, and all lie in the same vertical plane. A particle is released from rest

a Find the speed of the particle as it leaves the arc

.

b Find the time taken for the particle to hit the ground after it has left c Find the horizontal distance that the particle travels after it leaves the track before it hits the horizontal surface. X

P

3m 60°

OXY

Y 2m

5

A smooth piece of track

is constructed so that it is in the shape of a circular arc. The arc

has a radius of and subtends an angle of at its centre . The points and are on a line that is that is parallel to a horizontal surface, which is vertically below from the point parallel to a horizontal surface. released from rest at

, and all lie in the same vertical plane. A particle is

a Find the speed of the particle as it leaves the arc b Find the time taken for the particle to hit the ground after it has left c Find the horizontal distance the particle travels after it leaves the track before it hits the horizontal surface. X

P

0.5m

OXY 120°

Y 4m

6

A smooth piece of track is constructed so that it is in the shape of a circular arc. The arc has a radius of and subtends an angle of at its centre . The points and are on a line thatis parallel to a horizontal surface, which is vertically a horizontal surface. rest at

below from the point parallel to

, and all lie in the same vertical plane. A particle is released from

a Find the speed of the particle as it leaves the arc

.

b Find the time taken for the particle to hit the ground after it has left c Find the horizontal distance the particle travels after it leaves the track before it hits the horizontal surface. X

P

0.3m

OXY

Y 2m

7

A smooth solid hemisphere with centre and radius

is fixed with its flat surface in contact with

a horizontal plane. A particle is released from rest on the surface of the hemisphere, such that makes an angle of with the upward vertical. The particle leaves the hemisphere at Find the angle between 8

and the upward vertical.

A smooth sphere of centre and radius is fixed to a horizontal table. A particle of mass is released from rest at a point on the surface of the sphere such that the acute angle formed between and the line perpendicular to the horizontal surface going through is speed of the particle when it leaves the surface of the sphere.

9

. Find the

A light inextensible rope of length is attached at one end to a horizontal beam above the horizontal ground and at the other end to a seat. A horizontal platform above the horizontal ground has an acrobat of mass holding the rope taut while sitting on the seat attached to the rope. The acrobat is released from rest and follows a circular arc. After the acrobat has travelled through an angle of herself from the seat. a What angle does

, measured from the starting position, the acrobat releases

make with the downward vertical when held taut by the acrobat on the

platform? b Calculate the tension in the rope when the acrobat is directly below . c Calculate the speed and direction of the acrobat as she leaves the rope. 10 A marble of mass

is attached to one end of a light inextensible string of length

. The

other end of the string is attached to a fixed point such that hangs in equilibrium. The marble is set in motion with a horizontal speed of Let be the angle makes with the downward vertical at

If the string does not become slack:

a find an expression for the speed of b find an expression for the tension in the string in terms of c show that the marble does not make full circles. 11 A smooth rubbish chute is built in two sections, and The arc has a radius of and subtends an angle of radius

and subtends an angle of

in the same vertical plane with

at its centre and

a If a bag containing rubbish of mass the bag enters the arc

, each in the shape of an arc of a circle. at its centre . The arc has a

. The points



and

and all lie

on the same vertical line.

is released from rest at

calculate the speed at which

A large container to collect the rubbish bags from the chute is positioned

below

b Calculate the speed at which the rubbish bag reaches the large container below c Determine whether or not the rubbish bag of mass reaches

will lose contact with the chute before it

OAB 60° A B

C 45° OBC

Checklist of learning and understanding For a particle moving in a circular path of radius with varying angular speed, you can find the velocity of the particle at any point in the path using the principle of conservation of mechanical energy. This is assuming that the particle is only subject to weight and a central force. The acceleration is directed towards the centre of the circular motion and you can use this to find the force in the same direction. If a particle connected to a light inextensible string moving in vertical circles is to complete full circles, then the tension in the string must be greater than or equal to zero throughout its circular orbit. If a particle connected to a light rod moving in vertical circles is to complete full circles, then the speed of the particle must be greater than zero throughout its circular orbit. is the tangential (or transverse) component of the acceleration and

is the radial

component of the acceleration. When a particle loses contact with a surface, the normal reaction force between the particle and the surface becomes zero.

Mixed practice 9 Unless otherwise instructed, when a numerical value for the acceleration due to gravity is needed, use 1

A particle of mass is attached to one end of a light inextensible rope of length metres. The other end of the rope is attached to a fixed point The particle hangs in equilibrium with the rope taut at position . The particle is then set in motion with a horizontal speed of so that the particle moves in a vertical circle. Find the maximum height the particle reaches above .

2

A particle of mass is attached to one end of a light inextensible string of length . The other end of the string is attached to a fixed point The particle hangs in equilibrium with the string taut at position . The particle is then set in motion with a horizontal speed of so that the particle moves in a vertical circle. Find the tension in the string when the particle is in the same horizontal line as the fixed point

3

A smooth solid hemisphere with radius

and centre is resting on a horizontal table with

its flat face in contact with the table. A particle of mass is projected from the highest point parallel with the horizontal surface at an initial speed of . When makes an angle of with the horizontal, find the speed of the particle. 4

A particle of mass is attached to one end of a light rod of length The other end of the rod is attached to a fixed point The particle is hanging in equilibrium at its lowest point when it is projected with a horizontal speed of Using the principle of conservation of mechanical energy, find the speed of the particle when the angle is made with downward vertical .

5

One end of a light inextensible rope of length is attached to ball of mass and the other end is attached to a fixed point . The particle is hanging in equilibrium at when it is set in motion with a horizontal speed of Calculate the tension in the rope when makes an angle of with the downward vertical .

6

A hollow circular cylinder is fixed with its axis horizontal. The inner surface of the cylinder is smooth and has a radius of A particle of mass is projected horizontally with speed from the lowest point so that moves in a vertical circle centre which is perpendicular to the axis of the cylinder. The angle is .

O θ P

A 5 m s–1

7

i

While is in contact with the inner surface of the cylinder, the speed is equation for by using the principle of conservation of mechanical energy.

ii

For what value of will the particle leave the inner surface of the cylinder?

Find an

A light inextensible string of length of length has one end attached to a fixed point and the other end attached to a particle of mass . moves in a vertical circle with centre

and radius

. When is at the highest point on the circle it has a speed of

Determine

the tension in the string when is at its lowest point in the circular orbit. 8

A light rod of length

is freely hinged to a fixed point

while at the other end is attached

to a particle of mass . The particle starts at rest from a point vertically below and is projected horizontally with speed The particle moves in complete circles. Find the set of values of for which this happens. 9

A particle of mass is attached to a fixed point by a light inextensible string of length The particle is projected horizontally with speed from the point vertically below . The particle moves in a complete circle. Find the tension in the string when i

the string is horizontal,

ii

the particle is vertically above © OCR, GCE Mathematics, Paper 4730, January 2011

10 A smooth sphere of radius and centre has a particle of mass sitting at rest at the highest point of the sphere. The particle is projected horizontally with speed and the subsequent motion of is down the sphere. loses contact with the sphere when an angle of with the upward vertical. in terms of

makes

i

Find an expression for

and (the acceleration due to gravity).

ii

Determine the minimum value of in terms of and for which leaves the surface of the sphere the instant it is projected.

11 A light rod of length has a particle of mass attached at the point . The rod is free to rotate in a vertical plane about a fixed point The greatest force acting along the rod is . i

At which point in the particle’s circular orbit does the force along the rod reach this greatest value?

ii

Find the speed of the particle at the point where the force acting along the rod is greatest.

iii Find the magnitude of the force acting along the rod when the speed is

.

12 a O θ

v

R P

A hollow cylinder has internal radius . The cylinder is fixed with its axis horizontal. A particle of mass is at rest in contact with the smooth inner surface of the cylinder. is given a horizontal velocity in a vertical plane perpendicular to the axis of the cylinder, and begins to move in a vertical circle. While remains in contact with the surface, makes an angle of with the downward vertical, where is the centre of the circle. The speed of is and the magnitude of the force exerted on by the surface is (see diagram). i ii

Find in terms of

and and show that

Given that just reaches the highest point of the circle, find show that in this case the least value of is

.

in terms of and and

iii Given instead that oscillates between

radians, find

in terms of and

© OCR, GCE Mathematics, Paper 4730, June 2009 13

O θ°

2m P

A particle of mass

is attached to one end of a light inextensible string of length

.

The other end of the string is attached to a fixed point . With the string taut the particle is travelling in a circular path in a vertical plane. The angle between the string and the downward vertical is (see diagram). When the speed of is i

At the instant when the string is horizontal, find the speed of and the tension in the string.

ii

At the instant when the string becomes slack, find the value of . © OCR, GCE Mathematics, Paper 4730/01, January 2008

14

O

0.5m

3 m s–1

θ

P v m s–1

One end of a light inextensible string of length

is attached to a fixed point

A particle

of mass is attached to the other end of the string. With the string taut and horizontal, is projected with a velocity of vertically downwards. begins to move in a vertical circle with centre . While the string remains taut the angular displacement of is radians from its initial position, and the speed of is (see diagram). i

Show that

ii

Find, in terms of the radial and tangential components of acceleration of

iii Show that the tension in the string is and hence find the value of at the instant when the string becomes slack, giving your answer correct to decimal place. © OCR, GCE Mathematics, Paper 4730, January 2009 15 One end of a light inextensible string of length is attached to a fixed point A particle of mass is attached to the other end of the string. is projected horizontally from the point vertically below with speed . starts to move in a vertical circle with centre The speed of is when the string makes an angle with the downward vertical.

i

While the string remains taut, show that string in terms of .

and find the tension in the

ii

For the instant when the string becomes slack, find the value of and the value of .

iii Find, in either order, the speed of when it is at its greatest height after the string becomes slack, and the greatest height reached by above its point of projection. © OCR, GCE Mathematics, Paper 4730, June 2011 16 A particle is attached to a fixed point by a light inextensible string of length . A particle is in equilibrium suspended from by an identical string. With the string taut and horizontal, is projected vertically downwards with speed so that it strikes directly (see diagram). is brought to rest by the collision and starts to move with speed

6 m s–1

P

0.7 m

O

0.7 m

Q

i

Find the speed of immediately before the collision. Hence find the coefficient of restitution between and

ii

Given that the speed of is when makes an angle with the downward vertical, find an expression for in terms of , and show that the tension in the string is , where is the mass of .

iii Find the radial and transverse components of the acceleration of at the instant that the string

becomes slack.

iv Show that (after the string initial position.

, where is the speed of when it reaches its greatest height becomes slack). Hence find the greatest height reached by above its

© OCR, GCE Mathematics, Paper 4730, June 2010

Rewind Recall how to find the coefficient of restitution from Chapter 3.

10 Centres of mass 2 This chapter is for A Level students only. In this chapter you will learn how to: use integration to find centres of mass of rods of variable density, uniform laminas and uniform solids of revolution apply your knowledge of centres of mass to problems of equilibrium, including suspension of a lamina and toppling or sliding of a lamina acted on by several forces.

Before you start… A Level Mathematics Student Book 2

You should be able to integrate functions of the type: .

1 Evaluate:

A Level Mathematics Student Book 2

You should understand the moment of a force and be able to calculate moments.

2 Find the moment of a force of

You should be able to use the coefficient of friction and the inequality .

3 A particle rests on a horizontal rough surface. A horizontal force is applied such that the particle is on the point of sliding. Find the value of the frictional force, in terms of µ, given that the mass of the particle is .

A Level Mathematics Student Book 2

with position vector vector

acting at the point

about the point with position

.

Centres of mass, safety and stability An object placed on a horizontal plane may topple over. An object placed on a rough inclined plane may topple over or slide. If an external force is applied to an object on a rough plane (horizontal or inclined) the object may topple over or slide. By analysing the forces on an object, including its weight, and the moments of those forces, you can determine whether the object will remain stationary, topple over or slide. You can assess the stablility of objects, such as a tractor on a hillside, and consider the safety implications.

Did you know? Ships contain ballast below the waterline that lowers the centre of mass of the vessel. The ballast has to be adjusted carefully depending on the weight and distribution of the cargo, to ensure the ship is stable in a range of conditions at sea. The Hoegh Osaka left Southampton on 3 January

2015 but while still in the Solent it developed a severe list (tilt). The rudder and propeller were out of the water and the Captain could not control the ship. The crew were evacuated to safety and the tonne vessel was grounded on a sandbank to prevent capsize. The calculations that should have led to a safe distribution of ballast and cargo were inaccurate and the centre of mass of the loaded vessel was too high. This is a very practical example of the need for an accurate method for assessing the location of the centre of mass of a three-dimensional body.

Section 1: Centres of mass by integration The centre of mass of an object can be found by integration. When a rod has variable density that can be expressed as a function

you can use integration to find

the centre of mass. When a lamina has a shape that can be expressed as a function centre of mass.

you can use integration to find the

You can also use integration to find the centre of mass of a symmetrical solid of revolution defined by function . The centre of mass will lie on the axis of revolution. It is usually better to use standard results, if possible. If the shapes are non-standard, or the mass density is not uniform, integration is needed. You will also be expected to use integration to derive some of the standard results.

Tip You learned about density in GCSE. It is usually the mass per unit volume of a solid object, and is measured in . When working with a one-dimensional object such as a rod, density is mass per unit length, measured in .

Rewind You learned in Chapter 5, Section 1, how to find the centre of mass of a composite body in one dimension.

Centre of mass of a rod of variable density Rods may be designed to be non-uniform. For example, it may be desirable to reinforce a rod where it is expected to be subject to the greatest force.

Key point 10.1 The centre of mass of a rod of length metres with variable density function

is given by:

Tip Rods are assumed to be one-dimensional.

Tip is the mass of the rod.

WORKED EXAMPLE 10.1

A metal bar metres long is modelled as a rod with mass density function the distance of the centre of mass from the denser end.

. Find

Use the formula for centre of mass. Let be the position of the centre of mass from one end of the bar.

Substitute in the density function and the limits.

Integrate.

The rod is denser where , because but at the other end.

The centre of mass is from the denser end.

is

at this end

Centre of mass of a uniform lamina defined by a function If a lamina is not a standard shape, you may be able to model the shape using a mathematical function. y y = f(x)

R

O

a

x

In this example, the region represents a uniform lamina. is defined by the function

and the lines

.

Key point 10.2 The coordinates of the centre of mass of a uniform lamina defined by

and the lines

are given by integration:     

Tip Because the lamina is uniform, mass is proportional to area. You can work with area rather than mass.

Tip These formulae can be derived by imagining the region to be divided into many small rectangles and summing along the -axis. It is often convenient to use these rearrangements of the formulae:

where is the area of the lamina.

Tip The formula for is the same for a rod of variable density

and for a lamina defined by

.

WORKED EXAMPLE 10.2

A lamina has three straight edges along the lines The fourth edge is a curve modelled by

and

.

, as shown.

Find the -coordinate of the centre of mass of the lamina. y

8 f (x) = x2 + 4

4

O

x 2

Substitute for the function .

multiplying out the expression for

Integrate. Check that the value of looks sensible.

You can use integration to find the centre of mass of a right-angled isosceles triangle, as in Worked example 10.3. WORKED EXAMPLE 10.3

Use integration to find the centre of mass of the uniform triangular lamina shown. y

6

f (x) = 6 – x

O

6

x

The lamina is in the shape of an isosceles triangle so the centre of mass lies on its axis of symmetry. Use the formula for the centre of mass of a lamina by integration.

WORKED EXAMPLE 10.4

In the right-angled uniform triangular lamina

,

Ĉ

and

.

Use integration to find the centre of mass of the lamina. State the distances of the centre of mass from

and

.

A

Draw the triangle. Let

be the -axis and

be the -axis.

b

C

Gradient of

a

B

Find the equation of edge

.

Use

, with your equation for

Integrate from

to

Substitute limits to find

.

. .

You could use integration to find

, but in this case consider that you

could have chosen to be the -axis and would be swapped over.

the -axis, then and

Find using the formula for the area of a triangle. There is no need for integration here. You can check this result using the coordinates of the vertices.

The centre of mass lies from

and from

.

Rewind Recall from Chapter 5 that the centre of mass of a triangular lamina lies at the intersection of its medians.

WORKED EXAMPLE 10.5

Use integration, together with the inverse function , to find the -coordinate of the centre of mass of the uniform lamina defined by the straight lines , and .

y

8

O

f (x) = x 3

2

x The technique here is to find the -coordinate of the centre of mass of the uniform lamina defined by the inverse function bounded by , and .

y 2

O

3 f - 1 (x) = √ x

8

x

This will be the -coordinate of the centre of mass of empty figure defined by and

The denominator represents the area of the ‘inverse’ lamina which is

The -coordinate of the centre of mass of the required lamina can then be calculated by subtraction. The area of the bounding box is axis is and the area between

, the area between and the -axis is

and the .

You can check this answer using the formula given in Key point 10.2.

Rewind Finding inverse functions is covered in A Level Mathematics Student Book 2, Chapter 2.

Centre of mass of a uniform solid of revolution A solid of revolution is the solid three-dimensional shape formed by rotating a function about an axis. In the diagram, the -axis is the axis of revolution. The centre of mass lies on the axis of revolution. y 2 1 O

1

2

3

x

–1 –2

The formula can be derived by imagining the solid to be divided into many small discs and summing along the -axis.

Rewind Finding the volume of a solid of revolution is covered in Pure Core Student Book 2.

Key point 10.3 The position of the centre of mass of a uniform solid of revolution with radius defined by

where

is the volume of the solid of revolution.

It can be more convenient to use the formula in the form:

where is the volume of the solid.

is:

Tip Since the solid is uniform, you can work with volume rather than mass.

WORKED EXAMPLE 10.6

The region is bounded by the curve is rotated through

for

, the -axis and -axis.

radians about the -axis to produce a solid of revolution. Show by

integration that the centre of mass of the solid has -coordinate . y

The region is a quarter disc of radius , centre The solid of revolution is a hemisphere.

2 y = √4 – x2

The formula book gives a formula for the centre of mass of a hemisphere, but you are required to ‘show by integration’ so you cannot just quote this result.

R

O

2

x

Use the formula for the centre of mass of a volume of revolution by integration.

Cancel and expand the brackets in the numerator.

Write down the integrals and substitute the limits.

WORKED EXAMPLE 10.7

Use integration to show that the centre of mass of a uniform solid right circular cone of height lies from its base. Let be the radius of the cone. You can generate the solid cone by rotating a right-angled triangular region about the -axis.

y

r

O

x

h

Find the equation of the hypotenuse.

Use the formula for the centre of mass of a solid of revolution.

Cancel

Integrate and substitute limits.

Simplify. The centre of mass is

from the

vertex. The centre of mass is from the base of the cone. In Worked example 10.7, you could have used

to generate the cone as a solid of revolution.

This would have led directly to but the integration would have required more steps. y

r

O

r x y = r – h

h

x

WORKED EXAMPLE 10.8

The region is bounded by the line

for

the -axis and -axis.

y

4

2

R

O

2

is rotated through

4

x

radians about the -axis to produce a solid of revolution. Calculate the -

coordinate of the centre of mass of the solid. Use the formula for the centre of mass of a volume of revolution by integration. The solid formed is a frustum of a cone.

Cancel and expand the brackets in the numerator.

Write down the integrals and substitute the limits.

When integrating to find the centre of mass, you may need to use any of the integration techniques you have already learned. WORKED EXAMPLE 10.9

The region is bounded by the curve is rotated through

and

.

radians about the -axis. Find the -coordinate of the centre of mass of the

solid formed. Use the formula for the centre of mass of a volume of revolution. You need to express

Substitute for

using a double angle identity:

.

You need to integrate

by parts.

Integrate using integration by parts and being careful with and signs.

Substitute limits.

Check your solution using your calculator, if possible.

EXERCISE 10A 1

Use integration to find the centre of mass of a rod of length metres with mass density function: .

2

A javelin is modelled as a rod of length metres of mass density , where is measured from the tail end. Calculate the distance of the centre of mass of the javelin from its tail end.

3

A uniform triangular lamina is in the shape of a right-angled triangle

Â

and

. y

5

C

A O

B 8

x

a Find the equation of the line passing through

.

b Use integration to find the coordinates of the centre of mass of the lamina. 4

Use integration to find the centre of mass of a uniform lamina bounded by

5

Use integration to show that the distance of the centre of mass of a uniform solid cone of height metres and base radius metres is

6

and

.

from its vertex.

A uniform triangular lamina is bounded by the line

, and the positive - and -axes. Use

integration to find the centre of mass of the triangular lamina. 7

The shape of a solid toy can be modelled by rotating the graph of measured in , through about the -axis.

, where is

y 4

O

5

x

Assuming the solid formed is uniform, find the centre of mass of the toy. 8

The region bounded by the line and is rotated though about the -axis to form a truncated cone (frustum). Use integration to find the centre of mass of the frustum.

9

A uniform lamina is bounded by the curve

, the line

and the -axis. Find:

a the area of the lamina in terms of b the -coordinate of the centre of mass of the lamina c the -coordinate of the centre of mass of the lamina in terms of . 10 A uniform lamina is bounded by the curve of the centre of mass of the lamina.

and the -axis,

Find the -coordinate

11 A uniform lamina is defined by the positive -axis, the positive -axis and the curve with equation:

Use integration to find the position of the centre of mass of the lamina, in terms of . 12 The region bounded by the line and is rotated though radians about the axis to form a solid figure. Use integration to find the -coordinate of the centre of mass of the solid.



Section 2: Equilibrium of a rigid body A rigid body is a single or composite object consisting of particles, rods, wires, laminas and solids that is fixed in shape. It is in equilibrium if the resultant force acting on the body is zero and the resultant moment acting on the body is also zero.

Suspension of a lamina from a point The moment of the weight and reaction about the point of suspension are both zero, as the line of action of both forces passes through the point of suspension.

vertical

reaction

A point of suspension

G centre of mass

weight

Key point 10.4 If a rigid body is freely suspended, it will hang with its centre of mass vertically below the point of suspension. You can apply this principle to solve problems. WORKED EXAMPLE 10.10

A uniform lamina has its centre of mass at the point the lamina at . Find the angle between the line suspended from .

as shown. is a point on the edge of and the vertical when the lamina is freely

y A (2, 12) G (5, 8)

O

x The lamina will hang with the centre of mass vertically below . Draw a vertical line through and the line to rotate the diagram.

. There is no need

y A (2, 12)

θ

G (5, 8) ve r ti ca l

x

O

Let the angle between

and

be .

WORKED EXAMPLE 10.11

A composite lamina of mass is made from a uniform rectangular lamina and a uniform isosceles triangular lamina abutting as shown. D

E

13cm

10cm A

C 13cm 12cm

B

a Find the distance of the centre of mass from

and from

.

The lamina is freely suspended from . b Find the angle between and the vertical. The lamina remains suspended from but now has a point mass hangs with

attached at . The lamina now

horizontal.

c Find the exact value of . If a perpendicular is drawn from to triangles. a

it makes two

The mass of each component is directly proportional to area so you can work with area. Let the -axis pass through . Let vertex be

and the -axis pass through

.

Remember that the centre of mass of an isosceles triangular lamina is one-third the distance from the centre of the base to the opposite vertex. Draw in the vertical through , which passes through the centre of mass .

D

b

r ti

ca

l

θ

ve

5cm

G

2 cm 2 3

B

Calculate .

c

D

E 2 cm 2 3

A

12cm

C

B Mg

kMg

Take moments about the midpoint of value of .

, to work out the

As the lamina is hanging in equilibrium the resultant moment is zero. An object may be attached to a point or surface by a hinge. The hinge allows the object to rotate, like hinges connecting a door to its door frame. There is always a reaction force at the hinge, but this can often be eliminated from calculations by choosing a suitable point about which to take moments so that the moment of the reaction force is zero. If you need to calculate the reaction force at the hinge it can be useful to resolve the reaction force into perpendicular components.

Rewind Taking moments is covered in A Level Mathematics Student Book 2, Chapter 22.

WORKED EXAMPLE 10.12

A uniform rectangular lamina , , is smoothly pivoted at to a horizontal surface. It rests in equilibrium against a fixed smooth block of height . The mass of the lamina is . makes an angle of with the horizontal. Calculate: a the magnitude of the force exerted by the block on the lamina b the magnitude and direction of the reaction on the lamina at the hinge. Draw a diagram showing the forces acting on the lamina. The reaction of the hinge on the lamina has been resolved into components and , parallel to the edges of the lamina. The reaction, , of the smooth block on the lamina is normal to edge . The weight of the block has been resolved into

components parallel to the edges of the lamina.

C

12 c

c

m

m

4

os P

Q

10 cm

5°

4g

4

si n

c

D

B

4g

5°

18

R

45° A

The lamina is in equilibrium.

a

Take moments about , thus eliminating the reaction force from the calculation. Since the rod is in equilibrium the anticlockwise moment is equal to the clockwise moment. The centre of mass of the lamina is and from . b

from

There is zero resultant force parallel to

.

There is zero resultant force parallel to

.

Use Pythagoras’ theorem to calculate the magnitude of the reaction at the hinge.

The reaction force is at

to

.

Use trigonometry to find the direction of the reaction at the hinge.

WORKED EXAMPLE 10.13

A piece of uniform wire of length is bent to form three sides of a rectangle, , , and then freely suspended from a hinge at . Calculate the angle between and the vertical.

A

and

1cm

18cm

B

1cm

Let the midpoint of be and the centre of mass be at .

The wire is symmetrical so

.

Since the wire is uniform, mass is proportional to length, so you

can work with length to find the centre of mass. Calculate

A

using

Add a vertical line to the diagram from passing through .

1cm

Let be angle between

and vertical.

vertical

O

θ

G

B

WORKED EXAMPLE 10.14

A composite lamina of mass is made from two rectangular laminas joined as shown. There is an axis of symmetry passing between the midpoints of and . F

E D

10cm

7.5cm

A

B

12cm

8cm

C

a Find the distance of the centre of mass from and from The lamina is freely suspended from a hinge at .

.

b Find the angle between and the vertical. The lamina remains suspended from but now has a point mass hangs with horizontal.

attached at . The lamina now

c Find the exact value of . a Area of larger rectangular lamina Area of smaller rectangular lamina

The mass of each component is directly proportional to area so you can work with area. Work out the distances of the centres of mass of each rectangle from

and

, using

vertex as

.

F

E D

Gθ

7.5cm

ical

10cm

ve rt

b

A

12cm

B

8cm

C

The centre of mass will be vertically below when the lamina is suspended. Draw in the vertical through , which passes through .

Use trigonometry to calculate the required angle.

The weight of the lamina acts at . Take moments about the midpoint of , to work out the value of .

c

Tip In part a of Worked example 10.13 the location of the centre of mass in the -direction could have been found directly since it will lie on the line of symmetry.

Explore If a point on the rim of a lamina is attached to a fixed point, and the lamina is free to rotate, it will hang with its centre of mass vertically below the point of attachment. But what is there to stop it resting with its centre of mass vertically above the point of attachment? In this ‘upside-down’ situation forces and moments sum to zero, as required for equilibrium. Why, in practice, does the lamina hang with the centre of mass below the point of attachment?

Toppling of a lamina when there is sufficient friction to prevent sliding If a lamina is placed on a rough inclined plane, it may slide down the plane or it may topple over. A lamina will be in stable equilibrium if a vertical line through the centre of mass of the lamina lies within its line of contact (the line from the top-most point of contact to the bottom-most) with the plane. It will not topple.

G

stable equilibrium

A lamina will be in unstable (limiting) equilibrium if its centre of mass lies vertically above the end of its line of contact with the plane. It is about to topple.

G limiting equilibrium

A lamina will topple if its centre of mass lies vertically above a point on the inclined plane outside its line of contact with the plane.

G

toppling

Key point 10.5 If there is sufficient friction at the surface to prevent sliding, the lamina topples if its centre of mass lies vertically above a point on the inclined plane outside its line of contact with the plane.

Tip Use a diagram to find the position of the centre of mass of the lamina above its line of contact with the inclined plane.

WORKED EXAMPLE 10.15

A rectangular lamina measures

by

. It rests with one of its shorter sides on an inclined

plane. Friction between the lamina and the inclined plane is sufficiently large to prevent sliding. Find the maximum inclination of the plane to the horizontal that will allow the lamina to rest in equilibrium. Draw a sketch, showing the angle between the plane and the horizontal. When the lamina is in unstable equilibrium with its centre of mass vertically above its line of contact with the plane, the inclination of the plane is maximised.

G

5c m

θ

m 4c θ

horizontal

Use trigonometry to find an upper bound for .

The inclination of the plane must be no more than WORKED EXAMPLE 10.16

A rectangular lamina, measuring

by

, has a square removed from one corner as shown.

y 15cm 3.5cm 7.5cm

O

A

x

a Find the centre of mass of the lamina. The lamina is placed on a rough inclined plane and rests in limiting equilibrium on the point of toppling about point b Find the angle of inclination of the plane to the horizontal. a The centre of mass of the missing square lies at:

Find the area of the lamina by subtraction. Use of mass of the composite lamina.

, to find the centre

b

The centre of mass is resting above . The lamina is in unstable equilibrium.

an

e

Draw a sketch.

ne

d

pl

G in

c li

θ θ A

horizontal

Calculate the angle made by the plane with the horizontal using trigonometry.

Toppling or sliding of an object on an inclined plane If a force is applied to an object resting on a rough surface, the turning moment may cause the object to topple before it slides. In other cases, the resultant force may be sufficient to cause the object to slide before it topples. Consider the forces acting on an object resting in stable equilibrium on a rough inclined plane.

G al rm n o c ti o n re a

fr i c

ti o

n

weight

The weight acts vertically downwards through , the centre of mass. The normal reaction and friction are both considered to act at the point on the inclined plane that the weight passes through.

Tip If the object is about to topple, its centre of mass lies vertically above the end of its line of contact with the plane – so the normal reaction and frictional force both act at the point that the object would topple about.

Rewind You met the coefficient of friction in A Level Mathematics Student Book 2, Chapter 21. Remember: In the limiting case where the object is about to slide: µ .

WORKED EXAMPLE 10.17

WORKED EXAMPLE 10.17

A uniform solid cylinder is resting in equilibrium with its end on a rough plane inclined at a variable angle to the horizontal. The cylinder has diameter and height . a Assuming the plane is sufficiently rough to prevent sliding, find the maximum value of that would allow the cylinder to continue to rest in equilibrium. The coefficient of friction between the cylinder and the plane is . b As is increased, show that the cylinder will slide before it topples.

0.6m

Draw a diagram showing the forces. When the cylinder is in unstable equilibrium, its centre of mass lies directly above the outer edge of its base.

R

1.8m α

The centre of mass of the cylinder lies on its axis of symmetry at a height of above the base.

F α W a

b The cylinder slides down the plane if:

Consider the component of the weight acting parallel to the plane. The cylinder slides down the plane if the component of the weight acting down the plane exceeds the limiting value of friction, Resolve perpendicular to the plane to find the magnitude of in terms of , and use

The cylinder starts to slide when exceeds but does not topple until exceeds . An object resting on a horizontal plane may topple over if an applied force causes a resultant turning moment. WORKED EXAMPLE 10.18

A cardboard box in the form of a cuboid and its contents, with a combined mass of

, rests in

equilibrium on a rough horizontal plane. The contents of the box are evenly distributed and the centre of mass of the box lies at its geometric centre. The diagram shows a vertical cross-section through the centre of the box, .

T

Z

0.18m

Y

0.245m

W

X

A horizontal force of magnitude acts on a horizontal line through . The coefficient of friction between the box and the plane is µ . As gradually increases from zero, the box slides before it topples if µ

. Show that

.

Take moments about :

The box will topple about if the moment of exceeds the moment of the weight of the box. At the point of toppling the normal reaction from the surface acts through .

Resultantforce

The box will slide if there is a resultant force parallel to the plane surface.

to

The box will slide before it topples if:

WORKED EXAMPLE 10.19

A uniform solid cube, of side

and mass

, rests on a rough horizontal plane. The

diagram shows a vertical cross-section through the centre of mass of the cube. A force, , is applied at the midpoint of , acting at an angle of above the horizontal as shown such that .

A 10m

10m

D

H

5m θ 5m

B

C

a Assuming that the cube would not slide but is on the point of toppling about , find an expression for in terms of and . b Assuming that the cube would not topple but is on the point of sliding along the plane, show that: where µ is the coefficient of friction between the cube and the plane. c Find an inequality for µ if the cube is to slide before it topples. a Take moments about

The cube will topple about if the moment of exceeds the moment of the weight of the box. At the point of toppling the normal reaction from the surface acts through .

b Resolve perpendicular to the

The cube is on the point of sliding if the resultant force parallel to the plane surface is zero.

surface,and let the normal reaction at the surface be : Resolve // to surface:

Rearrange to make the subject: friction takes its limiting value, µ newtons. The cube will slide before it topples if

c

EXERCISE 10B In questions 1

the -axis lies in a horizontal plane and the -axis in a vertical plane.

A lamina with centre of mass at the point the angle between

2

and the line

A lamina with centre of mass at the point the angle between

and the line

. Find

is freely suspended from the point

. Find

is freely suspended from the point

. Find the

.

3

A lamina with centre of mass at the point angle between and the line .

4

A lamina with centre of mass at the point Find the angle between and the line

5

is freely suspended from the point

.

is freely suspended from the point

.

.

A uniform rectangular lamina with side lengths

and

is freely suspended from one

vertex. Find the angle between the longer side and the vertical. 6

A uniform rectangular lamina with side lengths

and

is freely suspended from one

vertex. Find the angle between the shorter side and the vertical. 7

A uniform lamina in the shape of an equilateral triangle triangle of side

of side

has an equilateral

removed from vertex as shown.

C

15cm

A

10cm

15cm

B

25cm

a Find the distance of the centre of mass from The lamina is freely suspended from b Find the angle between 8

and the vertical.

A rectangular lamina, measuring

by

, has a quarter disc removed from one corner as

shown. D

15cm

C

2.5cm E

7.5cm

A

10cm

B

a Find the distance of the centre of mass of the lamina from

and

.

The lamina is placed on an inclined plane so that its centre of mass is vertically above the point and rests in equilibrium. b Find the inclination of the plane to the horizontal. 9

A uniform rectangular lamina

has mass

. The side

measures

measures A uniform circular lamina, of mass , and radius rectangular lamina to form a sign. The centre of the circular lamina is shown. P

D

and the side

, is fixed to the from and

, as

C 20cm 40cm

10cm 20cm A

B

60cm

a Find the distances of the centre of mass from

and

The sign is freely suspended from , the midpoint of b Calculate the angle between

.

.

and the vertical when the sign hangs in equilibrium.

10 A uniform solid cylinder of height and radius rests with one of its plane faces on a rough plane inclined at to the horizontal. There is sufficient friction between the cylinder and the plane to prevent slipping. Calculate the value of if the cylinder is on the point of toppling. 11 A solid cone of base radius and height rests with its circular base on a rough plane inclined at to the horizontal. There is sufficient friction between the cone and the plane to prevent slipping. Calculate the value of if the cone is on the point of toppling. 12 A solid cone of base radius

and height metres rests with its circular base on a rough plane

inclined at to the horizontal. There is sufficient friction between the cone and the plane to prevent slipping. Calculate the exact value of , given the cone is on the point of toppling. 13 A piece of wire freely from .

of length

is bent to form an arc of a quarter-circle, and then suspended

a Calculate the angle between and the vertical. The arc of wire remains in suspension at when a force, , is applied to the arc at its midpoint along a tangent, upwards. The mass of the arc of wire is . b Find the magnitude of required to maintain the arc of wire in equilibrium with

, the axis of

symmetry, horizontal. 14 A uniform rectangular lamina of mass of mass rests with on horizontal ground, which is rough enough to prevent slipping. measures and measures . A force, , is applied to the lamina at at an angle of below the horizontal. Find the magnitude of , in terms of , if the lamina is on the point of toppling about . C

D

30° F

20cm

A

30cm

B

ground

15 A uniform rectangular lamina of mass rests with on a horizontal rough plane. measures and measures . The coefficient of friction between the plane and the block is A horizontal force, , is applied to the lamina at .

C

D

F

50cm

A

B

25cm

a Find the value of if the lamina is on the point of slipping along the plane. b Find the value of if the lamina is on the point of toppling about 16 a The region bounded by the -axis, the line

and the curve

, for

, is

occupied by a uniform lamina. Find, in exact form, the coordinates of the centre of mass of this lamina. y

1

0.5

O

1

ln 4

x

2

b The region bounded by the -axis, the line

and the curve

, for

, is

occupied by a second uniform lamina. By using your answer to part a, calculate, to significant figures, the -coordinate of the centre of mass of the second lamina.

y

2 ln 4 1

O 17 i

1 B

2

8 cm

x

C

6 cm

A

17 cm

D

Fig. 1

A uniform lamina

is in the form of a right-angled trapezium. (see Fig. 1). Taking and -axes along

respectively, find the coordinates of the centre of mass of the lamina.

and

C

ii

D

B 7 cm 30° A

Fig. 2

The lamina is smoothly pivoted at and it rests in a vertical plane in equilibrium against a fixed smooth block of height . The mass of the lamina is . makes an angle of with the horizontal (see Fig. 2). Calculate the magnitude of the force which the block exerts on the lamina. © OCR, GCE Mathematics, Paper 4729/01, June 2008

Checklist of learning and understanding The centre of mass of any symmetrical uniform lamina or solid lies on its axis or axes of symmetry. The centre of mass of a rod of variable density,

, can be found by integration:

The centre of mass of a uniform lamina with shape defined by

is found by integration:

The centre of mass of a uniform solid of revolution is found by integration:

If a rigid body is freely suspended, it will hang with its centre of mass vertically below the point of suspension. If there is sufficient friction at the surface to prevent sliding, the position of the centre of mass of a lamina above its line of contact with the inclined plane allows you to work out whether it will topple. You can use your knowledge of friction and moments to determine whether a body placed on a rough surface will slide or topple.

Mixed practice 10 1

A uniform rod of mass and length metres is suspended from two cables fixed to the rod at and . The rod hangs horizontally, with metres and . A packet, modelled as a point mass, is placed on the rod at . Calculate: a the distance

given is the centre of mass of the combined rod and packet

b the tensions in the cables at and . 2

The diagram shows a uniform lamina in the shape of two rectangles attached together, and , as shown. measures , measures , measured and measures

.

y

C

D F

E 7cm

4cm A

5cm

4cm

W

x

B

a Find the position of the centre of mass from

and from

.

The lamina is freely suspended from . b Find the angle between 3

and the vertical when the lamina is in equilibrium.

A uniform rectangular lamina has mass . is long and is long. A uniform square lamina of mass and side length is attached onto the rectangular lamina with one edge along the middle of the shorter side of the rectangular lamina. The diagram shows the system. 4cm

D

C

2cm

4cm

8cm

2cm A

10cm

B

a Explain why the centre of mass is

from

b Find the distance of the centre of mass from

. .

The composite body is freely suspended from . c Find the angle between 4

A straight rod

and the vertical when the body hangs in equilibrium.

has length . The rod has variable density, and at a distance from its

mass per unit length is given by

, where is a constant. Find the distance from

of the centre of mass of the rod. © OCR, GCE Mathematics, Paper 4731/01, June 2013

5

B

12cm

A

30°

9cm

15cm

C

A uniform right-angled triangular lamina with sides is freely suspended from a hinge at vertex . The lamina has mass

and is held in

equilibrium with horizontal by means of a string attached to . The string is at an angle of to the horizontal (see diagram). Calculate the tension in the string. © OCR, GCE Mathematics, Paper 4729, January 2009 6

A C θ° 4cm PN

B

A uniform semicircular arc

is freely pivoted at . The arc has mass

and is held in

equilibrium by a force of magnitude applied at . The line of action of this force lies in the same plane as the arc, and is perpendicular to . The diameter of has length and makes an angle of with the downward vertical (see diagram). i

Given that

, find the magnitude of the force acting on the arc at .

ii

Given instead that

, find the value of . © OCR, GCE Mathematics, Paper 4729/01, June 2013

7

A

O

6cm

C

B Fig. 1

i

A uniform piece of wire, , forms a semicircular arc of radius . is the midpoint of (see Fig. 1). Show that the distance from to the centre of mass of the wire is , correct to significant figures. A

B D

3grams

5grams C Fig. 2

ii

Two semicircular pieces of wire,

and

, are joined together at their ends to form a

circular hoop of radius . The mass of is and the mass of is . The hoop is freely suspended from (see Fig. 2). Calculate the angle which the diameter makes with the vertical, giving your answer correct to the nearest degree. © OCR, GCE Mathematics, Paper 4729, June 2010 8

A uniform solid is made of a hemisphere with centre and radius , and a cylinder of radius and height . The plane face of the hemisphere and a plane face of the cylinder coincide. (The formula for the volume of a sphere is

.)

i

Show that the distance of the centre of mass of the solid from is

.

ii

The solid is placed with the curved surface of the hemisphere on a rough horizontal surface and the axis inclined at to the horizontal. The equilibrium of the solid is maintained by a horizontal force of applied to the highest point on the circumference of its plane face (see diagram). Calculate a the mass of the solid, b the set of possible values of the coefficient of friction between the surface and the solid. 2N

0.6m O 0.6m 45°

© OCR, GCE Mathematics, Paper 4729, January 2011 9

A straight rod

has length . The rod has variable density, and at a distance from its

mass per unit length is , where is a constant. Find, in an exact form, the distance of the centre of mass of the rod from . © OCR, GCE Mathematics, Paper 4731, June 2011 10 The region bounded by the -axis, the -axis, the line , and the curve , is occupied by a uniform lamina. Find, in an exact form, the coordinates of the centre of mass of this lamina. © OCR, GCE Mathematics, Paper 4731, June 2010 11 a

The region bounded by the -axis, the line

and the curve

, is

occupied by a uniform lamina. Find, in exact form, the coordinates of the centre of mass of this lamina. y

1

O

2

x

b The region bounded by the -axis, the line and the curve , for , is occupied by a second uniform lamina. By using your answer to part a, calculate, to significant figures, the -coordinate of the centre of mass of the second lamina. y

2

1

O

12 i

x

1

A uniform semicircular lamina has radius . Show that the distance from its centre to its centre of mass is , correct to significant figures.

ii

20 cm

B

C F

8 cm

4 cm A

12 cm

E

O

G

D

Fig. 1

A model bridge is made from a uniform rectangular board, section, , removed. is the mid-point of . the radius of the semicircle is (see Fig. 1). a Show that the distance from significant figures. b Calculate the distance from iii

, with a semicircular and

to the centre of mass of the model is

, correct to

to the centre of mass of the model. C

B

D A

10°

Fig. 2 The model bridge is smoothly pivoted at and is supported in equilibrium by a vertical wire attached to . The weight of the model is and makes an angle of with the horizontal (see Fig. 2). Calculate the tension in the wire. © OCR, GCE Mathematics, Paper 4729/01, January 2008

FOCUS ON … PROOF 2

This section is for A Level students only. In Chapter 5, the formulae for the centres of mass of uniform triangular laminas were stated and used without proof. This Focus on … section explores why the centre of mass lies at the intersection of the medians. It then shows how you can set up proofs of the formulae, based on this fact and making use of vector arguments. Consider the statement: the centre of mass of a uniform triangular lamina lies at the intersection of the medians.

You learned that the centre of mass of a uniform lamina lies on any axis of symmetry. A uniform triangular lamina does not have an axis of symmetry unless it is isosceles. In the diagram the lamina has been divided into many trapezia parallel to . You can approximate these trapezia as uniform rods having centres of mass lying on , the median from to . Since the centres of mass of all of the rods lie on it follows that the centre of mass of the whole lamina lies on . The same argument would apply if you started from the other two vertices, so the centre of mass must lie at the intersection of the medians, known as the centroid.

Questions 1

Prove that the intersection of the medians of a triangle lies at a point that is two-thirds of the distance along the medians measured from the vertices. Vectors can be very useful in geometry and you can use them in proofs.

B

b N

X A M

O

a

Let the midpoint of be intersection of these be .

and the midpoint of

be . Consider the medians

and

. Let the

You can use vectors to find alternative expressions for the position of .

lies part-way along

and

; you use and to indicate this; you are trying to prove that

.

Use the vectors a and b, together with the constants and , to find two alternative vector expressions for

.

You may now equate coefficients of a and b in the two expressions to get two simultaneous equations. Solve these to find values for and . You should be able to prove that point is two-thirds along the medians from the vertices. 2

Prove that for a uniform triangular lamina:

Consider a triangle having vertices with position vectors Let the midpoint of the triangle lies on

and

.

be and the centre of mass be . Remember that the centre of mass, , of , such that is .

The position vector of the centre of mass, G, is then given by:

Find vector expressions for formula.

and

in terms of

and substitute them to prove the



FOCUS ON … PROBLEM SOLVING 2

This section is for A Level students only.

Alternative approaches Focus on … Problem solving 1 set out a step-by-step method for solving a Mechanics problem. You know that there are different methods for solving a quadratic equation such as factorisation, completing the square or the quadratic formula. You choose the method that takes you most quickly to the result(s) you need. Here, the focus is on two Mechanics problems that can be approached in different ways.

Problem 1 An elastic string of unstretched length has one end fixed at point . A bob is attached to the other end and dropped from . Find the maximum speed of the bob in the subsequent motion. Method 1 You can use conservation of energy to derive an expression for in terms of the extension of the elastic string, .

You need to be very careful with your definitions of distances. If you do this successfully you will derive the equation:

You can find the value of for which is maximised by completing the square:

When

, takes its maximum value of

.

In a simple question this approach may be ‘a sledgehammer to crack a nut’. A side benefit of this approach is that you can use the expression for to find the maximum extension of the string, by solving .

Method 2 When speed is maximised, acceleration is zero, likewise the resultant force. Resultant force:

This takes you straight to

.

You can substitute your value of into the energy equation to find :

This approach involves less algebra and quickly gets you to . To carry on and find the maximum extension of the string you would need to set

in the general

energy equation:

You will need to solve a quadratic equation.

The common theme is application of the principle of conservation of energy:

.

Problem 2 A small smooth sphere sliding across a smooth surface is acted on by an impulse directed towards the centre of the sphere. The speed of the sphere is increased from to and its direction of motion is diverted through angle . Values for , and are given. Find the magnitude, , and direction, , of the impulse. vm s–1 u m s–1

α β

I Ns

Method 1 You can use an impulse-momentum triangle:

mv I β

α mu

You can use the cosine rule to find the value of from values given for , and .

You will have an expression for in terms of , unless the value of is also given. You can use the sine rule in the vector triangle and hence find :

You must take care with your vector triangle. Make sure that the directions of the arrows on the sides make sense. Don’t forget to put momentums in the triangle rather than velocities to make the triangle dimensionally

consistent. Little algebraic rearrangement is necessary in this solution.

Method 2 You can resolve the final velocity into components and use conservation of momentum. to to These are simultaneous equations and you need to eliminate either or . You obtain:

Substitute the values for , and to obtain and hence . You must take care to resolve velocities and to apply conservation of momentum correctly. Simultaneous equations are a standard technique, but you need to take care with your rearrangements.

Questions 1

An elastic string is fixed at one end at the point . A bob of mass is attached to the other end of the string. The string has natural length and its modulus of elasticity is . The bob is held next to and dropped. Find: a the maximum speed of the bob as it descends b the maximum extension of the string. Solve this problem by both of the methods described for Problem 1 and compare your answers.

2

A smooth sphere of mass is sliding across a smooth horizontal floor with a speed when it receives an impulse acting towards its centre. The sphere slows to a speed of and its direction of motion is diverted through . Find the magnitude and direction of the impulse. Solve this problem by both of the methods described for Problem 2 and compare your answers.

FOCUS ON … MODELLING 2

The simple pendulum A simple pendulum consists of an inextensible string of length fixed at one end with a mass attached to the other end; you can call the mass the bob. The bob will hang in equilibrium immediately below the fixing unless it is displaced. If the bob is displaced out of the vertical and then released, it will swing back and forth along the arc of a circle.

O

θ

l

mg Assuming that any resistances to motion are negligible, you can derive an equation of motion of the pendulum bob. You need to consider the transverse acceleration of the pendulum bob, and apply Newton’s second law:

This equation of motion simplifies to:

Like free fall under gravity, the equation of motion of a pendulum bob is independent of mass. Thus the solution of the equation would be the same for a bob and a bob. It would be possible to solve

by numerical methods, but the theory of the simple pendulum

relies on some of your work on complex numbers in Pure Core Student Book 2. You also learned that so that, when close to , all terms are vanishingly small except the first. Thus, if is close to then

.

Now the equation of motion for the pendulum bob can be approximated to:

This is now similar in form, but not the same as, the equation for simple harmonic motion:

So the pendulum equation can be solved using your knowledge of simple harmonic motion.

Questions 1

Use the theory of simple harmonic motion to write down an expression for how varies with time, in terms of and , given that the pendulum bob is displaced so that the string makes an angle with the vertical and then released.

2

Work out the period of oscillation for a simple pendulum of length

3

Sketch a graph of angular displacement against time, i.e. against , given that a bob, attached to an inextensible string of length

4

Assume that

, is displaced to

, as a multiple of .

radians from the vertical and then released.

Differentiate to find an expression for . Also, derive an expression for the

transverse velocity from the angular velocity . 5

Find the maximum value for the transverse velocity to significant figures.

How accurate is the model? To get an idea of the accuracy of the model you can compare the velocity calculated using the simple harmonic motion approximation to the velocity calculated using conservation of mechanical energy.

Question 6

a With

and

, as previously, calculate

from the energy equation, to

significant figures. b Find the percentage error predicted in c Repeat Questions i ii

by the SHM model.

for pendulums with these different starting displacements:

CROSS-TOPIC REVIEW EXERCISE 2 1

A particle , of mass on by a force

, is moving in a straight line with velocity

at time . It is acted

, in the direction of the line of motion of , which is defined as:

for for

.

a Find an expression for in terms of for b If the velocity of is

when

c A particle, , of mass particle has velocity 2

A particle of mass magnitude

.

, find the velocity of when

, is acted on by a force . Find the velocity of when

.

for .

is moving in a straight line with speed

. When

, the

. An impulse of

deflects through an angle , and reduces its speed to

(see diagram).

By considering an impulse–momentum triangle: a Show that

.

b Find the angle that the impulse makes with the original direction of motion of .

3m s–1 P

θ

6.5m s–1 4N s 3

A small ball is moving across a horizontal plane floor when it strikes a smooth vertical wall. The coefficient of restitution between the ball and the wall is . Just before impact the direction of motion of the ball makes an angle of with the wall. Immediately after impact its direction of motion makes an angle of with the wall. a Find the fraction of the kinetic energy of the ball that is lost in the impact. b Find the value of .

4

A light rod of length metres has one end freely hinged at a fixed point and a particle of mass attached to the other end. The rod is set in motion and makes complete circles about . The speed of at the top of the circle is half of its speed at the bottom of the circle. Find the maximum speed of in terms of .

5

A region is defined by the curve , the line and the -axis. A uniform solid is formed by rotating through radians about the -axis. a Show that the volume of the solid is

.

b Show further that the -coordinate of the centre of mass of the solid is 6

A particle of mass is moving in a straight line on a smooth horizontal surface. A horizontal force then acts on the particle for . This force acts in the direction of motion of the particle and at time seconds has magnitude newtons. When , the velocity of the particle is . a Find the magnitude of the impulse of the force on the particle between b Hence find the velocity of the particle when

and

.

.

c Find the value of when the velocity of the particle is 7

.

.

Two smooth uniform spheres and of equal radius have masses and . They are moving on a smooth horizontal plane when they collide. Immediately before the collision the speed of is and the speed of is . When they collide, the line joining their

centres makes an angle with the direction of motion of and an angle with the direction of motion of , as shown in the diagram. It is given that

and

.

a Find the components of the velocities of and perpendicular and parallel to the lines of centres immediately before the collision. The coefficient of restitution between and is . b Find the speed of each sphere after the collision. 2m s–1 A

B β

α 1.5kg

0.75kg

3m s–1

8

A particle of mass

is attached to one end of a light inextensible string of length

metres. The other end of the string is attached to a fixed point . The particle is hanging in equilibrium at point , directly below , when it is given a horizontal speed of

. When

has turned through an angle and the string is still taut, the tension in the string is . a Find an expression for . b Find the height above at the instant when the string goes slack. c Find the maximum height above reached by before it starts to fall down again.

O

0.5m θ T

4.5m s–1

A 9

P

A smooth sphere lies at rest on a smooth horizontal plane. A second identical sphere , moving on the plane, collides with the sphere . Immediately before the collision the direction of motion of makes an angle with the line joining the centres of the spheres. Immediately after the collision the direction of motion of makes an angle with the line joining the centres of the spheres. The coefficient of restitution between the spheres is . Show that

.

R

S

α β

10 A uniform solid consists of a hemisphere of radius and a cylinder of base radius and height , fixed together so that the bases coincide. The solid can rest in equilibrium with any point on the curved surface of the hemisphere in contact with a horizontal plane. Find in terms of .

11 A particle is placed at the highest point on the outer surface of a fixed smooth hemisphere of radius and centre . The hemisphere is placed with its plane face on a horizontal surface. The particle is projected horizontally from with speed and initially moves along the surface of the sphere. The particle leaves the sphere at point , where makes an angle with the upward vertical through , as shown. a Find an expression for in terms of and . After the particle leaves the surface of the hemisphere, it strikes the horizontal surface with speed

.

b Find the value of . P Q a θ

a

O

12 A child’s toy is formed by joining two solid cones so that their circular bases coincide. The cones have the same uniform mass density and the same base radius. The heights of the cones are

and .

h

1.5 h

a Find the distance of the centre of mass of the toy from the vertex of the larger cone. The toy is now placed on horizontal ground with the sloping surface of the smaller cone in contact with the ground. The object rests in equilibrium but is on the point of toppling. b Find the radius of the base of the cones as an exact multiple of . 13 Tw uniform smooth spheres and , of equal radius, have masses and respectively. The spheres are moving on a horizontal surface when they collide. Before the collision, is moving with speed in a direction making an angle with the line of centres and is moving towards with speed in a direction making an angle with the line of centres (see diagram). After the collision, moves with velocity in a direction perpendicular to the line of centres and moves with velocity

in a direction making an angle of

with

the line of centres. The coefficient of restitution between and is . i

Show that

and find

ii Find the values of and . A 2m kg m kg α am s

–1

.

B β bm s–1

© OCR, GCE Mathematics, Paper 4730/01, June 2015

14

O θ

0.6m

P vm s–1

4m s–1

A hollow cylinder is fixed with its axis horizontal. The inner surface of the cylinder is smooth and has radius . A particle of mass is projected horizontally with speed from the lowest point of a vertical cross-section of the cylinder and moves in the plane of the cross-section, which is perpendicular to the axis of the cylinder. While remains in contact with the surface, its speed is when makes an angle with the downward vertical at , where is the centre of the cross-section (see diagram). The force exerted on by the surface is . Show that

i

and find an expression for in terms of .

ii Find the speed of at the instant when it leaves the surface. © OCR, GCE Mathematics, Paper 4730, June 2012 15

D

(a + 5)cm

C

5cm

A

5cm

B

The diagram shows the cross-section through the centre of mass of a uniform solid prism. The cross-section is a trapezium with and perpendicular to . The lengths of and are each and the length of is . i

Show the distance of the centre of mass of the prism from

The prism is placed with the face containing

is

in contact with a horizontal surface.

ii Find the greatest value of for which the prism does not topple. The prism is now placed on an inclined plane which makes an angle with the horizontal. lies along a line of greatest slope with higher than . iii Using the value for found in part ii, and assuming the prism does not slip down the plane, find the greatest value of for which the prism does not topple. © OCR, GCE Mathematics, Paper 4729, June 2012



AS LEVEL PRACTICE PAPER Time allowed: 1 hour and 15 minutes The total mark for this paper is 60. The acceleration due to gravity is denoted by is needed, use . 1

. Unless otherwise instructed, when a numerical value

A smooth bead is threaded onto a smooth circular wire with centre and radius metre.The circular wire is fixed in a vertical plane. The bead is projected from its lowest point on the wire with speed . Calculate the maximum vertical height, in metres, above the point of projection that the bead reaches.

2

[3 marks]

The string on a guitar is plucked, creating a wave. The velocity, , of the wave depends upon the mass of the string, its length, and the tension in the string so that . a Write down the dimensions of tension. b Find

3

[1 mark]

and given that is a dimensionless constant.

[3 marks]

c Comment on whether it was necessary to state that is a dimensionless constant in part b. [1 mark] A car of mass moves along a horizontal road against a constant resistive force of . a Find the maximum speed, in

, at which the car can move if the engine cannot exert more than [2 marks] A tow rope is now attached to the car, which pulls a trailer of mass The resistance to motion for the trailer is and the resistance to car’s motion has not changed. Let be the tension in the rope and, at the instant the speed is , the engine is working at .

b Find the driving force for the car.

[2 marks]

c What modelling assumptions have been made about the tow rope?

[2 marks]

d Use Newton’s second law to write an expression for the acceleration of the car in terms of . e Find the tension in the tow rope. [3 marks] [3 marks] 4

A particle is attached to one end of a string of length The other end of the string is attached to a fixed point A second particle of the same mass is attached to one end of another identical string, while the other end of the string is attached to the first particle The whole system moves with a constant angular speed of rad about the downward vertical through The upper string makes an angle of with the downward vertical through and the lower string makes an angle of with the downward vertical through .

O α P β

Q

a State two assumptions that you should make about the string in order to model the circular motion of and [1 mark] b By considering the forces and the circular motion at

show that

c By considering the forces and the circular motion at and part b show that

5

Two small spheres, and

of masses

and

[4 marks] [4 marks]

, respectively, are attached to opposite ends of a

light inextensible string of length . They are placed next to each other on a horizontal table and sphere is projected vertically upward with a speed of v m s–1 A

A

B

starting position

B after A has been projected vertically upward

a Find the speed of sphere at the instant the string becomes taut.

[2 marks]

b Comment on how the modelling assumption ‘the two spheres are placed next to each other  will affect your answer to part a.

[2 marks]

c Show, using that the law of conservation of momentum, that, at the instant immediately after the string becomes taut, the velocity of is

[2 marks]

d State the magnitude of the impulse in the string.

[1 mark]

e Calculate the maximum height that sphere reaches above the horizontal table when it first comes to instantaneous rest.

[3 marks]

f Calculate the loss in kinetic energy due to the tightening of the string. 6

Two smooth spheres, and

with masses

and

[3 marks]

respectively, are moving on a smooth

horizontal surface. The spheres are moving towards each other and before they collide each sphere is moving at a speed of . After the collision moves with a speed of in the opposite direction to its initial motion. 5 m s–1

X

5 m s–1

Y

a State the direction in which the sphere moves after the collision. b Calculate the coefficient of restitution between the two spheres. 7

A particle of mass from

to

slides down a slope at while sliding down

[1 mark] [5 marks]

to the horizontal. The particle increases its speed

of the slope.

a Calculate the gain in kinetic energy of the particle.

[2 marks]

b Calculate the work done against the resistance to motion. [4 marks] There is a constant frictional force between the particle and the slope, and this force is the only resistance to motion. c Calculate the coefficient of friction between the particle and the slope to significant figures. [3 marks] d For what value of the coefficient of friction (to significant figures) would the particle slide down the slope at a constant speed?

[3 marks]

A LEVEL PRACTICE PAPER Time allowed: 1 hour and 30 minutes The total mark for this paper is 75. The acceleration due to gravity is denoted by is needed, use . 1

. Unless otherwise instructed, when a numerical value

A ball is released from rest and falls from a height of metres above a horizontal table. Given that the ball rebounds to a height metres, find an expression for the coefficient of restitution in terms of and . [3 marks]

2

The force that two objects, of masses

and

, exert on each other is given by

, where

is the distance in metres between the two objects and is a constant. Find the dimensions of . [3 marks] 3

A light elastic spring with natural length rests on a smooth horizontal table. One end is attached to a fixed point while the other end is attached to a particle of mass . The particle is held away from . The modulus of elasticity of the spring is spring.

. Find the elastic potential energy in the [3 marks]

4

A particle of mass is attached to one end of a light elastic string of modulus of elasticity and natural length The other end of the string is attached to a fixed point on a smooth table so that the particle is moving in a horizontal circle with centre . a Find an expression for the force towards the centre of motion, if is the radius of the circular motion. [2 marks] b Given that the particle is moving at revolutions per second, find an expression for the radius of the circular motion. [4 marks] c For

metre, sketch a graph of against . [2 marks]

If the tension in the metre elastic string reaches

the string will break.

d Comment on your sketch from part c with reference to the values of that can be chosen for this model. [2 marks] 5

A particle is moving along the -axis, initially starting at the origin . At time seconds, the velocity of is in the direction of increasing, where: for

and

for

a Sketch a graph of against for

. When

, is at .

. [2 marks]

b Calculate the acceleration of at

. [2 marks]

c Calculate the total distance travelled by in the first

. [3 marks]

6

A light elastic string of modulus of elasticity

and natural length

is attached to two points,

and

which are

apart in a horizontal line. Two particles, and

each of mass

to fixed points of the string such that the unstretched lengths of and is in equilibrium with the angle equal to , both denoted by .

4m

A

are each

, are attached The system

B

α

α

X

Y

a Show that the tension in

is given by

and in

is given by

. [3 marks]

b Show that

. [6 marks]

c Verify that

is a solution to the equation found in part b. [1 mark]

7

A smooth sphere with centre and mass

is moving with speed

when it collides with a

second smooth sphere with centre and mass moving at a speed of . The velocities of the spheres immediately before impact are inclined at angles of and , respectively, to the direction at the moment of impact.

A

B

60°

v m s–1

2v m s–1

The coefficient of restitution between the spheres is a Calculate the speed and direction of the spheres after impact. [5 marks] b Calculate the loss of kinetic energy as a result of the impact. [2 marks] 8

An aeroplane of mass

travels vertically downwards while moving in a quarter circle and then

travels horizontally. 90m s–1 350m

350m

60m s–1

The circle has radius

and the linear speed of the plane reduces uniformly from

to

during the seconds it takes the plane to turn through the quarter circle. Let be the time after the plane enters the quarter turn. a Find expressions for the radial and tangential components of the acceleration of the plane during the turn. [4 marks]

b While the plane is in the turn, there is a constant resistive force. Calculate the energy lost due to resistances to motion in the turn, in . [2 marks] When the plane is halfway through the quarter circle, it releases a parcel. The plane is vertically above the horizontal ground. The particle is modelled as a projectile once it has been released. c Calculate the speed with which the parcel is released. [4 marks] d Calculate the time taken for the parcel to reach the ground. [3 marks] 9

A solid object is formed by rotating the area under the curve lines

and

with

around the . axis between the

.

a Show that the distance of the centre of mass from

is

. [5 marks]

A solid object is created in this way, with and The solid shape is placed on a rough inclined plane, at an angle of to the horizontal, with the largest flat face of the shape in contact with the inclined plane, and does not slide. b Find the angle at the point of toppling, to significant figures. [2 marks] c At what angle would the solid object slide down the inclined plane, if the coefficient of friction between the object and the inclined plane is [2 marks] d Given that the coefficient of friction between the inclined plane is

will the shape topple or slide

as the angle is increased? [1 mark] 10 A car of mass

starts at and moves along a horizontal road with a driving force

and a

variable resistance to motion force that is proportional to the square of the distance from . At the instant when is a distance of from the resultant force is . a If metres is the distance b Find the work done, in

, find an equation for the resultant force in terms of the distance . [2 marks]

, by the car when the car has travelled a distance of

metres. [3 marks]

c Given that the car starts from rest at , find the speed of the car when it is at a distance of metres from the point [4 marks]

FORMULAE

The following formulae will be given on the AS and A Level assessment papers: Kinematics Motion in a straight line

Motion in two dimensions

Newton’s experimental law Between two smooth spheres Between a smooth sphere with a fixed plane surface Motion in a circle Tangential velocity is Radial acceleration is

or

towards the centre

Tangential acceleration is Centres of mass Triangular lamina: along median from vertex Solid hemisphere, radius

from centre

Circular arc, radius , angle at centre Sector of circle, radius , angle at centre Solid cone or pyramid of height Conical shell of height

from centre from centre

above the base on the line from centre of base to vertex

above the base on the line from centre of base to vertex

Answers All answers are given to significant figures, where appropriate.

Chapter 1 Before you start… 1 2 3 4

Exercise 1A 1 2 3 4 5 6 a b 7 8

Exercise 1B 1 2 3 4 a b c 5 6 a b 7 8 9 10 11 Proof

12

Exercise 1C 1 2 3 4 5 a b c d 6 a b c d Model Anita as particle leaving the surface of the springboard and entering the water, no resistance to motion. 7 a b c d 8 9 a b 10 11 a b c

Exercise 1D 1 2 3 4 5 6 7 8

9 a b 10 a b c

Exercise 1E 1 2 3 4 5 6 7 8 9 a b c 10 11 12 13 14 a b

Mixed practice 1 1 2 3 4 i ii 5 i ii 6 i ii 7 i ii

8 a b c 9 i ii 10 i ii 11 i Proof ii 12 i ii 13 i Proof ii iii 14 i Proof ii iii 15 a b 16 i ii iii 17 i ii 18 i ii iii 19 i Proof ii 20 i ii 21 i ii iii iv

22 23 a b c

Chapter 2 Before you start… 1 a b c d 2 3 4 5 a b 6 7 8 9 10 11

if

Exercise 2A 1 a b c d e f Dimensionless g h i j 2 Dimensionless 3 Dimensionless 4 a

b Yes 5 a b The same 6 Dimensionless 7 8 a b 9 10

Exercise 2B 1 a b c 2 a b c d 3 4 a Consistent b Not consistent c Consistent d Consistent e Not consistent 5 a b Consistent 6 a b 7 a b

, not the same

c 8 a b The same 9 a b

10 a b Change the to

Work it out 2.1 Solution 2 is correct.

Exercise 2C 1 2 3 4 5 Proof 6 a Proof b Yes. As the angle must be dimensionless, must have the inverse dimension of , so the dimension of can be determined as . 7 8 a Dimensionless b 9 a b c 10

Exercise 2D 1 Quantity Time Mass Weight length (displacement) Area Volume Velocity Acceleration

Dimension SI unit

Acceleration due to gravity Force Kinetic energy Gravitational potential energy

Work done

Moment of a force

Power

newton metre

watt

Momentum

Impulse

Moment of inertia

Angular velocity

Density

Pressure

Periodic time (time for one complete cycle) Frequency

Surface tension

Mixed practice 2

newton second

1 a b 2 Proof 3 a b i Inconsistent,

is the incorrect term.

ii Consistent iii Inconsistent,

is the incorrect term.

4 a b 5 a b 6

and

7 a b

as the two masses must make a similar contribution to the attractive force.

c 8 a b c 9 a b c 10 a b 11 a b c 12 13 a b c d

if

Chapter 3 Before you start… 1 2 3

if

4 a b c 5 6

Exercise 3A 1 a b c d e 2 3 4 5 6

Positive direction Initial velocity

7 8 9 10

Final velocity





















Impulse

11

away from the wall

12

away from the wall

13 14

Work it out 3.1 Solution is the only possible answer.

Exercise 3B 1

2 3 4 5 a b 6

in

original direction of travel

7 8 9 10

. The balls being spherical, the balls being the same size, the impact being along the line of centres, the contact being smooth.

11 a b

Exercise 3C 1

i

ii

iii

iv

v

a b





c 2 Both change direction. Velocity of is

and of is

3 a b 4 a b c 5 a b c 6 a b c 7 8 a Velocity of is

and of is

, moving in opposite directions to each other.

b 9 10 a Velocity of A is

and of is

, both in the original direction of motion.

b 11 Velocity of is

and of is

, both in the original direction of motion.

and is

, both in the original direction of motion.

12 13 a Velocity of is b 14 a b c 15 16 17 18 19 a b

Mixed practice 3 1 a

b 2 a b 3

in the direction of the second roller skater

4

in the direction of initial motion

5 i a b ii 6 a b 7 i ii 8 i Proof ii iii 9 i

and has changed direction

ii a b 10 i ii As

there are no more collisions

11 i ii 12 i ii a b 13

and

away from each other

14 i ii Proof iii Proof iv Speed of is away from the wall and speed of is

towards the wall.

Chapter 4 Before you start… 1 2 3 4 5 Work it out 4.1 Solution 3 is correct.

Exercise 4A 1 a i ii b i ii c i ii d i ii 2 a i ii b i ii c i ii 3 4 a b 5 a b c 6 7 a b 8 a Its orbit is circular and its angular speed is constant.

b i ii

Exercise 4B 1 a i ii b i ii c i ii d i ii 2 a i ii b i ii c i ii 3 4 5 a b 6 Emily will feel an increase in force by a factor of . 7 a b No, it is not a sensible estimate; it is too small. This is a very low speed to drive around a bend of radius. 8 9 10 a Assume the only friction force acting on car is from road surface. Assume car is on point of slipping away from centre. b i ii c For asphalt, reduce max safe speed from For concrete, reduce max safe speed from

to to

when wet. when wet.

d Depends on where the road is to be built. For example, in a city with a low speed limit asphalt would be cheaper and suitable. 11 a Proof

b Proof c Proof d Proof e Proof

Exercise 4C 1 a i ii b i ii c i ii d i ii 2 a b c d 3 4 Proof 5 a b 6 Proof 7 a b c 8 a b c

, compression

9 a b 10 a b c 11

, away from the centre so the rod is in compression. ,

12 Max speed

, min speed

13 Proof

Mixed practice 4 1 a b c

mg N

2 a

F = 0.6mg N r

v2 a = r mg N

b 3 a b 4 5 Proof 6 7 a b c There was no frictional force acting. 8 a b i ii 9 i ii iii 10 i

in

and

in

in

and

in

ii 11 i Proof ii 12 i ii 13 a Proof b

c Proof 14 i µ ii

Chapter 5 Before you start… 1

Exercise 5A 1 2 3 4 5 6 7 8 9

Exercise 5B 1 2 3 4 a b c

d e 5 6 a b c d 7 8 9

Work it out 5.1 Solution 1 is correct.

Exercise 5C 1 a b c d e f g h 2 3

from

,

from

4 5 6 7

from

and

from

8 9 10

from

and

11 12 13 14 Proof

Mixed practice 5 1 2 3 4 5 6

from

7 8 9 10 11

Focus on … Proof 1 1 Proof 2 3

Focus on … Problem solving 1 1 2 3

Focus on … Modelling 1 1 2 3 Yes, the ball hits the bell. 4

Cross-topic review 1 1 2 a b 3 4 5 6 Proof 7 8

away from the wall

9 10 a b 11 12

13

Chapter 6 Before you start… 1 2 3 4

Exercise 6A 1 2 3 4 5 6 7 8 9 10 11 12 Work done

Work it out 6.1 Solution 2 is correct.

Exercise 6B 1 2 3 4 5 6 7 8 9 10

11 12 13 14 a b

upwards

15 Proof 16 17 18 a b

Exercise 6C 1 a b 2 a b 3 4 a b 5 6 7 8 a Proof b 9 a b 10 a Proof b

Exercise 6D 1 a i ii iii b c 2 a i ii

b i ii c 3 a i ii b c

or

4 5 6 7 8 9 a b c 10 a b c

Mixed practice 6 1 2 a b c 3 a b c 4 5 a b 6 i ii 7 i ii 8 i

upwards

ii 9 i ii 10 i Proof ii Proof 11 i Proof ii 12 13 14 15 i

, Proof

ii Proof 16 i ii iii 17 i Proof ii a b c 18 a

is the weight of a mass of variable weight model gives

b c 19 a b c d Proof

at sea level, where is the acceleration due to gravity. The as the weight of a mass of

at sea level.

Chapter 7 Before you start… 1 2 a

directed towards the origin

b where is a constant

3 4 a b 5

Exercise 7A In all of the following answers represents a constant. 1 a b 2 a b 3 a b 4 a b 5 a b 6 7 8 9 10 11 12 a b 13 a b 14

from origin in the negative -axis direction

15 a

directed in the negative -direction

b c As

,

16

so

, where is a constant

Exercise 7B 1 2 3

in the positive -direction

4 5

in the negative -direction

6

in the positive -direction

7 8 Proof 9 10 11 a b c d

Mixed practice 7 1 a b c d 2 Proof or

3 At

The particle is accelerating in the negative -direction. When

, so accelerating in the

positive -direction 4 a b

c d

in the positive -direction

5 i Proof ii Proof iii 6 i v

ii 5.76 1.92

O

7 Proof 8 i Proof ii 9 i Proof ii 10 i ii

12

24

t

Chapter 8 Before you start… 1 2 3

b

a

a + b 4 Horizontal 5

Exercise 8A 1 a b c d 2 a b c d 3 4 5 6 a b 7 a b 8 a i ii b 9 a i ii b 10

Exercise 8B

, vertical

1 a b 2 a b 3 a b c d

at

4

to the wall

at

to the wall

5 a b c d e f Proof 6 a

to the wall

b 7

at

to the positive i direction

8 9 10 a

at

b 11

speed

Exercise 8C 1 2 3

at

4

at

5

at

6

at

7 a b 8 a b

at to line of centres, at at

along the line of centres

9 a

at

to the string

b c

Mixed practice 8 1 2 3 i Proof ii 4 5 i

to the initial direction

ii 6 i ii 7 i ii 8

at an angle of

above the horizontal away from



in the opposite direction to

its initial motion. 9 i ii 10 i

to the right along the line of centres

ii 11 i ii

to the left along the line of centres

iii 12 i ii

to the left

iii iv 13 i Proof ii 14 a b

has below the line of centres. moves up perpendicular to line of centres.

Chapter 9 Before you start… 1 2 3 4

Work it out 9.1 Solution 3 is correct.

Exercise 9A 1 a i ii b i ii c i ii d i ii 2 a b c d 3 a No, maximum height b No, maximum height c No, maximum height 4 a b 5 6

for a full circle to occur.

7 8 a b c 9 a

when hangs vertically below

b c

Exercise 9B 1 a i Tangential

, radial

ii Tangential

, radial

b i Tangential ii Tangential

radial ,

c i ii

2 a Radial component

,

tangential component . b 3 4 5 6 Radial component

Work it out 9.2 Solution 2 is correct.

Exercise 9C 1 a i ii b i ii c i ii d i ii 2 a b

, tangential component is

.

3 a b c 4 a b c 5 a b c 6 a b c 7 8 9 a b c

in the direction

to the horizontal

10 a b c Proof 11 a b c It will lose contact with the chute.

Mixed practice 9 1 2 3 4 5 6 i ii 7 8 9 i ii

10 a b 11 a When is vertically below b c 12 i

, proof

ii

, proof

iii 13 i

and

ii 14 i Proof ii Radial iii

tangential

rads

15 i ii iii 16 i ii iii iv Proof,

, proof

Chapter 10 Before you start… 1 2

clockwise

3

Exercise 10A 1 2 3 a

b

4 5 Proof 6 7 8 9 a b c 10 11 12

Exercise 10B 1 2 3 4 5 6 7 a b 8 a b

from

,

from

9 a

from

,

from

b 10 11 12 13 a b 14 15 a b 16 a b 17 i ii

Mixed practice 10 1 a b 2 a

from

b 3 a Proof b c 4 5 6 i ii 7 i Proof ii 8 i Proof ii a b µ 9 10

,

from

11 a b 12 i Proof ii a Proof b iii

Focus on … Proof 2 1 Proof 2 Proof

Focus on … Problem solving 2 1 a b 2

at

to original direction of motion

Focus on … Modelling 2 1 2 θ

3 π 24

O

π

2π

t

–π 24

4 a

5 6 a b c i

ii

from SHM model. slower than the SHM model predicts).

from the energy equation (the bob is going

from SHM model. slower than the SHM model predicts).

from the energy equation (the bob is going

Cross-topic review 2 1 a

b c 2 a Proof b 3 a b 4 5 a Proof b Proof 6 a b c 7 a Sphere A:

Sphere B:

b Speed of 8 a b c 9 Proof 10 11 a b 12 a b 13 i ii

14 i Proof

, speed of

ii 15 i Proof ii iii

AS Level practice questions 1 2 a b c It was necessary since the dimensions of would affect the solution found in part b. 3 a b c Light and inextensible. d e 4 a Light and inextensible b Proof c Proof 5 a b It will affect how high above the horizontal can move before the string goes taut. The speed of will be larger if is not directly above . c Proof d e f 6 a

is moving in the opposite direction to its initial motion.

b 7 a b c d

A Level practice questions 1

2 3 4 a b

r 5

c

4 3 2 1 –1 O

2 k

1

d Asymptote at

so that the string does not break, so

.

5 a v 4

2

O

2

4

t

b c 6 a Proof b Proof c Proof 7 a Sphere b 8 a b c d 9 a Proof b c

Sphere :

to the direction

before the impact.

d it will slide. 10 a b c

Chapter 1 worked solutions 1 Work, energy and power 1 Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 1A 6

a b

7 8

EXERCISE 1B 4

a b c

5

6

a b

7

8

9

average force

10

11

12

EXERCISE 1C

EXERCISE 1C 5

a b c d

6

a b c d Model Anita as particle leaving the surface of the highboard and entering the water at a single instant, no resistance to motion. No account taken of bending of highboard.

7

a b c d

8



9

a b

10

11 a b c

EXERCISE 1D

EXERCISE 1D 8

9

a b

10 a b c

EXERCISE 1E 6

7

8

9

a b c

10

11

12

13

14 a

b

MIXED PRACTICE 1 1 2

, so

3 4

i ii

5

i ii

6

i

ii

7

i so ii

8

a

b c

9

i

ii 10 i

ii

11 i

ii

12 i

ii

13 i

ii

iii 14 i

ii iii

and

15 a

b

16 i

ii

iii 17 i

ii 18 i ii iii

19 i

ii

20 i

ii

21 i ii iii Constant speed:

iv

22

23 a

b

c

Chapter 2 worked solutions 2 Dimensional analysis Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 2A 4

a b

5

a b This is the same as the dimensions of so dimensionless.

6 7 8

a

b

9 Force

10

EXERCISE 2B 1

a b c

2

a

b

and

.

c

d

3

4

a

b

, So, not consistent.

c

d

e

5

so not consistent.

a

as is dimensionless.

b 6

a b

7

a

b

so not the same.

c Called moment of momentum as it is momentum length 8

a

b

9

a

,

b 10 a

So, dimensionless and therefore inconsistent. b Change the to

EXERCISE 2C

.

,

EXERCISE 2C ,

5

6

and

are dimensionless.

a b Yes. As the angle must be dimensionless, must have the inverse dimension of , so the dimension of can be determined as

.

7 8

a

is dimensionless. which is dimensionless.

b Let this be decibels 9

a

b

c 10

Index of

MIXED PRACTICE 2 1

a

b 2 because the difference of two lengths is also a length. Similarly,

3

so the equation is dimensionally consistent.

a b i

Inconsistent:

is inconsistent.

ii iii 4

a b Conversion table: CGS

MKS

Factor

has conversion factor

5

as cosine is dimensionless

a b

as cosine is dimensionless

6

7

a b

c

as the two masses must make the same contribution to the force.

,

,

,

,

8

a Angular acceleration



b If consistent the dimensions of all terms are the same.

c

9

a b

Equating indices:

Solving: c Substitute: 10 a b Equating indices:

11 a b



c

12 Equating indices:

13 a

b c

d



Chapter 3 worked solutions 3 Momentum and collisions 1 Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 3A 7

+ve

10N 600

0

t = 0

v

600 t 1 = 300

Change in momentum Change in momentum +ve

10N

8 600

8

t = 0

4

600 t = t

Change in momentum Impulse

or minutes

1.2

15ms–1

t = 0

+ve 1.2 t = 0

Impulse

1.2

v

t = 10

Impulse

10

+ve

0.6N

9

Change in momentum

0.6N 15ms–1

1.2

v

t = 10

Change in momentum

11

+ve 0.5

20ms–1 I

R

0.5

15ms–1

Change in momentum away from the wall. 12

+ve 2

15ms–1 I

2

10ms–1

change in momentum Impulse

away from the wall. +ve

13 250

25

250

t = 0

15

t = 20

Change in momentum Impulse Force is

in direction opposite to the motion. +ve

14 200

5

t = 0

200 t = 20

Change in momentum Impulse

EXERCISE 3B

20

+ve

2 1.5

3

A

B

1.5

0

v

2

A

B

+ve

3 2

10

8

A 6

2.5 B

2

2.5

v

1.5

0

+ve

4 3.5

20

v

5

5

0

2

a Collision 1: 0.45

0 3N s

0.45

v

b Collision 2: +ve 450

2 63

400

0

450

v

400

3

+ve

6 0.7

6

8

0.9

A

B

v

1.6

+ve

7 0.8

6

8

m

A

4

B

4

m

0.8

+ve

8 600

60/3.6

800

0

600

v

800

30/3.6

+ve

9 2

14

0.2

0

2

12

0.2

v

+ve

10

0.75v

8

v

0.38

0.38

0.42

0.75v

0.42

Assuming: the balls are spherical and the same size, the impact is along the line of centres, the contact is smooth. +ve

11 a m

3

5

3

A

B

v1

m + 3

Case 1

v2

Case 2

m + 3

Case 1: Case 2: b

Case 1

Case 2

EXERCISE 3C +ve

2 3kg

4

2kg

5

P

3

Q

v1

2

Momentum: Newton’s experimental law:

e = 1 2 v2

Subtract

and both change direction. Speed of is 3

. Speed of is

.

a Momentum: +ve 0.5

5

1

3

A

B

0.6

v

1

0.5

towards . b Newton’s experimental law: 4

+ve

1.5kg

12m s–1

16m s–1

1.5kg

a After the impulse let the velocity of the sphere be .

b Impulse change in momentum the wall. c Loss in 5

a Velocity

towards wall.

+ve J m v





or

away from

b Impulse

or

c Initial 6

Final +ve

a v

4

10

2.5

A

B

4

13.75

v

2.5

b c Initial Final Loss 7 Collision : +ve 10

0.2

wall v

0.2

Collision : +ve 0.4

0

6

0.2

B

v

A

0.4

Momentum

8

a Momentum:

0.2

0

away from wall.

+ve 0.4

6

8

0.6

A

B

v1

0.4

e = 0.8 v2

0.6

Newton’s experimental law:

Add

,

So velocity of is b Initial B

6

Final B

and velocity of is

0.6

0.6

4.08

Impulse 9 Energy:

loss

gain:

0.4

h

2m

u

gain

10 a

4m

v

loss:

2u

3u

m

A

4m

B

v1

m

e = 0.75 v2

; moving in opposite directions to each other.

Momentum: Newton’s experimental law:

Substitute in Both particles are moving in the original direction of . b Speed of

, Speed of

Initial Final Loss is +ve

11 m

3

m

6

A

B

m

v1

e= 0.8 v2

m

Momentum: Newton’s experimental law: Add

in original direction,

12

m

h1

h2 e

u

13 a

eu

loss

gain:

loss

gain:

: change in momentum impulse

+ve 0.4

6

3

0.4

A

B

v1

0.4

v2

0.4

0.96N

b Newton’s experimental law: +ve

14 a 0.04

5

0.04

3

A

2.2

B

v

0.04

0.04

b Newton’s experimental law: c Impulse on 15

0.5 0.5

2.5m

2m +ve

u

v

loss

gain:

loss

gain:

Impulse change in momentum

upward.

16

0.45

0.45

4 2m

2m

u

v

loss

gain:

gain

loss:

Positive direction

17 2m

0

2m

u

Impulse = 8m

Impulse change in momentum: Positive direction

2m Initial

4

3m

A 2m

0

B v1

3m

v2

Final

First collision: Using conservation of linear momentum:

Newton’s experimental law: Solving

and

and

Collision with wall: velocity of approaching wall is Velocity of leaving wall is 18 First collision:

so for to catch

Positive direction

m Initial

6

2m

P

0

Q

m

v1

2m

v2

Final

Conservation of linear momentum: Newton’s experimental law: Solving

and

simultaneously:

and

Second collision: Positive direction

2m Initial

3.4

7m

Q

0

R v3

2m

7m

v4

Final

Conservation of linear momentum: Newton’s experimental law: Solving

and

simultaneously:

For a second collision, must be moving away from and be moving faster than giving two conditions:

Combining

so

and

19 a Conservation of linear momentum: +ve 0

16 ms–1

3kg

5kg

A

B

3kg

5kg v

v

b Impulse change in momentum

MIXED PRACTICE 3 1

+ve

a 0.4

2.5

5

0.8

P

Q

1.5

v

0.8

0.4

Change in momentum of

in ’s final direction. b Momentum:

2

10

a

0 m kg

0.8kg P

Q

2

4

Momentum:

b

0.8

2

3

10

0.8

50kg

6

8

m

0

m

4

75kg

125

Momentum: So speed

in direction of the motion of heavier skater.

4

m

6

2.5

3m

P

Q

v

4m

Conservation of linear momentum:

5

i a Momentum: +ve 9 P

2 m

Q

0.8kg

v2 3.5

v1 3.5

b change in momentum of

ii Momentum: 0.8

6

3.5

0.4

1.2

v

i Momentum: 0.3

2.2

P

0.3

0.5

0.8

Q

1.1

ii 7

2.75

i Momentum:

0.5

v

0.4

1.5

3

0.6

P

ii

8

Q

0.4

0.1

0.6

v

0.4

–1

0.6

v

i Momentum: +ve

m

0

6

0.5

Q

P v

v + 1

ii Momentum: +ve

m

v

v + 1

Q

0.5 P

iii

9

i

2400

5

A

3

3600 B

v

v

has changed direction. +ve

ii a 2400

5

3

3600

A

v

B

2400

b change in momentum of

I = 0.9

10 i

v

3600

final momentum − initial momentum

I = 0.9

0.4

5

v

0.2

A

B

0.2

6

ii

0.2

6

B 0.2

0.1 C

v1

0.1

v2

Conservation of linear momentum for the second collisions:

NEL: Subtract As

, no more collisions.

11 i

0.2

3m 1.8m

+ve

I

u

v

loss

gain:

gain

loss:

ii +ve

12 i 0.18

m

3

2

m + 0.18

ii a

1.5

0

m

0.18

b Case 1:

Momentum: +ve

0.18

2

m + 0.18

Case 2: Momentum: 1.5

m + 0.18

m

3

v

1.5

13

2kg

8

2kg

v1

3kg

4

v2

3kg

Loss in Momentum: lost initial

–1

final

2

2

:

2

3

2 5

2.6

not possible

3

Possible answers are

and

or

and

collision as in diagram above. Speeds are

and

away from each other.

+ve

14 i

2m

u

A

2m

m

0

B

v

m

Momentum: ii Newton’s experimental law:

3v

which is not a possible result of the

iii

6u 5

m

v1

m

before impact

iv Momentum:

2m

2u 5

3u 5

m

A

2m

B

v1

m

e = 4 5 v2

NEL: Subtract

from

is away from wall, is

towards the wall.

Chapter 4 worked solutions 4 Circular motion 1 Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 4A 4

a

b

, consider proportion of angle in a full turn: in

5

a Angular change is

secs one circuit takes

, time taken is

secs.

secs.

Therefore angular speed is b c

and Therefore

6 For

and hence and

, therefore

As both and are connected they have the same linear velocity. Hence for 7

a

, hence

b 8

therefore

a Its orbit is circular and its angular speed is constant. b i

, therefore , Hence

ii

EXERCISE 4B 4

1500gN FR 35m

a

1500gN

and hence 5 a

2gN

T 0.3m

T

a

2gN

. At the object, b

hence At the object,

6

mgN

a F

E l 2

l 2 mgN

When halfway,

where

, hence

When she moves increases to and So

, so Emily will feel an increase in force by a factor of .

mgN

7 a

F = µ R 55m

a

mgN

hence Also,

, hence

, therefore

hence

, hence

b No it is not a sensible estimate, this is a very low speed to drive around a bend of 8

, rotating

radius.

, hence

mgN

a F

0.13m

mgN Also,



, hence

0.04gN

9

T T

T

A

F = µ R

0.15m

T

0.04gN

B 0.03gN ,

,

,

Towards the centre of motion,

at

.

Hence

mgN

10

F 25m

mgN a

Assume only friction force acting on car is from road surface. Assume car is on point of slipping away from centre.

b

Asphalt i As

, concrete , for asphalt

. and



ii For concrete

c

, hence

For asphalt

hence

Reduce max safe speed from For concrete

to

in wet conditions.

to

in wet conditions.

hence

Reduce max safe speed from d

Where the road is to be built could influence this decision. For a city with a lower speed limit asphalt would be cheaper and suitable. 11a Since and the coordinates of are you can write the position vector as . b Differentiating the position vector with respect to will give the velocity vector for : . c Differentiating the velocity vector with respect to will give the acceleration vector for : . d Since

. This shows that the acceleration is directed in the opposite

direction to the position vector and

.

e Looking at the scalar product of the vectors and you get:

. Hence the vectors are perpendicular and so the velocity vector is perpendicular to the radius.

EXERCISE 4C

2

a

A

θ 25m T

a

B

5g

ω = 4rads–1

unknown

Forces at



b





A

30° 1m T

a

B

0.2g

ω unknown

unknown, Forces at

and hence





c



A

θ 13m 12m

T

B

a

10g

ω unknown

unknown, unknown and hence and hence d

A

25° 1.2m T = 56

a

B

ω = 16rads–1

mg

unknown, 3

,

, hence

RN

4

mg θ

12m

a

and hence and hence

13

5

θ 12 hence Therefore 5

hence



as required.

a

30°

T T

0.8m

T 60°

T

At

0.5g



b

hence

hence

6 10000N

a ms–2

θ

mg Using

, you can find using a right-angled triangle that

Resolving forces perpendicularly you get

. .

. Using

directed towards the centre of the circular motion you get

Substituting the values in and rearranging, 7

a

A θ θ

4

5

1.44m

3

T1 1.80m T2

and so

.

sin θ = 3 5 cos θ = 4 5 tan θ = 3 4

P

θ

1.08m B

mg

Resolving forces vertically you get Using

.

towards the centre of the circular motion,

You can then solve the pair of simultaneous equations to find Using the diagram,

Multiplying equation

by

and equation

Adding the two equations together gives

by

gives

. (the tension in

).

b Given

, you can substitute this into

and rearrange to make

the subject: c For a lower bound for , you need the tension in

as

will be taut by the downward force of

the weight of the particle . Hence

8





a A

2m 30° T

2

B

Q T3

C

T1 2m 30° T 1 P

3g 2g

At : Resolving forces vertically gives Using

, so

towards the centre of the circular motion gives

Given that

,

So

and

Substituting

.

in from part a, you get

c At , you can use

The rod is therefore in compression.

a r r

3r 5

. .

directed towards the centre of the circular motion for .

.

a

.

.

b At : Resolving forces vertically gives

9

.

R

mgN

θ

r

3r 5

2

4r ⎧3 r ⎫ r 2 – = 5 5

↴

⎩ ⎭

b

hence

This gives











.

10 A

45° TA 1.3m TA

B

TB

P

TB ↵ω

3gN a At b and hence For it to be in tension

must be positive. Hence

Hence the lower bound is c When Hence

, away from the centre .

A

11

θ

T 2m T

B

3gN and hence , and hence



but no solution for Hence



)



R

12

F a

mg 22° 55m

When speed is at its maximum to prevent the car from slipping away from the centre friction acts inwards: hence

and

Hence

and

Hence , hence



R

F

a

mg 22° 55m

When speed is at its minimum to prevent the car from slipping in towards the centre friction acts outwards: and hence Hence and hence

R

13

mg θ rm

Assume that

the car can be modelled as a particle,

friction is preventing the car from slipping,

there are no other resistances to the movement. To work out whether the assumption is that the car is on the point of slipping inwards or outwards consider both cases and calculate which gives the coefficient of friction given.

R

F mg θ rm

Car is on point of slipping out. Hence

and





Hence



as required.

MIXED PRACTICE 4 1 a

2

b

and hence circumference

c

and hence



mgN

a F =µ R rm

a = rv

= 0.6mgN

2

mgN

b





3

a Using b Using

. .

4 Using towards the centre of circular motion of the car, the force keeping the car in circular motion is the frictional force. So

.

xl

5

A

(1 – x) l C

T

⟲ ωrads–1

T

mg

B

mg

Since the tension and angular speed are the same, you can use

at both and to give:

. So

and rearranging gives

. Hence

.

6 The tensions are equal, so you can resolve the forces vertically at to get at , directed towards the centre of circular motion of to give rearranging gives 7

a

and so

A

θ T 49L − 24L = 25L

7L

T

T B

T

R

24L a

mg



hence

b



Hence



c There was no frictional force acting. 8

R

a

mg 22° a

10m ↵ω



.

. You can then use .

and hence



b i





R

F = µ R

mg 22° a

10m

and hence

Hence





R

ii

F = µ R mg 22° 10m

and hence

Hence

9



5m

i

R

3m θ 4m

mg Resolving forces vertically you get

so

Using Pythagoras’ theorem you can find that

. and so you can calculate

.

5m

ii

R

3m

θ T 4m ϕ T 0.2g

0.1g Using

and resolving forces towards the centre of the circular motion you get: .

Using

you find that

.

iii Using the diagram you have

.

Resolving the forces at the attached hanging particle you get that the tension in the string is Resolving the forces vertically at you get: Using

and

gives

. .

Resolving the forces at towards the centre of the circular motion you find: . Using

and

gives

.

.

10 A

60°

60°

T

S

B

0.5g

i Resolving the forces vertically you get:

.

Resolving the forces horizontally towards the centre of the circular motion you get:

You now have two equations in and : and Solving these simultaneous equations gives ii When the string

and

is on the point of becoming slack,

Resolving the forces vertically,

. .

, thus

.

Resolving the forces horizontally towards the centre of the circular motion: . This gives 11 i

45°

.

R

T

45°

mg Resolving vertically: Resolving the forces horizontally towards the centre of the circular motion:

You now have two equations in and : and Adding the equations together gives: .

ii Looking at the greatest value for the angular speed this will occur when From the equations formed in part a with

.

:

This gives: Using 12 i

you get

.

A T θ θ

3.2m

ω = 4rads–1 S 2

1.6 B

2g

θ

Resolving the forces vertically you get

.

Resolving the forces horizontally towards the centre of the circular motion: You now have two equations to give and

and .

in the unknowns and . You can solve these simultaneously

ii The least possible speed will happen when

.

You can resolve the forces vertically and horizontally again to give: and

.

Dividing the first equation by the second: This gives

.

R

13

θ θ T

θ

mg

a Resolving forces vertically gives .

.

Resolving forces horizontally towards the centre of the circular motion:

Substituting

b Using

into the above equation gives:

from part a:

c For the motion of the particle, the string needs to be taut and thus must have tension. . This gives:

14

µR

R

mg i Resolving the forces horizontally towards the centre of the circular motion you get:

Resolving the forces vertically: Combining these two equations gives

.

ii A

T

0.5

0.3

0.4

θ

mg Resolving horizontally towards the centre of the circular motion:

Resolving the forces vertically you get:

Solving gives

and

.

Chapter 5 worked solutions 5 Centres of mass 1 Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 5B 9 Centre of mass for arc of angle

is given by:

Using Newton-Raphson method let

radians therefore angle of

radians

EXERCISE 5C 2 3

4 5 6 7 Total length of

The centre of mass is

8 For the triangular lamina:

For the composite body:

from

and

from

.

9

10 Total mass of rod and lamina

The centre of mass is 11 Cylinder:

. Measure distances left and down from . For rod:

metres from

and

metres from

, hemisphere:

Total mass 12 Radius of cone removed (similarity) Vol. of full cone: Vol. of cone removed:

13 Take measurements along the main axis of the hat. Conical shell: Curved surface area of hat Area of annulus

14 For composite body:

. Uniform solids so work with volumes.

Cancelling and rearranging:

MIXED PRACTICE 5 1 Moments about

, as required.

.

metres

metres 2

3

4 Area of triangular lamina

since perpendicular height of triangle is from

pythagorean triple. Area of

units squared

5 Centre of mass for quarter disc from

6

7 Centre of mass of arc of angle

Measuring from rod

8 Centre of mass of

9 The combined shape is a single uniform triangular lamina with vertices at the given lines. Therefore its centre of mass is:

10 Cancel , multiply by and substitute for and

11 Let vertex of conical shell be Using

relative to

Chapter 6 worked solutions 6 Work, energy and power 2 Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions.

EXERCISE 6A

EXERCISE 6A 5 6

7

8

9

10 Reduction in

work done by resistance

11

(reject the negative solution because of the context) 12 and, by trial and improvement,

EXERCISE 6B

EXERCISE 6B 5 6 7 8 9 10 Change in 11 12

13 Equilibrium extension, , such that 14 a Conservation of energy: Rearrange: b At the fullest extent: leading to 15 change in

change in tension mean extension

16 Using increase of

17 Let distance

loss of

be metres

18 a b

Conservation of energy:

EXERCISE 6C 1

a b

2

a

b 3 to nearest degree. 4

a b

5

6 Going up the plane: Going back down the plane:

7 Extension in longer string: Extension in shorter string: Let be the angle between the longer string and the vertical. Let be the angle between the shorter string and the vertical. The mass hangs in equilibrium:

8

a Conservation of energy:

b 9

a Acceleration resultant force mass

b

10 a Constant speed tractive force resistance resolved component of weight

b Using the same method:

EXERCISE 6D 4 5

6

7 8 9

a b c

10 a

b c

MIXED PRACTICE 6

1

2

a b c

so for

the force is

3

a b

c

4

5

a

b Integrate the expression for acceleration:

6

i

ii Starting

7

i

ii

8

i

ii

9

i

To calculate the max. speed at equilibrium extension: Using conservation of energy:

ii

in equation

: cancel

then:

Which simplifies to:

10 a Initially in equilibrium so

Conservation of energy for , speed just before collision:

Conservation of momentum, speed of combined particle: therefore

b Conservation of energy:

11 i Conservation of energy: loss of ²

²

ii

12 Work done by engine work done by resistance increase in KE

13

14

15 i

Use conservation of momentum when strikes

ii

16 i ii

iii

leading to

, a deceleration of

17 i ii a Max. extension when

.

b

c Differentiate equation

18 a

with respect to

is the weight of a mass of The variable weight model gives

at sea level, where is the acceleration due to gravity. as the weight of a mass

centre of the earth. Since the radius of the earth is metres sea level. b

c

19 a

b c

d

and

at a distance metres from the is the weight of a mass of

at

Chapter 7 worked solutions 7 Linear motion under variable force Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 7A 6





7

,

,

At



,



When

At

,

,

8







and









so ,

9

,





, is a constant Let



At

,

At

,



,

10







, :





is a constant. At

,

At

, ,

11

so

,





So,





Where is a constant. At

,





when



Direction for question:

12

a Since acceleration is positive for

the maximum velocity will occur at

b





At

,

,

13

.



,

a At

,





The particle is b

away from the origin in the negative -axis direction

At







At



Time taken from origin is

(since it takes

to travel to the origin from its starting

position). 14

direction for question:

at

,

therefore therefore when

where is constant ,

so

or

At 15

direction for question:

,

,

a At

,

so

Magnitude of acceleration is -direction. b

Using partial fractions:

and it is directed in the negative

Let

,

Let

so ,

so



At

,

:









As

c

,

so that

A bound for the distance travels from is 16 Maximum displacement is

.

, suggests sinusoidal function:



EXERCISE 7B 3

,

direction:



,





,



At

,

,



in the positive -direction.

,

4

At

,

At

,

5



,



direction:

(force in direction of motion),





,

,

where is a constant.

( is a constant) At

,

: so



, at

So has a speed of 6

,

and is travelling in the negative -direction.

,

direction for question:

when

.

and



when

:



When

:



travelling in the -direction. 7

,

direction:



,

,



,



At

,

:

8



direction:

,



and

At

,



:





9

,

direction:

At

,

:



Since

,

,

,









acceleration will always be positive and since at

positive. 10

,

direction:

,

,

,

,



velocity will always be

Use an integrating factor:



We need to integrate using by parts, twice.

.

11

,

.

when

direction



a

, where is the acceleration due to gravity.

At

,

.



:

where is a constant.



b At

,

:

, so



c





Let





So

giving

.

The other solution is unrealistic. If continued to increase with time, it would eventually exceed the weight, and the stone would fall back up! This possibility has been rejected. d Since the stone is falling from

it will be the first time the displacement in

.

MIXED PRACTICE 7 1

direction:

a b At



c At

,

:



, where is a constant

which is when

d 2

direction:

,

,

At 3

,

:





direction for question:

,

Instantaneously at rest when

,

,

:

At

,



, where is a constant.

:

so At

, hence:



or

,

The particle is accelerating in the negative -direction. At

,

, so the particle is accelerating in the positive -direction.

direction:

4 a



,

At

,

:





b (assuming positive displacements) when







c

d When

,



Magnitude of acceleration is

, direction is in the positive

-direction. 5

, i

,

ii

At

,

so:

,

,

,

, where is a constant.

iii

where is a constant. where is constant At

,

:

At



:

6

, i For



,

,

,

direction for question.

,



For the initial condition here we need the final condition from the first force acting. for the first force.

for At

,



When

,

So, at







for



At

:

, so the speed at

ii

for

is three times greater than at

. for

.

v

5.76

1.92

O

7

12

,

direction for question

24

t

,

,

,



At





:

When





:





8

, falls freely for

,

,

direction for question,

i First we need the initial velocity before it hits the resisting medium: ,

,

,

At



,



for the resisting medium.

In general for the resisting medium:





At

,





ii

distance travelled in

9

,

direction for question.

Deceleration:

i For engine Resulting force





.

ii

At

,

At 10



:



,

Direction for question,

,



i Using

:

Multiply by



:

ii



At

,

:

Chapter 8 worked solutions 8 Momentum and collisions 2 Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions.

EXERCISE 8A

EXERCISE 8A 4 Principle of conservation of linear momentum:

5 Principle of conservation of linear momentum:

Velocity of is 6

a b

7



a Principle of conservation of linear momentum:

b Initial Final Loss in 8

a i ii b

9

a i

ii

b Total impulse 10



but

EXERCISE 8B

no change in momentum so speed

1

a Direction – no change. +ve

⎧4⎫ ⎩7⎭

2

⎧a⎫ ⎩b⎭

2

before

after

to wall.



b Magnitude 2

+ve

a

⎧1⎫ ⎩6⎭

4

⎧a⎫ ⎩b⎭

4

before

after

i direction: no change

.

j direction: b



loss





loss 3

a Components Parallel to the wall: b Perpendicular to the wall: c After the collision: d

60°

θ v

5

If it rebounds with speed at to the wall,

4

30°

θ

5

v

There is no change in the component of velocity parallel to the wall.

Perpendicular to the wall

5

θ v1

v

a b c d e

f

6

Loss in

a

45°

θ v

4

Parallel to the wall: Perpendicular to the wall:

b Impulse on ball change in momentum perpendicular to the wall

7 Using the impulse/momentum triangle below: j

8 2

4 6Ns

By Pythagoras:

θ 6

mv



i

,

The minus sign indicates below the i direction. 8 Momentum–impulse triangle:

m u = 48kgms–1

48 θ I = 10N s

mv sketch

mv

10

triangle

Change in 9 Momentum–impulse triangle:

30° m u = 8

m v = 10

I

Remember to put

10 a

35°

and

coming from the same point to get the correct triangle.

θ v

5

Parallel to cushion, no impulse: Perpendicular to cushion:

b

11 Draw two diagrams showing the components of the velocities before and after each collision. Show the cushions at right angles. First collision: u cosθ

u cosθ

u sin θ

Second collision:

eu sin θ

eu sin θ

u cosθ eu cosθ ϕ

eu sin θ

EXERCISE 8C 1

2kg

3kg

⎧3⎫ ⎩2⎭

i

⎧2⎫ ⎩3⎭

No change in velocity in j direction. So velocity after collision Momentum in i direction: Newton’s experimental law:

:

Substitute in

:

2 No change in velocity in j direction.

In i direction momentum: NEL in i direction

:



Substitute in 3

:



3j 2i 3kg

Before

0 2kg

+ve 3j

0 xi

3kg

After

yi 2kg

After the collision:

in i direction.

Momentum: NEL:

:

Substitute in

:

4 Resolve into components:

,

5sinθ

0

A 3kg

Initial

B 1kg

5cosθ

0 e = 0.4

+ve 5sinθ

0

A 3kg

Final

B 1kg

v1

v2

Momentum along line of centres:

Newton’s experimental law:

:

Substitute into

A

4

1.95

:

0

B

3.15

5 Before

A

6sin30°= 3

6cos30° =3√3

+ve

A After

B

3sin40°

3cos40°

e = 0.3

B

2kg

1kg

3

v1

3sin40°

v2

Along the line of centres: Momentum



NEL:

:

Substitute into

:



,

3 θ 1.9486



3sin 40 θ1 4.1969

6 Along line of centres.

Before

+ve

6sin30= 3

P

6cos30 =3√3

After

10cos30 =5√3

3

0.5

10sin30= 5

Q

5

2

v1

e = 0.45

v2

Momentum: Newton’s experimental law:

:



3

5

P

  Q θ1

θ 6.28 3

2.68 3



7

a

5 45°

2kg B

v1

3kg v1 After

v2

Conservation of momentum parallel to the string:

b 8

Impulse change in momentum of (or )

a For before the collision: Impulse change in momentum

b

12 before 45° 2kg A v1

1kg B after

v1

v2

Conservation of momentum parallel to the string:

No impulse perpendicular to the string, so Final speed of 9

a

Before 6 60° 2m A v1 60° After

m B 60°

v1

v2

Conservation of momentum parallel to the string:

No impulse perpendicular to the string, so speed of Direction θ

to the line of the string 2

3 3

b

Impulse tension change in momentum of along the string

c Initial

Loss in

MIXED PRACTICE 8 1 Impulse–momentum triangle:

.

9.6N s

I

θ 7.2N s

2 Impulse–momentum triangle:

mv = 1.2

I

60°

θ mu = 1.6

3



i Impulse–momentum triangle: mv = 1.25

I = 2.6 α

θ m u = 3.15

ii



,

4 Impulse momentum triangle: 0.6 × 6 = 3.6

I =5

α 0.6 × 8 = 4.8



, so

5

i

Impulse momentum triangle: 15

mv

θ 11

ii 6

i

0.1 θ I

0.125

ii 2.5cosθ

2.5cosθ

2.5esinθ

2.5sinθ

Wall

7

i Along line of centres: v

0

Before 4

8

A 6kg

+ve

v

B 3kg

0

After v1

v2

Momentum: NEL:

:

the component of along the line of centres

ii

as 8

,

6

0

8 15

A 5kg

B 3kg

6

0

v1

v2

Initial components of velocity of are

vertically and

horizontally

Components of velocity vertical to line of centres unchanged by collision. Momentum along line of centres:

Newton’s experimental law:

at an angle of above the horizontal away from Speed of is 9

in the opposite direction to its initial motion.

i 0

0.6

Before 2.8

0.8 A 0.1kg

+ve

0

B 0.4kg 0.6

After 0

NEL:

v2

v

ii 0.6 α 1.5

Direction of

s motion before collision:

change in direction is 10 i 0.8

0 Before 3

0.6

A 0.1kg

+ve

B 0.2kg

0

0.8

After v2

v1

Momentum:

:

NEL: ii





B

θ

before

1

v

after

0.8 α 2.04

, Angle change

to the right.

0.3

0

11 i Before 0.3

0.3√3 A m

+ve

B m

0

0.3

v

After 60°

v1

v2

v

0.3

60°

v2

The velocity of perpendicular to the line of centres remains constant so

ii Momentum: iii Considering restitution 12 i 4

+ve

0

Before 3

2 A 2kg

B m kg

2

α

4

v1

Initial

After

0

0.8

of

, final

Energy loss ii Find : so to the left. iii Along line of centres: Momentum

iv Newton’s experimental law:

13 i

along line of centres. u sin α

u sin α

Before u cosα

u cosα

A 2m

B m

u sin α

u sin α

After v2

v1

Momentum:

 

NEL:

  :



speed of remains the same.

ii After the collision: A u sin α O

A moves up perpendicular to line of centres. u cos α α u sin α

has velocity at below the line of centres 14 i Let be the velocity (in the direction of

) at which begins to move when the string becomes

taut. By the principle of conservation of momentum parallel to

ii

starts to move along the length of the string with speed Impulsive tension change in momentum of (or )

:

.

Chapter 9 worked solutions 9 Circular motion 2 Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 9A 3

v ms–1

O θ

a

a ums–1

a

loss

When

gain

,

For a complete circle to occur





y

1 1 2

x

O

–1

If Hence

or for

or

so it will only reach a height of

coming back down again and repeating. b

again for a full circle

before

y

1 2 3

x

O

–1

when

or

, hence

for

or

.

Therefore the bead will not make a full circle. It will reach a height of before coming back down and repeating. c When

for a full circle



y

1

x

O

–1

When When A full circle will not occur for this bead as it will have velocity will come back down the same way it went up. 4

,

O

0.4m

A

mg = 0.5gN

loss

a

b



gain

by Newton’s second law:

at the top of the circle and

O 0.4m

a

T v = 2.8ms–1

0.5gN

Hence 5

v ms–1

O θ 0.4m

T u ms–1

A

loss

6

0.4m

gain



,

v ms–1

O θ

1.5m

1.5m

u ms–1

1.2g

1.2g

For a complete circle to occur loss

gain:

Minimum speed occurs when for a full circle to occur.



and hence

, hence

7

and

. For a complete circle to occur

.

u ms–1 TH

mg

r O

r

θ

r TL

mg N

2u ms–1

mg N

At the highest point loss



where

gain:

8







,

2m

O

P

0.5gN

a The maximum speed occurs when all below . gain

loss:

b

O

2m

u = √ 2g

θ h

2m T

0.5gN P

v ms–1 0.5gN

gain

loss:

has been converted to

so when hangs vertically







resolving and using , where

c

9

0.15m

O

a

60°

RN

0.15m

P

RN 0.025gN

0.025gN

gain

loss:

resolving and using

b



c Considering conservation of mechanical energy: loss work done due to friction

EXERCISE 9B 4

,

where

4 ms–1

v

v

When the particle is horizontally in line with the centre of the circular motion the tangential acceleration will be the acceleration due to gravity. When the particle is moving from the highest point the particle is moving towards the highest point

. Therefore the magnitude of the tangential

acceleration is 5

v O g ms–2

90°

5 ms–1

At the horizontal position you can find using conservation of energy: Before:

After:

We can find the tangential component of the acceleration by .

This gives: Using 6

, ,

P

T

O θ

4 ms–1 P v ms–1

Using conservation of mechanical energy:

, when

and radial acceleration

Before:

,

After:

,

The radial component of the acceleration is given by: The tangential component of the acceleration comes from a component of the acceleration due to gravity:

EXERCISE 9C 2

, u = Oms–1 0.5m

g

0.5g v ms–1 2m

a

gain

loss:

b

,

,

,

Hence , 3

,

,

,

a

gain

b



loss:



where

,

,



or

,

c

,

4

,

,

gain

a

loss:

b



where

,

c

,

5

, ,

X

P

Rm u = Oms–1

,

, Oxy θ Rm Y Dm

a

gain

loss:



,

(not possible)

b

where ,





c



:

6

, X

P

,

,

,

Rm u = Oms–1

Oxy θ Rm Y Dm

a

gain

loss:





b

:

,

,

c

where

,

,

7 The particle will lose contact with the surface when

.

RN

P

u = 0ms–1 RN X

5° mg θ O

At use

mg 2m

and resolve in the direction of acceleration: where

gain

Therefore, When

v = v ms–1

loss:

. We therefore need . with

8

,

. The particle leaves the surface of the sphere at . RN A u = Oms–1 25° 2gN

B

θ

RN

0.9m

O

v = v ms–1 2gN

gain

loss: hence



When the particle leaves the sphere at B, (

)

resolve in the direction of :

When

,

,

And calculating gives

When 9

: A

a

θ 2m

4m P

6m 4m

hence b ( ) Use

65g

.

and resolve in the direction of :

. Therefore we need .

A 4m a

4m

T

2m

P

vms–1

65g N

where gain

.

loss:

Therefore



and



, hence

c A

30°

60°

vms–1 30° P h

4m

65gN

gain

loss:



Height change Hence 10 a



loss

in the direction

to the horizontal.

gain:

O

v = v ms–1 θ

T 0.6m

T 0.03gN u = 3ms–1

0.03gN

given b Resolving the forces in the direction of and using Newton's second law, where

c The string becomes slack when

.

Therefore, since cosine is a decreasing function for . Hence the marble does not make full circles. 11 a

loss 3m

RN

θ

A

gain:

, hence

60° 3m

u = 0ms–1 RN

RN B

3gN 3gN 3gN

7m

45°

At

C 2m

θ OBC

7m



Hence b

gain

loss:

Hence

.

c For the bag to be in contact : For arc AB:

Use

where

and resolve in the direction of :

θ

60° 60°–θ

θ+ 30° 60°–θ θ+ 30° mgN

, the string will first become slack for

Hence ,



,



hence

For arc

At

, and contact is kept.

conservation of mechanical energy gives



, hence



gain

loss

where

Resolve in the direction of acceleration:





Hence y

17 21 y = cosx

O

36°,0 45°

x

Hence the bag will be in contact until

and then it will lose contact with the chute.

MIXED PRACTICE 9 1 When the rope makes angle with the downward vertical, by conservation of energy:

2 So

3 Angle made with upward vertical

4

so

.

,

O 60°

2m

v ms–1

2m

u = 7ms–1

P

0.4gN loss

gain where

Hence 5



metre, loss

,

.

gain

When the rope makes

degrees with the downward vertical we can use

the rope towards the centre of motion:

in the direction along

6

i

v ms–1

O 1.05m

θ

RN

1.05m RN

0.6gN 5ms–1

0.6gN

loss

gain where



Hence ii The particle will leave when (

) Use

and resolve in the direction of where

When 7



,

. At the lowest point:

u = 7ms–1 3m O

T

3m vms–1

0.4gN

Using where

Hence

and resolving in the direction of gives .

gain

loss

O

8

θ

a

v ms–1

a

P

u ms–1

mgN

mgN

loss

gain

where

Hence when the rod makes an angle with the downward vertical: For the particle to move in complete circles

for all values of .

Hence 9

i By considering conservation of mechanical energy from the initial starting position until the string is horizontal: loss gain. Taking the initial starting position as our zero

line:

Using

directed towards the centre O,

Thus

, where

.

.

ii By considering conservation of mechanical energy from the initial starting position until the string is highest point in the motion: loss gain Taking the initial starting position as our zero

Using

line:

directed towards the centre O:

, where

particle. Thus 10 i The particle loses contact with the sphere when u ms–1

RN

rm θ

mgN

vms–1

O

gain

loss

where Hence

.

.

and

is the weight of the

When the particle loses contact (

) Using

and resolving in the direction of gives: when OP makes angle with the vertical

ii If leaves the sphere the instant it is projected then

when

11 i When is vertically below as is the greatest value here. O

rm

a T

P mgN

ii Use

iii O θ

and resolve in the direction of

rm

vms–1 = 6 gr T

rm mgN u = 3 gr mgN

loss

When

gain





where

.

.

a

O

T mgN

120°

mgN

(

) Using

and resolving in the direction of

12 i By considering conservation of mechanical energy:

O am

v ms–1

am θ

R

R mg u ms–1 mg

loss

gain



where Hence acceleration (

) Using

and resolving in the direction of acceleration gives:

as required. ii ‘Just reaches’

when



.

Hence when Hence iii oscillates between





as required. hence when

13 i When the string is horizontal

O 2m

v ms–1

2m

θ

T

T 0.4g u = 7ms–1 0.4g

By considering conservation of mechanical energy: When

(

loss

where

,

) Use

and resolve in the direction of

ii Generally (

)

resolve in the direction of . where

When 14

,

, i Using the principle of conservation of mechanical energy:

0.5m

O

3 ms–1

θ

P v ms–1

This is the gain in kinetic energy:

ii Using

for radial acceleration: (towards the centre of the circular motion).

gain

θ

g

At the tangential component of acceleration is

.

iii Resolving forces along the radial direction:

When the string instantly becomes slack,

.

y

x

(in radians) and

(in radians) is the first positive value.

O

15 i

θ

0.8m

0.8m T

T

v ms–1

0.3gN u = 5.6ms–1

0.3gN

loss

gain where

(

) Using

and resolving in the direction of

ii The string becomes slack when

.

Hence iii



where

u = 2.29ms–1 180°–θ

g 3

2

θ –90° 5

Hence:



Velocity:

,

above the point of projection.



16 i Using the principle of conservation of mechanical energy for Before:

,

. After:

speed of before collision.

ii Using the principle of conservation of mechanical energy for Before:

Using

iii

when

. After:

in the direction towards the centre of the circular motion:

The radial acceleration Then transverse acceleration iv When reaches its greatest height there will be no vertical component of velocity. So we will only have the horizontal component of its velocity, which is: v

60°

120°

Using conservation of mechanical energy to find the additional height moves once the string is slack: Before:

30°

After:

1 0.7× = 0.35 2

120°

0.7m

Total height for



Chapter 10 worked solutions 10 Centres of mass 2 Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. EXERCISE 10A



1

Same answer can be seen by symmetry.

2





3

a b

4



5 As

, let



6

and



7

8

9

a

b



c





10

11 Hemisphere symmetry means

To find , integrating using the substitution

This result for is the area of a quadrant of a circle of radius .

and by symmetry,

and

12

EXERCISE 10B 7

a Height of trapezium Area of trapezium

b 8

a

.

the centre of mass of the missing quadrant lies at:

b 9

a Let be

, be

etc.

b Let be angle between

,

and vertical

10 11 12 13 a Let be the centre of arc and be the position of the centre of mass.

Radius of arc

and Let be mid-point of chord

Let be angle between

and vertical:

b 14 15 a Friction On point of sliding b

friction

16 a

(note that

)

and b Using

and considering the rectangle

and

:

for the inverse function 17 i uniform lamina mass area

ii Distance from to point of contact of lamina with block, Resolve weight into (parallel to ) and (parallel to ). Let the normal reaction at the point of contact of the lamina with the block be . Equilibrium equate moments about

MIXED PRACTICE 10 1

a b Take moments at Take moments at

2 Let be

,

lies on

,

lies on

.

a

b

3

, and

a Let be

,

lies on

,

lies on

.

because it is an axis of reflective symmetry.

b c

4

and 5 Take moments about But 6

i

ii

7

i

.

ii Let

be

: and

8

i Volume of solid Let the axis of symmetry be the -axis:

Cancel and ii a Take moments about contact point:

b 9

10

11 a

b Considering the function as two regions in the rectangle in part a:

The -coordinate for the inverse function, 12 i

ii uniform lamina mass area a

b

iii Let the tension in the wire be . Find the distance of the centre of mass from Angle between

and horizontal

Equilibrium equate moments about

Cross-topic review exercise 1 Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. 1 [F]=MLT−2∴MLT−2=[ c2][p]α[v]β[A]γ∴M1L1T−2=[ ML−3]α[ LT−1]β[ L2]γ M: 1=α, T:−2=−β, L:1=−3α+β+2γ=−3+2+2γ∴γ=1 D=c2pv2A∴v=2DcpA 2 a I=Ft=10(30)=30 kg N s−1 b (300)=2(x0)−2(10)∴2(x0)=(320)∴x=16 m s−1 3 15 revs per minute =15×2π radians per minute=1560×2π rads per second =π2rads s −1=ω∴F=mω2r=4×(π2)2×1=9.8696 N 4 Initial velocity=1.50.5=3 m s−1 Energy 12×0.5×9=0.5×10×h ∴h=920=0.45∴cos θ=11201∴θ=56.6∘ 5 Force down=mgsinθ+F=1500×10×0.1+300=1800 N Power=Fv=1800×20=36 000 W=36 kW 6

n=1. [ dxdt]=[ xt]=LT=LT−1 Assume truth of case n:[ dNxdtN]=LT−N Differentiate with respect to t:ddt(dNxdtN)=dN+1xdtN+1 ∴[ ddt(dNxdtN)]=1T[ dNxdtN]=1TLT−N=LT−N−1=LT−(N+1) ∴Truth of case N⇒truth of case N+1 and it is true for N=1∴True for all N.

7 Momentum(2×4)+(2×0)=(2×0)+(2×vi)∴vi=4 Time for B to get to edge of table=0.54=0.125 s s=ut+12gt2,u=0,g=10∴1.4=0+5t2∴t2=1.45 t=±1.45 but t cannot be negative∴Total time=0.125+1.45=0.654 seconds 8 CLM:  (4×1)+(3×2)=(4×v)∴v=2.5 NEL: v1=−ev=−0.8×2.5=2 m s−1 away from the wall.

y

9

–7

7

A

x

mBg mAg

Moments about O: mass of figure=40πρ0×49πρ−3×9πρ=40πρx∴−27=40x∴x= −0.675∴Coordinates (−0.675, 0) 10 a [ Force]=MLT−2 b [ 2P]=MLT−2,[ πD]=L and [ D2−d2]=L ∴[ D−D2−d2]=[ L−L]=L ∴[ BHN]=MLT−2L2=ML−1T−2 T

11

A

m v2/ r

x

T B 3g

T=mv2r,T=3g∴T=3v2r=3×3.52r=3g ∴r=3×3.5210×3=1.225∴x=2−1.225=0.775 m 12 θ T m v2 r

r 2g

Tcos θ=2g 1 Tsin θ=2×25r=50r 2 sin θ=r1.5 3 1÷2:cos θsin θ=2gr50=25r taking g = 10 m s−2 4 4÷3:cos θsin2 θ=2r5×1.5r=35 5cos θ=3sin2 θ=3(1−cos2 θ)=3−3cos2 θ 3cos θ+5cos θ−3=0 cos θ=−5±616=0.468 or−2.135 (invalid)∴θ=62.095∘ height of cone =1.5cos θ=0.702 , base radius of

cone =1.5sin θ=1.326 Volume of cone =13πr2h=1.29 m3 13 Mass of large square=m, mass of small square=m4×2=m2 Take moments about the x-axis: (2×m)+(3×m2)=3m2x∴7m2=3m2x∴x=73 By symmetry, the centre of mass lies on the line y=x ∴y=73

Cross-topic review exercise 2 Worked solutions are provided for all levelled practice and discussion questions, as well as Cross-topic review exercises. They are not provided for drill questions. 1

a

mdvdt=3t−1∴mv=3t22−t+c∴v=34t2−t2+c

b t=0 v=5 ∴c=5,v=34t2−t2+5 t=5:v=34×25−52+5=854=21.3 m s−1 c

Fm=(t−122)=a∴v=∫adt=(t22−12t+a2t+b) when t =0 v=(1−1)∴a=1 b=−1∴v=(t22−12t+12t−1) when t = 5 v=(252−52+110−1)=11i+19j

2

a Vector triangle and use of cosine rule: 42=2.42+5.22−2×2.4 ×5.2×cos θ ∴cos θ=3552 b

Sine rule: sin α2.4=sin θ4 and sin α= 2.44×1−(3552)2≈0.4437∴α≈26.3° ∴ the impulse is 180°−α=153.7° from the original direction of motion.

3

a

∥ to wall:mucos 60°=mvcos 45°  ∴u2=v22 ∴Loss of KE as fraction of starting KE =12m(u2−v2)12mu2 =u2−(u2)2u2 =12

b ⊥ to wall:mvsin 45°=e×mu sin60°∴v22× =e×u×32 But u=v×(2)∴v(2)=e×v(2)×(3)∴e=33 4 Maximum speed of P at the bottom of the circle. Conservation of energy: 12m×v2−12m× (v2)2=mg× 1.5 ∴v2− v24=3g∴v2=4g  and  v=2g 5

a Volume = π∫02.510x dx=π×[ 5x2 ]02.5 =125π4 as required. b

6

a

Ax¯ =π∫02.510x2dx=π× [ 10x33 ]02.5= 625π12 and =x¯=53 Impulse,J=∫04(5t−2)dt=[ 5t22−2t ]04=40−8=32  N s

b J=mv−mu⇒32=0.8 ×50−0.8×u ∴0.8u=8⇒u=10 m s−1 c

∫0t(5t−2)dt=[ 2.5t2−2t ]0T ∴2.5T2− 2T=0.8 ×70−0.8 ×10 ⇒2.5T2−2T=48 ⇒5T2−4T−96=0 ⇒(5T −24)(T+ 4)=0 T>0∴T= 245=4.8  seconds

7

a cos α=45, sin α=35, cos β=1213 and sin β=513 Sphere A:∥to lines of centres 3 cos α=125 m s−1

⊥ to line of centres 3 sin α=95 m s−1 Sphere B:∥to lines of centres −2 cos β=−2413 m s−1 ⊥ to line of centres −2 sin β=−1013 m s−1 b Let vA and vB be the final velocities of spheres A and B ∥ to line of centres. Newton's experimental law: (125+2413)e=vB−vA e=12 so this simplifies to 13865=vB−vA1 Conservation of momentum: (125×1.5−2413×0.75)=0.75vB+1.5vA Which can be simplified to: 9665=0.5vB+vA 2 Add equations 1 and 2: 23465=1.5vB and vB= 2.4 m s−1 and vA≈ 0.2769 m s−1 ∴Speed of A≈0.27692 + 1.82=1.82 m s−1 ∴Speed of B≈ 2.42 + (1013)2=2.52 m s−1 8

a Conservation of energy: 12 m× 4.52− ½ mv2=mg×0.5×(1−cos θ) Cancel m then simplify: v2=20.25−g+g cos θ and v2=10.45 +g cos θ ∴Force towards centre, mv20.5=T−mg cos  θ ∴T=mg cos θ+2mv2=mg cos θ+2m(10.45+g cos θ) ∴T= 3mgcos θ+20.9m b If T=0 cos  θ=20.93g so height=0.5 × (1 +20.93g)≈0.855 metres The horizontal speed when the string goes slack is vcos(180°−θ)=−vcosθ At max. ht. this will be all the speed it has (vertical speed =0). c KE lost = GPE gained: 12m×4.52−12mv2cos2 θ=mgh ∴ 4.52−3.483×(20.93g)2=2gh ∴h≈0.943 m

9 Let speed of R be u m s−1 before collision and v m s−1 afterwards. Let speed of S after collision be w m s−1. Conservation of momentum: ∥ to line of centres: u cos α=v cos β+w1 ⊥ to line of centres: u sin α=v sin β2 Newton’s experimental law: w−vcos β=e×ucos α3 Rearrange 3: e=w−vcos βucos α Substitute for w from 1: e=ucos α−2vcos βucos α =1−2vcos βucos α ∴1 −e=2vcos βucos α but from 2 vu=sin αsin β∴1−e=2sin αcos βsinβcos α=2tan αtanβ

∴(1−e)tan β=2 tan α as required. 10 Because the solid can rest anywhere on the surface of the hemisphere (with the centre of mass above the point of contact with the plane) the centre of mass lies at the centre of the disc joining the hemisphere and the cylinder. Take moments about the centre of mass: 3r8×43πr3=h2×πr2h Cancel πr2:3r28×43=h22and  h2=r2 and h=r 11 a Conservation of energy: 12mv2−12 mu2=mg(a−a cos θ) ∴v2−u2=2g(a− a cos θ) Force towards centre: mv2a=mgcos θ−(R) But R=0 as particle leaves the sphere∴v2=ag cos θ Substitute v2 in energy equation: ag cos θ−u2=2ga−2ga cos θ Rearrange: cos θ=u2+2ga3ga b Conservation of energy over remaining height to fall: 12m× 3ag2−12m×ag cos θ=mga cos θ 3ag−2ag cos θ= 4ag cos θ∴3=6 cosθ and θ=60° 12 a The centre of mass lies on the main axis of symmetry of the cone. Let the vertex of the larger cone be (0, 0) As the cones are uniform, work with volumes: (13πr2×1.5h+13πr2×h)x¯=13πr2×1.5h×1.125h+13πr2×h×1.75h Cancel  13πr2h, then 2.5x¯ =27h16+1.75h and x¯ =11h8 b Using trigonometry on a cross-section through the cones ∥ to axis of symmetry: hr=r1.5h−11h8∴r2=h28  and r=2h4 13 i Conservation of momentum ∥ line of centres: 2m×a cos α−m×b cos β=m× 2 cos 45°1 Newton’s experimental law: (a cos α+b cos β ) × 23=2 cos 45°2 From 1: b cos β=2a cos α−2 Substitute in 2: a cos α+2a cos α−2=322 ⇒3a cos α=522 ⇒a cos α=526, as required bcos β=2acos α−2=532−2=232 ii ⊥ line of centres: a sin α= 2 But a cos α=526 ⇒tan α= 1252⇒α=59.5° (3 s.f.)(59.49°) ∴a= 2sin 59.49∘=2.32 (3 s.f.) 14 i Conservation of ME: 12mv2=12m× 42−mg(0.6−0.6 cos θ)⇒v2=4.24+11.76 cos θ

Centripetal force: mv20.6 =R−mg cos θ⇒R=3.18 +13.23 cos θ ii R→ 0⇒cos θ≈−0.2404⇒v2≈ 4.24−11.76×0.2404⇒v=1.19 m s−1(3 s.f.) 15 i Let A be (0, 0) Square

Triangle

Whole trapezium

Area

25

5a2

a+102×5

x¯

25

5+a3

25×2.5+(5+a3)×12×5×a= x¯ (a+102)×5 Multiply by 6: 375+(15+a)×5a=15 (a+10)x¯ Divide by 5 and simplify: x¯ =a2+15a+753(a+10) ii a2+15a+753(a+10)< 5⇒a2< 75⇒a< 53 iii Let a=53 Rectangle

Triangle

Whole trapezium

Area

25

12 × 253

53+102×5

y¯

2.5

23×5

∴25×2.5+23×5×12× 25 ×3= y¯×(53+102×5) ⇒y¯≈2.89∴tan θ= 5y¯⇒θ≈60°

Worked solutions AS Level practice paper 1

O θ

G.P.E = O

√gms–1

Energy at start

Energy at end

So the height above the point of projection is 2

.

[3 marks]

a

[1 mark]

b



Therefore: for

; for

; and for [3 marks]

c It was necessary, otherwise we would not be able to solve the equations in , and since the dimensions of could involve , and . [1 mark] 3

a

300N

F ( ) direction

At the maximum speed the resistance force is equal to the driving force. Using

,

so

a

b 200N

.

[2 marks]

a T

T

F

300N ( ) direction

Using

for the car:

so

.

[2 marks]

c Light and inextensible.

[2 marks]

d Using Newton’s second law for the car in the direction of motion: [3 marks] e Using Newton’s second law for the trailer in the direction of motion:

Substituting your answer from part d above gives: .

[3 marks]

4

a Light and inextensible.

[1 mark]

b O

[4 marks]

α

T1 P

T2

T2

β

Q

mg

mg

c Resolving forces vertically at we get Using

towards the centre of the circular motion for :

Dividing one equation by the other we get: Resolving forces vertically at we get Using

towards the centre of the circular motion for :

Substituting

into

Substituting

and

Using part a we get:

gives: into

gives:

[4 marks] 5

a Using the equations of motion under constant acceleration and choosing the direction of motion as vertically upwards: This gives:

[2 marks]

b If the two spheres are not next to each other we cannot assume that A moves a distance of before the string becomes taut. The answer to part a would be larger, since the height that A would rise would be smaller. [2 marks] c Using conservation of momentum: Before

After

[2 marks] [1 mark]

d Impulse at e As a whole system, all the kinetic energy is converted to gravitational potential energy.

height above the table for (Total height above the table.) f

Kinetic energy before

[3 marks] Kinetic energy after

[3 marks] 6

a

is moving in the opposite direction to its original motion before the collision.

[1 mark]

b Using conservation of momentum in the direction of motion of :

[5 marks]

7

a Using conservation of energy: Kinetic energy before

Gain in kinetic energy b Using conservation of energy:

Kinetic energy after

[2 marks]

start 8m 8sin30 = 4m 30°

end

G.P.E = O

Energy at start

Energy at end



[4 marks]

R

c

30° 30°

25g

Frictional force at a maximum is given by

, where is the normal reaction force. and [3 marks]

d For a constant speed, the work done against friction must be equal to the loss in gravitational potential energy. If the particle travels metres down the slope, then: [3 marks]

Worked solutions A Level practice paper 1 Take the downward vertical as the positive direction. You can find the speed at which the ball hits the ground using

with

This gives

and

.

.

Therefore the velocity before impact is and, from the question, the rebound velocity is using (this time with and . Using

(since the ball changes direction):

.

2 You can use dimensional analysis where: Therefore,

.

. This gives

.

3 You can calculate the stored elastic potential energy using

[3 marks] . [3 marks]

Converting each of the units for distance into metres: 4

a

and natural Let

[3 marks]

.

be the extension in the string and let the tension in the string be given by [2 marks]

. b

is

.

Using Newton’s second law towards the centre of the circular motion:

[4 marks] c You need to let

and look, sketching the graph of

with

, as you can consider

as the angular speed. r 5 4 3 2 1

–1

O

1

d Asymptote at

2 k

.

[2 marks]

Given that

, if the string breaks when the tension is and so

then:

.

Given that the string has natural length , you have a bound for

.

This restricts the graph from part c. Looking at the values for that give this bound: 5

[2 marks]

.

a v

[2 marks]

4 2

O

2

4

6

t

8

b giving

.

[2 marks]

c You can use integration to find the total distance travelled:

This gives distance as 6

a

4m

A

.

[3 marks]

B

α

α

X

Y

The diagram is symmetrical about the perpendicular bisector of

and thus you can consider the

equilibrium of the particle at . You have three forces acting at : the tension tension in the string ; the weight of . Resolving the forces at horizontally gives Rearranging gives

and vertically gives

in the string

; the

.

and substituting gives .

[3 marks]

b Let be the extension of the part of the string .

and let be the extension of the part of the string

Using Hooke’s law

.

From part a you obtained

and

, with the two equations above you have

obtained four equations in the unknowns and . You need a fifth equation if you are going to be able to find the equation you are looking for. Using the diagram and also that the length of is you get . After substituting

and

into the equations you found using Hooke’s law, you

can rearrange to make and the subject: You can then use equation

. to find:

[6 marks]

Rearranging: c When 7

[1 mark]



a A

B

60°

v m s–1

2v m s–1

Using the conservation of momentum in the horizontal direction:

Using the law of restitution in the horizontal direction:

Solving this pair of simultaneous equations gives. Vertical components of velocity unchanged. Therefore, the speed of the sphere directly after impact is given by: and the direction is given by:

For sphere you have a speed of

and at an angle of

. [5 marks]

b The loss in kinetic energy is given by: kinetic energy before − kinetic energy after

[2 marks]

. 8

a

90m s–1 350m

350m

60m s–1 For the tangential component of the acceleration you need to find Since the plane decreases its speed by To find the radial acceleration, You can integrate to find ,

in

seconds then

, you must first find . .

. .

This gives

. At

and so

This means the radial acceleration is

. [4 marks]

.

b You can look at the conservation of energy, taking the gravitational potential energy level to be zero at the end of the quarter circle. Energy at start

Energy at end

Work done against friction at end so

.

[2 marks]

c You can look at the conservation of energy, taking the gravitational potential energy level to be zero when the plane releases the parcel. You need to find the speed with which the parcel is released. Energy at start

Energy at end

(work done against friction) Energy at start energy at end So This gives a speed of

.

[4 marks]

d To calculate the time taken to reach the ground you need to look at the vertical component of the velocity of the parcel. You will take the direction for the movement as vertically downwards. Using trigonometry, the parcel is moving . Since you can model the parcel as a projectile, you can use the equations of motion under a constant acceleration to find the time . With and you can find using and the quadratic formula:

This gives

9

a

.

[3 marks]

This is the coordinate of the centre of mass on the -axis. To find the distance from you must subtract :

[5 marks] b For

and

, the centre of mass lies on the axis of symmetry of the object and is units from its base. The solid shape will be on the point of toppling when

the perpendicular from the centre of mass passes through a corner on the base of the solid shape. Half the width of the base measures . At the toppling point the angle is given by: [2 marks]

. c Given that the coefficient of friction is inclined plane to find .

you can resolve the forces parallel and perpendicular to the

Parallel: Perpendicular: This gives

and thus

.

d The shape will slide first.

[2 marks] [1 mark]

N

10 a

R

2000N

1250g

Resolving forces in the direction of the motion gives: , where is the resistance force. , since is directly proportional to the square of the distance from and is a constant of proportionality. At

, therefore:

[2 marks] b Using

equals work done for a variable force:

[3 marks]

c From part b,

. Using Newton’s second law:

.

At

, therefore

. At

.

[4 marks]

Glossary Acceleration vector: The rate of change of the velocity vector of an object. Angular speed: The rate at which an object is rotating, measured as the angle in radians turned through in unit time. Centre of mass: A single point at which the mass of an object can be considered to be located. Centripetal force: The force directed towards the centre of a circle that hold a moving object in a circular path. Composite body: An object made from a combination of shapes. Conical pendulum: A particle attached to the end of a string or rod suspended from a pivot point; the particle moves in a horizontal circle with constant angular speed, with the string or rod tracing out the curved surface of a cone. Dimensions: These describe what type of quantity you are measuring. In mechanics the usual dimensions used to describe a quantity are combinations of mass ( ), length ( ) and time ( ). Displacement: Distance moved in a particular direction. Elastic limit: When an elastic spring or string is stretched beyond its elastic limit it does not return to its original length when the force is removed. Elastic potential energy: Energy stored in an elastic string when stretched or an elastic spring when stretched or compressed. Elastic spring: A spring that can be stretched or compressed when a force is applied and will return to its original length when the force is removed. Elastic string: A string that can be stretched when a force is applied and will return to its original length when the force is removed. Energy: The capacity of a physical system to do work. Gravitational potential energy: energy possessed by an object because of its height above an arbitrary fixed level. Impulse: The product of the force on an object and the time for which the force acts, which results in a change in the object’s momentum. Kinetic energy: Energy that a body possesses by virtue of being in motion. Lamina: A two-dimensional object. Mechanical energy: The sum of the kinetic energy and gravitational potential energy of an object. Modulus of elasticity: Theoretically the force required to double the length of an elastic string or spring, or to compress the length of an elastic spring to zero. Momentum: The product of mass and velocity of a moving object. Oblique impact: A collision at an angle to the line joining the centres of two colliding objects. Perfectly elastic collision: A collision in which there is no loss of kinetic energy. Power: The rate at which energy is transferred when a force does work. Propulsive forces: forces that promote movement.

Radial acceleration: Acceleration towards the centre of circular motion. Radial direction: The direction along a radius of a circle. Resistive forces: forces that oppose movement. Scalar quantity: A quantity that has only magnitude but not direction. SI: The international system of units based on seven base units, including the kilogram, metre and second. Stiffness: The force required for unit extension of an elastic string or spring or for unit compression of an elastic spring. Tangential direction: The direction along a tangent at a point of a curve. Tension: A force that is transmitted through a string or rod. Thrust (in a spring): A force that opposes a compressive force. Tractive force: The driving force of an engine. Uniform lamina: A two-dimensional object that has constant mass per unit area. Vector quantity: A quantity that has both magnitude and direction. Velocity vector: The rate of change of the position of an object. The magnitude gives the speed and the vector direction gives the direction of the motion. Work done: A force does work when it moves an object. Work–energy principle: An essential idea in mechanics that enables us to calculate the work necessary to cause a change in energy.

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