Physics for the IB Diploma 6th Edition (PDF)

Physics for the IB Diploma Sixth Edition K. A. Tsokos Cambridge University Press’s mission is to advance learning, knowledge and research worldwide. Our IB Diploma resources aim to:  encourage learners to explore concepts, ideas and topics that have local and global significance  help st...

Physics for the IB Diploma Sixth Edition K. A. Tsokos Cambridge University Press’s mission is to advance learning, knowledge and research worldwide. Our IB Diploma resources aim to:  encourage learners to explore concepts, ideas and topics that have local and global significance  help students develop a positive attitude to learning in preparation for higher education  assist students in approaching complex questions, applying critical-thinking skills and forming reasoned answers. University Printing House, Cambridge CB2 8BS, United Kingdom Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org First, second and third editions © K. A. Tsokos 1998, 1999, 2001 Fourth, fifth, fifth (full colour) and sixth editions © Cambridge University Press 2005, 2008, 2010, 2014 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1998 Second edition 1999 Third edition 2001 Fourth edition published by Cambridge University Press 2005 Fifth edition 2008 Fifth edition (full colour version) 2010 Sixth edition 2014 Printed in the United Kingdom by Latimer Trend A catalogue record for this publication is available from the British Library isbn 978-1-107-62819-9 Paperback Additional resources for this publication at education.cambridge.org/ibsciences Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables, and other factual information given in this work is correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter. The material has been developed independently by the publisher and the content is in no way connected with nor endorsed by the International Baccalaureate Organization. notice to teachers in the uk It is illegal to reproduce any part of this book in material form (including photocopying and electronic storage) except under the following circumstances: (i) where you are abiding by a licence granted to your school or institution by the Copyright Licensing Agency; (ii) where no such licence exists, or where you wish to exceed the terms of a licence, and you have gained the written permission of Cambridge University Press; (iii) where you are allowed to reproduce without permission under the provisions of Chapter 3 of the Copyright, Designs and Patents Act 1988, which covers, for example, the reproduction of short passages within certain types of educational anthology and reproduction for the purposes of setting examination questions. The website accompanying this book contains further resources to support your IB Physics studies.Visit education.cambridge.org/ibsciences and register for access. Separate website terms and conditions apply.  Contents Introduction v 7 Atomic, nuclear and particle Note from the author vi physics 270 7.1 Discrete energy and radioactivity 270 1 Measurements and uncertainties 1 7.2 Nuclear reactions 285 1.1 Measurement in physics 1 7.3 The structure of matter 295 1.2 Uncertainties and errors 7 Exam-style questions 309 1.3 Vectors and scalars 21 Exam-style questions 32 8 Energy production 314 8.1 Energy sources 314 2 Mechanics 35 8.2 Thermal energy transfer 329 2.1 Motion 35 Exam-style questions 340 2.2 Forces 57 2.3 Work, energy and power 78 9 Wave phenomena (HL) 346 2.4 Momentum and impulse 98 9.1 Simple harmonic motion 346 Exam-style questions 110 9.2 Single-slit diffraction 361 9.3 Interference 365 3 Thermal physics 116 9.4 Resolution 376 3.1 Thermal concepts 116 9.5 The Doppler effect 381 3.2 Modelling a gas 126 Exam-style questions 390 Exam-style questions 142 10 Fields (HL) 396 4 Waves 146 10.1 Describing fields 396 4.1 Oscillations 146 10.2 Fields at work 415 4.2 Travelling waves 153 Exam-style questions 428 4.3 Wave characteristics 162 4.4 Wave behaviour 172 11 Electromagnetic 4.5 Standing waves 182 induction (HL) 434 Exam-style questions 190 11.1 Electromagnetic induction 434 11.2 Transmission of power 444 5 Electricity and magnetism 196 11.3 Capacitance 457 5.1 Electric fields 196 Exam-style questions 473 5.2 Heating effect of electric currents 207 5.3 Electric cells 227 12 Quantum and nuclear 5.4 Magnetic fields 232 physics (HL) 481 Exam-style questions 243 12.1 The interaction of matter with radiation 481 6 Circular motion and gravitation 249 12.2 Nuclear physics 505 6.1 Circular motion 249 Exam-style questions 517 6.2 The law of gravitation 259 Exam-style questions 265 III  Appendices 524 Answers to Test yourself questions 528 1 Physical constants 524 2 Masses of elements and selected isotopes 525 Glossary 544 3 Some important mathematical results 527 Index 551 Credits 560 Free online material The website accompanying this book contains further resources to support your IB Physics studies. Visit education.cambridge.org/ibsciences and register to access these resources:r7 Options Self-test questions Option A Relativity Assessment guidance Option B Engineering physics Model exam papers Option C Imaging Nature of Science Option D Astrophysics Answers to exam-style questions Additional Topic questions to Answers to Options questions accompany coursebook Answers to additional Topic questions Detailed answers to all coursebook Options glossary test yourself questions Appendices A Astronomical data B Nobel prize winners in physics IV  Introduction This sixth edition of Physics for the IB Diploma is fully updated to cover the content of the IB Physics Diploma syllabus that will be examined in the years 2016–2022. Physics may be studied at Standard Level (SL) or Higher Level (HL). Both share a common core, which is covered in Topics 1–8. At HL the core is extended to include Topics 9–12. In addition, at both levels, students then choose one Option to complete their studies. Each option consists of common core and additional Higher Level material.You can identify the HL content in this book by ‘HL’ included in the topic title (or section title in the Options), and by the red page border. The four Options are included in the free online material that is accessible using education.cambridge.org/ibsciences. The structure of this book follows the structure of the IB Physics syllabus. Each topic in the book matches a syllabus topic, and the sections within each topic mirror the sections in the syllabus. Each section begins with learning objectives as starting and reference points. Worked examples are included in each section; understanding these examples is crucial to performing well in the exam. A large number of test yourself questions are included at the end of each section and each topic ends with exam- style questions. The reader is strongly encouraged to do as many of these questions as possible. Numerical answers to the test yourself questions are provided at the end of the book; detailed solutions to all questions are available on the website. Some topics have additional questions online; these are indicated with the online symbol, shown here. Theory of Knowledge (TOK) provides a cross-curricular link between different subjects. It stimulates thought about critical thinking and how we can say we know what we claim to know. Throughout this book, TOK features highlight concepts in Physics that can be considered from a TOK perspective. These are indicated by the ‘TOK’ logo, shown here. 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. Throughout this book, International- Mindedness features highlight international concerns in Physics. These are indicated by the ‘International-Mindedness’ logo, shown here. Nature of science is an overarching theme of the Physics course. The theme examines the processes and concepts that are central to scientific endeavour, and how science serves and connects with the wider community. At the end of each section in this book, there is a ‘Nature of science’ paragraph that discusses a particular concept or discovery from the point of view of one or more aspects of Nature of science. A chapter giving a general introduction to the Nature of science theme is available in the free online material. INTRODUCTION V  Free online material Additional material to support the IB Physics Diploma course is available online.Visit education.cambridge.org/ibsciences and register to access these resources. Besides the Options and Nature of science chapter, you will find a collection of resources to help with revision and exam preparation. This includes guidance on the assessments, additional Topic questions, interactive self-test questions and model examination papers and mark schemes. Additionally, answers to the exam-style questions in this book and to all the questions in the Options are available. Note from the author This book is dedicated to Alexios and Alkeos and to the memory of my parents. I have received help from a number of students at ACS Athens in preparing some of the questions included in this book. These include Konstantinos Damianakis, Philip Minaretzis, George Nikolakoudis, Katayoon Khoshragham, Kyriakos Petrakos, Majdi Samad, Stavroula Stathopoulou, Constantine Tragakes and Rim Versteeg. I sincerely thank them all for the invaluable help. I owe an enormous debt of gratitude to Anne Trevillion, the editor of the book, for her patience, her attention to detail and for the very many suggestions she made that have improved the book substantially. Her involvement with this book exceeded the duties one ordinarily expects from an editor of a book and I thank her from my heart. I also wish to thank her for her additional work of contributing to the Nature of science themes throughout the book. Finally, I wish to thank my wife, Ellie Tragakes, for her patience with me during the completion of this book. K.A. Tsokos VI  Measurement and uncertainties 1 1.1 Measurement in physics Learning objectives Physics is an experimental science in which measurements made must be expressed in units. In the international system of units used throughout State the fundamental units of the SI system. this book, the SI system, there are seven fundamental units, which are defined in this section. All quantities are expressed in terms of these units Be able to express numbers in scientific notation. directly, or as a combination of them. Appreciate the order of magnitude of various quantities. The SI system Perform simple order-of- The SI system (short for Système International d’Unités) has seven magnitude calculations mentally. fundamental units (it is quite amazing that only seven are required). Express results of calculations to These are: the correct number of significant 1 The metre (m). This is the unit of distance. It is the distance travelled figures. 1 by light in a vacuum in a time of seconds. 299 792 458 2 The kilogram (kg). This is the unit of mass. It is the mass of a certain quantity of a platinum–iridium alloy kept at the Bureau International des Poids et Mesures in France. 3 The second (s). This is the unit of time. A second is the duration of 9 192 631 770 full oscillations of the electromagnetic radiation emitted in a transition between the two hyperfine energy levels in the ground state of a caesium-133 atom. 4 The ampere (A). This is the unit of electric current. It is defined as that current which, when flowing in two parallel conductors 1 m apart, produces a force of 2 × 107 N on a length of 1 m of the conductors. 1 5 The kelvin (K). This is the unit of temperature. It is of the 273.16 thermodynamic temperature of the triple point of water. 6 The mole (mol). One mole of a substance contains as many particles as there are atoms in 12 g of carbon-12. This special number of particles is called Avogadro’s number and is approximately 6.02 × 1023. 7 The candela (cd). This is a unit of luminous intensity. It is the intensity 1 of a source of frequency 5.40 × 1014 Hz emitting W per steradian. 683 You do not need to memorise the details of these definitions. In this book we will use all of the basic units except the last one. Physical quantities other than those above have units that are combinations of the seven fundamental units. They have derived units. For example, speed has units of distance over time, metres per second (i.e. m/s or, preferably, m s−1). Acceleration has units of metres per second squared (i.e. m/s2, which we write as m s−2 ). Similarly, the unit of force is the newton (N). It equals the combination kg m s−2. Energy, a very important quantity in physics, has the joule (J) as its unit. The joule is the combination N m and so equals (kg m s−2 m), or kg m2 s−2. The quantity 1 MEASUREMENT AND UNCERTAINTIES 1  power has units of energy per unit of time, and so is measured in J s−1. This combination is called a watt. Thus: 1 W = (1 N m s−1) = (1 kg m s−2 m s−1) = 1 kg m2 s−3 Metric multipliers Small or large quantities can be expressed in terms of units that are related to the basic ones by powers of 10. Thus, a nanometre (nm) is 10−9 m, a microgram (µg) is 10−6 g = 10−9 kg, a gigaelectron volt (GeV) equals 109 eV, etc. The most common prefixes are given in Table 1.1. Power Prefix Symbol Power Prefix Symbol 10−18 atto- A 101 deka- da −15 2 10 femto- F 10 hecto- h −12 3 10 pico- p 10 kilo- k −9 6 10 nano- n 10 mega- M 10 −6 micro- μ 10 9 giga- G −3 12 10 milli- m 10 tera- T 10−2 centi- c 1015 peta- P 10−1 deci- d 1018 exa- E Table 1.1 Common prefixes in the SI system. Orders of magnitude and estimates Expressing a quantity as a plain power of 10 gives what is called the order of magnitude of that quantity. Thus, the mass of the universe has an order of magnitude of 1053 kg and the mass of the Milky Way galaxy has an order of magnitude of 1041 kg. The ratio of the two masses is then simply 1012. Tables 1.2, 1.3 and 1.4 give examples of distances, masses and times, given as orders of magnitude. Length / m distance to edge of observable universe 1026 distance to the Andromeda galaxy 1022 diameter of the Milky Way galaxy 1021 distance to nearest star 1016 diameter of the solar system 1013 distance to the Sun 1011 radius of the Earth 107 size of a cell 10−5 size of a hydrogen atom 10−10 size of an A = 50 nucleus 10−15 size of a proton 10−15 Planck length 10−35 Table 1.2 Some interesting distances. 2  Mass / kg Time / s the universe 1053 age of the universe 1017 41 the Milky Way galaxy 10 age of the Earth 1017 the Sun 1030 time of travel by light to nearby star 108 the Earth 1024 one year 107 Boeing 747 (empty) 105 one day 105 an apple 0.2 period of a heartbeat 1 −6 a raindrop 10 lifetime of a pion 10–8 a bacterium 10−15 lifetime of the omega particle 10–10 smallest virus 10−21 time of passage of light across a proton 10–24 a hydrogen atom 10−27 Table 1.4 Some interesting times. −30 an electron 10 Table 1.3 Some interesting masses. Worked examples 1.1 Estimate how many grains of sand are required to fill the volume of the Earth. (This is a classic problem that goes back to Aristotle. The radius of the Earth is about 6 × 106 m.) The volume of the Earth is: 4 3 4 6 3 20 21 3 3πR ≈ 3 × 3 × (6 × 10 ) ≈ 8 × 10 ≈ 10 m The diameter of a grain of sand varies of course, but we will take 1 mm as a fair estimate. The volume of a grain of sand is about (1 × 10−3)3 m3. Then the number of grains of sand required to fill the Earth is: 1021 ≈ 1030 (1 × 10−3)3 1.2 Estimate the speed with which human hair grows. I have my hair cut every two months and the barber cuts a length of about 2 cm. The speed is therefore: 2 × 10−2 10−2 m s–1 ≈ 2 × 30 × 24 × 60 × 60 3 × 2 × 36 × 104 10−6 10−6 ≈ = 6 × 40 240 ≈ 4 × 10–9 m s–1 1 MEASUREMENT AND UNCERTAINTIES 3  1.3 Estimate how long the line would be if all the people on Earth were to hold hands in a straight line. Calculate how many times it would wrap around the Earth at the equator. (The radius of the Earth is about 6 × 106 m.) Assume that each person has his or her hands stretched out to a distance of 1.5 m and that the population of Earth is 7 × 109 people. Then the length of the line of people would be 7 × 109 × 1.5 m = 1010 m. The circumference of the Earth is 2πR ≈ 6 × 6 × 106 m ≈ 4 × 107 m. 1010 So the line would wrap ≈ 250 times around the equator. 4 × 107 1.4 Estimate how many apples it takes to have a combined mass equal to that of an ordinary family car. Assume that an apple has a mass of 0.2 kg and a car has a mass of 1400 kg. 1400 Then the number of apples is = 7 × 103. 0.2 1.5 Estimate the time it takes light to arrive at Earth from the Sun. (The Earth–Sun distance is 1.5 × 1011 m.) distance 1.5 × 1011 The time taken is = ≈ 0.5 × 104 = 500 s ≈ 8 min speed 3 × 108 Significant figures The number of digits used to express a number carries information about how precisely the number is known. A stopwatch reading of 3.2 s (two significant figures, s.f.) is less precise than a reading of 3.23 s (three s.f.). If you are told what your salary is going to be, you would like that number to be known as precisely as possible. It is less satisfying to be told that your salary will be ‘about 1000’ (1 s.f.) euro a month compared to a salary of ‘about 1250’ (3 s.f.) euro a month. Not because 1250 is larger than 1000 but because the number of ‘about 1000’ could mean anything from a low of 500 to a high of 1500.You could be lucky and get the 1500 but you cannot be sure. With a salary of ‘about 1250’ your actual salary could be anything from 1200 to 1300, so you have a pretty good idea of what it will be. How to find the number of significant figures in a number is illustrated in Table 1.5. 4  Number Number of s.f. Reason Scientific notation 504 3 in an integer all digits count (if last digit is 5.04 × 102 not zero) 608 000 3 zeros at the end of an integer do not count 6.08 × 105 200 1 zeros at the end of an integer do not count 2 × 102 0.000 305 3 zeros in front do not count 3.05 × 10−4 0.005 900 4 zeros at the end of a decimal count, those 5.900 × 10−3 in front do not Table 1.5 Rules for significant figures. Scientific notation means writing a number in the form a × 10b, where a is decimal such that 1 ≤ a < 10 and b is a positive or negative integer. The number of digits in a is the number of significant figures in the number. In multiplication or division (or in raising a number to a power or taking a root), the result must have as many significant figures as the least precisely known number entering the calculation. So we have that: 4  × 578=13 23  294 ≈ 1.3  × 10  (the least number of s.f. is shown in red) 2 s.f. 3 s.f. 2 s.f. 6.244  4 s.f. 0 =4.9952… ≈ 5.00  × 10  =5.00 1.25  3 s.f. 3 s.f. 3 3  =1860.867… ≈ 1.86 12.3  × 10  3 s.f. 3 s.f. 2    = 242.6932… ≈ 2.43 58900   × 10  3 s.f. 3 s.f. In adding and subtracting, the number of decimal digits in the answer must be equal to the least number of decimal places in the numbers added or subtracted. Thus: 3.21  =7.32 ≈ 7.3  + 4.1  (the least number of d.p. is shown in red) 2 d.p. 1 d.p. 1 d.p.  − 3.15=9.217 12.367  ≈ 9.22  3 d.p. 2 d.p. 2 d.p. Use the rules for rounding when writing values to the correct number of decimal places or significant figures. For example, the number 542.48 = 5.4248 × 102 rounded to 2, 3 and 4 s.f. becomes: 5.4|248 × 102 ≈ 5.4 × 102 rounded to 2 s.f. 5.42|48 × 102 ≈ 5.42 × 102 rounded to 3 s.f. 5.424|8 × 102 ≈ 5.425 × 102 rounded to 4 s.f. There is a special rule for rounding when the last digit to be dropped is 5 and it is followed only by zeros, or not followed by any other digit. 1 MEASUREMENT AND UNCERTAINTIES 5  This is the odd–even rounding rule. For example, consider the number 3.250 000 0… where the zeros continue indefinitely. How does this number round to 2 s.f.? Because the digit before the 5 is even we do not round up, so 3.250 000 0… becomes 3.2. But 3.350 000 0… rounds up to 3.4 because the digit before the 5 is odd. Nature of science Early work on electricity and magnetism was hampered by the use of different systems of units in different parts of the world. Scientists realised they needed to have a common system of units in order to learn from each other’s work and reproduce experimental results described by others. Following an international review of units that began in 1948, the SI system was introduced in 1960. At that time there were six base units. In 1971 the mole was added, bringing the number of base units to the seven in use today. As the instruments used to measure quantities have developed, the definitions of standard units have been refined to reflect the greater precision possible. Using the transition of the caesium-133 atom to measure time has meant that smaller intervals of time can be measured accurately. The SI system continues to evolve to meet the demands of scientists across the world. Increasing precision in measurement allows scientists to notice smaller differences between results, but there is always uncertainty in any experimental result. There are no ‘exact’ answers. ? Test yourself 1 How long does light take to travel across a proton? 9 Give an order-of-magnitude estimate of the 2 How many hydrogen atoms does it take to make density of a proton. up the mass of the Earth? 10 How long does light take to traverse the 3 What is the age of the universe expressed in diameter of the solar system? units of the Planck time? 11 An electron volt (eV) is a unit of energy equal to 4 How many heartbeats are there in the lifetime of 1.6 × 10−19 J. An electron has a kinetic energy of a person (75 years)? 2.5 eV. 5 What is the mass of our galaxy in terms of a solar a How many joules is that? mass? b What is the energy in eV of an electron that 6 What is the diameter of our galaxy in terms of has an energy of 8.6 × 10−18 J? the astronomical unit, i.e. the distance between 12 What is the volume in cubic metres of a cube of the Earth and the Sun (1 AU = 1.5 × 1011 m)? side 2.8 cm? 7 The molar mass of water is 18 g mol−1. How 13 What is the side in metres of a cube that has a many molecules of water are there in a glass of volume of 588 cubic millimetres? water (mass of water 300 g)? 14 Give an order-of-magnitude estimate for the 8 Assuming that the mass of a person is made up mass of: entirely of water, how many molecules are there a an apple in a human body (of mass 60 kg)? b this physics book c a soccer ball. 6  15 A white dwarf star has a mass about that of the 20 A block of mass 1.2 kg is raised a vertical distance Sun and a radius about that of the Earth. Give an of 5.55 m in 2.450 s. Calculate the power order-of-magnitude estimate of the density of a mgh delivered. (P = and g = 9.81 m s−2 ) white dwarf. t 21 Find the kinetic energy (EK = 12mv2 ) of a block of 16 A sports car accelerates from rest to 100 km per mass 5.00 kg moving at a speed of 12. 5 m s−1. hour in 4.0 s. What fraction of the acceleration 22 Without using a calculator, estimate the value due to gravity is the car’s acceleration? of the following expressions. Then compare 17 Give an order-of-magnitude estimate for the your estimate with the exact value found using a number of electrons in your body. calculator. 18 Give an order-of-magnitude estimate for the 243 ratio of the electric force between two electrons a 43 1 m apart to the gravitational force between the electrons. b 2.80 × 1.90 19 The frequency f of oscillation (a quantity with 480 c 312 × units of inverse seconds) of a mass m attached 160 to a spring of spring constant k (a quantity with 8.99 × 109 × 7 × 10−16 × 7 × 10−6 d units of force per length) is related to m and k. (8 × 102 )2 By writing f = cmxk y and matching units 6.6 × 10−11 × 6 × 1024 k e on both sides, show that f = c , where c is a (6.4 × 106)2 m dimensionless constant. 1.2 Uncertainties and errors Learning objectives This section introduces the basic methods of dealing with experimental Distinguish between random error and uncertainty in measured physical quantities. Physics is an and systematic uncertainties. experimental science and often the experimenter will perform an Work with absolute, fractional experiment to test the prediction of a given theory. No measurement will and percentage uncertainties. ever be completely accurate, however, and so the result of the experiment Use error bars in graphs. will be presented with an experimental error. Calculate the uncertainty in a gradient or an intercept. Types of uncertainty There are two main types of uncertainty or error in a measurement. They can be grouped into systematic and random, although in many cases it is not possible to distinguish clearly between the two. We may say that random uncertainties are almost always the fault of the observer, whereas systematic errors are due to both the observer and the instrument being used. In practice, all uncertainties are a combination of the two. Systematic errors A systematic error biases measurements in the same direction; the measurements are always too large or too small. If you use a metal ruler to measure length on a very hot day, all your length measurements will be too small because the metre ruler expanded in the hot weather. If you use an ammeter that shows a current of 0.1 A even before it is connected to 1 MEASUREMENT AND UNCERTAINTIES 7  a circuit, every measurement of current made with this ammeter will be larger than the true value of the current by 0.1 A. Suppose you are investigating Newton’s second law by measuring the acceleration of a cart as it is being pulled by a falling weight of mass m (Figure 1.1). Almost certainly there is a frictional force f between the cart and the table surface. If you forget to take this force into account, you would expect the cart’s acceleration a to be: m mg a= M Figure 1.1 The falling block accelerates the where M is the constant combined mass of the cart and the falling block. cart. The graph of the acceleration versus m would be a straight line through the origin, as shown by the red line in Figure 1.2. If you actually do the experiment, you will find that you do get a straight line, but not through the origin (blue line in Figure 1.2). There is a negative intercept on the vertical axis. a /m s–2 2.0 1.5 1.0 0.5 0.0 0.1 0.2 0.3 0.4 m / kg –0.5 –1.0 Figure 1.2 The variation of acceleration with falling mass with (blue) and without (red) frictional forces. This is because with the frictional force present, Newton’s second law predicts that: mg f a= − M M So a graph of acceleration a versus mass m would give a straight line with a negative intercept on the vertical axis. Systematic errors can result from the technique used to make a measurement. There will be a systematic error in measuring the volume of a liquid inside a graduated cylinder if the tube is not exactly vertical. The measured values will always be larger or smaller than the true value, depending on which side of the cylinder you look at (Figure 1.3a). There will also be a systematic error if your eyes are not aligned with the liquid level in the cylinder (Figure 1.3b). Similarly, a systematic error will arise if you do not look at an analogue meter directly from above (Figure 1.3c). Systematic errors are hard to detect and take into account. 8 a b c Figure 1.3 Parallax errors in measurements. Random uncertainties The presence of random uncertainty is revealed when repeated measurements of the same quantity show a spread of values, some too large some too small. Unlike systematic errors, which are always biased to be in the same direction, random uncertainties are unbiased. Suppose you ask ten people to use stopwatches to measure the time it takes an athlete to run a distance of 100 m. They stand by the finish line and start their stopwatches when the starting pistol fires.You will most likely get ten different values for the time. This is because some people will start/stop the stopwatches too early and some too late.You would expect that if you took an average of the ten times you would get a better estimate for the time than any of the individual measurements: the measurements fluctuate about some 21 22 23 24 25 26 27 value. Averaging a large number of measurements gives a more accurate estimate of the result. (See the section on accuracy and precision, overleaf.) We include within random uncertainties, reading uncertainties (which really is a different type of error altogether). These have to do with the precision with which we can read an instrument. Suppose we use a ruler to record the position of the right end of an object, Figure 1.4. 30 31 32 33 34 35 36 The first ruler has graduations separated by 0.2 cm. We are confident that the position of the right end is greater than 23.2 cm and smaller than 23.4 cm. The true value is somewhere between these bounds. The Figure 1.4 Two rulers with different average of the lower and upper bounds is 23.3 cm and so we quote the graduations. The top has a width between measurement as (23.3 ± 0.1) cm. Notice that the uncertainty of ± 0.1 cm graduations of 0.2 cm and the other 0.1 cm. is half the smallest width on the ruler. This is the conservative way of doing things and not everyone agrees with this. What if you scanned the diagram in Figure 1.4 on your computer, enlarged it and used your computer to draw further lines in between the graduations of the ruler. Then you could certainly read the position to better precision than the ± 0.1 cm. Others might claim that they can do this even without a computer or a scanner! They might say that the right end is definitely short of the 23.3 cm point. We will not discuss this any further – it is an endless discussion and, at this level, pointless. Now let us use a ruler with a finer scale. We are again confident that the position of the right end is greater than 32.3 cm and smaller than 32.4 cm. The true value is somewhere between these bounds. The average of the bounds is 32.35 cm so we quote a measurement of (32.35 ± 0.05) cm. Notice 1 MEASUREMENT AND UNCERTAINTIES 9  again that the uncertainty of ± 0.05 cm is half the smallest width on the ruler. This gives the general rule for analogue instruments: The uncertainty in reading an instrument is ± half of the smallest width of the graduations on the instrument. For digital instruments, we may take the reading error to be the smallest Instrument Reading error division that the instrument can read. So a stopwatch that reads time to ruler ± 0.5 mm two decimal places, e.g. 25.38 s, will have a reading error of ± 0.01 s, and a vernier calipers ± 0.05 mm weighing scale that records a mass as 184.5 g will have a reading error of micrometer ± 0.005 mm ± 0.1 g. Typical reading errors for some common instruments are listed in electronic weighing ± 0.1 g Table 1.6. scale stopwatch ± 0.01 s Accuracy and precision In physics, a measurement is said to be accurate if the systematic error Table 1.6 Reading errors for some common instruments. in the measurement is small. This means in practice that the measured value is very close to the accepted value for that quantity (assuming that this is known – it is not always). A measurement is said to be precise if the random uncertainty is small. This means in practice that when the measurement was repeated many times, the individual values were close to each other. We normally illustrate the concepts of accuracy and precision with the diagrams in Figure 1.5: the red stars indicate individual measurements. The ‘true’ value is represented by the common centre of the three circles, the ‘bull’s-eye’. Measurements are precise if they are clustered together. They are accurate if they are close to the centre. The descriptions of three of the diagrams are obvious; the bottom right clearly shows results that are not precise because they are not clustered together. But they are accurate because their average value is roughly in the centre. not accurate and not precise accurate and precise not accurate and not precise accurate and precise not accurate but precise accurate but not precise not accurate but precise accurate but not precise Figure 1.5 The meaning of accurate and precise measurements. Four different sets of four measurements each are shown. 10 Averages In an experiment a measurement must be repeated many times, if at all possible. If it is repeated N times and the results of the measurements are x1, x2, …, xN, we calculate the mean or the average of these values (x– ) using: x + x + … + xN x– = 1 2 N This average is the best estimate for the quantity x based on the N measurements. What about the uncertainty? The best way is to get the standard deviation of the N numbers using your calculator. Standard deviation will not be examined but you may need to use it for your Internal Assessment, so it is good idea to learn it – you will learn it in your mathematics class anyway. The standard deviation σ of the N measurements is given by the formula (the calculator finds this very easily): (x1 – x– )2 + (x2 – x– )2 + … + (xN – x– )2 σ= N–1 A very simple rule (not entirely satisfactory but acceptable for this course) is to use as an estimate of the uncertainty the quantity: xmax − xmin ∆x = 2 i.e. half of the difference between the largest and the smallest value. For example, suppose we measure the period of a pendulum (in seconds) ten times: 1.20, 1.25, 1.30, 1.13, 1.25, 1.17, 1.41, 1.32, 1.29, 1.30 We calculate the mean: t + t + … + t10 t– = 1 2 = 1.2620 s 10 and the uncertainty: tmax − tmin 1.41 − 1.13 ∆t = = = 0.140 s 2 2 Exam tip How many significant figures do we use for uncertainties? The general There is some case to be made rule is just one figure. So here we have ∆t = 0.1 s. The uncertainty is in the for using two significant figures first decimal place. The value of the average period must also be in the uncertainty when the expressed to the same precision as the uncertainty, i.e. here to one first digit in the uncertainty decimal place, t– = 1.3 s. We then state that: is 1. So in this example, since ∆t = 0.140 s does begin period = (1.3 ± 0.1) s with the digit 1, we should state ∆t = 0.14 s and quote (Notice that each of the ten measurements of the period is subject to a the result for the period as reading error. Since these values were given to two decimal places, it is ‘period = (1.26 ± 0.14) s’. implied that the reading error is in the second decimal place, say ± 0.01 s. 1 MEASUREMENT AND UNCERTAINTIES 11  This is much smaller than the uncertainty found above so we ignore the reading error here. If instead the reading error were greater than the error due to the spread of values, we would have to include it instead. We will not deal with cases when the two errors are comparable.) You will often see uncertainties with 2 s.f. in the scientific literature. For example, the charge of the electron is quoted as e = (1.602 176 565 ± 0.000 000 035) × 10−19 C and the mass of the electron as me = (9.109 382 91 ± 0.000 000 40) × 10−31 kg. This is perfectly all right and reflects the experimenter’s level of confidence in his/her results. Expressing the uncertainty to 2 s.f. implies a more sophisticated statistical analysis of the data than is normally done in a high school physics course. With a lot of data, the measured values of e form a normal distribution with a given mean (1.602 176 565 × 10−19 C) and standard deviation (0.000 000 035 × 10−19 C). The experimenter is then 68% confident that the measured value of e lies within the interval [1.602 176 530 × 10−19 C, 1.602 176 600 × 10−19 C]. Worked example 1.6 The diameter of a steel ball is to be measured using a micrometer caliper. The following are sources of error: 1 The ball is not centred between the jaws of the caliper. 2 The jaws of the caliper are tightened too much. 3 The temperature of the ball may change during the measurement. 4 The ball may not be perfectly round. Determine which of these are random and which are systematic sources of error. Sources 3 and 4 lead to unpredictable results, so they are random errors. Source 2 means that the measurement of diameter is always smaller since the calipers are tightened too much, so this is a systematic source of error. Source 1 certainly leads to unpredictable results depending on how the ball is centred, so it is a random source of error. But since the ball is not centred the ‘diameter’ measured is always smaller than the true diameter, so this is also a source of systematic error. Propagation of uncertainties A measurement of a length may be quoted as L = (28.3 ± 0.4) cm. The value 28.3 is called the best estimate or the mean value of the measurement and the 0.4 cm is called the absolute uncertainty in the measurement. The ratio of absolute uncertainty to mean value is called the fractional uncertainty. Multiplying the fractional uncertainty by 100% gives the percentage uncertainty. So, for L = (28.3 ± 0.4) cm we have that: absolute uncertainty = 0.4 cm 0.4 fractional uncertainty = 28.3 = 0.0141 percentage uncertainty = 0.0141 × 100% = 1.41% 12 In general, if a = a0 ± ∆a, we have: The subscript 0 indicates the mean absolute uncertainty = ∆a ∆a value, so a0 is the mean value of a. fractional uncertainty = a0 ∆a percentage uncertainty = a0 × 100% Suppose that three quantities are measured in an experiment: a = a0 ± ∆a, b = b0 ± ∆b, c = c0 ± ∆c. We now wish to calculate a quantity Q in terms of a, b, c. For example, if a, b, c are the sides of a rectangular block we may want to find Q = ab, which is the area of the base, or Q = 2a + 2b, which is the perimeter of the base, or Q = abc, which is the volume of the block. Because of the uncertainties in a, b, c there will be an uncertainty in the calculated quantities as well. How do we calculate this uncertainty? There are three cases to consider. We will give the results without proof. Addition and subtraction The first case involves the operations of addition and/or subtraction. For example, we might have Q = a + b or Q = a − b or Q = a + b − c. Then, in all cases the absolute uncertainty in Q is the sum of the absolute uncertainties in a, b and c. Exam tip Q=a+b ⇒ ∆Q = ∆a + ∆b In addition and subtraction, Q=a−b ⇒ ∆Q = ∆a + ∆b we always add the absolute Q=a+b−c ⇒ ∆Q = ∆a + ∆b + ∆c uncertainties, never subtract. Worked examples 1.7 The side a of a square, is measured to be (12.4 ± 0.1) cm. Find the perimeter P of the square including the uncertainty. Because P = a + a + a + a, the perimeter is 49.6 cm. The absolute uncertainty in P is: ∆P = ∆a + ∆a + ∆a + ∆a ∆P = 4∆a ∆P = 0.4 cm Thus, P = (49.6 ± 0.4) cm. 1.8 Find the percentage uncertainty in the quantity Q = a − b, where a = 538.7 ± 0.3 and b = 537.3 ± 0.5. Comment on the answer. The calculated value is 1.7 and the absolute uncertainty is 0.3 + 0.5 = 0.8. So Q = 1.4 ± 0.8. 0.8 The fractional uncertainty is = 0.57, so the percentage uncertainty is 57%. 1.4 The fractional uncertainty in the quantities a and b is quite small. But the numbers are close to each other so their difference is very small. This makes the fractional uncertainty in the difference unacceptably large. 1 MEASUREMENT AND UNCERTAINTIES 13  Multiplication and division The second case involves the operations of multiplication and division. Here the fractional uncertainty of the result is the sum of the fractional uncertainties of the quantities involved: ∆Q ∆a ∆b Q = ab ⇒ = + Q0 a0 b0 a ∆Q ∆a ∆b Q= ⇒ = + b Q0 a0 b0 ab ∆Q ∆a ∆b ∆c Q= ⇒ = + + c Q0 a0 b0 c0 Powers and roots The third case involves calculations where quantities are raised to powers or roots. Here the fractional uncertainty of the result is the fractional uncertainty of the quantity multiplied by the absolute value of the power: ∆Q ∆a Q = an ⇒ = |n| Q0 a0 n ∆Q 1 ∆a Q = √a ⇒ = Q0 n a0 Worked examples 1.9 The sides of a rectangle are measured to be a = 2.5 cm ± 0.1 cm and b = 5.0 cm ± 0.1 cm. Find the area A of the rectangle. The fractional uncertainty in a is: ∆a 0.1 = = 0.04 or 4% a 2.5 The fractional uncertainty in b is: ∆b 0.1 = = 0.02 or 2% b 5.0 Thus, the fractional uncertainty in the area is 0.04 + 0.02 = 0.06 or 6%. The area A0 is: A0 = 2.5 × 5.0 = 12.5 cm2 ∆A and = 0.06 A0 ⇒ ∆A = 0.06 ×12.5 = 0.75 cm2 Hence A = 12.5 cm2 ± 0.8 cm2 (the final absolute uncertainty is quoted to 1 s.f.). 14 1.10 A mass is measured to be m = 4.4 ± 0.2 kg and its speed v is measured to be 18 ± 2 m s−1. Find the kinetic energy of the mass. The kinetic energy is E = 12mv2, so the mean value of the kinetic energy, E0, is: E0 = 12 × 4.4 × 182 = 712.8 J Using: ∆E ∆m 2× ∆v = + E0 m0 v0 because of the square we find: ∆E0.2 2 = 2 × = 0.267 = 712.8 4.4 18 So: ∆E = 712.8 × 0.2677 = 190.8 J Exam tip To one significant figure, the uncertainty The final absolute uncertainty must be expressed to one is ∆E = 200 = 2 × 102 J; that is E = (7 ± 2) × 102 J. significant figure. This limits the precision of the quoted value for energy. 1.11 The length of a simple pendulum is increased by 4%. What is the fractional increase in the pendulum’s period? L The period T is related to the length L through T = 2π. g Because this relationship has a square root, the fractional uncertainties are related by: ∆T 1 × ∆L = T0 2 L0 because of the square root ∆L We are told that = 4%. This means we have : L0 ∆T 1 = × 4% = 2% T0 2 1 MEASUREMENT AND UNCERTAINTIES 15  1 1.12 A quantity Q is measured to be Q = 3.4 ± 0.5. Calculate the uncertainty in a and b Q2. Q 1 1 a = = 0.294 118 Q 3.4 ∆(1/Q) ∆Q = 1/Q Q ∆Q 0.5 ⇒ ∆(1/Q) = 2 = = 0.043 25 Q 3.42 1 Hence: = 0.29 ± 0.04 Q b Q2 = 3.42 = 11.5600 ∆(Q2 ) ∆Q =2× Q 2 Q ⇒ ∆(Q2 ) = 2Q × ∆Q = 2 × 3.4 × 0.5 = 3.4 Hence: Q2 = 12 ± 3 1.13 The volume of a cylinder of base radius r and height h is given by V = πr2h. The volume is measured with an uncertainty of 4% and the height with with an uncertainty of 2%. Determine the uncertainty in the radius. V We must first solve for the radius to get r =. The uncertainty is then: πh ∆r r × 100% = 2 V ( 1 ∆V ∆h + h 1 ) × 100% = (4 + 2) × 100% = 3% 2 Best-fit lines In mathematics, plotting a point on a set of axes is straightforward. In physics, it is slightly more involved because the point consists of measured or calculated values and so is subject to uncertainty. So the point (x0 ± ∆x, y0 ± ∆y) is plotted as shown in Figure 1.6. The uncertainties are y 2∆ x y0 + ∆ y y0 2∆ y y0 – ∆ y 0 x0 – ∆ x x0 x0 + ∆ x x Figure 1.6 A point plotted along with its error bars. 16 represented by error bars. To ‘go through the error bars’ a best-fit line can go through the area shaded grey. In a physics experiment we usually try to plot quantities that will give straight-line graphs. The graph in Figure 1.7 shows the variation with extension x of the tension T in a spring. The points and their error bars are plotted. The blue line is the best-fit line. It has been drawn by eye by trying to minimise the distance of the points from the line – this means that some points are above and some are below the best-fit line. The gradient (slope) of the best-fit line is found by using two points on the best-fit line as far from each other as possible. We use (0, 0) and (0.0390, 7.88). The gradient is then: ∆F gradient = ∆x 7.88 − 0 gradient = 0.0390 – 0 gradient = 202 N m−1 The best-fit line has equation F = 202x. (The vertical intercept is essentially zero; in this equation x is in metres and F in newtons.) F/N 8 7 6 5 4 ∆F 3 2 1 0 1 2 3 4 ∆x x /cm Figure 1.7 Data points plotted together with uncertainties in the values for the tension. To find the gradient, use two points on the best-fit line far apart from each other. 1 MEASUREMENT AND UNCERTAINTIES 17  On the other hand it is perfectly possible to obtain data that cannot be easily manipulated to give a straight line. In that case a smooth curve passing through all the error bars is the best-fit line (Figure 1.8). From the graph the maximum power is 4.1 W, and it occurs when R = 2.2 Ω. The estimated uncertainty in R is about the length of a square, i.e. ± 0.1 Ω. Similarly, for the power the estimated uncertainty is ± 0.1 W. P/W 5 4 3 2 1 0 0 1 2 3 4 R /Ω Figure 1.8 The best-fit line can be a curve. Uncertainties in the gradient and intercept When the best-fit line is a straight line we can easily obtain uncertainties in the gradient and the vertical intercept. The idea is to draw lines of maximum and minimum gradient in such a way that they go through all the error bars (not just the ‘first’ and the ‘last’ points). Figure 1.9 shows the best-fit line (in blue) and the lines of maximum and minimum gradient. The green line is the line through all error bars of greatest gradient. The red line is the line through all error bars with smallest gradient. All lines are drawn by eye. The blue line has gradient kmax = 210 N m−1 and intercept −0.18 N. The red line has gradient kmin = 193 N m−1 and intercept +0.13 N. So we can find the uncertainty in the gradient as: kmax − kmin 210 − 193 ∆k = = = 8.5 ≈ 8 Nm−1 2 2 18  F/N 8 7 6 5 4 3 2 1 0 1 2 3 4 x /cm Figure 1.9 The best-fit line, along with lines of maximum and minimum gradient. The uncertainty in the vertical intercept is similarly: 0.13 − (−0.18) ∆intercept = = 0.155 ≈ 0.2 N 2 We saw earlier that the line of best fit has gradient 202 N m−1 and zero intercept. So we quote the results as k = (2.02 ± 0.08) ×102 and intercept = 0.0 ± 0.2 N. Nature of science A key part of the scientific method is recognising the errors that are present in the experimental technique being used, and working to reduce these as much as possible. In this section you have learned how to calculate errors in quantities that are combined in different ways and how to estimate errors from graphs.You have also learned how to recognise systematic and random errors. No matter how much care is taken, scientists know that their results are uncertain. But they need to distinguish between inaccuracy and uncertainty, and to know how confident they can be about the validity of their results. The search to gain more accurate results pushes scientists to try new ideas and refine their techniques. There is always the possibility that a new result may confirm a hypothesis for the present, or it may overturn current theory and open a new area of research. Being aware of doubt and uncertainty are key to driving science forward. 1 MEASUREMENT AND UNCERTAINTIES 19  ? Test yourself 23 The magnitudes of two forces are measured to 31 In a similar experiment to that in question 30, be 120 ± 5 N and 60 ± 3 N. Find the sum and the following data was collected for current difference of the two magnitudes, giving the and voltage: (V, I ) = {(0.1, 27), (0.2, 44), (0.3, uncertainty in each case. 60), (0.4, 78)} with an uncertainty of ± 4 mA in 24 The quantity Q depends on the measured values the current. Plot the current versus the voltage a and b in the following ways: and draw the best-fit line. Suggest whether the a current is proportional to the voltage. a Q = , a = 20 ± 1, b = 10 ± 1 b 32 A circle and a square have the same perimeter. b Q = 2a + 3b, a = 20 ± 2, b = 15 ± 3 Which shape has the larger area? c Q = a − 2b, a = 50 ± 1, b = 24 ± 1 33 The graph shows the natural logarithm of d Q = a2, a = 10.0 ± 0.3 the voltage across a capacitor of capacitance a2 e Q = 2 , a = 100 ± 5, b = 20 ± 2 C = 5.0 µF as a function of time. The voltage is b given by the equation V = V0 e−t/RC, where R is In each case, find the value of Q and its the resistance of the circuit. Find: uncertainty. 2 a the initial voltage 25 The centripetal force is given by F = mvr. The b the time for the voltage to be reduced to half mass is measured to be 2.8 ± 0.1 kg, the velocity its initial value 14 ± 2 m s−1 and the radius 8.0 ± 0.2 m; find the c the resistance of the circuit. force on the mass, including the uncertainty. 26 The radius r of a circle is measured to be ln V 4.0 2.4 cm ± 0.1 cm. Find the uncertainty in: 3.5 a the area of the circle 3.0 b the circumference of the circle. 27 The sides of a rectangle are measured as 2.5 4.4 ± 0.2 cm and 8.5 ± 0.3 cm. Find the area and 2.0 perimeter of the rectangle. 0 5 10 15 20 28 The length L of a pendulum is increased by 2%. t /s Find the percentage increase in the period T. 34 The table shows the mass M of several stars and L their corresponding luminosity L (power emitted). T = 2π g a Plot L against M and draw the best-fit line. 29 The volume of a cone of base radius R and 2 b Plot the logarithm of L against the logarithm height h is given by V = πR3 h. The uncertainty of M. Use your graph to find the relationship in the radius and in the height is 4%. Find the between these quantities, assuming a power percentage uncertainty in the volume. law of the kind L = kMα. Give the numerical 30 In an experiment to measure current and voltage value of the parameter α. across a device, the following data was collected: (V, I ) = {(0.1, 26), (0.2, 48), (0.3, 65), (0.4, 90)}. Mass M (in solar Luminosity L (in terms The current was measured in mA and the masses) of the Sun’s luminosity) voltage in mV. The uncertainty in the current 1.0 ± 0.1 1±0 was ± 4 mA. Plot the current versus the voltage 3.0 ± 0.3 42 ± 4 and draw the best-fit line through the points. Suggest whether the current is proportional to 5.0 ± 0.5 230 ± 20 the voltage. 12 ± 1 4700 ± 50 20 ± 2 26 500 ± 300 20 1.3 Vectors and scalars Learning objectives Quantities in physics are either scalars (i.e. they just have magnitude) or vectors (i.e. they have magnitude and direction). This section provides the Distinguish between vector and scalar quantities. tools you need for dealing with vectors. Resolve a vector into its components. Vectors Reconstruct a vector from its Some quantities in physics, such as time, distance, mass, speed and components. temperature, just need one number to specify them. These are called Carry out operations with scalar quantities. For example, it is sufficient to say that the mass of a vectors. body is 64 kg or that the temperature is −5.0 °C. On the other hand, many quantities are fully specified only if, in addition to a number, a direction is needed. Saying that you will leave Paris now, in a train moving Vectors Scalars at 220 km/h, does not tell us where you will be in 30 minutes because we displacement distance do not know the direction in which you will travel. Quantities that need velocity speed a direction in addition to magnitude are called vector quantities. Table 1.7 gives some examples of vector and scalars. acceleration mass A vector is represented by a straight arrow, as shown in Figure 1.10a. force time The direction of the arrow represents the direction of the vector and the weight density length of the arrow represents the magnitude of the vector. To say that electric field electric potential two vectors are the same means that both magnitude and direction are magnetic field electric charge the same. The vectors in Figure 1.10b are all equal to each other. In other gravitational field gravitational words, vectors do not have to start from the same point to be equal. potential We write vectors as italic boldface a. The magnitude is written as |a| momentum temperature or just a. area volume angular velocity work/energy/power Table 1.7 Examples of vectors and scalars. a b Figure 1.10 a Representation of vectors by arrows. b These three vectors are equal to each other. Multiplication of a vector by a scalar A vector can be multiplied by a number. The vector a multiplied by the 2a number 2 gives a vector in the same direction as a but 2 times longer. The vector a multiplied by −0.5 is opposite to a in direction and half as long –0.5a (Figure 1.11). The vector −a has the same magnitude as a but is opposite a in direction. Figure1.11 Multiplication of vectors by a scalar. 1 MEASUREMENT AND UNCERTAINTIES 21  Addition of vectors Figure 1.12a shows vectors d and e. We want to find the vector that equals d + e. Figure 1.12b shows one method of adding two vectors. e e d+e d+e O e d d d a b c Figure 1.12 a Vectors d and e. b Adding two vectors involves shifting one of them parallel to itself so as to form a parallelogram with the two vectors as the two sides. The diagonal represents the sum. c An equivalent way to add vectors. To add two vectors: 1 Draw them so they start at a common point O. 2 Complete the parallelogram whose sides are d and e. 3 Draw the diagonal of this parallelogram starting at O. This is the vector d + e. Equivalently, you can draw the vector e so that it starts where the vector d stops and then join the beginning of d to the end of e, as shown in Figure 1.12c. Exam tip Figure 1.13 Vectors (with arrows pointing in the same sense) forming closed polygons add up to zero. Subtraction of vectors Exam tip Figure 1.14 shows vectors d and e. We want to find the vector that equals The change in a quantity, and d − e. in particular the change in a To subtract two vectors: vector quantity, will follow us 1 Draw them so they start at a common point O. through this entire course.You 2 The vector from the tip of e to the tip of d is the vector d − e. need to learn this well. (Notice that is equivalent to adding d to −e.) e d–e e d O e d –e –e d–e d a b c Figure 1.14 Subtraction of vectors. 22 Worked examples 1.14 Copy the diagram in Figure 1.15a. Use the diagram to draw the third force that will keep the point P in equilibrium. P a b Figure 1.15 We find the sum of the two given forces using the parallelogram rule and then draw the opposite of that vector, as shown in Figure 1.15b. 1.15 A velocity vector of magnitude 1.2 m s−1 is horizontal. A second velocity vector of magnitude 2.0 m s−1 must be added to the first so that the sum is vertical in direction. Find the direction of the second vector and the magnitude of the sum of the two vectors. We need to draw a scale diagram, as shown in Figure 1.16. Representing 1.0 m s−1 by 2.0 cm, we see that the 1.2 m s−1 corresponds to 2.4 cm and 2.0 m s−1 to 4.0 cm. First draw the horizontal vector. Then mark the vertical direction from O. Using a compass (or a ruler), mark a distance of 4.0 cm from A, which intersects the vertical line at B. AB must be one of the sides of the parallelogram we are looking for. Now measure a distance of 2.4 cm horizontally from B to C and join O to C. This is the direction in which the second velocity vector must be pointing. Measuring the diagonal OB (i.e. the vector representing the sum), we find 3.2 cm, which represents 1.6 m s−1. Using a protractor, we find that the 2.0 m s−1 velocity vector makes an angle of about 37° with the vertical. C B A O Figure 1.16 Using a scale diagram to solve a vector problem. 1 MEASUREMENT AND UNCERTAINTIES 23  1.16 A person walks 5.0 km east, followed by 3.0 km north and then another 4.0 km east. Find their final position. The walk consists of three steps. We may represent each one by a vector (Figure 1.17). The first step is a vector of magnitude 5.0 km directed east (OA). The second is a vector of magnitude 3.0 km directed north (AB). Vectors corresponding to line The last step is represented by a vector of 4.0 km directed east (BC). segments are shown as bold The person will end up at a place that is given by the vector sum of capital letters, for example these three vectors, that is OA + AB + BC, which equals the vector OC. OA. The magnitude of the By measurement from a scale drawing, or by simple geometry, the distance vector is the length OA from O to C is 9.5 km and the angle to the horizontal is 18.4°. and the direction is from O towards A. B 4 km C 3 km O 5 km A Figure 1.17 Scale drawing using 1 cm = 1 km. 1.17 A body moves in a circle of radius 3.0 m with a constant speed of 6.0 m s−1. The velocity vector is at all times tangent to the circle. The body starts at A, proceeds to B and then to C. Find the change in the velocity vector between A and B and between B and C (Figure 1.18). vA A B vC C vB Figure 1.18 For the velocity change from A to B we have to find the difference vB − vA. and for the velocity change from B to C we need to find vC − vB. The vectors are shown in Figure 1.19. vA vC vB vB vB – vA vC – vB Figure 1.19 24  The vector vB − vA is directed south-west and its magnitude is (by the Pythagorean theorem): vA2 + vB2 = 62 + 62 = √72 = 8.49 m s−1 The vector vC − vB has the same magnitude as vB − vA but is directed north-west. Components of a vector Suppose that we use perpendicular axes x and y and draw vectors on this x–y plane. We take the origin of the axes as the starting point of the vector. (Other vectors whose beginning points are not at the origin can be shifted parallel to themselves until they, too, begin at the origin.) Given a vector a we define its components along the axes as follows. From the tip of the vector draw lines parallel to the axes and mark the point on each axis where the lines intersect the axes (Figure 1.20). y y y y A A θ θ y-component θ θ φ 0 0 x x x x 0 0 φ φ x-component A A Figure 1.20 The components of a vector A and the angle needed to calculate the components. The angle θ is measured counter-clockwise from the positive x-axis. The x- and y-components of A are called Ax and Ay. They are given by: Exam tip Ax = A cos θ The formulas given for the components of a vector can Ay = A sin θ always be used, but the angle where A is the magnitude of the vector and θ is the angle between the must be the one defined in vector and the positive x-axis. These formulas and the angle θ defined Figure 1.20, which is sometimes as shown in Figure 1.20 always give the correct components with the awkward.You can use other correct signs. But the angle θ is not always the most convenient. A more more convenient angles, but convenient angle to work with is φ, but when using this angle the signs then the formulas for the have to be put in by hand. This is shown in Worked example 1.18. components may change. 1 MEASUREMENT AND UNCERTAINTIES 25  Worked examples 1.18 Find the components of the vectors in Figure 1.21. The magnitude of a is 12.0 units and that of b is 24.0 units. y x 45° 30° a b Figure 1.21 Taking the angle from the positive x-axis, the angle for a is θ = 180° + 45° = 225° and that for b is θ = 270° + 60° = 330°. Thus: ax = 12.0 cos 225° bx = 24.0 cos 330° ax = −8.49 bx = 20.8 ay = 12.0 sin 225° by = 24.0 sin 330° ay = −8.49 by = −12.0 But we do not have to use the awkward angles of 225° and 330°. For vector a it is better to use the angle of φ = 45°. In that case simple trigonometry gives: ax = −12.0 cos 45° = −8.49 and ay = −12.0 sin 45° = −8.49 ↑ ↑ put in by hand put in by hand For vector b it is convenient to use the angle of φ = 30°, which is the angle the vector makes with the x-axis. But in this case: bx = 24.0 cos 30° = 20.8 and by = −24.0 sin 30° = −12.0 ↑ put in by hand 26  1.19 Find the components of the vector W along the axes shown in Figure 1.22. θ W Figure 1.22 See Figure 1.23. Notice that the angle between the vector W and the negative y-axis is θ. y-axis x-axis Then by simple trigonometry Wx = −W sin θ (Wx is opposite the angle θ so the sine is used) Wx Wy = −W cos θ (Wy is adjacent to the angle θ so the cosine is used) Wy θ θ (Both components are along the negative axes, so a minus sign has been put in by hand.) W Figure 1.23 Reconstructing a vector from its components Knowing the components of a vector allows us to reconstruct it (i.e. to find the magnitude and direction of the vector). Suppose that we are given that the x- and y-components of a vector are Fx and Fy. We need to find the magnitude of the vector F and the angle (θ ) it makes with the x-axis (Figure 1.24). The magnitude is found by using the Pythagorean theorem and the angle by using the definition of tangent. Fy F = Fx2 + Fy2, θ = arctan Fx As an example, consider the vector whose components are Fx = 4.0 and Fy = 3.0. The magnitude of F is: F = Fx2 + Fy2 = 4.02 + 3.02 = 25 = 5.0 y y Fy F θ x x Fx Figure 1.24 Given the components of a vector we can find its magnitude and direction. 1 MEASUREMENT AND UNCERTAINTIES 27  y y and the direction is found from: Fy 3 Fx Fx θ = arctan = arctan = 36.87° ≈ 37° x x Fx 4 63.43° Here is another example. We need to find the magnitude and direction of Fy Fy the vector with components Fx = −2.0 and Fy = −4.0. The vector lies in the third quadrant, as shown in Figure 1.25. Figure 1.25 The vector is in the third The magnitude is: quadrant. F = Fx2 + Fy2 = (−2.0)2 + (−4.0)2 = 20 = 4.47 ≈ 4.5 The direction is found from: Fy −4 φ = arctan = arctan = arctan 2