Physics for the IB Diploma Coursebook PDF

Physics for the IB Diploma COURSEBOOK K. A. Tsokos  Contents How to use this series vi How to use this book vii Unit A Space, time and motion 1 1 Kinematics 2 1.1 Displacemen...

Physics for the IB Diploma COURSEBOOK K. A. Tsokos  Contents How to use this series vi How to use this book vii Unit A Space, time and motion 1 1 Kinematics 2 1.1 Displacement, distance, speed and velocity 3 1.2 Uniformly accelerated motion: the equations of kinematics 7 1.3 Graphs of motion 16 1.4 Projectile motion 20 2 Forces and Newton’s laws 29 2.1 Forces and their direction 30 2.2 Newton’s laws of motion 39 2.3 Circular motion 52 3 Work, energy and power 61 3.1 Work 62 3.2 Conservation of energy 70 3.3 Power and efficiency 78 3.4 Energy transfers 80 4 Linear momentum 84 4.1 Newton’s second law in terms of momentum 85  4.2 Impulse and force–time graphs 87 4.3 Conservation of momentum 90 4.4 Kinetic energy and momentum 93 4.5 Two-dimensional collisions 97 5 Rigid body mechanics 102 5.1 Kinematics of rotational motion 103 5.2 Rotational equilibrium and Newton’s second law 106 5.3 Angular momentum 119 6 Relativity 125 6.1 Reference frames and Lorentz transformations 126 6.2 Effects of relativity 134 6.3 Spacetime diagrams 143 Unit B The particulate nature of matter 155 7 Thermal energy transfers 156 7.1 Particles, temperature and energy 157 7.2 Specific heat capacity and change of phase 161 7.3 Thermal energy transfer 168 8 The greenhouse effect 177 8.1 Radiation from real bodies 178 8.2 Energy balance of the earth 180 9 The gas laws 188  9.1 Moles, molar mass and the Avogadro constant 189 9.2 Ideal gases 191 9.3 The Boltzmann equation 199 10 Thermodynamics 205 10.1 Internal energy 206 10.2 The first law of thermodynamics 211 10.3 The second law of thermodynamics 218 10.4 Heat engines 224 11 Current and circuits 231 11.1 Potential difference, current and resistance 232 11.2 Voltage, power and emf 238 11.3 Resistors in electrical circuits 241 11.4 Terminal potential difference and the potential divider 254 Unit C Wave behaviour 263 12 Simple harmonic motion 264 12.1 Simple harmonic oscillations 265 12.2 Details of simple harmonic motion 273 12.3 Energy in simple harmonic motion 279 12.4 More about energy in SHM 281 13 The wave model 286 13.1 Mechanical pulses and waves 287 13.2 Transverse and longitudinal waves 290  13.3 Electromagnetic waves 298 13.4 Waves extension 299 14 Wave phenomena 301 14.1 Reflection and refraction 302 14.2 The principle of superposition 308 14.3 Diffraction and interference 311 14.4 Single-slit diffraction 318 14.5 Multiple slits 322 15 Standing waves and resonance 329 15.1 Standing waves 330 15.2 Standing waves on strings 331 15.3 Standing waves in pipes 334 15.4 Resonance and damping 340 16 The Doppler effect 346 16.1 The Doppler effect at low speeds 347 16.2 The Doppler effect for sound 351 Unit D Fields 357 17 Gravitation 358 17.1 Newton’s law of gravitation 359 17.2 Gravitational potential and energy 366 17.3 Motion in a gravitational field 373 18 Electric and magnetic fields 384 18.1 Electric charge, force and field385 18.2 Magnetic field and force 395  18.3 Electric potential and electric potential energy 406 19 Motion in electric and magnetic fields 416 19.1 Motion in an electric field 417 19.2 Motion in a magnetic field 421 20 Electromagnetic induction 428 20.1 Electromagnetic induction 429 20.2 Generators and alternating current 443 Unit E Nuclear and quantum physics 449 21 Atomic physics 450 21.1 The structure of the atom 451 21.2 Quantisation of angular momentum 457 22 Quantum physics 462 22.1 Photons and the photoelectric effect 463 22.2 Matter waves 473 23 Nuclear physics 478 23.1 Mass defect and binding energy 479 23.2 Radioactivity 486 23.3 Nuclear properties and the radioactive decay law 494 24 Nuclear fission 506 24.1 Nuclear fission 507 25 Nuclear fusion and stars 514  25.1 Nuclear fusion 515 25.2 Stellar properties and the Hertzsprung– Russell diagram 516 25.3 Stellar evolution extension 523 Acknowledgements 551 How to use this series  How to use this book Throughout this book, you will find lots of different features that will help your learning. These are explained below. UNIT INTRODUCTION A unit is made up of a number of chapters. The key concepts for each unit are covered throughout the chapters. LEARNING OBJECTIVES Each chapter in the book begins with a list of learning objectives. These set the scene for each chapter, help with navigation through the coursebook and indicate the important concepts in each topic. A bulleted list at the beginning of each section clearly shows the learning objectives for the section. GUIDING QUESTIONS This feature contains questions and activities on subject knowledge you will need before starting this chapter. Links These are a mix of questions and explanation that refer to other chapters or sections of the book. The content in this book is divided into Standard and Higher Level material. A vertical line runs down the margin of all Higher Level material, allowing you to easily identify Higher Level from Standard material. Key terms are highlighted in orange bold font at their first appearance in the book so you can immediately recognise them. At the end of the book, there is a glossary that defines all the key terms. KEY POINTS This feature contains important key learning points (facts) and/or equations to reinforce your understanding and engagement. EXAM TIPS These short hints contain useful information that will help you tackle the tasks in the exam. SCIENCE IN CONTEXT This feature presents real-world examples and applications of the content in a chapter, encouraging you to look further into topics. You will note that some of these features end with questions intended to stimulate further thinking, prompting you to consider some of the benefits and problems of these applications. NATURE OF SCIENCE Nature of Science is an overarching theme of the IB Physics Diploma course. The theme examines the processes and concepts that are central to scientific endeavour, and how science serves and connects with the wider community. Throughout the book, there are ‘Nature of Science’ features that discuss particular concepts or discoveries from the point of view of one or more aspects of Nature of Science. THEORY OF KNOWLEDGE This section stimulates thought about critical thinking and how we can say we know what we claim to know. You will note that some of these features end with questions intended to get you thinking and discussing these important Theory of Knowledge issues. INTERNATIONAL MINDEDNESS Throughout this Physics for the IB Diploma course, the international mindedness feature highlights international concerns. Science is a truly international endeavour, being practised across all continents, frequently in international or even global partnerships. Many problems that science aims to solve are international and will require globally implemented solutions. CHECK YOURSELF These appear throughout the text so you can check your progress and become familiar with the important points of a section. Answers can be found at the back of the book. TEST YOUR UNDERSTANDING These questions appear within each chapter and help you develop your understanding. The questions can be used as the basis for class discussions or homework assignments. If you can answer these questions, it means you have understood the important points of a section. WORKED EXAMPLE Many worked examples appear throughout the text to help you understand how to tackle different types of questions. REFLECTION These questions appear at the end of each chapter. The purpose is for you as a learner to reflect on the development of your skills proficiency and your progress against the objectives. The reflection questions are intended to encourage your critical thinking and inquiry-based learning. EXAM-STYLE QUESTIONS  Exam-style questions at the end of each chapter provide essential practice and self-assessment. These are signposted in the print coursebook and can be found in the digital version of the coursebook. SELF-EVALUATION CHECKLIST These appear at the end of each chapter/section as a series of statements. You might find it helpful to rate how confident you are for each of these statements when you are revising. You should revisit any topics that you rated ‘Needs more work’ or ‘Almost there’. Free online material Additional material to support the Physics for the IB Diploma course is available online. This includes Assessment guidance—a dedicated chapter in the digital coursebook helps teachers and students unpack the new assessment and model exam specimen papers. Additionally, answers to the Exam- style question and Test your understanding are also available. Visit Cambridge GO and register to access these resources.  Unit A Space, time and motion INTRODUCTION This unit deals with Classical Mechanics. The basic concepts that we will use include position in space, displacement (change in position), mass, velocity, acceleration, force, momentum, energy and of course time. These concepts, the relations between them and the laws they give rise to are discussed in the first five chapters. This incredible structure that began with Newton's work detailed in his Principia is also called Newtonian Mechanics. The Newtonian view of the world has passed every conceivable experimental test both on a local terrestrial scale as well as on a much larger scale when it is applied to the motion of celestial bodies. It is the theory upon which much of engineering is based with daily practical applications. Of course, no theory of Physics can be considered “correct” no matter how many experimental tests it passes. The possibility always exists that new phenomena, new observations and new experiments may lead to discrepancies with the theory. In that case it may be necessary to modify the theory or even abandon it completely in favour of a new theory that explains the old as well as the new phenomena. This is the case, too, with Newtonian Mechanics. In chapter 6 of this unit we will see that the Newtonian concepts of space and time need to be revised in situations where the speeds involved approach the speed of light. This is not to say that Newtonian Mechanics is useless; the theory of relativity that replaces it, does becomes Newtonian Mechanics in the limit of speeds that are small compared to that of light. Laws that have been derived with Newtonian Mechanics such as the conservation of energy, the conservation of momentum and the conservation of angular momentum also hold in the theories that replace Newtonian Mechanics. There is another limit in which Newtonian Mechanics is unable to describe observed phenomena. This is the physics on a very small, atomic and nuclear scale. At these scales Newtonian Mechanics fails completely to describe the observed phenomena and needs to be replaced by a new theory, Quantum Mechanics.  Chapter 1 Kinematics LEARNING OBJECTIVES In this chapter you will: learn the difference between displacement and distance learn the difference between speed and velocity learn the concept of acceleration learn how to analyse graphs describing motion learn how to solve motion problems using the equations for constant acceleration  learn how to describe the motion of a projectile gain a qualitative understanding of the effects of a fluid resistance force on motion gain an understanding of the concept of terminal speed. GUIDING QUESTIONS Which equations are used to describe the motion of an object? How does graphical analysis help us to describe motion? Introduction This chapter introduces the basic concepts used to describe motion. First, we consider motion in a straight line with constant velocity. We then discuss motion with constant acceleration. Knowledge of uniformly accelerated motion allows us to analyse more complicated motions, such as the motion of projectiles. We use graphical analysis when acceleration is not constant. 1.1 Displacement, distance, speed and velocity Straight line motion in one dimension means that the particle that moves is constrained to move along a straight line. The position of the particle is then given by its coordinate on the straight line (Figure 1.1). Position is a vector quantity—this is important when discussing projectile motion but, while discussing motion on a straight line, we can just express position as a positive or a negative number since here the vector can only point in one direction or the opposite. If the line is horizontal, we use the symbol x to represent the coordinate and position. If the line is vertical, we use the symbol y. In general, for an arbitrary line, we use a generic symbol, s, for position. So, in Figure 1.1, x = 6 m, y = −4 m, s = 0 and y = 4 m. It is up to us to decide which side of zero we call positive and which side of zero we call negative; the decision is arbitrary. Figure 1.1: The position of a particle is determined by the coordinate on the number line. The change in position is called displacement, Δs = sfinal − sinitial. Displacement is a vector. Table 1.1 shows four different motions. Make sure you understand how to calculate the displacement and that you understand the direction of motion. Initial Final Displacement Direction of position position motion 12 m 28m +16 m Towards increasing s −6 m −14 m −8 m Towards decreasing s 10 m −5 m −15 m Towards decreasing s −20 m −15 +5 m Towards increasing s Table 1.1: Four different motions. Consider the two motions shown in Figure 1.2. In the first motion, the particle leaves its initial position at −4 m and continues to its final position at 16 m. The displacement is: Δs = sfinal − sinitial = 16 − (−4) = 20 m The distance travelled is the actual length of the path followed (20 m). In the second motion, the particle leaves its initial position at 12 m, arrives at position 20 m and then comes back to its final position at 4 m. Figure 1.2 The second motion is an example of motion with changing direction. The change in the position of this particle (its displacement) is: Δs = sfinal − sinitial = 4 − 12 = −8 m But the distance travelled by the particle (the length of the path) is 8 m in the outward trip and 16 m on the return trip, making a total distance of 24 m. So, we must be careful to distinguish distance from displacement: Distance is a scalar quantity but displacement is a vectorquantity. Numerically, they are different if there is a change of direction, as in Figure 1.2. As the particle moves on the straight line its position changes. In uniform motion in equal intervals of time, the position changes by the same amount. For uniform motion the velocity, v, of the particle is the displacement divided by the time to achieve that Δs displacement: v =. Δt The average speed is the total distance travelled divided by the time taken. Assume that the first motion in Figure 1.2 took 4.0 s to 20 complete. The velocity is −1 = 5.0 m s. The 4.0 average speed is the same since the distance travelled is also 20 m. Assume the second motion took 6.0 s to complete. The −8.0 velocity is = −1.3 m s −1 and the average speed 6.0 24 is = 4.0 m s −1. So, in uniform motion, average 6.0 speed and velocity are not the same when there is change in direction. This implies that for uniform motion: s − sinitial v = t − 0 which can be re-arranged to give: s = sinitial + vt (Notice how we use s for final position, rather than sf or sfinal, and we will write si for sinitial for simplicity). So for motion with constant velocity: s = si + vt EXAM TIP This formula can only be used when the velocity is constant. This formula gives, in uniform motion, the position s of the moving object t seconds after time zero, given that the constant velocity is v and the initial position is si. This means that a graph of position against time is a straight line and the graph of velocity against time is a horizontal straight line (Figure 1.3). Figure 1.3: In uniform motion the graph of position against time is a straight line with non-zero gradient. The graph of velocity against time is a horizontal straight line. Positive velocity means that the position s is increasing. Negative velocity means that s is decreasing. Observe that the area under the v versus t graph from t = 0 to time t is vt. From s = si + vt we deduce that s − si = vt and so the area in a velocity-against-time graph is the displacement. CHECK YOURSELF 1 An object moves from A to B at speed 15 m s−1 and returns from B to A at speed 30 m s−1. What is the average speed for the round trip? WORKED EXAMPLE 1.1 Two cyclists, A and B, start moving at the same time. The initial position of A is 0 km and her velocity is +20 km h−1. The initial position of B is 150 km away from A and he cycles at a velocity of −30 km h−1. a Determine the time and position at which they will meet. b What is the displacement of each cyclist when they meet? c On another occasion the same experiment is performed but this time B starts 1 h after A. When will they meet? Answer a The position of A is given by the formula sA = 0 + 20t. The position of B is given by the formula sB = 150 − 30t. They will meet when they are at the same position, i.e. when sA = sB. This implies:  20t = 150 − 30t 50t = 150 t = 3.0 h The common position is found from either sA = 20 × 3.0 = 60 km or sB = 150 − 30 × 3.0 = 60 km. b The displacement of A is 60 km − 0 = 60 km. That of B is 60 km − 150 km = −90 km. c sA = 0 + 20t as before. When t h go by, B will have been moving for only t − 1 h. Hence sB = 150 − 30(t − 1). They will meet when sA = sB: 20t = 150 − 30(t − 1) 50t = 180 t = 3.6 h TEST YOUR UNDERSTANDING 1 A car must be driven a distance of 120 km in 2.5 h. During the first 1.5 h the average speed was 70 km h−1. Calculate the average speed for the remainder of the journey. 2 Find the constant velocity for each motion whose position–time graphs are shown in Figure 1.4.  Figure 1.4 3 Draw the position–time graphs for an object moving in a straight line with velocity–time graphs as shown in Figure 1.5. The initial position is zero. In each case, state the displacement at 4 s. Figure 1.5 4 Two cyclists, A and B, have displacements 0 km and 70 km, respectively. At t = 0 they begin to cycle towards each other with velocities 15 km h−1 and 20 km h−1, respectively. At the same time, a fly that was sitting on cyclist A starts flying towards cyclist B with a velocity of 30 km h−1. As soon as the fly reaches cyclist B it immediately turns around and flies towards cyclist A, and so on until cyclist A and cyclist B meet. a Find the position of the two cyclists and the fly when all three meet. b Determine the distance travelled by the fly. 1.2 Uniformly accelerated motion: the equations of kinematics Defining velocity in non-uniform motion How is velocity defined when it is not constant? We have to refine what we did in Section 1.1. We now define the average velocity as: Δs v̄ = Δt where Δs is the total displacement for the motion and Δt the total time taken. We would like to have a concept of velocity at an instant of time, the (instantaneous) velocity. We need to make the time interval Δt very, very small. The instantaneous velocity is defined as: Δs v = lim Δt→0 Δt In other words, instantaneous velocity is the rate of change of position. This definition implies that velocity is the gradient of a position-against-time graph. Consider Figure 1.6a. Choose a point on this curve. Draw a tangent to the curve at the point. The gradient of the tangent line Δs is the meaning of v = lim and, therefore, also of velocity. Δt→0 Δt Figure 1.6: a In uniformly accelerated motion the graph of position against time is a curve. b The gradient (slope) of the tangent at a particular point gives the velocity at that point. In Figure 1.6b the tangent is drawn at t = 3.0 s. We can use this to find the instantaneous velocity at t = 3.0 s. The gradient of this tangent line is: 25 − 1.0 −1 = 6.0 m s 5.0 − 1.0 To find the instantaneous velocity at some other instant of time we must take another tangent and we will find a different instantaneous velocity. At the point at t = 0 it is particularly easy to find the velocity: the tangent is horizontal and so the velocity is zero. From now on we drop the word instantaneous and refer just to velocity. The magnitude of the velocity is known as the (instantaneous) speed. When the velocity changes we say that we have acceleration. The Δv average acceleration is defined as ā =. So for a body that Δt accelerates from a velocity of 2.0 m s−1 to a velocity of 8.0 m s−1 in a time of 3.0 s the average acceleration is 8.0 − 2.0 ā = = 2.0 m s −2.. 3.0 To define instantaneous acceleration or just simply acceleration, we use the same idea as for velocity. We let the time interval Δt get very small and define acceleration as: Δv Acceleration is the rate of change of velocity. a = lim. It is Δt→0 Δt the gradient of a velocity-against-time graph. Acceleration is a vector. Uniformly accelerated motion means that the acceleration is constant: the graph of velocity against time is a non-horizontal straight line (Figure 1.7). In equal intervals of time the velocity changes by the same amount. Figure 1.7: In uniformly accelerated motion the graph of velocity against time is a straight line with non-zero slope. Figure 1.8: This fighter jet is accelerating. When the acceleration is positive, the velocity is increasing (Figure 1.8). Negative acceleration means that v is decreasing. For constant acceleration there is no difference between instantaneous acceleration and average acceleration. Suppose we choose a time interval from t = 0 to some arbitrary time t later. Let the velocity at t = 0 (the initial velocity) be u and the velocity at time t be v. Then: Δv v − u a = = Δt t − 0 which can be re-arranged to: v = u + at For uniformly accelerated motion, this formula gives the velocity v of the moving object t seconds after time zero, given that the initial velocity is u and the acceleration is a. WORKED EXAMPLE 1.2 A particle has an initial velocity 12 m s−1 and moves with a constant acceleration of −3.0 m s−2. Determine the time at which the particle stops instantaneously. Answer  At some point it will stop instantaneously; that is, its velocity v will be zero. We know that v = u + at. Substituting values gives: 0 = 12 + (−3.0) × t 3.0t = 12 Hence t = 4.0 s. Consider the graph of velocity against time in Figure 1.9. Imagine approximating the straight line with a staircase. The area under the staircase is the change in position since at each step the velocity is constant. If we make the steps of the staircase smaller and smaller, the area under the line and the area under the staircase will be indistinguishable and so we have the general result that: KEY POINTS The area under the curve in a velocity against time graph is the change in position; in other words, the displacement. From Figure 1.9 this area is (the shape is a trapezoid): u + v Δs = ( )t 2 Figure 1.9: The straight-line graph may be approximated by a staircase. But v = u + at, so this becomes: u + u + at 1 2 Δs = ( ) t = ut + at 2 2 So we have two formulas for position in the case of uniformly accelerated motion (recall that Δs = s − si): u + v u + v s = si + ( ) t or Δs = ( )t 2 2 1 1 2 2 s = si + ut + at or Δs = ut + at 2 2 1 We get a final formula if we combine s = si + ut + at 2 with v 2 = u + at. v − u From the second equation, we find t =. a Substituting in the first equation we get: 2 2 2 2 v − u 1 v − u uv u 1 v uv 1 u s − si = u + a( ) = − + − + a 2 a a a 2 a a 2 a 2 2 v − u = 2a This becomes: 2 2 2 2 v = u + 2a (s − si ) or v = u + 2aΔs Usually si = 0 so this last equation is usually written as v2 = u2 + 2as. This formula is useful to solve problems in which no information about time is available. In summary, for motion along a straight line with constant acceleration: v = u + at u + v Δs = ( )t 2 1 2 Δs = ut + at 2 2 2 v = u + 2aΔs Graphs of position against time for uniformly accelerated motion are parabolas (Figure 1.10). If the parabola ‘holds water’ (concave up) the acceleration is positive. If its concave down, the acceleration is negative. Figure 1.10: Graphs of position s against time t for uniformly accelerated motion. a Positive acceleration. b Negative acceleration. EXAM TIP Table 1.2 summarises the meaning of the gradient and area for the different motion graphs. Graph of … Gradient Area position against velocity time velocity against acceleration change in position, i.e. time displacement acceleration against change in velocity time Table 1.2: Information that can be derived from motion graphs The equations of kinematics can be used only for motion on a straight line with constant acceleration. (If the initial position is zero, Δs may be replaced by just s.) CHECK YOURSELF 2 The graph in Figure 1.11 shows the variation of position with time. Figure 1.11 Indicate the intervals of times for which the acceleration is: a positive b negative c zero. WORKED EXAMPLE 1.3 A particle has an initial velocity 2.00 m s−1. Its acceleration is a = 4.00 m s−2. Find its displacement after 10.0 s. Answer Displacement is the change of position, i.e. ∆s = s − si. We use the equation: 1 2 Δs = ut + at 2 2 = 2.00 × 10.0 + 1/2 × 4.00 × 10.0 = 220 m WORKED EXAMPLE 1.4 A car has an initial velocity of u = 5.0 m s−1. After a displacement of 20 m, its velocity becomes 7.0 m s−1. Find the acceleration of the car. Answer Here, ∆s = s − si = 20 m. So, use v2 = u2 + 2a∆s to find a. 2 2 7.0 = 5.0 + 2a × 20 24 = 40a Therefore, a = 0.60 m s−2. WORKED EXAMPLE 1.5 A body has an initial velocity of 4.0 m s−1. After 6.0 s the velocity is 12 m s−1. Determine the displacement of the body in the 6.0 s. Answer We know u, v and t. We can use: v + u Δs = ( )t 2 to get:  12 + 4.0 Δs = ( ) × 6.0 2 Δs = 48 m A slower method would be to use v = u + at to find the acceleration: 12 = 4.0 + 6.0a −2 ⇒ a = 1.333 m s Then use the value of a to find ∆s: 1 2 Δs = ut + at 2 1 = 4.0 × 6.0 + × 1.333 × 36 2 = 48 m WORKED EXAMPLE 1.6 A ball, X, starts moving to the right from rest with a constant acceleration of 2.0 m s−2. 2.0 s later, a second ball, Y, starts moving to the right with a constant velocity of 9.0 m s−1. Both balls start from the same position. Determine the position of the balls when they meet. Describe what is going on. Answer Let the two balls meet t s after X starts moving. The position of X is: 1 2 1 2 2 s X = at = × 2.0 × t = t 2 2 The position of Y is: s Y = 9(t − 2) (The factor t – 2 must be considered because, after t s, Y has actually been moving for only t − 2 seconds.) These two positions are equal when the two balls meet, and so: 2 t = 9(t − 2) Solving the quadratic equation we find t = 3.0 s and 6.0 s. At these times the positions are 9.0 m and 36 m. At 3.0 s Y caught up with X and passed it. At 6.0 s X caught up with Y and passed it. (Making position–time graphs here is instructive.) WORKED EXAMPLE 1.7 A particle starts out from the origin with velocity 10 m s−1 and continues moving at this velocity for 5 s. The velocity is then abruptly reversed to −5 m s−1 and the object moves at this velocity for 10 s. For this motion find: a the change in position, i.e. the displacement b the total distance travelled c the average speed d the average velocity Answer The problem is best solved using the velocity–time graph, which is shown in Figure 1.12. Figure 1.12 a The initial position is zero. So, after 5.0 s, the position is 10 × 5.0 m = 50 m (the area under the first part of the graph). In the next 10 s, the displacement changes by −5.0 × 10 = −50 m (the area under the second part of the graph). So the change in position (the displacement) = 50 − 50 = 0 m. b Take the initial velocity as moving to the right. The object moved towards the right, stopped and returned to its starting position. (We know this because the displacement was 0.) The distance travelled is 50 m moving to the right and 50 m coming back, giving a total distance travelled of 100 m. 100 c The average speed is = 6.7 m s −1. 15 d The average velocity is zero, since the displacement is zero. WORKED EXAMPLE 1.8 An object with an initial velocity 20 m s−1 and initial position of −75 m experiences a constant acceleration of −2 m s−2. Sketch the position–time graph for this motion for the first 20 s. Answer  1 Use the equation s = si + ut + 2 at. Substituting the 2 values we know, the position is given by s = −75 + 20t − t2. This is the function we must graph. The result is shown in Figure 1.13. Figure 1.13 At 5 s the object reaches the origin and overshoots it. It returns to the origin 10 s later (t = 15 s). The furthest it gets from the origin is 25 m. The velocity at 5 s is 10 m s−1 and at 15 s it is −10 m s−1. At 10 s the velocity is zero. A special acceleration Assuming that we can neglect air resistance and other frictional forces, an object thrown into the air will experience the acceleration of free fall while in the air. This is an acceleration caused by the attraction between the earth and the body. The magnitude of this acceleration is denoted by g. Near the surface of the earth g = 9.8 m s−2. The direction of this acceleration is always vertically downwards. (We will sometimes approximate g to 10 m s−2.) WORKED EXAMPLE 1.9 An object is thrown vertically upwards with an initial velocity of 20 m s−1 from the edge of a cliff that is 25 m from the sea below, as shown in Figure 1.14. Determine: a the ball’s maximum height b the time taken for the ball to reach its maximum height c the time to hit the sea d the speed with which it hits the sea (You may approximate g to 10 m s−2.) Figure 1.14: A ball is thrown upwards from the edge of a cliff. Answer We have motion on a vertical line so we will use the symbol y for position (Figure 1.15a). We take the direction up to be positive. The zero for position is the ball’s initial position. Figure 1.15: Diagrams for solving the ball’s motion. a Displacement upwards is positive. b The highest point is the zero of position. Here we take displacement downwards to be positive. a The quickest way to get the answer to this part is to use v2 = u2 − 2gy. (The acceleration is a = −g.) At the highest point v = 0, and so: 2 0 = 20 − 2 × 10y ⇒ y = 20 m b At the highest point the object’s velocity is zero. Using v = 0 in v = u − gt gives: 0 = 20 − 10 × t 20 t = = 2.0 s 10 c There are many ways to do this. One is to use the displacement arrow shown in blue in Figure 1.15a. Then, when the ball hits the sea, y = −25 m. Now use the 1 formula y = ut − gt 2 to find an equation that only 2 has the variable t: 2 −25 = 20 × t − 5 × t 2 t − 4t − 5 = 0 This is a quadratic equation. Using your calculator you can find the two roots as −1.0 s and 5.0 s. Choose the positive root to find the answer t = 5.0 s. Another way of looking at this is shown in Figure 1.15b. Here we start at the highest point and take the direction down to be positive. Then, at the top y = 0, at the sea y = +45 m and a = +10 m s−2. Now, the initial velocity is zero because we take our initial point to be at the top. 1 Using y = ut + gt 2 , we find: 2 2 45 = 5t ⇒ t = 3.0 s This is the time to fall to the sea. It took 2.0 s to reach the highest point, so the total time from launch to hitting the sea is: 2.0 + 3.0 = 5.0 s. d Use v = u − gt and t = 5.0 s to get v = 20 − 10 × 5.0 = −30 m s−1. The speed is then 30 m s−1. (If you prefer the diagram in Figure 1.15b for working out part c and you want to continue this method for part d, then you would write v = u + gt with t = 3.0 s and u = 0 to get v = 10 × 3.0 = +30 m s−1.) CHECK YOURSELF 3 A ball is thrown vertically upwards with a speed of 20 m s−1 and another vertically downwards with the same speed and from the same height. What interval separates the landing times of the two balls? (Take g = 10 m s−2.) TEST YOUR UNDERSTANDING 5 The initial velocity of a car moving on a straight road is 2.0 m s−1. It becomes 8.0 m s−1 after travelling for 2.0 s under constant acceleration. Find the acceleration. 6 A car accelerates from rest to 28 m s−1 in 9.0 s. Find the distance it travels. 7 A particle has an initial velocity of 12 m s−1 and is brought to rest over a distance of 45 m. Find the acceleration of the particle. 8 A particle at the origin has an initial velocity of −6.0 m s−1 and moves with an acceleration of 2.0 m s−2. Determine when its position will become 16 m. 9 A plane starting from rest takes 15.0 s to take off after speeding over a distance of 450 m on the runway with constant acceleration. Find the take-off velocity. 10 A particle starts from rest with constant acceleration. After travelling a distance d the speed is v. What was the speed d when the particle had travelled a distance ? 2 11 A car is travelling at 40.0 m s−1. The driver sees an emergency ahead and 0.50 s later slams on the brakes. The deceleration of the car is 4.0 m s−2. a Find the distance travelled before the car stops. b Calculate the stopping distance if the driver could apply the brakes instantaneously without a reaction time. c Calculate the difference in your answers to a and b. d Assume now that the car was travelling at 30.0 m s−1 instead. Without performing any calculations, state whether the answer to c would now be less than, equal to or larger than before. Explain your answer. 12 Two balls are dropped from rest from the same height. One of the balls is dropped 1.00 s after the other. Air resistance is ignored. a Find the distance that separates the two balls 2.00 s after the second ball is dropped. b Explain what happens to the distance separating the balls as time goes on. 13 A particle moves in a straight line with an acceleration that varies with time as shown in Figure 1.16. Initially the velocity of the object is 2.00 m s−1. Figure 1.16  a Find the maximum velocity reached in the first 6.00 s of this motion. b Draw a graph of the velocity against time. 14 The graph in Figure 1.17 shows the variation of velocity with time of an object. Find the acceleration at 2.0 s. Figure 1.17 15 Your brand-new convertible is parked 15 m from its garage when it begins to rain. You do not have time to get the keys, so you begin to push the car towards the garage. The maximum acceleration you can give the car is 2.0 m s−2 by pushing and 3.0 m s−2 by pulling back on the car. Find the least time it takes to put the car in the garage. (Assume that the car and the garage are point objects.) 16 A stone is thrown vertically up from the edge of a cliff 35.0 m from the sea (Figure 1.18). The initial velocity of the stone is 8.00 m s−1.  Figure 1.18 Determine: a the maximum height of the stone b the time when it hits the sea c the velocity just before hitting the sea d the distance the stone covers e the average speed and the average velocity for this motion 17 A ball is thrown upwards from the edge of a cliff with velocity 20.0 m s−1. It reaches the bottom of the cliff 6.0 s later. Take g = 10 m s−2. a Determine the height of the cliff. b Calculate the speed of the ball as it hits the ground. 18 A toy rocket is launched vertically upwards with acceleration 3.0 m s−2. The fuel runs out when the rocket reaches a height of 85 m. a Calculate the velocity of the rocket when the fuel runs out. b Determine the maximum height reached by the rocket. c Draw a graph to show the variation of i the velocity of the rocket ii the position of the rocket from launch until it reaches its maximum height d Calculate the time the rocket takes to fall to the ground from its maximum height. 1.3 Graphs of motion If the acceleration is constant the equations derived in Section 1.2 hold and we can use them to analyse a motion problem. But if acceleration is not constant these equations do not apply and cannot be used. But we can still analyse a motion using graphs. Our main tools will be those provided in Table 1.1. Consider the graph of position against time shown in Figure 1.19. Figure 1.19: Position against time for accelerated motion. We would like to predict the shape of the graph of velocity against time. We use the fact that velocity is the gradient of the position-against-time graph. We see that at t = 0 the gradient is positive. As time increases the gradient is still positive but gets smaller and smaller and at t = 0.25 s it becomes zero. It then becomes negative. It gets more and more negative, reaching its most negative value at t = 0.5 s. It remains negative past 0.5 s but the curve gets less steep and the gradient becomes zero again at 0.75 s. From then it becomes positive and increases until 1.0 s. These observations lead us to the velocity-against-time graph shown in Figure 1.20. Note that it is the shape we are after, not detailed numerical values of velocity; hence there is no need to put numbers on the vertical axis. Figure 1.20: The velocity graph corresponding to the graph of Figure 1.19. Working in exactly the same way we can find the shape of the acceleration-against-time graph by examining the gradient of the velocity graph we just obtained. The result is Figure 1.21. Figure 1.21: Acceleration graph corresponding to the velocity graph of Figure 1.20. CHECK YOURSELF 4 Look at the motion in Figure 1.22. Figure 1.22 a State the time(s) at which the acceleration is zero. b State the time(s) at which the acceleration is negative. c At what time is the acceleration most negative? d At 2.3 s, do you expect the displacement to be positive or negative and why? TEST YOUR UNDERSTANDING 19 The graph in Figure 1.23 shows the variation of the position of a moving object with time. Draw the graph showing the variation of the velocity of the object with time. Figure 1.23 20 The graph in Figure 1.24 shows the variation of the position of a moving object with time. Draw the graph showing the variation of the velocity of the object with time.  Figure 1.24 21 The graph in Figure 1.25 shows the variation of the position of a moving object with time. Draw the graph showing the variation of the velocity of the object with time. Figure 1.25 22 The graph in Figure 1.26 shows the variation of the velocity of a moving object with time. Draw the graph showing the variation of the position of the object with time. The initial position is zero.  Figure 1.26 23 The graph in Figure 1.27 shows the variation of the velocity of a moving object with time. Draw the graph showing the variation of the position of the object with time (assuming the initial position to be zero). Figure 1.27 24 The graph in Figure 1.28 shows the variation of the velocity of a moving object with time. Draw the graph showing the variation of the acceleration of the object with time.  Figure 1.28 25 The graph in Figure 1.29 shows how acceleration varies with time. Figure 1.29 Draw a graph to show how velocity varies with time. The initial velocity is zero. 26 The graph in Figure 1.30 shows how acceleration varies with time.  Figure 1.30 Draw a graph to show how velocity varies with time. The initial velocity is zero. 27 The graph in Figure 1.31 shows the position against time of an object moving in a straight line. Four points on this graph have been selected. Figure 1.31 a Is the velocity between A and B positive, zero or negative? b What can you say about the velocity between B and C? c Is the acceleration between A and B positive, zero or negative?  d Is the acceleration between C and D positive, zero or negative? 28 Sketch velocity–time plots (no numbers are necessary on the axes) for the following motions. a A ball is dropped from a certain height and bounces off a hard floor. The speed just before each impact with the floor is the same as the speed just after impact. Assume that the time of contact with the floor is negligibly small. b A cart slides with negligible friction along a horizontal air track. When the cart hits the ends of the air track it reverses direction with the same speed it had right before impact. Assume the time of contact of the cart and the ends of the air track are negligibly small. c A person jumps from a hovering helicopter. After a few seconds she opens a parachute. Eventually she will reach a terminal speed and will then land. 1.4 Projectile motion Figure 1.32 shows the positions of two objects every 0.2 s: the first was simply allowed to drop vertically from rest; the other was launched horizontally with no vertical component of velocity. We see that in the vertical direction, both objects fall the same distance in the same time and so will hit the ground at the same time. Figure 1.32: A body dropped from rest and one launched horizontally cover the same vertical distance in the same time. THEORY OF KNOWLEDGE How do we understand this fact? Consider Figure 1.33, in which a black ball is projected horizontally with velocity v. A blue ball is allowed to drop vertically from the same height.  Figure 1.33a shows the situation when the balls are released as seen by an observer X at rest on the ground. But suppose there is an observer Y, who v moves to the right with velocity with respect to the 2 ground. What does Y see? Observer Y sees the black v ball moving to the right with velocity and the blue 2 v ball approaching with velocity − Figure 1.18b). The 2 motions of the two balls are therefore identical (except for direction). So this observer will determine that the two bodies reach the ground at the same time. Since time is absolute in Newtonian physics, the two bodies must reach the ground at the same time as far as any other observer is concerned as well. Figure 1.33: a A ball projected horizontally and one simply dropped from rest from the point of view of observer X. Observer Y is moving to the right with v velocity with respect to the ground. b From the 2 point of view of observer Y, the black and the blue balls have identical motions. The motion of a ball that is projected at some angle can be analysed by separately looking at the horizontal and the vertical directions. All we have to do is consider two motions, one in the horizontal direction in which there is no acceleration and another in the vertical direction in which we have an acceleration, g. Consider Figure 1.34, where a projectile is launched at an angle θ to the horizontal with speed u. The components of the initial velocity vector are ux = u cos θ and uy = u sin θ. At some later time t the components of velocity are vx and vy. In the x-direction there is zero acceleration and in the y-direction, the acceleration is −g and so: Horizontal direction Vertical direction vx = u cos θ vy = u sin θ − gt Figure 1.34: A projectile is launched at an angle θ to the horizontal with speed u. The green vector in Figure 1.35a shows the position of the projectile t seconds after launch. The red arrows in Figure 1.35b show the velocity vectors and their components. Here the true vector nature of what we have been calling position comes into its own. The position of the projectile is determined by the → position vector r , a vector from the origin to the position of the projectile. The components of this vector are x and y. Figure 1.35: a The position of the particle is determined if we know the x- and y-components of the position vector. b The velocity vectors for projectile motion are tangents to the parabolic path. EXAM TIP All that we are doing is using the formulas from the previous section for velocity and position – v = u + at 1 and s = ut + 2 at – and rewriting them separately 2  for each direction x and y. In the x-direction there is zero acceleration, and in the y-direction there is an acceleration –g. We would like to know the x- and y-components of the position vector. We now use the formula for position. In the x-direction: x = uxt and so x = ut cos θ 1 In the y-direction: y = uy t − 2 gt and 2 1 so y = ut sin θ − 2 gt. 2 In summary: Horizontal direction Vertical direction vx = u cos θ vy = u sin θ − gt x = ut cos θ 1 2 y = ut sin θ − gt 2 The equation with ‘squares of speeds’ is a bit trickier (carefully review the following steps). It is: 2 2 v = u − 2gy Since v2 = vx2 + vy2 and u2 = ux2 + uy2, and in addition vx2 = ux2, this is also equivalent to: 2 2 vx = uy − 2gy EXAM TIP Always choose your x- and y-axes so that the origin is the point where the launch takes place. CHECK YOURSELF 5 A projectile is launched with kinetic energy K at 60° to the horizontal. What is the kinetic energy of the projectile at the highest point of its path? (Kinetic 1 energy is equal to mv 2.) 2 WORKED EXAMPLE 1.10 A body is launched with a speed of 18.0 m s−1 at the following angles: a 30° to the horizontal b 0° to the horizontal c 90° to the horizontal Find the x- and y-components of the initial velocity in each case. Answer a vx = u cos θ vy = u sin θ vx = 18.0 × cos 30° vy = 18.0 × sin 30° −1 −1 vx = 15.6 m s vy = 9.00 m s b vx = 18.0 m s −1 vy = 0 m s −1 c vx = 0 vy = 18.0 m s −1 WORKED EXAMPLE 1.11 For a projectile launched at some angle above the horizontal, sketch graphs to show the variation with time of: a the horizontal component of velocity b the vertical component of velocity Answer The graphs are shown in Figure 1.36. Figure 1.36: Answer to worked example 1.11 WORKED EXAMPLE 1.12 An object is launched horizontally from a height of 20 m above the ground with speed 15 m s−1. Determine: a the time at which it will hit the ground b the horizontal distance travelled c the speed with which it hits the ground (Take g = 10 m s−2.) Answer a The launch is horizontal, i.e. θ = 0°, and so the formula for vertical position is just 1 y = − gt 2. 2 The object will hit the ground when y = −20 m. Substituting the values, we find: 2 −20 = −5t ⇒ t = 2.0 s b The horizontal position is found from x = ut. Substituting values: x = 15 × 2.0 = 30 m (Remember that θ = 0°.) c Use v2 = u2 − 2gy to get: 2 2 v = 15 − 2 × 10 × (−20) −1 v = 25 m s WORKED EXAMPLE 1.13  An object is launched horizontally with a velocity of 12 m s−1. Determine: a the vertical component of velocity after 4.0 s b the x- and y-components of the position vector of the object after 4.0 s. Answer a The launch is again horizontal, i.e. θ = 0°, so substitute this value in the formulas. The horizontal component of velocity is 12 m s−1 at all times. From vy = −gt, the vertical component after 4.0 s is vy = −40 m s−1. b The coordinates after time t are: 1 2 x = ut y = − gt 2 x = 12.0 × 4.0 y = −5 × 16 x = 48 m and y = −80 m EXAM TIP Worked example 1.13 is a basic problem— you must know how to do this! Figure 1.37 shows an object thrown at an angle of θ = 30° to the horizontal with initial speed 20 m s−1. The position of the object is shown every 0.2 s. Note how the dots get closer together as the object rises (the speed is decreasing) and how they move apart on the way down (the speed is increasing). It reaches a maximum height of 5.1 m and travels a horizontal distance of 35 m. The photo in Figure 1.38 shows an example of projectile motion. Figure 1.37: A launch at of θ = 30° to the horizontal with initial speed 20 m s−1. At what point in time does the vertical velocity component become zero? Setting vy = 0 we find: 0 = u sin θ − gt u sin θ ⇒ t = g Figure 1.38: A real example of projectile motion. The time when the vertical velocity becomes zero is, of course, the time when the object attains its maximum height. What is this height? Going back to the equation for the vertical component of position, we find that when: u sin θ t = g y is given by: 2 u sin θ 1 u sin θ ymax = u sin θ − g( ) g 2 g 2 2 u sin θ ymax = 2g What about the maximum displacement in the horizontal direction (called the range)? At this point y is zero. We can find the time by setting y = 0 in the formula for y but it is easier to notice that since the path is symmetrical the time to cover the range is double the 2u sin θ time to get to the top, so t =. g EXAM TIP You should not remember these formulas by heart. You should be able to derive them quickly. Therefore the range is: 2 2u sin θ cos θ x = g A bit of trigonometry allows us to rewrite this as: 2 u sin 2θ x = g One of the identities in trigonometry is 2 sinθ cosθ = sin2θ. The maximum value of sin 2θ is 1, and this happens when 2θ = 90° (θ = 45°). In other words, we obtain the maximum range with a launch angle of 45°. This equation also says that there are two different angles of launch that give the same range for the same initial speed. These two angles add up to a right angle. (Can you see why?) CHECK YOURSELF 6 On earth the maximum height and range of a projectile are H and R, respectively. The projectile is launched with the same velocity on a planet where the acceleration of free fall is 2g. What are the maximum height and range of the projectile on the planet? WORKED EXAMPLE 1.14 A projectile is launched at 32.0° to the horizontal with initial speed 25.0 m s−1. Determine the maximum height reached. (Take g = 9.81 m s−2.) Answer The vertical velocity is given by vy = u sin θ − gt and becomes zero at the highest point. Thus: u sin θ t = g 25.0 × sin 32.0° t = 9.81 t = 1.35 s Substituting in the formula for y, 1 y = ut sin θ − 2 gt , we get: 2 1 2 y = 25 × sin 32.0° × 1.35 − × 9.81 × 1.35 2 y = 8.95 m Equivalently, 2 (25.0 × sin 32.0°) 2 0 = (u sin θ) − 2gy ⇒ y = = 8.95 m 2 × 9.81 WORKED EXAMPLE 1.15 A projectile is launched horizontally from a height of 42 m above the ground. As it hits the ground, the velocity makes an angle of 55° to the horizontal. Find the initial velocity of launch. (Take g = 9.8 m s−2.) Answer The time it takes to hit the ground is found from 1 y = − gt 2. (Here θ = 0° since the launch is 2 horizontal.) The ground is at y = −42 m and so: 1 2 −42 = − × 9.8t 2 ⇒ t = 2.928 s Using vy = u sin θ − gt, when the projectile hits the ground: vy = 0 − 9.8 × 2.928 −1 vy = −28.69 m s We know the angle the final velocity makes with the ground (Figure 1.39). Hence:  ∣ vy ∣ tan 55° = ∣ ∣ ∣ vx ∣ 28.69 ⇒ vx = tan 55° −1 vx = 20.09 ≈ 20 m s Figure 1.39 Links In projectile motion we have a single constant force whose direction is always vertical. This force is the weight of the body. When the body is projected at an angle to the vertical the result is motion along a path that is parabolic. There is one other instance where again a force that is constant in magnitude and direction acts on a body. This is the case of motion of an electric charge in a uniform electric field. So we expect that a charge projected at an angle in a region of uniform electric field will also result in motion along a parabolic path and what we have learned in this chapter will apply to that case as well. See Chapter 19. Fluid resistance The discussion of the previous sections has neglected air resistance forces. In general, whenever a body moves through a fluid (gas or liquid) it experiences a fluid resistance force that is directed opposite to the velocity. Typically F = kv for low speeds and F = kv2 for high speeds (where k is a constant). The magnitude of this force increases with increasing speed. Imagine dropping a body of mass m from some height. Assume that the force of air resistance on this body is F = kv. Initially, the only force on the body is its weight, which accelerates it downwards. As the speed increases, the force of air resistance also increases. Eventually, this force will become equal to the weight and so the acceleration will become zero: the body will then move at constant speed, called terminal speed, vT. This speed can be found from: mg = kvT which leads to: mg vT = k Figure 1.40 shows how the speed and acceleration vary for motion with an air resistance force that is proportional to speed. The speed eventually becomes the terminal speed, and the acceleration becomes zero. The initial acceleration is g. Figure 1.40: a The variation with time of a speed. b Acceleration in motion with an air resistance force proportional to speed. The effect of air resistance forces on projectiles is very pronounced. Figure 1.41 shows the positions of a projectile with (red) and without (blue) air resistance forces. With air resistance forces the range and maximum height are smaller and the shape is no longer symmetrical. The projectile hits the ground at a steeper angle. Figure 1.41: The effect of air resistance on projectile motion. WORKED EXAMPLE 1.16 The force of air resistance in the motion described by Figure 1.40 is given by F = 0.653v. Determine the mass of the projectile. Answer The terminal speed is 30 m s−1 and is given by mg vT =. Hence: k 0.653 × 30 m = 9.8 m ≈ 2.0 kg NATURE OF SCIENCE The simple and the complex Careful observation of motion in the natural world led to the equations for motion with uniform acceleration along a straight line that we have used in this section. Thinking about what causes an object to move links to the idea of forces. However, although the material in this section is perhaps some of the ‘easiest’ material in your physics course, it does not enable you to understand the falling of a leaf off a tree. The falling leaf is complicated because it is acted upon by several forces: its weight, but also air resistance forces that constantly vary as the orientation and speed of the leaf change. In addition, there is wind to consider as well as the fact that turbulence in air greatly affects the motion of the leaf. So the physics of the falling leaf is far away from the physics of motion along a straight line at constant acceleration. But learning the principles of physics in a simpler context allows its application in more involved situations. TEST YOUR UNDERSTANDING 29 A ball rolls off a table with a horizontal speed of 2.0 m s−1. The table is 1.3 m high. Calculate how far from the table the ball will land. 30 Two particles are on the same vertical line. They are thrown horizontally with the same speed, 4.0 m s−1, from heights of 4.0 m and 8.0 m. a Calculate the distance that will separate the two objects when both land on the ground. b The particle at the 4.0 m height is now launched with horizontal speed u such that it  lands at the same place as the particle launched from 8.0 m. Calculate u. 31 For an object thrown at an angle of 40° to the horizontal at a speed of 20 m s−1, draw graphs of: a horizontal velocity against time b vertical velocity against time c acceleration against time. 32 Determine the maximum height reached by an object thrown with speed 24 m s−1 at 40° to the horizontal. 33 An object is thrown with speed 20.0 m s−1 at an angle of 50° to the horizontal. Draw graphs to show the variation with time of: a the horizontal position b the vertical position 34 A cruel man takes aim horizontally at a chimp that is hanging from the branch of a tree, as shown in Figure 1.42. The chimp lets go of the branch as soon as the hunter pulls the trigger. Treating the chimp and the bullet as point particles, determine if the bullet will hit the chimp.  Figure 1.42 35 A ball is launched from the surface of a planet. Air resistance and other frictional forces are neglected. The graph in Figure 1.43 shows the position of the ball every 0.20 s. Figure 1.43 a Use this graph to determine:  i the components of the initial velocity of the ball, ii the angle to the horizontal the ball was launched at, iii the acceleration of free fall on this planet. b Make a copy of the graph and draw two arrows to represent the velocity and the acceleration vectors of the ball at t = 1.0 s. c The ball is now launched under identical conditions from the surface of a different planet where the acceleration due to gravity is twice as large. Draw the path of the ball on your graph. 36 A stone is thrown with a speed of 20.0 m s−1 at an angle of 48° to the horizontal from the edge of a cliff 60.0 m above the surface of the sea. a Calculate the velocity with which the stone hits the sea. b Discuss qualitatively the effect of air resistance on your answer to a. 37 a State what is meant by terminal speed. b A ball is dropped from rest. The force of air resistance on the ball is proportional to the ball’s speed. Explain why the ball will reach terminal speed. 38 A projectile is launched with speed u at 45° to the horizontal. The projectile is at height h at two different times. a Show that the horizontal distance separating u those points is √u2 − 4gh. g b Deduce, using the result in part a, the maximum height reached by this projectile. 39 A projectile is launched with some speed at some angle to the horizontal. At 1.0 s and at 5.0 s the height of the projectile from the ground is the same. What is the maximum height reached by this projectile? (Take g = 10 m s–2.) SELF-ASSESSMENT CHECKLIST Ready Nearly to I am able to... Section Not yet there move on define and 1.1 distinguish between the concepts of position, displacement, average and instantaneous velocity and average and instantaneous acceleration solve problems 1.2 using the equations of kinematics analyse motion 1.3 through graphs solve problems 1.4 with projectile motion describe the 1.4 effect of air resistance force  on projectile motion REFLECTION Do you understand the difference between distance and displacement? Do you understand the difference between speed and velocity? Are you confident using the equations of kinematics? Do you know what information a graph of position versus time gives? Do you know what information a graph of velocity versus time gives? Can you explain why a body reaches terminal speed if it is acted upon by a speed dependent resistance force? Do you understand how to solve problems with projectile motion? Can you describe the effect of air resistance on the path of a projectile? EXAM-STYLE QUESTIONS You can find questions in the style of IB exams online in the digital coursebook. CHECK YOURSELF ANSWERS 1 Let d be the distance between A and B and t1, t2 the travel times back and forth. Then d = 15 t1 = 30 t2, i.e. t1 = 2 t2. The average speed is then 2d 2 × 30t2 −1 v̄ = = = 20 m s t1 + t2 3t2 2 Positive for 1.0 s on, because curve is concave up, negative for 0 s to 1.0 s and zero at 1.0 s. 3 It takes 2 s to get to the maximum height and 2 to return so the required time is 4 s. 4 a Zero at about 0.4 s and 1.6 s. b Negative from 0.4 s to 1.6 s. c Most negative at 1.0 s. d The displacement from 0 s to 1.0 s is about the same in magnitude as that from 1.0 s to 2.0 s so the displacement at 2.3 s will be positive. 1 5 K = mu 2. At the highest point, 2 1 1 1 K KE = mu 2 cos 2 60° = mu 2 × =. 2 2 4 4 6 Both H and R are inversely proportional to the acceleration of free fall and so both are halved.  Chapter 2 Forces and Newton’s laws LEARNING OBJECTIVES In this chapter you will learn how to: treat bodies as point particles construct and interpret free-body force diagrams apply the equilibrium condition, ΣF = 0 apply Newton’s three laws of motion solve problems involving frictional forces explain why we have acceleration in circular motion  apply Newton’s laws to circular motion identify centripetal forces. GUIDING QUESTIONS What is the connection between force and motion? How does knowledge of the forces on a body allow a prediction of the state of motion of a body? Introduction This chapter introduces Newton’s laws of motion. A lot of classical physics is based on these laws. It was once thought that knowledge of the present state of a system and all forces acting on it would enable the complete prediction of the state of that system in the future. We will learn how to correctly identify the forces that act on a system, find the net force on a system, and apply Newton’s laws to describe the motion that may take place. 2.1 Forces and their direction A force, which we commonly associate with a pull or a push, is a vector quantity. We represent forces with arrows whose length is proportional to the magnitude of the force. It is important that we are able to correctly identify the direction of forces. Forces In this chapter we will deal with the following forces: weight, string tension, spring tension, normal contact forces, drag forces, buoyant forces (upthrust) and frictional forces. Weight Weight is the gravitational force between a body and the planet the body is on. The weight of a body is given by the formula W = mg where m is the mass of the body and g is the gravitational field strength of the planet. The unit of g is newton per kilogram, N kg−1. The gravitational field strength is also known as ‘the acceleration due to gravity’ or the ‘acceleration of free fall’. Therefore the unit of g is also m s−2. If m is in kg and g in N kg−1 or m s−2 then W is in newtons, N. On the surface of the earth, g = 9.81 N kg−1 —a number that we will sometime approximate by the more convenient 10 N kg−1. (We almost always do this for multiple choice exam questions.) This force is always directed vertically downward, as shown in Figure 2.1. Figure 2.1: The weight of an object is always directed vertically downward. The mass of an object is the same everywhere in the universe, but its weight depends on the location of the body. For example, a mass of 70 kg has a weight of 687 N on the surface of the earth (g = 9.81 N kg−1) and a weight of 635 N at a height of 250 km from the earth’s surface (where g = 9.07 N kg−1). However, on the surface of Venus, where the gravitational field strength is only 8.9 N kg−1, the weight is 623 N. Tension The force that arises in any body when it is stretched or compressed is called tension. A string that is taut is said to be under tension. This force is the result of interactions between the molecules of the material making up the string and is electrical in nature. A tension force in a string is created when two forces are applied in opposite directions at the ends of the string (see Figure 2.2). Figure 2.2: A tension force in a string. To say that there is tension in a string means that an arbitrary point P on the string is acted upon by two forces (the tension T) as shown in Figure 2.3. If the string hangs from a ceiling and a mass m is tied at the other end, tension develops in the string. At the point of support at the ceiling, the tension force pulls down on the ceiling and at the point where the mass is tied the tension acts upwards on the mass. Figure 2.3: The tension is directed along the string. In most cases we will idealize the string by assuming it is massless. This does not mean that the string really is massless, but rather that its mass is so small compared with any other masses in the problem that we can neglect it. In that case, the tension T is the same at all points on the string. The direction of the tension force is along the string. Further examples of tension forces in a string are given in Figure 2.4. A string or rope that is not taut has zero tension in it. Figure 2.4: More examples of tension forces. Springs A spring that is pulled so that its length increases will develop a tension force inside the spring that will tend to bring the length back to its original value. Similarly, if it is compressed a tension force will again try to restore the length of the spring, Figure 2.5. Experiments show that for a range of extensions of the spring, the tension force is proportional to the extension, T = kx, where k is known as the spring constant. This relation between tension and extension is known as Hooke’s law. Figure 2.5: Tension is proportional and opposite to the extension or compression. CHECK YOURSELF 1 A block is attached to a vertical spring. Will the extension of the spring be different on the earth and on the moon? Normal contact forces If a body touches another body, there is a normal contact force between the two bodies. This force is perpendicular to the surface of the body exerting the force, hence the name normal. Like tension, the origin of this force is also electrical. In Figure 2.6 we show the normal force R on several bodies. Figure 2.6: Examples of normal forces, R. We can understand the existence of contact forces in a simple model in which atoms are connected by springs. The block pushes down on the atoms of the table, compressing the springs under the block. This creates the normal force on the block, Figure 2.7. Figure 2.7: A simple model of contact forces. Drag forces Drag forces are forces that oppose the motion of a body through a fluid (a gas or a liquid). Typical examples are the air resistance force experienced by a car (see Figure 2.8) or plane, or the resistance force experienced by a steel marble dropped into a jar of honey. Drag forces are directed opposite to the velocity of the body and in general depend on the speed and shape of the body. The higher the speed, the higher the drag force. Figure 2.8: Drag forces oppose motion. For the case of a spherical body of radius r falling through a fluid of density ρ with speed v the drag force is given by Stoke’s law: FD = 6πηrv where η is a quantity known as the viscosity of the fluid. (This formula applies to spherical, small objects moving through a fluid smoothly.) Viscosity is a measure of the forces between layers of the fluid that are moving relative to each other; water has a viscosity of about 10−3 Pa s, motor oil 0.5 Pa s and honey 1.5 Pa s. The unit is Pa s where the pascal (Pa) is the unit of pressure and equals N m−2. Viscosity decreases with increasing temperature. Buoyant forces (upthrust) Any object placed in a fluid (liquid or gas) experiences an upward force called the buoyant force (see Figure 2.9). This force is given by FB = ρgVimm where Vimm is the volume of the body that is immersed in the fluid. If the buoyant force equals the weight of the body, the body will float in the fluid. If the buoyant force is less than the weight, the body will sink. The buoyant force is caused by the pressure that the fluid exerts on the body. The pressure is greater at the bottom compared to that at the top. Figure 2.9: Buoyant force. Frictional forces Frictional forces generally oppose the motion of a body. They are also electrical in origin. (See Figure 2.10.) Figure 2.10: Examples of frictional forces, f. In a there is motion to the right, which is opposed by a single frictional force that will eventually stop the body. In b the force accelerating the body is opposed by a frictional force. In c the body does not move, but it does have a tendency to move down the plane, and so a frictional force directed up the plane opposes this tendency, keeping the body in equilibrium. Friction arises whenever one body slides over another. In this case we have dynamic friction. Friction also arises whenever there is a tendency for motion, not necessarily motion itself. For example, a block that rests on an inclined plane has a tendency to slide down the plane, so there is a force of friction up the plane. Similarly, if you pull on a block on a level rough road with a small force the block will not move. This is because a force of friction develops that is equal and opposite to the pulling force. In this case we have static friction. In the simple model of matter consisting of atoms connected by springs, pushing the block to the right results in springs stretching and compressing. The net result is a force opposing the motion, friction, Figure 2.11. Figure 2.11: Friction in the simple atoms-and-springs model of matter. A more realistic model involves irregularities (called asperities) in the surfaces which interlock, opposing sliding, Figure 2.12. Figure 2.12: Exaggerated view of how asperities oppose the sliding of one surface over the other. Frictional forces are still not very well understood, and there is no theory of friction that follows directly from the fundamental laws of physics. However, a number of simple, empirical ‘laws’ of friction have been discovered. These are not always applicable and are only approximately true, but they are useful in describing frictional forces in general terms. The ‘friction laws’ may be summarised as follows: 1 the area of contact between the two surfaces does not affect the frictional force. 2 the force of dynamic friction is equal to fd = μd R, where R is the normal force between the surfaces and μd is the coefficient of dynamic friction. 3 the force of dynamic friction does not depend on the speed of sliding. 4 the maximum force of static friction that can develop between two surfaces is given by fs = μs R, where R is the normal force between the surfaces and μs is the coefficient of static friction with μs > μd. EXAM TIP One of the most common mistakes is to think that μsR is the formula that gives the static friction force. This is not correct. This formula gives the maximum possible static friction force that can develop between two surfaces. Figure 2.13 shows how the frictional force f varies with a pulling force F. The force F pulls on a body on a horizontal rough surface. Initially the static frictional force matches the pulling force and we have no motion, fs = F. When the pulling force exceeds the maximum possible static friction force, μs R, the frictional force drops abruptly to the dynamic value of μd R and stays at that constant value as the body accelerates. This is a well-known phenomenon of everyday life: it takes a lot of force to get a heavy piece of furniture to start moving (you must exceed the maximum value of the static friction force), but once you get it moving, pushing it along becomes easier (you are now opposed by the smaller dynamic friction force). Figure 2.13: The variation of the frictional force f between surfaces with the pulling force F. CHECK YOURSELF 2 A block of weight 5.0 N rests on a rough horizontal table. The static coefficient of friction between the block and the table is 0.60. A horizontal force of 2.2 N is applied to the block. What is the frictional force on the block? Other forces In later chapters we will deal with other forces, such as the electric force, the magnetic force, and the nuclear force. Free-body force diagrams A free-body force diagram is a diagram showing the magnitude and direction of all the forces acting on a chosen body. The body is shown on its own, free of its surroundings and of any other bodies it may be in contact with. In Figure 2.14 we show three situations in which forces are acting; below each is the corresponding free-body force diagram for the shaded bodies. Figure 2.14: Free-body diagrams. In any mechanics problem, it is important to be able to draw correctly the free-body force diagrams for all the bodies of interest. It is also important that the length of the arrow representing a given force is proportional to the magnitude of the force. WORKED EXAMPLE 2.1 a Draw the free-body diagram for a pendulum when the bob is at the extreme left position. b Draw the free-body diagram for a ball that moves on the inside surface of a vertical circle at the three positions shown in Figure 2.15. Figure 2.15: For worked example 2.1. Answer a See Figure 2.16. Figure 2.16: The two forces on the bob. b See Figure 2.17. The size of the normal force here will be understood when we move to Newton’s second law. For now make sure you understand the direction.  Figure 2.17: The forces on the ball. Equilibrium Equilibrium of a point particle means that the resultant or net force on the particle is zero. The net force on a particle is the one single force whose effect is the same as the combined effect of individual forces acting on the particle. We denote it by ∑F. Finding the net force is easy when the forces are in the same or opposite directions. In Figure 2.18a, the net force is (we take the direction to the right to be positive) ∑ F = 12 + 6.0 − 8.0 = 10 N. This is positive, indicating a direction to the right. In Figure 2.18b it is (we take the direction upward to be positive) ∑ F = 5.0 + 6.0 − 4.0 − 8.0 = −1.0 N The negative sign indicates a direction vertically down. Figure 2.18: The net force is found by plain addition and/or subtraction when the forces are in the same or opposite direction. WORKED EXAMPLE 2.2 Determine the magnitude of the force F given that the block (Figure 2.19) is in equilibrium. Figure 2.19: For worked example 2.2. Answer For equilibrium, ∑F = 0 and so 6.0 + F + 6.0 – 15 = 0. This gives F = 3.0 N. WORKED EXAMPLE 2.3 A brick of weight 50 N rests on a horizontal surface. The coefficient of static friction between the brick and the surface is 0.60, and the coefficient of dynamic friction is 0.20. A horizontal force F is applied to the brick, its magnitude increasing uniformly from zero. Once the brick starts moving the pulling force no longer increases. Estimate the net force on the moving brick. Answer The maximum frictional force that can develop between the brick and the surface is fs = μs R, which evaluates to 0.60 × 50 = 30 N. So motion takes place when the pulling force is just barely larger than 30 N. Once motion starts the frictional force will be equal to μd R, i.e., 0.20 × 50 = 10 N. The net force on the brick in that case will be just larger than 30 – 10 = 20 N. CHECK YOURSELF 3 An ice cube floats in water in a glass. What happens to the level of the water when the ice cube melts? WORKED EXAMPLE 2.4 An iceberg floats in seawater. The density of ice is 920 kg m−3 and that of seawater is 1020 kg m−3. Estimate the fraction of the volume of the iceberg that is under water. Answer The weight of the iceberg is balanced by the buoyant force. The weight is given by W = mg = ρice gV and the buoyant force is FB = ρwater gVimm. The weight and the buoyant forces are equal and so ρice gV = ρwater gVimm Vimm ρice 920 = = = 0.90 V ρwater 1020 WORKED EXAMPLE 2.5 A small sphere of radius r and density ρbody falls vertically through a viscous fluid of viscosity η and density ρfluidwith constant speed v. (As we will be discussing extensively later on, motion with constant velocity is equivalent to equilibrium, and the net force is zero.) Show that the speed v is given by 2 2 (ρbody − ρfluid ) r g v =. 9 η Answer The forces on the falling sphere are the weight vertically downward and the drag force and the buoyant force upward. Hence W = B + FD. Now W = mg = ρbody gV, B = ρfluid gV and FD = 6πηrv so that: ρbody gV = ρfluid gV + 6πηrv. Now use the formula for the volume of a sphere, 3 4πr V = , to get 3 3 3 4πr 4πr ρbody g = ρfluid g + 6πηrv 3 3 3 4πr g 6πηrv = (ρbody − ρfluid ) 3 2 2 (ρbody − ρfluid ) r g v = 9 η CHECK YOURSELF 4 A car moves on a straight horizontal road with constant speed. The force due to the engine pushing the car forward is F. Is the total resistive force on the car less than F, equal to F, or greater than F? TEST YOUR UNDERSTANDING 1 Figure 2.20 shows a block at rest on a rough table. The block is connected by a string that goes over a pulley to a second hanging block. Draw the forces on each body. Figure 2.20: For question 1. 2 a A bead is at rest at the top of a sphere, Figure 2.21. Draw the forces on the bead. b The bead is given a small push, and at the right-most position shown, it is just about to lose contact with the sphere. Draw the forces on the bead. c Draw the forces on the bead in the intermediate position. Figure 2.21: For question 2. 3 Look at Figure 2.22. Compare the tension in the string in the two cases. Figure 2.22: For question 3. 4 A spring is compressed by a certain distance and a mass is attached to its right end, as shown in  Figure 2.23. The mass is at rest on a rough table. On a copy of the diagram, draw the forces acting on the mass. Figure 2.23: For question 4. 5 A mass hangs attached to three strings, as shown in Figure 2.24. On a copy of the diagram, draw the forces on: a the hanging mass b the point where the strings join. Figure 2.24: For question 5. 6 A force of 12 N extends a spring by 3 cm. What force will extend the spring by 4 cm? 7 Two identical springs, each with a spring constant of k = 220 N m−1, are connected to a trolley and  fixed supports as shown in Figure 2.25. a When the trolley is in equilibrium the springs have their natural length. Calculate the net force on the trolley when it is moved 2.0 cm to the right. b When the trolley is in equilibrium each spring is extended by 4.0 cm. Calculate the net force on the trolley when it is moved 2.0 cm to the right. Figure 2.25: For question 7. 8 Find the net force on each of the bodies, A–F, shown in Figure 2.26. The only forces acting are the ones shown. Indicate direction by ‘right’, ‘left’, ‘up’ and ‘down’.  Figure 2.26: For question 8. 9 A small ball falls through a liquid at constant speed. Copy Figure 2.27, then draw and label the forces acting on the ball. Figure 2.27: For question 9. 10 Explain why it is impossible for a mass to hang attached to two horizontal strings as shown in Figure 2.28. Figure 2.28: For question 10. 11 A block X rests on top of another block Y (see Figure 2.29). Both blocks are on a frictionless, horizontal table. There is friction between the two blocks. In a, a horizontal force is applied to block Y. In b, the force is applied to block X. In both cases, the two blocks move together without sliding on each other. In each case draw the horizontal forces acting on each block. Figure 2.29: In a the force is applied to Y. In b it is applied to X. The blocks move together. 12 A mass of 2.00 kg rests on a rough horizontal table. The coefficient of static friction between the block and the table is 0.60. The block is  attached to a hanging mass by a string that goes over a smooth pulley, as shown in Figure 2.30. Determine the largest mass that can hang in this way without forcing the block to slide. Figure 2.30: For question 12. 13 A boy tries to lift a suitcase of weight 220 N by pulling upward on it with a force of 140 N. The suitcase does not move. Calculate the normal force from the floor on the suitcase. 14 A block of weight 15 N rests on a horizontal table. A man pushes the block downward with a force of 12 N. What is the normal force of the table on the block? 2.2 Newton’s laws of motion Newton’s first law of motion Suppose you have two identical train carriages. Both are equipped with all the apparatus you need to do physics experiments. One train carriage is at rest at the train station. The other moves in a straight line with constant speed—the ride is perfectly smooth, there are no bumps, there is no noise and there are no windows to look outside. Every physics experiment conducted in the train at rest will give identical results to similar experiments made in the moving train. We have no way of determining whether a carriage is ‘really at rest’ or ‘really moving’. We find it perfectly natural to believe, correctly, that no net force is present in the case of the carriage at rest. Therefore no net force is required in the case of the carriage moving in a straight line with constant speed. Newton’s first law of motion states that: KEY POINT When the net force on a body is zero, the body will move with constant velocity (which may be zero); in other words, it will move on a straight line with constant speed (which may be zero). In effect, Newton’s first law defines what a force is. A force is what changes a body’s velocity. A force is not what is required to keep something moving as Aristotle thought. Using the law in reverse allows us to conclude that if a body is not moving with constant velocity (which may mean not moving in a straight line, or not moving with constant speed or both) then a force must be acting on the body. So, since the earth revolves around the sun we know that a force must be acting on the earth. Newton’s first law is also called the law of inertia. Inertia is what keeps the body in the same state of motion when no forces act on the body. When a car accelerates forward, the passengers are thrown back into their seats because their original state of motion was motion with low speed. If a car brakes abruptly, the passengers are thrown forward, Figure 2.31. This implies that a mass tends to stay in the state of motion it was in before the force acted on it. The reaction of a body to a change in its state of motion (acceleration) is inertia. Figure 2.31: The car was originally travelling at high speed. When it hits the wall the car stops, but the passenger wants to stay in the original high speed state of motion. This results in him hitting the steering wheel and the windshield (which is why it is a good idea to have safety belts and air bags). Newton’s third law of motion Newton’s third law of motion states that: KEY POINT If body X exerts a force on body Y, then body Y will exert an equal and opposite force on body X. Make sure you understand that these equal and opposite forces act on different bodies. Thus, you cannot use this law to claim that it is impossible to ever have a net force on a body because for every force on it there is also an equal and opposite force. Here are four examples of this law: 1 You stand on roller-skates facing a wall. You push on the wall, and you move away from it. This is because you exerted a force on the wall, and in turn the wall exerted an equal and opposite force on you, making you move away (Figure 2.32). Figure 2.32: The girl pushes on the wall, so the wall pushes on her in the opposite direction. 2 You step on a scale. The scale exerts an upward force on you, and so you exert a downward force on the scale. This is the force that the scale reads (Figure 2.33).  Figure 2.33: The familiar bathroom scale does not measure mass. It measures the force that you exert on it. This force is the weight only when the scale is at rest. 3 A helicopter hovers in air, Figure 2.34. Its rotors exert a force downward on the air. Thus, the air exerts the upward force on the helicopter that keeps it from falling. Figure 2.34: The upward force on the rotor is due to the force the rotor exerts on the air downward. 4 A book of mass 2 kg is allowed to fall feely. The earth exerts a force on the book, namely the weight of the book of about 20 N. Thus, the book exerts an equal and opposite force on the earth—a force upward equal to 20 N. You must be careful with situations with two forces that are equal and opposite; they do not always have to do with the third law. For example, a block of mass 3 kg resting on a horizontal table has two forces acting on it. Its weight of about 30 N, and the normal force from the table that is also 30 N. These two forces are equal and opposite, but they are acting on the same body and so have nothing to do with Newton’s third law. (The force that pairs with the weight of the block is an upward force on the earth, and the one that pairs with the normal force is a downward force on the table.) Newton’s third law also applies to cases where there is no contact between the bodies. Examples are the electric force between two electrically charged particles or the gravitational force between any two massive particles. These forces must be equal and opposite. (See Figure 2.35.) Figure 2.35: The two charges and the two masses are different, but the forces they exert on each other are equal and opposite. SCIENCE IN CONTEXT Helicopters, planes and rockets function because of Newton’s third law. Equilibrium When there are angles between the various forces, solving equilibrium problems is a bit more cumbersome and will involve getting components of forces using vector methods. We choose a set of axes whose origin is the body in question and find the components of all the forces on the body. Figure 2.36 shows three forces acting at the same point. We have equilibrium, which means the net force acting at the point is zero. We need to find the unknown magnitude and direction of force F1. This situation could represent three people pulling on three ropes that are tied at a point. Figure 2.36: Force diagram of three forces in equilibrium pulling a common point. Notice that the three vectors representing the three forces form a triangle. Getting compo