SQA Advanced Higher Physics Unit 1: Mechanics

Transcription

1 SCHOLAR Study Guide SQA Advanced Higher Physics Unit 1: Mechanics Andrew Tookey Heriot-Watt University Campbell White Tynecastle High School Heriot-Watt University Edinburgh EH14 4AS, United Kingdom.

2 First published 2001 by Heriot-Watt University. This edition published in 2013 by Heriot-Watt University SCHOLAR. Copyright 2013 Heriot-Watt University. Members of the SCHOLAR Forum may reproduce this publication in whole or in part for educational purposes within their establishment providing that no profit accrues at any stage, Any other use of the materials is governed by the general copyright statement that follows. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, without written permission from the publisher. Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the information contained in this study guide. Distributed by Heriot-Watt University. SCHOLAR Study Guide Unit 1: SQA Advanced Higher Physics 1. SQA Advanced Higher Physics ISBN Printed and bound in Great Britain by Graphic and Printing Services, Heriot-Watt University, Edinburgh.

3 Acknowledgements Thanks are due to the members of Heriot-Watt University's SCHOLAR team who planned and created these materials, and to the many colleagues who reviewed the content. We would like to acknowledge the assistance of the education authorities, colleges, teachers and students who contributed to the SCHOLAR programme and who evaluated these materials. Grateful acknowledgement is made for permission to use the following material in the SCHOLAR programme: The Scottish Qualifications Authority for permission to use Past Papers assessments. The Scottish Government for financial support. All brand names, product names, logos and related devices are used for identification purposes only and are trademarks, registered trademarks or service marks of their respective holders.

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5 i Contents 1 Kinematic relationships Introduction Derivation of kinematic relationships Motion in one dimension Motion in two dimensions Summary End of topic test Relativistic motion Introduction Relativistic dynamics Relativistic energy Other relativistic effects Summary End of topic test Angular velocity and acceleration Introduction Angular displacement and radians Angular velocity and acceleration Kinematic relationships for angular motion Tangential speed and angular velocity Summary End of topic test Centripetal force Introduction Centripetal acceleration Centripetal force Applications Summary End of topic test Rotational dynamics Introduction Torque and moment Newton's laws applied to rotational dynamics Angular momentum and kinetic energy Summary End of topic test

6 ii CONTENTS 6 Gravitational force and field Introduction Newton's law of gravitation Weight Gravitational fields Summary End of topic test Gravitational potential and satellite motion Introduction Gravitational potential and potential energy Satellite motion Escape velocity Summary End of topic test Simple harmonic motion Introduction Defining SHM Equations of motion in SHM Energy in SHM Applications and examples Damping Summary End of topic test Wave-particle duality Introduction Wave-particle duality of waves Wave-particle duality of particles Summary End of topic test Introduction to quantum mechanics Introduction Atomic models Quantum mechanics Summary End of topic test Mechanics end-of-unit assessment 165 Glossary 168 Hints for activities 172 Answers to questions and activities Kinematic relationships Relativistic motion Angular velocity and acceleration Centripetal force

7 CONTENTS iii 5 Rotational dynamics Gravitational force and field Gravitational potential and satellite motion Simple harmonic motion Wave-particle duality Introduction to quantum mechanics Mechanics end-of-unit assessment

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9 1 Topic 1 Kinematic relationships Contents 1.1 Introduction Derivation of kinematic relationships Motion in one dimension Horizontal motion Vertical motion Motion in two dimensions Summary End of topic test Prerequisite knowledge Knowledge of the difference between vector and scalar quantities. Calculus - definite integrals. Some familiarity with the kinematic relationships would be useful. Learning Objectives By the end of this topic, you should be able to: derive the three kinematic relationships from the calculus definitions of acceleration and velocity; apply these equations to describe the motion of a particle with uniform acceleration moving in one dimension; find the horizontal and vertical components of the velocity of a body moving in two dimensions; apply the kinematic relationships to describe the motion of a particle with uniform acceleration moving in two dimensions.

10 2 TOPIC 1. KINEMATIC RELATIONSHIPS 1.1 Introduction This topic deals with the motion of particles moving with a uniform acceleration. You should already be familiar with the kinematic relationships describing this motion, and with the vector nature of the quantities displacement, velocity and acceleration. For an object moving in a straight line with uniform acceleration, the average velocity over a period of time is given by (u + v) 2 where u and v are the initial and final velocities, respectively. The displacement s after a time t is then given by (u + v) s = t 2 This relationship is, of course, equivalent to the scalar equation d = v t, where d = distance and v = average speed. There are three more kinematic relationships used to describe motion with uniform acceleration. We will begin by using calculus to derive these relationships, starting from the definitions of instantaneous acceleration and velocity. These relationships will then be applied to motion in one and two dimensions. 1.2 Derivation of kinematic relationships Learning Objective To derive the three kinematic relationships from the calculus definitions of acceleration and velocity. The equations describing motion with a constant acceleration can be derived from the definitions of acceleration and velocity. We will start with the definition of instantaneous acceleration: acceleration is the rate of change of velocity a = dv dt We are looking at motion where a is a constant. To find the velocity after time t, we will integrate this expression over the time interval from t =0tot = t. Carrying out this integration: v u dv = t t=0 t adt = a dt t=0

11 TOPIC 1. KINEMATIC RELATIONSHIPS 3 [v] v u =a [t]t 0 v u =at v =u + at (1.1) Equation 1.1 gives us the velocity v after time t, in terms of the acceleration a and the initial velocity u. Velocity is defined as the rate of change of displacement. We will now use this definition to derive the second of the kinematic relationships: v = ds dt We can substitute for v in Equation 1.1 using this expression. ds dt = u + at Again, we will integrate over the time interval from 0 to t. s t ds = (u + at)dt s=0 t=0 [ [s] s 0 = ut + 1 ] t 2 at2 0 s = ut at2 Equation 1.2 gives us the displacement s after time t, in terms of the acceleration and the initial velocity. To obtain the third kinematic relationship, we first rearrange Equation 1.1. t = v u a (1.2) Substituting this expression for t into Equation 1.2 gives us ( ) v u s =u + 1 ( ) v u 2 a 2 a 2as =2vu 2u 2 + v 2 2vu + u 2 2as = 2u 2 + v 2 + u 2 v 2 =u 2 +2as (1.3)

12 4 TOPIC 1. KINEMATIC RELATIONSHIPS We have obtained three equations relating displacement s, time elapsed t, acceleration a and the initial and final velocities u and v. Note that with the exception of t, these are all vector quantities. v = u + at s = ut at2 v 2 = u 2 +2as Although we are dealing with vector quantities, we are only considering the special case of motion in a straight line. It is vital to ensure we assign the correct +ve or -ve sign to each of the quantities s, u, v and a. We can show graphically how the acceleration, velocity and displacement all vary with time. With these three equations, along with the equation in the Introduction relating displacement to average velocity and time, we can solve all the problems we will encounter during this topic. Acceleration, velocity and displacement 20 min At this stage there is an online activity. This simulation shows how the quantities acceleration a, velocity v and displacement s vary as functions of time. The program allows you to set values for a and the initial velocity u, and then calculates how a, v, and s evolve in time. You should be able to explain how the velocity-time graph relates to the equation v = u + at and how the displacement-time graph relates to the equation s = ut at Motion in one dimension Learning Objective To apply the kinematic relationships to describe the motion of a particle moving in one dimension. Both horizontal and vertical motion are covered in this section Horizontal motion The kinematic relationships can be used to solve problems of motion in one dimension with constant acceleration. In all cases we will be ignoring the effects of air resistance. Example A car accelerates from rest at a rate of 4.0 m s What is its velocity after 10 s?

13 TOPIC 1. KINEMATIC RELATIONSHIPS 5 2. How long does it take to travel 72 m? 3. How far has it travelled after 8.0 s? We can list the data given to us in the question: u =0ms -1 (the car starts from rest) a = 4.0 m s We are told that the time elapsed t = 10 s and we wish to find v. So with u, a and t known and v unknown, we will use the equation v = u + at. L L =? K =0 J =10 J v = u + at v =0+(4.0 10) v =40ms 1 2. In part 2 we know u, a and s, and t is the unknown, so we use s = ut at2. u =0ms -1 a = 4.0 m s -2 s =72m t =? I I =72 J =? J

14 6 TOPIC 1. KINEMATIC RELATIONSHIPS s = ut at2 72 = 0 + ( t2) 72 = 2t 2 t 2 =36 t =6.0s 3. Finally in part 3 we are asked to find s, given u, a and t, so we will again use s = ut at2, this time using different data. u =0ms -1 a =4.0ms -2 t= 8.0 s s =? s = ut at2 s =0+ ( ) s = 128 m The same strategy should be used in solving all of the problems in this topic. Firstly, list the data given to you in the question. This will ensure that you use the correct values when you perform any calculations, and should also make it clear to you which of the kinematic relationships to use. It is often useful to make a sketch diagram with arrows, to ensure that any vector quantities are being measured in the correct direction. Horizontal Motion 20 min Suppose a car is being driven at a velocity of 12.0 m s -1 towards a set of traffic lights, which are changing to red. The car driver applies her brakes when the car is 30.0 m from the stop line. What is the minimum uniform deceleration needed to ensure the car stops at the line? At this stage there is an online activity where you try it out. Always use the same procedure to solve a kinematics problem in one dimension - sketch a diagram, list the data and select the appropriate kinematic relationship Vertical motion When dealing with freely-falling bodies, the acceleration of the body is the acceleration due to gravity, g =9.8ms -2. If any force other than gravity is acting in the vertical plane, the body is no longer in free-fall, and the acceleration will take a different value. Problems should be solved using exactly the same method we used to solve horizontal motion problems.

15 TOPIC 1. KINEMATIC RELATIONSHIPS 7 Example A student drops a stone from a second floor window, 15 m above the ground. 1. How long does it take for the stone to reach the ground? 2. With what velocity does it hit the ground? When dealing with motion under gravity, we must take care with the direction we choose as the positive direction. Here, if we take a as a positive acceleration, then v and s will also be positive in the downward direction. L I L =? I =15 K =0 J J =? J We are told that u =0ms -1 a = g = 9.8 m s -2 s =15m 1. To find t, we use s = ut at2. Putting in the appropriate values s = ut at2 15 = 0 + ( t2) 15 = 4.9t 2 t 2 = t 2 =3.06 t =1.7s 2. To find v, we use v 2 = u 2 +2as. Again, we put the appropriate values into this kinematic relationship v 2 = u 2 +2as v 2 =0+( ) v 2 = 294 v =17ms 1

16 8 TOPIC 1. KINEMATIC RELATIONSHIPS Difficulties sometimes occur when the initial velocity is directed upwards, for example if an object is being thrown upwards from the ground. To solve such a problem, it is usual to take the vertical displacement as being positive in the upwards direction. The velocity vector is then positive when the object is travelling upwards, and negative when it is returning to the ground. In this situation the acceleration is always negative, as it is always directed towards the ground. Quiz 1 Motion in one dimension 20 min First try the questions. If you get a question wrong or do not understand a question, there are 'Hints'. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: acceleration due to gravity g 9.8ms -2 Q1: An object is moving with a uniform acceleration of 5 m s -2. A displacement-time graph showing the motion of this object has a gradient which a) increases with time. b) decreases with time. c) equals 5 m s -1. d) equals 5 m s -2. e) equals 0. Q2: A car accelerates from rest with a uniform acceleration of 2.50 m s -2. How far does it travel in 6.00 s? a) 7.50 m b) 15.0 m c) m d) 45.0 m e) m Q3: Neglecting air resistance, a stone dropped from the top of a building 125 m high hits the ground after a) 1.25 s. b) 4.00 s. c) 5.05 s. d) 12.7 s. e) 25.5 s.

17 TOPIC 1. KINEMATIC RELATIONSHIPS 9 Q4: A diver jumps upwards with an initial vertical velocity of 3.00 m s -1 from a diving board which is 8.00 m above the swimming pool. With what vertical velocity does he enter the pool? a) 3.0 m s -1 b) 3.6 m s -1 c) 9.0 m s -1 d) 13 m s -1 e) 24 m s -1 Q5: A stone is dropped from a window 28.0 m above the ground. What is the velocity of the stone when it is 10.0 m above the ground? a) 10.0 m s -1 b) 14.0 m s -1 c) 18.8 m s -1 d) 23.4 m s -1 e) 98.1 m s Motion in two dimensions Learning Objective To apply the kinematic relationships to an object moving in two dimensions. Because we are dealing with vector quantities, we can also solve problems of motion in two dimensions. Usually this involves separating the motion into x- and y-components, which can be treated independently. Most problems we come across involve projectile motion, where horizontal and vertical components are studied separately. For example, consider a parcel dropped from a train window 3 m above the ground, when the train is travelling at 40 m s -1 along a straight track. Neglecting air resistance, there is no horizontal force acting on the parcel, and hence no acceleration. The parcel will continue to move at the same horizontal velocity in the same direction as the train. In the vertical plane, the parcel is falling under gravity, and will accelerate downwards at a rate of g ms -2. We can calculate the time taken for the parcel to hit the ground by considering the vertical motion only. Once we know this time, we can find out the distance between the point when the parcel was released and where it hits the ground by considering the horizontal motion only. Vertically u y =0ms -1, a y = g = 9.8 m s -2, s y =3m,t =?

18 10 TOPIC 1. KINEMATIC RELATIONSHIPS s y = u y t a yt 2 3=0+ ( t2) t 2 = t 2 =0.612 t =0.782 s Since there is no acceleration horizontally, u x = v x =40ms -1, and we can use the simple equation s x = v x t. s x = v x t s x = s x =31m Example Consider another example - a golf ball struck with an initial velocity of 24 m s -1 at an angle of 30 to the ground. How far does it travel horizontally before striking the ground again? We must first separate the initial velocity into its orthogonal components u x and u y. K K O θ K N Using the trigonometric relationships, we can see that u x = u cos θ and u y = u sin θ The initial velocity of the particle is given by u 2 = u 2 x + u2 y in the direction given by the angle θ. Now we can study the x- and y-components of the motion separately, and as in the previous example, we need to work out the time-of-flight from the vertical motion in order to calculate the horizontal range.

19 TOPIC 1. KINEMATIC RELATIONSHIPS 11 Vertically, we want to know the time taken until s y returns to zero. Taking upwards as the positive direction, we have u y =24 sin30 = 12 m s -1 a y =-g =-9.8ms -2 s y =0m t =? s = ut at2 0=(24 sin 30 t)+ ( t2) 0=12t 4.9t 2 One solution to this equation, of course, is t = 0 s, since the displacement is 0 when t = 0 s. Ignoring this solution, we can divide both sides of the above equation by t to give us 0=12 4.9t t = t =2.45 s Knowing this time-of-flight, we can use the simple s x = v x t relation for the horizontal motion to calculate s x s x = v x t s x =(24 cos 30) 2.45 s x = s x =51m There are a couple of important points brought up in this example. First, make sure you understand the sign convention that has been used for the vertical motion. We have taken upwards as the positive direction, so the initial velocity u y is a positive vector quantity, and the acceleration a y is a negative vector quantity. The second point to note is that when the golf ball has returned to Earth, its vertical displacement s y is zero. The displacement tells us how far, and in what direction, an object is from its starting point, not the total distance it has travelled. In this case, when the ball hits the ground its vertical displacement is exactly the same as it was at t =0,sos y =0. By considering the x- and y-components, we can work out the angle of trajectory required to give the maximum horizontal range for a given initial projectile speed. To solve this problem, we first consider the horizontal motion. s x = u cos θ t So for a given u, the range depends on θ and t. Let us look now at the vertical motion. The value of t that we are considering is the time taken for the projectile to return to Earth. This means we require s y, the vertical displacement, to return to zero.

20 12 TOPIC 1. KINEMATIC RELATIONSHIPS s y =u y t a yt 2 0=u sin θt 1 2 gt2 Ignoring the t = 0 solution, we can divide by t and then rearrange this equation in terms of t. 2u sin θ t = g We can substitute this expression for t into our equation for horizontal motion. s x =u cos θ t 2u sin θ s x =u cos θ g s x = u2 2sinθcos θ g Now we can use the trig relation sin 2θ =2sinθcos θ in the above equation to give us s x = u2 sin 2θ g So for any value of u, the maximum value of s x is obtained when sin2θ is at a maximum. This occurs when sin2θ = 1, and hence 2θ =90. Therefore the maximum range is obtained when θ =45. Putting this value of θ back into the above equation, we can see that s x,max = u2 g (1.4) Motion in two dimensions 25 min A motorcycle stunt rider is trying to jump a 41.0 m gap. The take-off ramp launches him at an angle of 20.0, and the landing ramp is at the same height as the take-off ramp. Use the kinematic relationships to calculate the minimum speed at take-off that will get the rider safely to the other side. You can test your result out in the online activity. In all problems which involve motion in two dimensions, the motion must be split into x- and y-components.

21 TOPIC 1. KINEMATIC RELATIONSHIPS 13 Quiz 2 Motion in two dimensions First try the questions. If you get a question wrong or do not understand a question, there are 'Hints'. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. 20 min Useful data: acceleration due to gravity g 9.8ms -2 Q6: A truck is travelling at 20 m s -1 on a straight horizontal road. A man in the truck fires a gun straight up in the air. Ignoring air resistance, where will the bullet land? a) In front of the truck. b) Behind the truck. c) On the truck. d) Depends on the vertical speed of the bullet. e) Depends on the value of g. Q7: What is the maximum horizontal range of a football kicked with an initial velocity of 12.0 m s -1, if air resistance is negligible? a) 1.73 m b) 3.83 m c) 8.49 m d) 12.0 m e) 14.7 m Q8: An arrow is fired from a bow at 24.0 m s -1 towards a target 30.0 m away. If the bow and the bull's-eye of the target are at the same height, what is the minimum angle of elevation needed to ensure the arrow hits the bull's-eye? a) 15.3 b) 30.7 c) 45.0 d) 53.1 e) 74.6 Q9: A boy kicks a ball at an angle of elevation of 75 and velocity 8.00 m s -1. With what velocity must he run over level ground to catch the ball before it bounces? a) 0.11 m s -1 b) 2.07 m s -1 c) 4.00 m s -1 d) 7.73 m s -1 e) 8.00 m s -1

22 14 TOPIC 1. KINEMATIC RELATIONSHIPS Q10: An aeroplane flying horizontally at an altitude of 900 m releases a package when it is a horizontal distance of 1600 m from its target. How fast is the plane travelling, if the package lands on its target? a) 1.78 m s -1 b) 118 m s -1 c) 144 m s -1 d) 900 m s -1 e) 1600 m s Summary The material covered in this topic should have added to your understanding of motion of bodies moving with uniform acceleration. By the end of this topic you should be able to: derive the three kinematic relationships v = u + at s = ut at2 v 2 = u 2 +2as from the calculus definitions of acceleration and velocity. apply these equations to describe the motion of a particle with uniform acceleration moving in one dimension. find the horizontal and vertical components of the velocity of a body moving in two dimensions. apply the kinematic relationships to describe the motion of a particle with uniform acceleration moving in two dimensions. 1.6 End of topic test Learning Objective topic End of topic test 30 min At this stage there is an end of topic test available online.. If however you do not have access to the internet you may try the questions which follow. The following data should be used when required: acceleration due to gravity g = 9.8 m s -2 Q11: A car accelerates uniformly from rest, travelling 55 m in the first 8.5s. Calculate its acceleration in m s -2.

23 TOPIC 1. KINEMATIC RELATIONSHIPS 15 Q12: The brakes on a car can provide a deceleration of up to 4.9 m s -2. Calculate the minimum stopping distance (in m) if the brakes are applied when the car is initially travelling at 20 m s -1. Q13: A toy rocket is fired vertically upwards with an initial velocity of 16 m s -1. Calculate how long (in s) the rocket stays in the air before it returns to the ground. Q14: A piano is being raised to the third floor of a building using a rope and pulley. The piano is 15 m above the ground, moving upwards at 0.25 m s -1, when the rope snaps. Calculate how much time (in s) elapses before the piano hits the ground. Q15: A cricket ball is thrown with an initial velocity 17 m s -1 at an angle of elevation 48 above the horizontal. Calculate how much time (in s) passes before the ball hits the ground again. Q16: In a film stunt, a car is driven off a cliff with a horizontal velocity of 14 m s -1. The cliff face is vertical, and the cliff is 46 m high. Calculate the distance in m from the base of the cliff to the point where the car strikes the ground. Q17: A golfer strikes a ball with an initial velocity 30 m s -1. The ball lands on the green, which is on the same horizontal plane 45 m away. Calculate the minimum angle of elevation θ at which the golf ball should be struck in order for it to land on the green.

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25 17 Topic 2 Relativistic motion Contents 2.1 Introduction Relativistic dynamics Relativistic energy Other relativistic effects Summary End of topic test Prerequisite knowledge Knowledge of Newtonian mechanics, such as the equation for the kinetic energy of an object. Learning Objectives By the end of this topic, you should be able to: Understand that the properties of objects moving at very large speeds are subject to relativistic phenomena; State that the speed of light in a vacuum is constant, and that no object can travel faster than the speed of light; Calculate the relativistic mass and energy of an object.

26 18 TOPIC 2. RELATIVISTIC MOTION 2.1 Introduction Probably the most well-known of all physics equations is E = mc 2. Almost everyone has heard or seen this equation, even if most people don't understand its meaning or implications. In this Topic we will be studying relativity. By the end of it you should understand what this equation means and when it should be applied. The concept of relativity was first proposed by Albert Einstein in 1905 in his Special Theory of Relativity. It is something that we are not often aware of in everyday life, and in some ways requires a new way of looking at things. This Topic only scratches the surface of the subject, as many of its harder concepts require a lot of advanced mathematics. One consequence of the principle of relativity is that two people measuring the same thing may not necessarily get the same result if they are moving with respect to each other. The greater their relative speed, the greater the discrepancy in their measurements. The laws of mechanics that you are used to applying are called Newtonian mechanics, since they are based on Newton's laws of motion. In this Topic we will see that when an object is moving with a speed close to the speed of light, Newtonian mechanics can no longer be used to describe its motion. 2.2 Relativistic dynamics Learning Objective To state that the relativistic mass of an object increases as its speed approaches the speed of light. Consider an object of mass m travelling at speed v. We could use a balance to find the value of m, a ruler to find how far it has travelled and a stopwatch to find how long it takes to travel this distance. We could also apply Newtonian mechanics to calculate the momentum and kinetic energy of the object, and the time it takes to travel between any two points. When an object is moving at a very high speed, the laws of Newtonian mechanics are no longer valid. To an observer who is stationary compared to the moving object, the mass, distance travelled, time taken and other related properties are subject to relativistic mechanics and are calculated using a different set of equations. Since relativity is such a large topic we will concentrate on calculating relativistic mass and kinetic energy. For completeness, relativistic effects on time and length are also discussed at the end of the topic. The relativistic mass m of an object moving with speed v is given by the equation m = m 0 1 v2/ c 2 (2.1)

27 TOPIC 2. RELATIVISTIC MOTION 19 In Equation 2.1 m 0 is the rest mass of the object (its mass when it is stationary) and c is the speed of light in a vacuum ( ms -1 ). So to an observer who is stationary, the mass of the object is m rather than m 0. Example An electron (rest mass kg) is accelerated to a velocity of ms -1. What is the relativistic mass of the electron? Putting the values of rest mass and speed into Equation 2.1, m = m 0 1 v2/ c 2 m = m = m = m = ( ) 2 ( ) ( ) ( ) m = m = kg The mass of the electron has increased to kg, for a stationary observer making measurements or calculations on a moving electron. This example shows that even at a speed of ms -1, the relativistic increase in mass is still only around 1% of the rest mass. As the speed of the object gets closer and closer to c, so the relativistic mass gets larger and larger. There are a couple of important points to be aware of when calculating the relativistic mass of an object. Firstly, relativistic effects only become noticeable when an object is moving with a speed of around 0.1c or greater. If the speed is less than this, the factor v2/ c2 is extremely small, and the value of the square root term in Equation 2.1 is very close to 1, hence m m 0. A second point to note is that an object's mass would become infinite if it were travelling at the speed of light. If v = c in Equation 2.1, the denominator becomes zero and hence m becomes infinite. To accelerate an object to the speed of light would therefore require an infinite-sized force to be applied to the object, which is impossible. This means that the speed of light is "nature's speed limit" as it is impossible for an object with non-zero rest mass to reach this speed. (It can also be shown that an object with zero rest mass, such as a photon, travels at speed c.)

28 20 TOPIC 2. RELATIVISTIC MOTION Quiz 1 Relativistic dynamics 20 min First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint ans still do not understand then ask your tutor. All references in the hints are to online materials. Useful data: rest mass of an electron m e kg rest mass of a proton m p kg speed of light in a vacuum c ms -1 Q1: A proton is accelerated to a speed of ms -1. What is the relativistic mass of the proton? a) kg b) kg c) kg d) kg e) kg Q2: Einstein's theory of relativity tells us that a) electrons can be accelerated to the speed of light. b) the relativistic mass of an object is less than its rest mass. c) objects moving at very high speeds do not obey the laws of Newtonian mechanics. d) any particle can travel with a speed greater than c. e) only objects with a non-zero rest mass can travel at the speed of light. Q3: At approximately what speed is an object travelling when relativistic effects become important? a) 340 m s -1 (speed of sound) b) 0.1 c c) c d) 10c e) c 2 Q4: The relativistic mass of a particle is equal to twice its rest mass. At what speed is the particle travelling? a) ms -1 b) ms -1 c) ms -1 d) ms -1 e) ms -1

29 TOPIC 2. RELATIVISTIC MOTION 21 Q5: When the value of v<<c, the value of a) 0 b) 1 c) v d) c e) 1 v2/ c2 is approximately 2.3 Relativistic energy Learning Objective To state the relativistic energy of an object, and calculate the rest energy, kinetic energy and total energy of an object. One important consequence of Einstein's work on special relativity is the principle of mass-energy equivalence. This principle states that the mass of an object is a measure of its total energy. This is summed up in the famous equation E = mc 2 (2.2) Here E is the total energy of an object, and m is its relativistic mass. A stationary object therefore has energy E 0 = m 0 c 2 (2.3) E 0 is called the rest energy of the object. A moving object has kinetic energy E k in addition to its rest energy, so E = E k + E 0 E k = E E 0 (2.4) We can substitute for the values of E and E 0 in Equation 2.4 to find the kinetic energy of an object moving at a relativistic speed E k = E E 0 E k = mc 2 m 0 c 2 m 0 E k = 1 v2/ c 2 m 0 c 2 c 2 E k = m 0 c v2/ 1 c 2 (2.5)

30 22 TOPIC 2. RELATIVISTIC MOTION This equation looks very different from the equation for kinetic energy that you will have seen before, E k = 1 2 mv2. However it can be shown that when v is much smaller than c, and relativistic effects are negligible, Equation 2.5 does give us the values we would expect. Relativistic and Newtonian kinetic energy (optional) 20 min This is an OPTIONAL activity for students wishing to show that the relativistic expression for kinetic energy is the same as the Newtonian expression for an object moving at low speed. Equation 2.5 gives the kinetic energy of an object moving at relativistic speeds. Does this mean that the kinetic energy equation derived from Newtonian mechanics is incorrect? Well luckily it doesn't, unless we are moving at very high speeds. In this exercise we will show that the two expressions are equivalent at low speeds. To do this we will make a binomial expansion of the relativistic expression for kinetic energy. The binomial theory allows us to expand any expression which is of the general form (1+x) n, (1 + x) n n (n 1) x2 n (n 1) (n 2) x3 =1+nx (2.6) 2! 3! We need to rearrange Equation 2.5 to allow us to expand the term in brackets as a binomial, E k = m 0 c v2/ 1 c 2 ( ( E k = m 0 c 2 1 v2/ ) 1 ) c (2.7) Comparing the (1 v2/ ) 1 c 2 2 term in Equation 2.7 to Equation 2.6, we have a binomial with x = v2/ c 2 and n = 1 2. Performing this expansion we obtain, (1 + x) n =1+ nx + ) 1 ( ) ( (1 v2 2 1 =1+ 2 v 2 ) c 2 c 2 + ) 1 (1 v2 2 v 2 =1+ c 2 2c 2 + n (n 1) x 2 2! ( ) ( ) v 4 2c v 4 8c n (n 1) (n 2) x !

31 TOPIC 2. RELATIVISTIC MOTION 23 When v is much less than c, all terms in the expansion become extremely small, except the first two. So under these conditions, ) 1 (1 v2 2 v 2 1+ c 2 2c 2 If we substitute this expression back into Equation 2.7, we obtain (( ) ) E k = m 0 c 2 1+ v2 2c 2 1 E k = m 0 c 2 v2 2c 2 E k = 1 2 m 0v 2 This is the Newtonian expression for kinetic energy. So as long as v is much less than c, we can use the standard expression for kinetic energy. Using a binomial expansion it can be shown that the equations for Newtonian and relativistic kinetic energy are equivalent for an object moving at low speed. Relativistic Mass and Energy At this stage there is an online activity. This simulation plots the relativistic mass and energy of an object as its speed increases from ms min The mass of an object increases the closer to the speed of light c that the object is travelling at. Kinetic energy also increases with increasing speed, but the relativistic kinetic energy becomes much larger than the Newtonian kinetic energy as the speed becomes closer to c. Quiz 2 Relativistic energy First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint ans still do not understand then ask your tutor. All references in the hints are to online materials. 20 min Useful data: rest mass of an electron m e kg rest mass of a proton m p kg speed of light in a vacuum c ms -1

32 24 TOPIC 2. RELATIVISTIC MOTION Q6: A stationary nucleus has mass kg. What is the rest energy of this nucleus? a) J b) J c) J d) J e) J Q7: The total relativistic energy of a moving object is made up of a) kinetic energy only. b) rest energy and light energy. c) kinetic energy and light energy. d) rest energy and kinetic energy. e) rest energy only. Q8: A particle with a rest mass of kg emerges from a particle accelerator, travelling at such a speed that its relativistic mass is kg. What is the total energy of the particle? a) J b) J c) J d) J e) J Q9: How does the kinetic energy of an object moving at a speed close to the speed of light, calculated using a relativistic calculation, compare with the kinetic energy calculated using a Newtonian calculation? a) The relativistic calculation gives a lower value of kinetic energy. b) The relativistic calculation gives a higher value of kinetic energy. c) Both calculations give the same value of kinetic energy. d) Impossible to say without knowing the exact speed of the object. e) Impossible to say without knowing the rest mass of the object. Q10: A proton is accelerated to a speed of ms -1. Using a relativistic calculation, find the kinetic energy of the proton. a) J b) J c) J d) J e) J

33 TOPIC 2. RELATIVISTIC MOTION Other relativistic effects Learning Objective To describe some of the other concepts of relativity The theory of relativity does not just deal with the effects on mass and energy of travelling at very high speeds. Other effects become apparent too. For example, length contraction occurs, which means that to a stationary observer, an object travelling at high speeds appears to be shorter than it does when it is at rest. Another relativistic effect, time dilation, means that time passes more slowly for an object moving at high speeds than it does for a stationary object. An example of this is the creation of muons in the upper atmosphere by cosmic radiation. Muons are subatomic particles with extremely short lifetimes ( 10-6 s) and are typically created at an altitude of over 10 km above the Earth's surface. Even travelling at close to the speed of light, these particles would disintegrate before reaching the Earth's surface, yet detectors on Earth can detect muons. The reason that they survive long enough to be detected is because the time dilation increases their lifetime. Time dilation At this stage there is an online activity. A spaceship is flying a distance of 5 light hours, for example from Earth to the planet Pluto. The speed can be regulated with the upper buttons. The animation demonstrates that the clock in the spaceship goes more slowly than the two clocks of the system in which Earth and Pluto are motionless. 15 min Time dilation occurs for an object travelling at close to the speed of light, meaning that a stationary clock on Earth would run faster than an identical clock placed in a rocket ship travelling at high speed. 2.5 Summary A stationary observer will find that measurements made on the motion of an object moving at a very high speed will not be consistent with the laws of Newtonian mechanics. An object moving at a speed close to the speed of light will undergo relativistic effects. In this topic we have concentrated on the relativistic mass and energy of an object. The relativistic mass of an object is greater than its rest mass, and increases with increasing speed. The total energy of such an object consists of rest energy and kinetic energy, where the kinetic energy is larger than that calculated using Newtonian mechanics.

34 26 TOPIC 2. RELATIVISTIC MOTION By the end of this topic you should be able to: use the equation m = m 0 1 v2/ c 2 to calculate the relativistic mass of an object of rest mass m 0 ; state that the speed of light in a vacuum is the maximum speed in nature, and that an object with non-zero rest mass cannot attain this speed; state that the rest energy of an object is given by the equation E = m 0 c 2 and the relativistic energy of a particle is given by the equation E = mc 2 ; use the equation E k = m 0 c v2/ 1 c 2 to find the kinetic energy of a particle travelling at relativistic speed. 2.6 End of topic test End of topic test 30 min At this stage there is an end of topic test available online. If however you do not have access to the internet you may try the questions which follow. the following data should be used when required. speed of light in a vacuum c ms -1 rest mass of an electron m e rest mass of a proton m p kg kg Q11: Calculate the relativistic mass (in kg) of a proton moving at speed ms -1. Q12: A electron, rest mass m e, has been accelerated to such a speed that its relativistic mass is 2.1 m e. At what speed, in m s -1, is the electron travelling?. Q13: The nucleus of an atom has rest mass kg. Calculate the rest energy (in J) of this nucleus. Q14: An electron is travelling at speed ms -1. Calculate the relativistic energy, measured in J, of the electron.

35 TOPIC 2. RELATIVISTIC MOTION 27 Q15: A proton travelling at high speed has relativistic mass kg. Calculate the kinetic energy of the proton, measured in J. Q16: An electron is travelling at a speed of c. Calculate the total relativistic energy (in J) of the electron.

36 28 TOPIC 2. RELATIVISTIC MOTION

37 29 Topic 3 Angular velocity and acceleration Contents 3.1 Introduction Angular displacement and radians Angular velocity and acceleration Kinematic relationships for angular motion Tangential speed and angular velocity Summary End of topic test Prerequisite knowledge Kinematic relationships (Mechanics Topic 1). Calculus - definite integrals. Learning Objectives By the end of this topic, you should be able to: express angles and angular displacement in radians; calculate the angular velocity, periodic time and angular acceleration of an object moving in a circle; derive the kinematic relationships for angular motion, and use the relationships to solve problems; relate angular velocity and acceleration to the tangential speed and acceleration.

38 30 TOPIC 3. ANGULAR VELOCITY AND ACCELERATION 3.1 Introduction In the Kinematics topic we studied the motion of objects travelling in a straight line. Now we will move on to study circular motion, which has a wide range of applications. These include the motion of planets around the Sun, the disc spinning in a CD player and a car taking a corner. We begin here by introducing angular displacement, angular velocity and angular acceleration, and introducing the units in which they are measured. We will use these quantities to describe the motion of a point object moving in a circular path, or a solid object rotating about an axis. We have already used kinematic relationships involving linear displacement, velocity and acceleration. We will derive the angular equivalent of these relationships, and look at the relationship between angular and linear quantities. 3.2 Angular displacement and radians Learning Objective To use radians to measure angles and angular displacement. In the first Topic of the course we investigated motion in one and two dimensions. We are going to apply many of the ideas we met in that Topic to describe the motion of an object moving in a circle. We will see in the next Section that instead of using velocity and acceleration vectors, we will be using angular velocity and angular acceleration. Firstly we will look at angular displacement, which replaces the linear displacement we are used to dealing with. Imagine a disc spinning about a central axis, as shown in Figure 3.1. We can draw a reference line along the radius of the disc. The angular displacement after time t is the angle through which this line has swept in time t. Figure 3.1: Angular displacement At J =0 After time interval J θ

39 TOPIC 3. ANGULAR VELOCITY AND ACCELERATION 31 The angular displacement is given the symbol θ, and is measured in radians (rad). Throughout this Topic, radians will be used to measure angles and angular displacement. A brief explanation of radian measurement, and how radians and degrees are related, will be given next. Figure 3.2: Radian measurement θ r s With reference to Figure 3.2, the angle θ, measured in radians, is equal to s / r. Since the radian is defined as being one distance (s) divided by another (r) then strictly speaking it is a dimensionless quantity, however the radian is regarded as a supplementary SI unit. To compare radians with degrees, consider an angular displacement of one complete circle, equivalent to a rotation of 360. In this case, the distance s in Figure 3.2 is equal to the circumference of the circle. Hence θ = s r θ = 2πr r θ =2πrad (3.1) So 360 is equivalent to 2π rad, and this relationship can be used to convert from radians to degrees, and vice versa. It is useful to remember that π rad is equivalent to 180 and π/2 rad is equivalent to 90. For the sake of neatness and clarity, it is common to leave an angle as a multiple of π rather than as a decimal, so the equivalent of 30 is usually expressed as π/6 rad rather than rad.

40 32 TOPIC 3. ANGULAR VELOCITY AND ACCELERATION Quiz 1 Radian measurement 15 min First try the questions. If you get a question wrong or do not understand a question, there are Hints. The hints are in the same order as the questions. If you read the hint and still do not understand then ask your tutor. All references in the hints are to online materials. Q1: Convert the angle 145 into radians. a) rad b) rad c) 2.53 rad d) 23.1 rad e) 911 rad Q2: What is the equivalent in degrees ( ) to 1.20 radians? a) 3.76 b) 7.54 c) d) 138 e) 432 Q3: Express 120 in radians. a) π /120 rad b) π /4 rad c) π /3 rad d) 2π /3 rad e) 3π /2 rad Q4: An object moves through 3 complete rotations about an axis. What is its total angular displacement? a) b) π /6 rad π /3 rad c) 3π rad d) 6π rad e) 2π 3 rad Q5: An object moves 5.00 cm around the circumference of a circle of radius 24.0 cm. What is the angular displacement of the object? a) rad b) 0.208π rad c) 4.80 rad

41 TOPIC 3. ANGULAR VELOCITY AND ACCELERATION 33 d) 4.80π rad e) 9.60π rad 3.3 Angular velocity and acceleration Learning Objective To define the angular velocity, angular acceleration and periodic time of an object moving in a circle, and to calculate these quantities. Now that we have introduced the concept of angular displacement, it follows that the angular velocity ω is equal to the rate of change of angular displacement, just as the linear velocity v is the rate of change of linear displacement s. ω = dθ dt v = ds dt ω is measured in radians per second (rad/s or rad s -1 ). The average angular velocity over a period of time t is the total angular displacement in time t divided by t. Example It takes the Moon 27.3 days to complete one orbit of the Earth. Assuming the Moon travels in a circular orbit at constant angular velocity, what is the angular velocity of the Moon? The Moon moves through 2π rad in 27.3 days. Now, 27.3 days is equal to ( ) = hours which is equal to ( ) = minutes which is equal to ( ) = s So the angular velocity ω is given by angular displacement ω = time elapsed 2π ω = ω = rad s 1 You should also be aware of two other useful ways of describing the rate at which a body is moving in circular motion. One is to use the periodic time (or period) T, which

42 34 TOPIC 3. ANGULAR VELOCITY AND ACCELERATION is the time taken for one complete rotation. The other is to express the rate in terms of revolutions per second, which is the inverse of the periodic time. In the example above, the periodic time T for the Moon orbiting the Earth is 27.3 days, or s in SI units. The rate of rotation is equal to 1 1 / T = = revolutions per second. The relationship between periodic time and angular velocity is ω = 2π T (3.2) Orbits of the planets 20 min Following on from the above example, calculate the periodic times and angular velocities of the motion of Mercury, Venus and the Earth around the Sun. Fill in the gaps in the table below. Planet Orbit radius (m) Period (days) Period (s) Mercury Venus Earth Angular velocity (rad s -1 ) The angular velocity and periodic time of an object moving in a circle are related by the equation ω = 2π / T. Having defined angular displacement θ and angular velocity ω, it should be clear that if ω is changing then we have an angular acceleration. The instantaneous angular acceleration α is the rate of change of angular velocity, measured in rad s -2. α = dω dt (3.3) The average angular acceleration over time t is the total change in angular velocity divided by t.