Cambridge International AS & A Level Physics

Special thanks to Sohaib kindly remember him in your prayers  Physics for Cambridge International AS & A Level Coursebook David Sang, Graham Jones, Gurinder Chadha & Richard Woodside  Teachers play an important part in shaping futur...

Special thanks to Sohaib kindly remember him in your prayers  Physics for Cambridge International AS & A Level Coursebook David Sang, Graham Jones, Gurinder Chadha & Richard Woodside  Teachers play an important part in shaping futures. Our Dedicated Teacher Awards recognise the hard work that teachers put in every day. Thank you to everyone who nominated this year, we have been inspired and moved by all of your stories. Well done to all of our nominees for your dedication to learning and for inspiring the next generation of thinkers, leaders and innovators. Congratulations to our incredible winner and finalists For more information about our dedicated teachers and their stories, go to dedicatedteacher.cambridge.org 〉 Contents Introduction How to use this series How to use this book Resource index 1 Kinematics 1.1 Speed 1.2 Distance and displacement, scalar and vector 1.3 Speed and velocity 1.4 Displacement–time graphs 1.5 Combining displacements 1.6 Combining velocities 1.7 Subtracting vectors 1.8 Other examples of scalar and vector quantities 2 Accelerated motion 2.1 The meaning of acceleration 2.2 Calculating acceleration 2.3 Units of acceleration 2.4 Deducing acceleration 2.5 Deducing displacement 2.6 Measuring velocity and acceleration 2.7 Determining velocity and acceleration in the laboratory 2.8 The equations of motion 2.9 Deriving the equations of motion 2.10 Uniform and non-uniform acceleration 2.11 Acceleration caused by gravity 2.12 Determining g 2.13 Motion in two dimensions: projectiles 2.14 Understanding projectiles 3 Dynamics 3.1 Force, mass and acceleration 3.2 Identifying forces 3.3 Weight, friction and gravity 3.4 Mass and inertia 3.5 Moving through fluids 3.6 Newton’s third law of motion 3.7 Understanding SI units 4 Forces 4.1 Combining forces 4.2 Components of vectors 4.3 Centre of gravity 4.4 The turning effect of a force 4.5 The torque of a couple 5 Work, energy and power 5.1 Doing work, transferring energy 5.2 Gravitational potential energy 5.3 Kinetic energy 5.4 Gravitational potential to kinetic energy transformations  5.5 Down, up, down: energy changes 5.6 Energy transfers 5.7 Power 6 Momentum 6.1 The idea of momentum 6.2 Modelling collisions 6.3 Understanding collisions 6.4 Explosions and crash-landings 6.5 Collisions in two dimensions 6.6 Momentum and Newton’s laws 6.7 Understanding motion 7 Matter and materials 7.1 Density 7.2 Pressure 7.3 Archimedes’ principle 7.4 Compressive and tensile forces 7.5 Stretching materials 7.6 Elastic potential energy 8 Electric current 8.1 Circuit symbols and diagrams 8.2 Electric current 8.3 An equation for current 8.4 The meaning of voltage 8.5 Electrical resistance 8.6 Electrical power 9 Kirchhoff’s laws 9.1 Kirchhoff’s first law 9.2 Kirchhoff’s second law 9.3 Applying Kirchhoff’s laws 9.4 Resistor combinations 10 Resistance and resistivity 10.1 The I–V characteristic for a metallic conductor 10.2 Ohm’s law 10.3 Resistance and temperature 10.4 Resistivity 11 Practical circuits 11.1 Internal resistance 11.2 Potential dividers 11.3 Sensors 11.4 Potentiometer circuits 12 Waves 12.1 Describing waves 12.2 Longitudinal and transverse waves 12.3 Wave energy 12.4 Wave speed 12.5 The Doppler effect for sound waves 12.6 Electromagnetic waves 12.7 Electromagnetic radiation 12.8 Orders of magnitude 12.9 The nature of electromagnetic waves 12.10 Polarisation 13 Superposition of waves 13.1 The principle of superposition of waves 13.2 Diffraction of waves 13.3 Interference 13.4 The Young double-slit experiment 13.5 Diffraction gratings 14 Stationary waves 14.1 From moving to stationary 14.2 Nodes and antinodes 14.3 Formation of stationary waves 14.4 Determining the wavelength and speed of sound 15 Atomic structure 15.1 Looking inside the atom 15.2 Alpha-particle scattering and the nucleus 15.3 A simple model of the atom 15.4 Nucleons and electrons 15.5 Forces in the nucleus 15.6 Discovering radioactivity 15.7 Radiation from radioactive substances 15.8 Energies in α and β decay 15.9 Equations of radioactive decay 15.10 Fundamental particles 15.11 Families of particles 15.12 Another look at β decay 15.13 Another nuclear force P1 Practical skills at AS Level P1.1 Practical work in physics P1.2 Using apparatus and following instructions P1.3 Gathering evidence P1.4 Precision, accuracy, errors and uncertainties P1.5 Finding the value of an uncertainty P1.6 Percentage uncertainty P1.7 Recording results P1.8 Analysing results P1.9 Testing a relationship P1.10 Combining uncertainties P1.11 Identifying limitations in procedures and suggesting improvements 16 Circular motion 16.1 Describing circular motion 16.2 Angles in radians 16.3 Steady speed, changing velocity 16.4 Angular speed 16.5 Centripetal forces 16.6 Calculating acceleration and force 16.7 The origins of centripetal forces 17 Gravitational fields 17.1 Representing a gravitational field 17.2 Gravitational field strength g 17.3 Energy in a gravitational field 17.4 Gravitational potential 17.5 Orbiting under gravity 17.6 The orbital period 17.7 Orbiting the Earth 18 Oscillations 18.1 Free and forced oscillations 18.2 Observing oscillations 18.3 Describing oscillations 18.4 Simple harmonic motion 18.5 Representing s.h.m. graphically 18.6 Frequency and angular frequency 18.7 Equations of s.h.m. 18.8 Energy changes in s.h.m. 18.9 Damped oscillations 18.10 Resonance 19 Thermal physics 19.1 Changes of state 19.2 Energy changes 19.3 Internal energy 19.4 The meaning of temperature 19.5 Thermometers 19.6 Calculating energy changes 20 Ideal gases 20.1 Particles of a gas 20.2 Explaining pressure 20.3 Measuring gases 20.4 Boyle’s law 20.5 Changing temperature 20.6 Ideal gas equation 20.7 Modelling gases: the kinetic model 20.8 Temperature and molecular kinetic energy 21 Uniform electric fields 21.1 Attraction and repulsion 21.2 The concept of an electric field 21.3 Electric field strength 21.4 Force on a charge 22 Coulomb’s law 22.1 Electric fields 22.2 Coulomb’s law 22.3 Electric field strength for a radial field 22.4 Electric potential 22.5 Gravitational and electric fields 23 Capacitance 23.1 Capacitors in use 23.2 Energy stored in a capacitor 23.3 Capacitors in parallel 23.4 Capacitors in series 23.5 Comparing capacitors and resistors 23.6 Capacitor networks 23.7 Charge and discharge of capacitors 24 Magnetic fields and electromagnetism 24.1 Producing and representing magnetic fields 24.2 Magnetic force 24.3 Magnetic flux density 24.4 Measuring magnetic flux density 24.5 Currents crossing fields  24.6 Forces between currents 24.7 Relating SI units 24.8 Comparing forces in magnetic, electric and gravitational fields 25 Motion of charged particles 25.1 Observing the force 25.2 Orbiting charged particles 25.3 Electric and magnetic fields 25.4 The Hall effect 25.5 Discovering the electron 26 Electromagnetic induction 26.1 Observing induction 26.2 Explaining electromagnetic induction 26.3 Faraday’s law of electromagnetic induction 26.4 Lenz’s law 26.5 Everyday examples of electromagnetic induction 27 Alternating currents 27.1 Sinusoidal current 27.2 Alternating voltages 27.3 Power and alternating current 27.4 Rectification 28 Quantum physics 28.1 Modelling with particles and waves 28.2 Particulate nature of light 28.3 The photoelectric effect 28.4 Threshold frequency and wavelength 28.5 Photons have momentum too 28.6 Line spectra 28.7 Explaining the origin of line spectra 28.8 Photon energies 28.9 The nature of light: waves or particles? 28.10 Electron waves 28.11 Revisiting photons 29 Nuclear physics 29.1 Balanced equations 29.2 Mass and energy 29.3 Energy released in radioactive decay 29.4 Binding energy and stability 29.5 Randomness and radioactive decay 29.6 The mathematics of radioactive decay 29.7 Decay graphs and equations 29.8 Decay constant λ and half-life t½ 30 Medical imaging 30.1 The nature and production of X-rays 30.2 X-ray attenuation 30.3 Improving X-ray images 30.4 Computerised axial tomography 30.5 Using ultrasound in medicine 30.6 Echo sounding 30.7 Ultrasound scanning 30.8 Positron Emission Tomography 31 Astronomy and cosmology 31.1 Standard candles  31.2 Luminosity and radiant flux intensity 31.3 Stellar radii 31.4 The expanding Universe P2 Practical skills at A Level P2.1 Planning and analysis P2.2 Planning P2.3 Analysis of the data P2.4 Treatment of uncertainties P2.5 Conclusions and evaluation of results Appendix 1: Physical quantities and units Appendix 2: Data and formulae Appendix 3: Mathematical equations and conversion factors Appendix 4: The Periodic Table Acknowledgements 〉 Introduction This book covers the entire syllabus of Cambridge International AS & A Level Physics (9702) for examination from 2022. This book is in three parts: Chapters 1–15 and P1: the AS Level content, covered in the first year of the course, including a chapter (P1) dedicated to the development of your practical skills Chapters 16–31 and P2: the A Level content, including a chapter (P2) dedicated to developing your ability to plan, analyse and evaluate practical investigations Appendices of useful formulae, a Glossary and an Index. The main tasks of a textbook like this are to explain the various concepts of physics that you need to understand, and to provide you with questions that will help you to test your understanding and develop the key skills you need to succeed on this course. You will find a visual guide to the structure of each chapter and the features of this book on the next two pages. In your study of physics, you will find that certain key concepts are repeated, and that these concepts form ‘themes’ that link the different areas of physics together. It will help you to progress and gain confidence in your understanding of physics if you take note of these themes. For this Coursebook, these key concepts include: models of physical systems testing predictions against evidence mathematics as a language and problem-solving tool matter, energy and waves forces and fields. In this Coursebook, the mathematics has been kept to the minimum required by the Cambridge International AS & A Level Physics syllabus (9702). If you are also studying mathematics, you may find that more advanced techniques such as calculus will help you with many aspects of physics. Studying physics is a stimulating and worthwhile experience. It is an international subject; no single country has a monopoly on the development of the ideas. It can be a rewarding exercise to discover how men and women from many countries have contributed to our knowledge and well-being, through their research into and application of the concepts of physics. We hope not only that this book will help you to succeed in your future studies and career, but also that it will stimulate your curiosity and fire your imagination. Today’s students become the next generation of physicists and engineers, and we hope that you will learn from the past to take physics to ever-greater heights. 〉 How to use this series  CAMBRIDGE._,una,mRUI Physics for Cambridge International AS & A Level N4-Hilli·ii21 iJS:\-1-iM I I 〉 How to use this book Throughout this book, you will notice lots of different features that will help your learning. These are explained below. LEARNING INTENTIONS These set the scene for each chapter, help with navigation through the Coursebook and indicate the important concepts in each topic. BEFORE YOU START This contains questions and activities on subject knowledge you will need before starting this chapter. SCIENCE IN CONTEXT This feature presents real-world examples and applications of the content in a chapter, encouraging you to look further into topics. There are discussion questions at the end that look at some of the benefits and problems of these applications. PRACTICAL ACTIVITIES This book does not contain detailed instructions for doing particular experiments, but you will find background information about the practical work you need to do in these boxes. There are also two chapters, P1 and P2, which provide detailed information about the practical skills you need to develop during the course. Questions Appearing throughout the text, questions give you a chance to check that you have understood the topic you have just read about. You can find the answers to these questions in the digital Coursebook. KEY EQUATIONS Key equations are highlighted in the text when an equation is first introduced. Definitions for the equation and further information are given in the margin. KEY WORDS Key vocabulary is highlighted in the text when it is first introduced. If you hover your cursor over the word, the definition will appear. COMMAND WORDS Command words that appear in the syllabus and might be used in exams are highlighted in the exam-style questions when they are first introduced. If you hover your cursor over the word, the Cambridge International definition will appear. WORKED EXAMPLES Wherever you need to know how to use a formula to carry out a calculation, there are worked examples boxes to show you how to do this. KEY IDEAS Important scientific concepts, facts and tips are given in these boxes. REFLECTION These activities ask you to look back on the topics covered in the chapter and test how well you understand these topics and encourage you to reflect on your learning. SUMMARY There is a summary of key points at the end of each chapter. EXAM-STYLE QUESTIONS Questions at the end of each chapter provide more demanding exam-style questions, some of which may require use of knowledge from previous chapters. Answers to these questions can be found in the digital Coursebook. SELF-EVALUATION CHECKLIST The summary checklists are followed by ‘I can’ statements that match the Learning intentions at the beginning of the chapter. 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’. I can See topic… Needs more Almost there Ready to work move on Resource index The resource index is a convenient place for you to download all answer files for this resource. 〉 Chapter 1 Kinematics: describing motion LEARNING INTENTIONS In this chapter you will learn how to: define and use displacement, speed and velocity draw and interpret displacement–time graphs describe laboratory methods for determining speed understand the differences between scalar and vector quantities and give examples of each use vector addition to add and subtract vectors that are in the same plane. BEFORE YOU START Do you know how to rearrange an equation that involves fractions? Choose an equation that you know 2 from your previous physics course, such as P = VR , and rearrange it to make R or V the subject of the formula. Can you write down a direction using compass bearings, for example, as 014°, N14°E or 14° east of north? DESCRIBING MOVEMENT Our eyes are good at detecting movement. We notice even quite small movements out of the corners of our eyes. It’s important for us to be able to judge movement – think about crossing the road, cycling or driving, or catching a ball. Figure 1.1 shows a way in which movement can be recorded on a photograph. This is a stroboscopic photograph of a boy juggling three balls. As he juggles, a bright lamp flashes several times a second so that the camera records the positions of the balls at equal intervals of time. How can the photograph be used to calculate the speed of the upper ball horizontally and vertically as it moves through the air? What other apparatus is needed? You can discuss this with someone else. Figure 1.1: This boy is juggling three balls. A stroboscopic lamp flashes at regular intervals; the camera is moved to one side at a steady rate to show separate images of the boy. 1.1 Speed We can calculate the average speed of something moving if we know the distance it moves and the time it takes: distance average speed = time In symbols, this is written as: d v= t where v is the average speed and d is the distance travelled in time t. If an object is moving at a constant speed, this equation will give us its speed during the time taken. If its speed is changing, then the equation gives us its average speed. Average speed is calculated over a period of time. If you look at the speedometer in a car, it doesn’t tell you the car’s average speed; rather, it tells you its speed at the instant when you look at it. This is the car’s instantaneous speed. KEY EQUATION distance average speed = time d v = t Question 1 Look at Figure 1.2. The runner has just run 10 000 m in a time of 27 minutes 5.17 s. Calculate his average speed during the race. Figure 1.2: England’s Mo Farah winning his second gold medal at the Rio Olympics in 2016. Units In the Système Internationale d’Unités (the SI system), distance is measured in metres (m) and time in seconds (s). Therefore, speed is in metres per second. This is written as m s−1 (or as m/s). Here, s−1 is the same as 1/s, or ‘per second’. There are many other units used for speed. The choice of unit depends on the situation. You would probably give the speed of a snail in different units from the speed of a racing car. Table 1.1 includes some alternative units of speed. Note that in many calculations it is necessary to work in SI units (m s−1). m s−1 metres per second cm s−1 centimetres per second km s−1 kilometres per second km h−1 or km/h kilometres per hour mph miles per hour Table 1.1: Units of speed. Questions 2 Here are some units of speed: m s−1 mm s−1 km s−1 km h−1 Which of these units would be appropriate when stating the speed of each of the following? a a tortoise b a car on a long journey c light d a sprinter. 3 A snail crawls 12 cm in one minute. What is its average speed in mm s−1? Determining speed You can find the speed of something moving by measuring the time it takes to travel between two fixed points. For example, some motorways have emergency telephones every 2000 m. Using a stopwatch you can time a car over this distance. Note that this can only tell you the car’s average speed between the two points. You cannot tell whether it was increasing its speed, slowing down or moving at a constant speed. PRACTICAL ACTIVITY 1.1 Laboratory measurements of speed Here we describe four different ways to measure the speed of a trolley in the laboratory as it travels along a straight line. Each can be adapted to measure the speed of other moving objects, such as a glider on an air track or a falling mass. Measuring speed using two light gates The leading edge of the card in Figure 1.3 breaks the light beam as it passes the first light gate. This starts the timer. The timer stops when the front of the card breaks the second beam. The trolley’s speed is calculated from the time interval and the distance between the light gates. Figure 1.3: Using two light gates to find the average speed of a trolley. Measuring speed using one light gate The timer in Figure 1.4 starts when the leading edge of the card breaks the light beam. It stops when the trailing edge passes through. In this case, the time shown is the time taken for the trolley to travel a distance equal to the length of the card. The computer software can calculate the speed directly by dividing the distance by the time taken. Figure 1.4: Using a single light gate to find the average speed of a trolley. Measuring speed using a ticker-timer The ticker-timer (Figure 1.5) marks dots on the tape at regular intervals, usually s (i.e. 0.02 s). (This is because it works with alternating current, and in most countries the frequency of the alternating mains is 50 Hz.) The pattern of dots acts as a record of the trolley’s movement. Figure 1.5: Using a ticker-timer to investigate the motion of a trolley. Start by inspecting the tape. This will give you a description of the trolley’s movement. Identify the start of the tape. Then, look at the spacing of the dots: even spacing – constant speed increasing spacing – increasing speed. Now you can make some measurements. Measure the distance of every fifth dot from the start of the tape. This will give you the trolley’s distance at intervals of 0.10 s. Put the measurements in a table and draw a distance–time graph. Measuring speed using a motion sensor The motion sensor (Figure 1.6) transmits regular pulses of ultrasound at the trolley. It detects the reflected waves and determines the time they took for the trip to the trolley and back. From this, the computer can deduce the distance to the trolley from the motion sensor. It can generate a distance–time graph. You can determine the speed of the trolley from this graph. Figure 1.6: Using a motion sensor to investigate the motion of a trolley. Choosing the best method  Each of these methods for finding the speed of a trolley has its merits. In choosing a method, you might think about the following points: Does the method give an average value of speed or can it be used to give the speed of the trolley at different points along its journey? How precisely does the method measure time–to the nearest millisecond? How simple and convenient is the method to set up in the laboratory? Questions 4 A trolley with a 5.0 cm long card passed through a single light gate. The time recorded by a digital timer was 0.40 s. What was the average speed of the trolley in m s−1? 5 Figure 1.7 shows two ticker-tapes. Describe the motion of the trolleys that produced them. Figure 1.7: Two ticker-tapes. For Question 5. 6 Four methods for determining the speed of a moving trolley have been described. Each could be adapted to investigate the motion of a falling mass. Choose two methods that you think would be suitable, and write a paragraph for each to say how you would adapt it for this purpose. 1.2 Distance and displacement, scalar and vector In physics, we are often concerned with the distance moved by an object in a particular direction. This is called its displacement. Figure 1.8 illustrates the difference between distance and displacement. It shows the route followed by walkers as they went from town A to town C. Figure 1.8: If you go on a long walk, the distance you travel will be greater than your displacement. In this example, the walkers travel a distance of 15 km, but their displacement is only 10 km, because this is the distance from the start to the finish of their walk. Their winding route took them through town B, so that they covered a total distance of 15 km. However, their displacement was much less than this. Their finishing position was just 10 km from where they started. To give a complete statement of their displacement, we need to give both distance and direction: displacement = 10 km at 030° or 30° E of N Displacement is an example of a vector quantity. A vector quantity has both magnitude (size) and direction. Distance, on the other hand, is a scalar quantity. Scalar quantities have magnitude only. 1.3 Speed and velocity It is often important to know both the speed of an object and the direction in which it is moving. Speed and direction are combined in another quantity, called velocity. The velocity of an object can be thought of as its speed in a particular direction. So, like displacement, velocity is a vector quantity. Speed is the corresponding scalar quantity, because it does not have a direction. So, to give the velocity of something, we have to state the direction in which it is moving. For example, ‘an aircraft flies with a velocity of 300 m s−1 due north’. Since velocity is a vector quantity, it is defined in terms of displacement: change in displacement velocity = time taken We can write the equation for velocity in symbols: s v= t KEY EQUATION change in displacement velocity = time taken Alternatively, we can say that velocity is the rate of change of an object’s displacement: Δs v= Δt where the symbol Δ (the Greek letter delta) means ‘change in’. It does not represent a quantity (in the way that s and t do). Another way to write Δs would be s2 − s1, but this is more time-consuming and less clear. From now on, you need to be clear about the distinction between velocity and speed, and between displacement and distance. Table 1.2 shows the standard symbols and units for these quantities. Quantity Symbol for quantity Symbol for unit distance d m displacement s, x m time t s speed, velocity v m s−1 Table 1.2: Standard symbols and units. (Take care not to confuse italic s for displacement with s for seconds. Notice also that v is used for both speed and velocity.) Question 7 Do these statements describe speed, velocity, distance or displacement? (Look back at the definitions of these quantities.) a The ship sailed south-west for 200 miles. b I averaged 7 mph during the marathon.  c The snail crawled at 2 mm s−1 along the straight edge of a bench. d The sales representative’s round trip was 420 km. Speed and velocity calculations Δs The equation for velocity, v = Δt , can be rearranged as follows, depending on which quantity we want to determine: change in displacement Δs = v × Δt Δs change in time Δt = v Note that each of these equations is balanced in terms of units. For example, consider the equation for displacement. The units on the right-hand side are m s−1 × s, which simplifies to m, the correct unit for displacement. We can also rearrange the equation to find distance s and time t: Δs = v × t Δs t= v WORKED EXAMPLES 1 A car is travelling at 15 m s−1. How far will it travel in 1 hour? Step 1 It is helpful to start by writing down what you know and what you want to know: v = 15 m s−1 t = 1 h = 3600 s s =? Step 2 Choose the appropriate version of the equation and substitute in the values. Remember to include the units: s = v×t = 15 × 3600 = 5.4 × 104 m = 54 km The car will travel 54 km in 1 hour. 2 The Earth orbits the Sun at a distance of 150 000 000 km. How long does it take light from the Sun to reach the Earth? (Speed of light in space = 3.0 × 108 m s−1.) Step 1 Start by writing what you know. Take care with units; it is best to work in m and s. You need to be able to express numbers in scientific notation (using powers of 10) and to work with these on your calculator. v = 3.0 × 108 m s−1 s = 150 000 000 km = 150 000 000 m = 1.5 × 1011 m Step 2 Substitute the values in the equation for time:  s t = v 11 1.5×10 = 3.0×10 8 = 500 s Light takes 500 s (about 8.3 minutes) to travel from the Sun to the Earth. Hint: When using a calculator, to calculate the time t, you press the buttons in the following sequence: [1.5] [10n] [÷] [10n] Making the most of units In Worked example 1 and Worked example 2, units have been omitted in intermediate steps in the calculations. However, at times it can be helpful to include units as this can be a way of checking that you have used the correct equation; for example, that you have not divided one quantity by another when you should have multiplied them. The units of an equation must be balanced, just as the numerical values on each side of the equation must be equal. If you take care with units, you should be able to carry out calculations in non-SI units, such as kilometres per hour, without having to convert to metres and seconds. For example, how far does a spacecraft travelling at 40 000 km h−1 travel in one day? Since there are 24 hours in one day, we have: distance travelled = 40 000 km h−1 × 24 h = 960 000 km Questions 8 A submarine uses sonar to measure the depth of water below it. Reflected sound waves are detected 0.40 s after they are transmitted. How deep is the water? (Speed of sound in water = 1500 m s−1.) 9 The Earth takes one year to orbit the Sun at a distance of 1.5 × 1011 m. Calculate its speed. Explain why this is its average speed and not its velocity. 1.4 Displacement–time graphs We can represent the changing position of a moving object by drawing a displacement–time graph. The gradient (slope) of the graph is equal to its velocity (Figure 1.9). The steeper the slope, the greater the velocity. A graph like this can also tell us if an object is moving forwards or backwards. If the gradient is negative, the object’s velocity is negative – it is moving backwards. Deducing velocity from a displacement–time graph A toy car moves along a straight track. Its displacement at different times is shown in Table 1.3. This data can be used to draw a displacement–time graph from which we can deduce the car’s velocity. Displacement 1.0 3.0 5.0 7.0 7.0 7.0 s/m Time t / s 0.0 1.0 2.0 3.0 4.0 5.0 Table 1.3: Displacement s and time t data for a toy car. It is useful to look at the data first, to see the pattern of the car’s movement. In this case, the displacement increases steadily at first, but after 3.0 s it becomes constant. In other words, initially the car is moving at a steady velocity, but then it stops. Figure 1.9: The slope of a displacement–time (s–t) graph tells us how fast an object is moving. Now we can plot the displacement–time graph (Figure 1.10). We want to work out the velocity of the car over the first 3.0 seconds. We can do this by working out the gradient of the graph, because: velocity = gradient of displacement−time graph We draw a right-angled triangle as shown. To find the car’s velocity, we divide the change in displacement by the change in time. These are given by the two sides of the triangle labelled Δs and Δt. Figure 1.10: Displacement–time graph for a toy car; data as shown in Table 1.3. change in displacement velocity = time taken Δs = Δt (7.0−1.0) = (3.0−0) 6.0 = 3.0 = 2.0 ms−1 If you are used to finding the gradient of a graph, you may be able to reduce the number of steps in this calculation. Questions 10 The displacement–time sketch graph in Figure 1.11 represents the journey of a bus. What does the graph tell you about the journey?  Figure 1.11: For Question 10. 11 Sketch a displacement–time graph to show your motion for the following event. You are walking at a constant speed across a field after jumping off a gate. Suddenly you see a horse and stop. Your friend says there’s no danger, so you walk on at a reduced constant speed. The horse neighs, and you run back to the gate. Explain how each section of the walk relates to a section of your graph. 12 Table 1.4 shows the displacement of a racing car at different times as it travels along a straight track during a speed trial. a Determine the car’s velocity. b Draw a displacement–time graph and use it to find the car’s velocity. Displacement / m 0 85 170 255 340 Time / s 0 1.0 2.0 3.0 4.0 Table 1.4: Displacement s and time t data for Question 12. 13 An old car travels due south. The distance it travels at hourly intervals is shown in Table 1.5. a Draw a distance–time graph to represent the car’s journey. b From the graph, deduce the car’s speed in km h−1 during the first three hours of the journey. c What is the car’s average speed in km h−1 during the whole journey? Time / h 0 1 2 3 4 Distance / km 0 23 46 69 84 Table 1.5: Data for Question 13. 1.5 Combining displacements The walkers shown in Figure 1.12 are crossing difficult ground. They navigate from one prominent point to the next, travelling in a series of straight lines. From the map, they can work out the distance that they travel and their displacement from their starting point: distance travelled = 25 km Figure 1.12: In rough terrain, walkers head straight for a prominent landmark. (Lay thread along route on map; measure thread against map scale.) displacement = 15 km in the direction 045°, N45° E or north-east (Join starting and finishing points with straight line; measure line against scale.) A map is a scale drawing. You can find your displacement by measuring the map. But how can you calculate your displacement? You need to use ideas from geometry and trigonometry. Worked examples 3 and 4 show how. WORKED EXAMPLES 3 A spider runs along two sides of a table (Figure 1.13). Calculate its final displacement.  Figure 1.13: The spider runs a distance of 2.0 m. For Worked example 3. Step 1 Because the two sections of the spider’s run (OA and AB) are at right angles, we can add the two displacements using Pythagoras’s theorem: OB 2 = OA 2 + AB 2 = 0.82 + 1.22 = 2.08 −−−− OB = √ 2.08 = 1.44 m ≈ 1.4 m Step 2 Displacement is a vector. We have found the magnitude of this vector, but now we have to find its direction. The angle θ is given by: opp 0.8 tan θ = adj = 1.2 = 0.667 θ = tan −1 (0.667) = 33.7° ≈ 34° So the spider’s displacement is 1.4 m at 056° or N56°E or at an angle of 34° north of east. 4 An aircraft flies 30 km due east and then 50 km north-east (Figure 1.14). Calculate the final displacement of the aircraft. Figure 1.14: For Worked example 4. Here, the two displacements are not at 90° to one another, so we can’t use Pythagoras’s theorem. We can solve this problem by making a scale drawing, and measuring the final displacement. (However, you could solve the same problem using trigonometry.) Step 1 Choose a suitable scale. Your diagram should be reasonably large; in this case, a scale of 1 cm to represent 5 km is reasonable.  Step 2 Draw a line to represent the first vector. North is at the top of the page. The line is 6 cm long, towards the east (right). Step 3 Draw a line to represent the second vector, starting at the end of the first vector. The line is 10 cm long, and at an angle of 45° (Figure 1.15). Figure 1.15: Scale drawing for Worked example 4. Using graph paper can help you to show the vectors in the correct directions. Step 4 To find the final displacement, join the start to the finish. You have created a vector triangle. Measure this displacement vector, and use the scale to convert back to kilometres: length of vector = 14.8 cm final displacement = 14.8 × 5 = 74 km Step 5 Measure the angle of the final displacement vector: angle = 28° N of E Therefore the aircraft’s final displacement is 74 km at 28° north of east, 062° or N62°E. Questions 14 You walk 3.0 km due north, and then 4.0 km due east. a Calculate the total distance in km you have travelled. b Make a scale drawing of your walk, and use it to find your final displacement. Remember to give both the magnitude and the direction. c Check your answer to part b by calculating your displacement. 15 A student walks 8.0 km south-east and then 12 km due west. a Draw a vector diagram showing the route. Use your diagram to find the total displacement. Remember to give the scale on your diagram and to give the direction as well as the magnitude of your answer. b Calculate the resultant displacement. Show your working clearly. This process of adding two displacements together (or two or more of any type of vector) is known as vector addition. When two or more vectors are added together, their combined effect is known as the resultant of the vectors. 1.6 Combining velocities Velocity is a vector quantity and so two velocities can be combined by vector addition in the same way that we have seen for two or more displacements. Imagine that you are attempting to swim across a river. You want to swim directly across to the opposite bank, but the current moves you sideways at the same time as you are swimming forwards. The outcome is that you will end up on the opposite bank, but downstream of your intended landing point. In effect, you have two velocities: the velocity due to your swimming, which is directed straight across the river the velocity due to the current, which is directed downstream, at right angles to your swimming velocity. These combine to give a resultant (or net) velocity, which will be diagonally downstream. In order to swim directly across the river, you would have to aim upstream. Then your resultant velocity could be directly across the river. WORKED EXAMPLE 5 An aircraft is flying due north with a velocity of 200 m s−1. A side wind of velocity 50 m s−1 is blowing due east. What is the aircraft’s resultant velocity (give the magnitude and direction)? Here, the two velocities are at 90°. A sketch diagram and Pythagoras’s theorem are enough to solve the problem. Step 1 Draw a sketch of the situation – this is shown in Figure 1.16a. Step 2 Now sketch a vector triangle. Remember that the second vector starts where the first one ends. This is shown in Figure 1.16b. Figure 1.16: Finding the resultant of two velocities. For Worked example 5. Step 3 Join the start and end points to complete the triangle. Step 4 Calculate the magnitude of the resultant vector v (the hypotenuse of the right-angled triangle). v2 = 2002 + 502 = 40 000 + 2500 = 42 500 −−−−− v = √ 42 500≈ 206m s−1 Step 5 Calculate the angle θ:  50 tan θ = 200 = 0.25 θ = tan−1 (0.25) ≈ 14° So the aircraft’s resultant velocity is 206 m s−1 at 14° east of north, 076° or N76°E. Questions 16 A swimmer can swim at 2.0 m s−1 in still water. She aims to swim directly across a river that is flowing at 0.80 m s−1. Calculate her resultant velocity. (You must give both the magnitude and the direction.) 17 A stone is thrown from a cliff and strikes the surface of the sea with a vertical velocity of 18 m s−1 and a horizontal velocity v. The resultant of these two velocities is 25 m s−1. a Draw a vector diagram showing the two velocities and the resultant. b Use your diagram to find the value of v. c Use your diagram to find the angle between the stone and the vertical as it strikes the water. 1.7 Subtracting vectors Sometimes, vectors need to be subtracted rather than added. For example, if you are in a car moving at 2.0 m s−1 and another car on the same road is moving in the same direction at 5.0 m s−1, then you approach the car at 5.0 – 2.0 = 3.0 m s−1. You are subtracting two velocity vectors. Subtraction of vectors can be done using the formula: A − B = A + (− B) where A and B are vectors. KEY IDEA To subtract a vector, add on the vector to be subtracted in the opposite direction. So, to subtract, just add the negative vector. But first you have to understand what the negative of vector B means. The negative of vector B is another vector of the same size as B but in the opposite direction. This is straightforward if the velocities are in the same direction. For example, to subtract a velocity of 4 m s−1 north from a velocity of 10 m s−1 north, you start by drawing a vector 10 m s−1 north and then add a vector of 4 m s−1 south. The answer is 6 m s−1 north. It is less straightforward if the velocities are in the opposite direction. For example, to subtract a velocity of 4 m s−1 south from a velocity of 10 m s−1 north, you start by drawing a vector 10 m s−1 north and then add a vector of 4 m s−1 north. The answer is 14 m s−1 north. The example in Figure 1.17 shows how to find A − B and A + B when the vectors are along different directions. Question 18 A velocity of 5.0 m s−1 is due north. Subtract from this velocity another velocity that is: a 5.0 m s−1 due south b 5.0 m s−1 due north Figure 1.17: Subtracting and adding two vectors A and B in different directions. c 5.0 m s−1 due west d 5.0 m s−1 due east (You can do a scale drawing or make a calculation but remember to give the direction of your answers as well as their size.) 1.8 Other examples of scalar and vector quantities Direction matters when vectors are combined. You can use this to decide whether a quantity is a vector or a scalar. For example, if you walk for 3 minutes north and then 3 minutes in another direction, the total time taken is 6 minutes whatever direction you choose. A vector of 3 units added to another vector of 3 units can have any value between 0 and 6 but two scalars of 3 units added together always make six units. So, time is a scalar. Mass and density are also both scalar quantities. Force and acceleration, as you will see in later chapters, are both vector quantities. This is because, if an object is pushed with the same force in two opposite directions, the forces cancel out. Work and pressure, which you will also study in later chapters, both involve force. However, work and pressure are both scalar quantities. For example, if you pull a heavy case along the floor north and then the same distance south, the total work done is clearly not zero. You just add scalar quantities even if they are in the opposite direction. REFLECTION Write down anything that you found interesting or challenging in this chapter. Look at your notes later when you revise this topic. SUMMARY Displacement is the distance travelled in a particular direction. Velocity is defined by the word equation: change in displacement velocity = time taken The gradient of a displacement–time graph is equal to velocity: Δs v= Δt Distance, speed, mass and time are scalar quantities. A scalar quantity has only magnitude. Displacement and velocity are vector quantities. A vector quantity has both magnitude and direction. Vector quantities may be combined by vector addition to find their resultant. The second vector can be subtracted from the first by adding the negative of the second vector, which acts in the opposite direction. EXAM-STYLE QUESTIONS 1 Which of the following pairs contains one vector and one scalar quantity? A displacement : mass B displacement : velocity C distance : speed D speed : time 2 A vector P of magnitude 3.0 N acts towards the right and a vector Q of magnitude 4.0 N acts upwards. What is the magnitude and direction of the vector (P − Q)? A 1.0 N at an angle of 53° downwards to the direction of P B 1.0 N at an angle of 53° upwards to the direction of P C 5.0 N at an angle of 53° downwards to the direction of P D 5.0 N at an angle of 53° upwards to the direction of P 3 A car travels one complete lap around a circular track at a constant speed of 120 km h−1. a If one lap takes 2.0 minutes, show that the length of the track is 4.0 km. b Explain why values for the average speed and average velocity are different. c Determine the magnitude of the displacement of the car in a time of 1.0 min. (The circumference of a circle = 2πR, where R is the radius of the circle.) [Total: 5] 4 A boat leaves point A and travels in a straight line to point B. The journey takes 60 s. Figure 1.18 Calculate: a the distance travelled by the boat b the total displacement of the boat c the average velocity of the boat. Remember that each vector quantity must be given a direction as well as a magnitude. [Total: 6] 5 A boat travels at 2.0 m s−1east towards a port, 2.2 km away. When the boat reaches the port, the passengers travel in a car due north for 15 minutes at 60 km h−1.  Calculate: a the total distance travelled b the total displacement c the total time taken d the average speed in m s−1 e the magnitude of the average velocity. [Total: 11] 6 A river flows from west to east with a constant velocity of 1.0 m s−1. A boat leaves the south bank heading due north at 2.4 m s−1. Find the resultant velocity of the boat. 7 a Define displacement. b Use the definition of displacement to explain how it is possible for an athlete to run round a track yet have no displacement. [Total: 6] 8 A girl is riding a bicycle at a constant velocity of 3.0 m s−1 along a straight road. At time t = 0, she passes her brother sitting on a stationary bicycle. At time t = 0, the boy sets off to catch up with his sister. His velocity increases from time t = 0 until t = 5.0 s, when he has covered a distance of 10 m. He then continues at a constant velocity of 4.0 m s−1. a Draw the displacement–time graph for the girl from t = 0 to t = 12 s. b On the same graph axes, draw the displacement–time graph for the boy. c Using your graph, determine the value of t when the boy catches up with his sister. [Total: 4] 9 A student drops a small black sphere alongside a vertical scale marked in centimetres. A number of flash photographs of the sphere are taken at 0.10 s intervals: Figure 1.19 The first photograph is taken with the sphere at the top at time t = 0 s. a Explain how Figure 1.19 shows that the sphere reaches a constant speed. b Determine the constant speed reached by the sphere. c Determine the distance that the sphere has fallen when t = 0.80 s. d In a real photograph, each image of the sphere appears slightly blurred because each flash is not instantaneous and takes a time of 0.0010 s. Determine the absolute uncertainty that this gives in the position of each position of the black sphere when it is travelling at the final constant speed. Suggest whether this should be observable on the diagram. [Total: 8] 10 a State one difference between a scalar quantity and a vector quantity and give an example of each. b A plane has an air speed of 500 km h−1 due north. A wind blows at 100 km h−1 from east to west. Draw a vector diagram to calculate the resultant velocity of the plane. Give the direction of travel of the plane with respect to north. c The plane flies for 15 minutes. Calculate the displacement of the plane in this time. [Total: 8] 11 A small aircraft for one person is used on a short horizontal flight. On its journey from A to B, the resultant velocity of the aircraft is 15 m s−1 in a direction 60° east of north and the wind velocity is 7.5 m s−1 due north. Figure 1.20 a Show that for the aircraft to travel from A to B it should be pointed due east. b After flying 5 km from A to B, the aircraft returns along the same path from B to A with a resultant velocity of 13.5 m s−1. Assuming that the time spent at B is negligible, calculate the average speed for the complete journey from A to B and back to A. [Total: 5] SELF-EVALUATION CHECKLIST After studying the chapter, complete a table like this: I can See topic… Needs more Almost there Ready to work move on define and use displacement, speed and 1.1, 1.2, 1.3 velocity draw and interpret displacement–time 1.4 graphs describe laboratory methods for 1.1 determining speed understand the differences between 1.2 scalar and vector quantities and give examples of each use vector addition to add and subtract 1.6, 1.7 vectors that are in the same plane. 〉 Chapter 2 Accelerated motion LEARNING INTENTIONS In this chapter you will learn how to: define acceleration draw and interpret graphs of speed, velocity and acceleration calculate displacement from the area under a velocity–time graph calculate velocity and acceleration using gradients of a displacement–time graph and a velocity–time graph derive and use the equations of uniformly accelerated motion describe an experiment to measure the acceleration of free fall, g use perpendicular components to represent a vector explain projectile motion in terms of uniform velocity and uniform acceleration. BEFORE YOU START Write down definitions of speed and velocity. Write a list of all the vectors that you know. Why are some quantities classed as vectors? QUICK OFF THE MARK The cheetah (Figure 2.1) has a maximum speed of more than 30 m s−1 (108 km/h). A cheetah can reach 20 m s−1 from a standing start in just three or four strides, taking only two seconds. A car cannot increase its speed as rapidly but on a long straight road it can easily travel faster than a cheetah. How do you think such measurements can be made? What apparatus is needed? Figure 2.1: The cheetah is the world’s fastest land animal. Its acceleration is impressive, too. 2.1 The meaning of acceleration In everyday language, the term accelerating means ‘speeding up’. Anything whose speed is increasing is accelerating. Anything whose speed is decreasing is decelerating. To be more precise in our definition of acceleration, we should think of it as changing velocity. Any object whose speed is changing or which is changing its direction has acceleration. Because acceleration is linked to velocity in this way, it follows that it is a vector quantity. Some examples of objects accelerating are shown in Figure 2.2. Figure 2.2: Examples of objects accelerating. 2.2 Calculating acceleration The acceleration of something indicates the rate at which its velocity is changing. Language can get awkward here. Looking at the sprinter in Figure 2.3, we might say, ‘The sprinter accelerates faster than the car.’ However, ‘faster’ really means ‘greater speed’. It is better to say, ‘The sprinter has a greater acceleration than the car.’ Acceleration is defined as follows: acceleration = rate of change of velocity change in velocity average acceleration = time taken So to calculate acceleration a, we need to know two quantities – the change in velocity Δv and the time taken Δt: Δv a= Δt Sometimes this equation is written differently. We write u for the initial velocity and v for the final velocity (because u comes before v in the alphabet). The moving object accelerates from u to v in a time t (this is the same as the time represented by Δt in the equation). Then the acceleration is given by the equation: v−u a= t Figure 2.3: The sprinter has a greater acceleration than the car, but her top speed is less. You must learn the definition of acceleration. It can be put in words or symbols. If you use symbols you must state what those symbols mean. 2.3 Units of acceleration The unit of acceleration is m s−2 (metres per second squared). The sprinter might have an acceleration of 5 m s−2; her velocity increases by 5 m s−1 every second. You could express acceleration in other units. For example, an advertisement might claim that a car accelerates from 0 to 60 miles per hour (mph) in 10 s. Its acceleration would then be 6 mph s−1 (6 miles per hour per second). However, mixing together hours and seconds is not a good idea, and so acceleration is almost always given in the standard SI unit of m s−2. WORKED EXAMPLES 1 Leaving a bus stop, a bus reaches a velocity of 8.0 m s−1 after 10 s. Calculate the acceleration of the bus. Step 1 Note that the bus’s initial velocity is 0 m s−1. Therefore: change in velocity Δv = (8.0 − 0) m s−1 time taken Δt = 10 s Step 2 Substitute these values in the equation for acceleration: Δv acceleration = Δt 8.0 = 10 = 0.80m s−2 2 A sprinter starting from rest has an acceleration of 5.0 m s−2 during the first 2.0 s of a race. Calculate her velocity after 2.0 s. v−u Step 1 Rearranging the equation a = t gives: v = u + at Step 2 Substituting the values and calculating gives: v = 0 + (5.0 × 2.0) = 10 m s−1 3 A train slows down from 60 m s−1 to 20 m s−1 in 50 s. Calculate the magnitude of the deceleration of the train. Step 1 Write what you know: u = 60 m s−1 v = 20 m s−1 t = 50 s Step 2 Take care! Here the train’s final velocity is less than its initial velocity. To ensure that we arrive at the correct answer, we will use the alternative form of the equation to calculate a. v−u a = t 20−60 −40 = 50 = 50 −2 = −0.80 m s The minus sign (negative acceleration) indicates that the train is slowing down. It is decelerating. The magnitude of the deceleration is 0.80 m s−2. Questions 1 A car accelerates from a standing start and reaches a velocity of 18 m s−1 after 6.0 s. Calculate its acceleration. 2 A car driver brakes gently. Her car slows down from 23 m s−1 to 11 m s−1 in 20 s. Calculate the magnitude (size) of her deceleration. (Note that, because she is slowing down, her acceleration is negative.) 3 A stone is dropped from the top of a cliff. Its acceleration is 9.81 m s−2. How fast is it moving: a after 1.0 s? b after 3.0 s? 2.4 Deducing acceleration The gradient of a velocity–time graph tells us whether the object’s velocity has been changing at a high rate or a low rate, or not at all (Figure 2.4). We can deduce the value of the acceleration from the gradient of the graph: acceleration = gradient of velocity–time graph KEY IDEA acceleration = gradient of velocity–time graph The graph (Figure 2.5) shows how the velocity of a cyclist changed during the start of a sprint race. We can find his acceleration during the first section of the graph (where the line is straight) using the triangle as shown. The change in velocity Δv is given by the vertical side of the triangle. The time taken Δt is given by the horizontal side. change in velocity acceleration = time taken 25−0 = 5 −2 = 4.0 m s A more complex example where the velocity–time graph is curved is shown in Figure 2.18. Figure 2.4: The gradient of a velocity–time graph is equal to acceleration. Figure 2.5: Deducing acceleration from a velocity–time graph. 2.5 Deducing displacement We can also find the displacement of a moving object from its velocity–time graph. This is given by the area under the graph: displacement = area under velocity–time graph KEY IDEA displacement = area under velocity–time graph It is easy to see why this is the case for an object moving at a constant velocity. The displacement is simply velocity × time, which is the area of the shaded rectangle (Figure 2.6a). For changing velocity, again the area under the graph gives displacement (Figure 2.6b). Figure 2.6: The area under the velocity–time graph is equal to the displacement of the object. So, for this simple case in which the area is a triangle, we have: 1 displacement = 2 × base × height 1 = 2 × 5.0 × 10 = 25 m It is easy to confuse displacement–time graphs and velocity–time graphs. Check by looking at the quantity marked on the vertical axis. For more complex graphs, you may have to use other techniques such as counting squares to deduce the area, but this is still equal to the displacement. (Take care when counting squares: it is easiest when the sides of the squares stand for one unit. Check the axes, as the sides may represent 2 units, 5 units or some other number.) Questions 4 A lorry driver is travelling at the speed limit on a motorway. Ahead, he sees hazard lights and gradually slows down. He sees that an accident has occurred, and brakes suddenly to a halt. Sketch a velocity–time graph to represent the motion of this lorry. 5 Table 2.1 shows how the velocity of a motorcyclist changed during a speed trial along a straight road. a Draw a velocity–time graph for this motion. b From the table, deduce the motorcyclist’s acceleration during the first 10 s. c Check your answer by finding the gradient of the graph during the first 10 s. d Determine the motorcyclist’s acceleration during the last 15 s. e Use the graph to find the total distance travelled during the speed trial. Velocity / m s−1 0 15 30 30 20 10 0 Time / s 0 5 10 15 20 25 30 Table 2.1: Data for a motorcyclist. 2.6 Measuring velocity and acceleration In a car crash, the occupants of the car may undergo a very rapid deceleration. This can cause them serious injury, but can be avoided if an air-bag is inflated within a fraction of a second. Figure 2.7 shows the tiny accelerometer at the heart of the system, which detects large accelerations and decelerations. The acceleration sensor consists of two rows of interlocking teeth. In the event of a crash, these move relative to one another, and this generates a voltage that triggers the release of the air-bag. Figure 2.7: A micro-mechanical acceleration sensor is used to detect sudden accelerations and decelerations as a vehicle travels along the road. This electron microscope image shows the device magnified about 1000 times. At the top of the photograph (Figure 2.7), you can see a second sensor that detects sideways accelerations. This is important in the case of a side impact. These sensors can also be used to detect when a car swerves or skids, perhaps on an icy road. In this case, they activate the car’s stability-control systems. 2.7 Determining velocity and acceleration in the laboratory In Chapter 1, we looked at ways of finding the velocity of a trolley moving in a straight line. These involved measuring distance and time, and deducing velocity. Practical Activity 2.1 shows how these techniques can be extended to find the acceleration of a trolley. PRACTICAL ACTIVITY 2.1: LABORATORY MEASUREMENTS OF ACCELERATION Measurements using light gates The computer records the time for the first ‘interrupt’ section of the card to pass through the light beam of the light gate (Figure 2.8). Given the length of the interrupt, it can work out the trolley’s initial velocity u. This is repeated for the second interrupt to give final velocity v. The computer also records the time interval t3 − t1 between these two velocity measurements. Now it can calculate the acceleration a as shown: l1 u= t 2 −t 1 (l1 = length of first section of the interrupt card) and l2 v= t 4 −t 3 (l2 = length of second section of the interrupt card) Therefore: change in velocity a = time taken v−u = t 3 −t 1 (Note that this calculation gives only an approximate value for a. This is because u and v are average speeds over a period of time; for an accurate answer we would need to know the speeds at times t1 and t3.) Sometimes two light gates are used with a card of length l. The computer can still record the times as shown and calculate the acceleration in the same way, with l1 = l2 = l. Figure 2.8: Determining acceleration using a single light gate. Measurements using a ticker-timer The practical arrangement is the same as for measuring velocity. Now we have to think about how to interpret the tape produced by an accelerating trolley (Figure 2.9). Figure 2.9: Ticker-tape for an accelerating trolley. The tape is divided into sections, as before, every five dots. Remember that the time interval between adjacent dots is 0.02 s. Each section represents 0.10 s. By placing the sections of tape side by side, you can picture the velocity–time graph. The length of each section gives the trolley’s displacement in 0.10 s, from which the average velocity during this time can be found. This can be repeated for each section of the tape, and a velocity–time graph drawn. The gradient of this graph is equal to the acceleration. Table 2.2 and Figure 2.10 show some typical results. The acceleration is calculated to be: Δv a = Δt 0.93 = 0.20 ≈ 4.7 ms−2 Section of tape Time at start / s Time interval / s Length of section / Velocity / m s−1 cm 1 0.0 0.10 2.3 0.23 2 0.10 0.10 7.0 0.70 3 0.20 0.10 11.6 1.16 Table 2.2: Data for Figure 2.10.  Figure 2.10: Deducing acceleration from measurements of a ticker-tape. Measurements using a motion sensor The computer software that handles the data provided by the motion sensor can calculate the acceleration of a trolley. However, because it deduces velocity from measurements of position, and then calculates acceleration from values of velocity, its precision is relatively poor. Questions 6 Sketch a section of ticker-tape for a trolley that travels at a steady velocity and then decelerates. 7 Figure 2.11 shows the dimensions of an interrupt card, together with the times recorded as it passed through a light gate. Use these measurements to calculate the acceleration of the card. (Follow the steps outlined in Practical Activity 2.1.) Figure 2.11: For Question 7. 8 Two adjacent five-dot sections of a ticker-tape measure 10 cm and 16 cm, respectively. The interval between dots is 0.02 s. Deduce the acceleration of the trolley that produced the tape. 2.8 The equations of motion As a space rocket rises from the ground, its velocity steadily increases. It is accelerating (Figure 2.12). Eventually, it will reach a speed of several kilometres per second. Any astronauts aboard find themselves pushed back into their seats while the rocket is accelerating. Figure 2.12: A rocket accelerates as it lifts off from the ground. The engineers who planned the mission must be able to calculate how fast the rocket will be travelling and where it will be at any point in its journey. They have sophisticated computers to do this, using more elaborate versions of the four equations of motion. There is a set of equations that allows us to calculate the quantities involved when an object is moving with a constant acceleration. The quantities we are concerned with are: s displacement u initial velocity v final velocity a acceleration t time taken The four equations of motion are shown above. Take care using the equations of motion. They can only be used for: motion in a straight line  an object with constant acceleration. KEY EQUATIONS The four equations of motion: equation 1: v = u + at (u+v) equation 2: s = ×t 2 equation 3: s = ut + 12 at 2 equation 4: v2 = u 2 + 2as To get a feel for how to use these equations, we will consider some worked examples. In each example, we will follow the same procedure: Step 1 We write down the quantities that we know, and the quantity we want to find. Step 2 Then we choose the equation that links these quantities, and substitute in the values. Step 3 Finally, we calculate the unknown quantity. We will look at where these equations come from in the next topic, ‘Deriving the equations of motion’. WORKED EXAMPLES 4 The rocket shown in Figure 2.12 lifts off from rest with an acceleration of 20 m s−2. Calculate its velocity after 50 s. Step 1 What we know: u = 0 m s−1 a = 20 m s−2 t = 50 s and what we want to know: v = ? Step 2 The equation linking u, a, t and v is equation 1: v = u + at Substituting gives: v = 0 + (20 × 50) Step 3 Calculation then gives: v = 1000 m s−1 So the rocket will be travelling at 1000 m s−1 after 50 s. This makes sense, since its velocity increases by 20 m s−1 every second, for 50 s. You could use the same equation to work out how long the rocket would take to reach a velocity of 2000 m s−1, or the acceleration it must have to reach a speed of 1000 m s−1 in 40 s and so on. 5 The car shown in Figure 2.13 is travelling along a straight road at 8.0 m s−1. It accelerates at 1.0 m s−2 for a distance of 18 m. How fast is it then travelling?  Figure 2.13: For Worked example 5. This car accelerates for a short distance as it travels along the road. In this case, we will have to use a different equation, because we know the distance during which the car accelerates, not the time. Step 1 What we know: u = 8.0 m s−1 a = 1.0 m s−2 s = 18 m and what we want to know: v = ? Step 2 The equation we need is equation 4: v2 = u2 + 2as Substituting gives: v2 = 8.02 + (2 × 1.0 × 18) Step 3 Calculation then gives: v2 = 64 + 36 = 100 m2 s−2 v = 10 m s−1 So the car will be travelling at 10 m s−1 when it stops accelerating. (You may find it easier to carry out these calculations without including the units of quantities when you substitute in the equation. However, including the units can help to ensure that you end up with the correct units for the final answer.) 6 A train (Figure 2.14) travelling at 20 m s−1 accelerates at 0.50 m s−2 for 30 s. Calculate the distance travelled by the train in this time.  Figure 2.14: For Worked example 6. This train accelerates for 30 s. Step 1 What we know: u = 20 m s−1 t = 30 s a = 0.50 m s−2 and what we want to know: s = ? Step 2 The equation we need is equation 3: s = ut + 12 at 2 Substituting gives: s = (20 × 30) + 1 2 × 0.5 × (30)2 Step 3 Calculation then gives: s = 600 + 225 = 825 m So the train will travel 825 m while it is accelerating. 7 The cyclist in Figure 2.15 is travelling at 15 m s−1. She brakes so that she doesn’t collide with the wall. Calculate the magnitude of her deceleration. Figure 2.15: For Worked example 7. The cyclist brakes to stop herself colliding with the wall. This example shows that it is sometimes necessary to rearrange an equation, to make the unknown quantity its subject. It is easiest to do this before substituting in the values. Step 1 What we know: u = 15 m s−1 v = 0 m s−1 s = 18 m and what we want to know: a = ? Step 2 The equation we need is equation 4:  v2 = u2 + 2as Rearranging gives: v 2 −u 2 a = 2s 02 −152 a = 2×18 −225 = 36 Step 3 Calculation then gives: a = −6.25 m s−2 ≈ −6.3 m s−2 So the cyclist will have to brake hard to achieve a deceleration of magnitude 6.3 m s−2. The minus sign shows that her acceleration is negative; in other words, a deceleration. Questions 9 A car is initially stationary. It has a constant acceleration of 2.0 m s−2. a Calculate the velocity of the car after 10 s. b Calculate the distance travelled by the car at the end of 10 s. c Calculate the time taken by the car to reach a velocity of 24 m s−1. 10 A train accelerates steadily from 4.0 m s−1 to 20 m s−1 in 100 s. a Calculate the acceleration of the train. b From its initial and final velocities, calculate the average velocity of the train. c Calculate the distance travelled by the train in this time of 100 s. 11 A car is moving at 8.0 m s−1. The driver makes it accelerate at 1.0 m s−2 for a distance of 18 m. What is the final velocity of the car? 2.9 Deriving the equations of motion We have seen how to make use of the equations of motion. But where do these equations come from? They arise from the definitions of velocity and acceleration. We can find the first two equations from the velocity–time graph shown in Figure 2.16. The graph represents the motion of an object. Its initial velocity is u. After time t, its final velocity is v. Figure 2.16: This graph shows the variation of velocity of an object with time. The object has constant acceleration. Equation 1 The graph of Figure 2.16 is a straight line, therefore the object’s acceleration a is constant. The gradient (slope) of the line is equal to acceleration. The acceleration is defined as: (v−u) a= t which is the gradient of the line. Rearranging this gives the first equation of motion: v = u + at (equation 1) Equation 2 Displacement is given by the area under the velocity–time graph. Figure 2.17 shows that the object’s average velocity is half-way between u and v. So the object’s average velocity, calculated by averaging its initial and final velocities, is given by: (u+v) 2 The object’s displacement is the shaded area in Figure 2.17. This is a rectangle, and so we have: displacement = average velocity × time taken and hence: (u+v) s= 2 ×t (equation 2) Figure 2.17: The average velocity is half-way between u and v. Equation 3 From equations 1 and 2, we can derive equation 3: v = u + at (equation 1) (u+v) s= 2 ×t (equation 2) Substituting v from equation 1 gives: (u+u+at) s = 2 ×t 2ut at 2 = 2 + 2 So s = ut + 12 at 2 (equation 3) Looking at Figure 2.16, you can see that the two terms on the right of the equation correspond to the areas of the rectangle and the triangle that make up the area under the graph. Of course, this is the same area as the rectangle in Figure 2.17. Equation 4 Equation 4 is also derived from equations 1 and 2: v = u + at (equation 1) (equation 2)  (u+v) s= 2 ×t Substituting for time t from equation 1 gives: (u+v) (v−u) s= 2 × a Rearranging this gives: 2as = (u + v) (v − u) = v2 − u 2 or simply: v2 = u2 + 2as (equation 4) Investigating road traffic accidents The police frequently have to investigate road traffic accidents. They make use of many aspects of physics, including the equations of motion. The next two questions will help you to apply what you have learned to situations where police investigators have used evidence from skid marks on the road. Questions 12 Trials on the surface of a new road show that, when a car skids to a halt, its acceleration is −7.0 m s−2. Estimate the skid-to-stop distance of a car travelling at a speed limit of 30 m s−1 (approximately 110 km h−1 or 70 mph). 13 At the scene of an accident on a country road, police find skid marks stretching for 50 m. Tests on the road surface show that a skidding car decelerates at 6.5 m s−2. Was the car that skidded exceeding the speed limit of 25 m s−1 (90 km h−1) on this road? 2.10 Uniform and nonuniform acceleration It is important to note that the equations of motion only apply to an object that is moving with a constant acceleration. If the acceleration a was changing, you wouldn’t know what value to put in the equations. Constant acceleration is often referred to as uniform acceleration. The velocity–time graph in Figure 2.18 shows non-uniform acceleration. It is not a straight line; its gradient is changing (in this case, decreasing). Figure 2.18: This curved velocity–time graph cannot be analysed using the equations of motion. The acceleration at any instant in time is given by the gradient of the velocity–time graph. The triangle in Figure 2.18 shows how to find the acceleration at t = 20 seconds: At the time of interest, mark a point on the graph. Draw a tangent to the curve at that point. Make a large right-angled triangle, and use it to find the gradient. You can find the change in displacement of the body as it accelerates by determining the area under the velocity–time graph. To find the displacement of the object in Figure 2.18 between t = 0 and t = 20 s, the most straightforward, but lengthy, method is just to count the number of small squares. In this case, up to t = 20 s, there are about 250 small squares. This is tedious to count but you can save yourself a lot of time by drawing a line from the origin to the point at 20 s. The area of the triangle is easy to find (200 small squares) and then you only have to count the number of small squares between the line you have drawn and the curve on the graph (about 50 squares) In this case, each square is 1 m s−1 on the y-axis by 1 s on the x-axis, so the area of each square is 1 × 1 = 1 m and the displacement is 250 m. In other cases, note carefully the value of each side of the square you have chosen. Questions 14 The graph in Figure 2.19 represents the motion of an object moving with varying acceleration. Lay your ruler on the diagram so that it is tangential to the graph at point P. a What are the values of time and velocity at this point? b Estimate the object’s acceleration at this point. Figure 2.19: For Question 14. 15 The velocity–time graph (Figure 2.20) represents the motion of a car along a straight road for a period of 30 s. a Describe the motion of the car. b From the graph, determine the car’s initial and final velocities over the time of 30 s. c Determine the acceleration of the car. d By calculating the area under the graph, determine the displacement of the car. e Check your answer to part d by calculating the car’s displacement using s = ut 12 at 2. Figure 2.20: For Question 15. 2.11 Acceleration caused by gravity If you drop a ball or stone, it falls to the ground. Figure 2.21, based on a multiflash photograph, shows the ball at equal intervals of time. You can see that the ball’s velocity increases as it falls because the spaces between the images of the ball increase steadily. The ball is accelerating. A multiflash photograph is useful to demonstrate that the ball accelerates as it falls. Usually, objects fall too quickly for our eyes to be able to observe them speeding up. It is easy to imagine that the ball moves quickly as soon as you let it go, and falls at a steady speed to the ground. Figure 2.21 shows that this is not the case. If we measure the acceleration of a freely falling object on the surface of the Earth, we find a value of about 9.81 m s−2. This is known as the acceleration of free fall, and is given the symbol g: Figure 2.21: This diagram of a falling ball, based on a multiflash photo, clearly shows that the ball’s velocity increases as it falls. acceleration of free fall g = 9.81 m s−2 The value of g depends on where you are on the Earth’s surface, but we usually take g = 9.81 m s−2. If we drop an object, its initial velocity u = 0. How far will it fall in time t? Substituting in s = ut 12 at 2 gives displacement s:  1 s = 2 × 9. 81 × t 2 = 4.9 × t 2 Hence, by timing a falling object, we can determine g. Questions 16 If you drop a stone from the edge of a cliff, its initial velocity u = 0, and it falls with acceleration g = 9.81 m s−2. You can calculate the distance s it falls in a given time t using an equation of motion. a Copy and complete Table 2.3, which shows how s depends on t. b Draw a graph of s against t. c Use your graph to find the distance fallen by the stone in 2.5 s. d Use your graph to find how long it will take the stone to fall to the bottom of a cliff 40 m high. Check your answer using the equations of motion. Time / s 0 1.0 2.0 3.0 4.0 Displacement / m 0 4.9 Table 2.3: Time t and displacement s data fo 17 An egg falls off a table. The floor is 0.8 m from the table-top. a Calculate the time taken to reach the ground. b Calculate the velocity of impact with the ground. 2.12 Determining g One way to measure the acceleration of free fall g would be to try bungee-jumping (Figure 2.22). You would need to carry a stopwatch, and measure the time between jumping from the platform and the moment when the elastic rope begins to slow your fall. If you knew the length of the unstretched rope, you could calculate g. There are easier methods for finding g that can be used in the laboratory. These are described in Practical Activity 2.2. Figure 2.22: A bungee-jumper falls with initial acceleration g. PRACTICAL ACTIVITY 2.2: LABORATORY MEASUREMENTS OF g Measuring g using an electronic timer In this method, a steel ball-bearing is held by an electromagnet (Figure 2.23). When the current to the magnet is switched off, the ball begins to fall and an electronic timer starts. The ball falls through a trapdoor, and this breaks a circuit to stop the timer. This tells us the time taken for the ball to fall from rest through the distance h between the bottom of the ball and the trapdoor. Here is how we can use one of the equations of motion to find g: displacement s = h time taken = t initial velocity u = 0 acceleration a = g Substituting in s = ut + 12 at 2 gives: h = 12 gt 2 and for any values of h and t we can calculate a value for g. Figure 2.23: The timer records the time for the ball to fall through the distance h. A more satisfactory procedure is to take measurements of t for several different values of h. The height of the ball bearing above the trapdoor is varied systematically, and the time of fall measured several times to calculate an average for each height. Table 2.4 and Figure 2.24 show some typical results. We can deduce g from the gradient of the graph of h against t2. The equation for a straight line through the origin is: y = mx In our experiment we have: h/m t/s t2 / s2 0.27 0.25 0.063 0.39 0.30 0.090 0.56 0.36 0.130 0.70 0.41 0.168 0.90 0.46 0.212 Table 2.4: Data for Figure 2.24. These are mean values. g The gradient of the straight line of a graph of h against t2 is equal to. 2 Figure 2.24: The acceleration of free fall can be determined from the gradient. Therefore: g gradient = 2 0.84 = 0.20 = 4.2 g = 4.2 × 2 = 8.4 m s−2 Sources of uncertainty The electromagnet may retain some magnetism when it is switched off, and this may tend to slow the ball’s fall. Consequently, the time t recorded by the timer may be longer than if the ball were to fall completely freely. From h = 12 gt 2 , it follows that, if t is too great, the experimental value of g will be too small. This is an example of a systematic error – all the results are systematically distorted so that they are too great (or too small) as a consequence of the experimental design. Measuring the height h is awkward. You can probably only find the value of h to within ±1 mm at best. So there is a random error in the value of h, and this will result in a slight scatter of the points on the graph, and a degree of uncertainty in the final value of g. If you just have one value for h and the corresponding value for t you can use the uncertainty in h and t to find the uncertainty in g. The percentage uncertainty in g = percentage uncertainty in h + 2 × percentage uncertainty in t. For more about errors and combining uncertainties, see Chapter P1. Measuring g using a ticker-timer Figure 2.25 shows a weight falling. As it falls, it pulls a tape through a ticker-timer. The spacing of the dots on the tape increases steadily, showing that the weight is accelerating. You can analyse the tape to find the acceleration, as discussed in Practical Activity 2.1. Figure 2.25: A falling weight pulls a tape through a ticker-timer. This is not a very satisfactory method of measuring g. The main problem arises from friction between the tape and the ticker-timer. This slows the fall of the weight and so its acceleration is less than g. (This is another example of a systematic error.) The effect of friction is less of a problem for a large weight, which falls more freely. If measurements are made for increasing weights, the value of acceleration gets closer and closer to the true value of g. Measuring g using a light gate Figure 2.26 shows how a weight can be attached to a card ‘interrupt’. The card is designed to break the light beam twice as the weight falls. The computer can then calculate the velocity of the weight twice as it falls, and hence find its acceleration: x initial velocity u = t 2 −t 1 x final velocity v = t 4 −t 3 Therefore: v−u acceleration a = t 3 −t 1 The weight can be dropped from different heights above the light gate. This allows you to find out whether its acceleration is the same at different points in its fall. This is an advantage over Method 1, which can only measure the acceleration from a stationary start.  Figure 2.26: The weight accelerates as it falls. The upper section of the card falls more quickly through the light gate. WORKED EXAMPLE 8 To get a rough value for g, a student dropped a stone from the top of a cliff. A second student timed the stone’s fall using a stopwatch. Here are their results: estimated height of cliff = 30 m time of fall = 2.6 s Use the results to estimate a value for g. Step 1 Calculate the average speed of the stone: 30 average speed of stone during fall = 2.6 = 11.5 m s−1 Step 2 Find the values of v and u: final speed v = 2 × 11.5 m s−1 = 23.0 m s−1 initial speed u = 0 m s−1 Step 3 Substitute these values into the equation for acceleration: v−u a = t 23.0 = 2.6 = 8.8 m s−2 Note that you can reach the same result more directly using s = ut + 12 at 2 , but you may find it easier to follow what is going on using the method given here. We should briefly consider why the answer is less than the expected value of g = 9.81 m s−2. It might be that the cliff was higher than the student’s estimate. The timer may not have been accurate in switching the stopwatch on and off. There will have been air resistance that slowed the stone’s fall. Questions 18 A steel ball falls from rest through a height of 2.10 m. An electronic timer records a time of 0.67 s for the fall. a Calculate the average acceleration of the ball as it falls. b Suggest reasons why the answer is not exactly 9.81 m s−2.  c Suppose the height is measured accurately but the time is measured to an uncertainty of ±0.02 s. Calculate the percentage uncertainty in the time and the percentage uncertainty in the average acceleration. You can do this by repeating the calculation for g using a time of 0.65 s. You can find out more about uncertainty in Chapter P1. 19 In an experiment to determine the acceleration due to gravity, a ball was timed electronically as it fell from rest through a height h. The times t shown in Table 2.5 were obtained. a Plot a graph of h against t2. b From the graph, determine the acceleration of free fall g. c Comment on your answer. Height h / m 0.70 1.03 1.25 1.60 1.99 Time t / s 0.99 1.13 1.28 1.42 1.60 Table 2.5: Height h and time t data for Question 19. 20 In Chapter 1, we looked at how to use a motion sensor to measure the speed and position of a moving object. Suggest how a motion sensor could be used to determine g. 2.13 Motion in two dimensions: projectiles A curved trajectory A multiflash photograph can reveal details of the path, or trajectory, of a projectile. Figure 2.27 shows the trajectories of a projectile – a bouncing ball. Once the ball has left the child’s hand and is moving through the air, the only force acting on it is its weight. The ball has been thrown at an angle to the horizontal. It speeds up as it falls – you can see that the images of the ball become further and further apart. At the same time, it moves steadily to the right. You can see this from the even spacing of the images across the picture. The ball’s path has a mathematical shape known as a parabola. After it bounces, the ball is moving more slowly. It slows down, or decelerates, as it rises – the images get closer and closer together. We interpret this picture as follows. The vertical motion of the ball is affected by the force of gravity, that is, its weight. When it rises it has a vertical deceleration of magnitude g, which slows it down, and when it falls it has an acceleration of g, which speeds it up. The ball’s horizontal motion is unaffected by gravity. In the absence of air resistance, the ball has a constant velocity in the horizontal direction. We can treat the ball’s vertical and horizontal motions separately, because they are independent of one another. Figure 2.27: A bouncing ball is an example of a projectile. This multiflash photograph shows details of its motion that would escape the eye of an observer. Components of a vector In order to understand how to treat the velocity in the vertical and horizontal directions separately we start by cons