O X F O R D I B P R E P A R E D PH YSI C S I B D I P L O M A David Homer P R O G R A M M E O X F O R D I B P R E P A R E D PH YSI C S I B D I P L O M A P R O G R A M M E David Homer 1 M16 SL P3 Sa mple 3 T Z1 SL P3 st udent Great Clarendon Street, Oxford, OX2 6DP, United Press is a department of the University of Oxford. the University’s objective of excellence in research, education by publishing worldwide. 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Sto c k M m a xer/Shut ter sto c k ; Shut ter sto c k ; p167: a n s wer, T Z2 M12 a n s wer, M11 HL Q20, c r ed it s: C over a n s wer, P3 st udent Q10 (a),( b); st udent a n s wer, Q19; P rac t ice 3, Department, Ac k nowledgement s Photo p172: HL M12 P rac t ice p231: Q2 4, Printed a n s wer, Q15 (a),( b);Sa mple E1( b),(c); You Q4; HL P3 C .4. 2, p2 0 0: M18 p19 9: Q21(a),( b); a n s wer, T Z2 HL Sa mple P3 p213: Q2 4 (a),( b); M16 pap er 1, Q17, Sa mple p228: 3 No permission permitted Ad apted P3 2019 reserved. prior Sa mple st udent f r om Ad apted All T Z2 Sa mple st udent 978 SL SL 2019 HL ISBN Q6; a n s wer, M17 st udent First P3 st udent Ad apted The st udent M16 p145: countries C . 2 . 2, © 1, M16 mark Q11; of Sa mple p156: p148: scholarship, p177: and pr oblem a n s wer, It st udent furthers Q2 (a); a n s wer, T Z2 Sa mple University P rac t ice Kingdom M17 Oxford Q2; st udent P2 P1 2, M10 P rac t ice p239: P rac t ice p232: T Z2 pap er T Z2 pap er M10 p229: P rac t ice M17 P rac t ice p238: Q28, Q29; Q7, A6 ( b); p23 5: 1, T Z2 3, P3 P3 3, P3 M12 Q16, I3; E2; T Z1 pap er T Z2 pap er Q5; Q21, M10 3, P3 p237: P rac t ice p23 5: pap er M10 T Z2 pap er M12 P rac t ice P1 Q3, A 5 ( b); Q3, P rac t ice p238: 2, T Z2 pap er 3, 3, P3 C ontents Introduction 1 iv Measurements and uncertainties 1.1 Measurements in physics 2 1.2 Uncer tainties and errors 5 1.3 Vectors and scalars 7 10 Fields (AHL) 10.1 Describing elds 10 6 10.2 Fields at work 109 11 Electromagnetic induction (AHL) 2 11.1 Mechanics 2.1 Motion Electromagnetic induction 116 11.2 Power generation and transmission 118 11.3 Capacitance 122 10 2.2 Forces 13 2.3 Work , energy and power 17 2.4 Momentum 21 12 Quantum and nuclear physics (AHL) 3 Thermal physics 3.1 Temperature and energy changes 25 3.2 Modelling a gas 28 12.1 The interaction of matter with radiation 128 12.2 Nuclear physics 133 13 Data-based and practical questions (Section A ) 4 A 4.1 Oscillations 34 4.2 Travelling waves 36 4.3 4.4 4.5 5 Wave characteristics Wave behaviour Standing waves 40 43 47 Relativity A.1 Beginnings of relativity 146 A.2 Lorentz transformations 148 A.3 Spacetime diagrams 152 A.4 Relativistic mechanics (AHL) 156 A.5 General relativity (AHL) 158 Electricity and magnetism B 5.1 Electric elds 52 5.2 Heating eect of an electric current 55 5.3 Electric cells 59 5.4 Magnetic eects of electric currents 61 6 140 Oscillations and waves Engineering physics B.1 Rigid bodies and rotational dynamics 164 B.2 Thermodynamics 168 B.3 Fluids and uid dynamics (AHL) B.4 Forced vibrations and resonance (AHL) 174 178 Circular motion and gravity C 6.1 Circular motion 66 6.2 Newton’s law of gravitation 68 7 C.1 Introduction to imaging 182 C.2 Imaging instrumentation 188 C.3 Fibre optics 193 C.4 Medical imaging (AHL) 19 6 Atomic, nuclear and par ticle physics 7.1 Discrete energy and radioactivity 72 7.2 Nuclear reactions 76 7.3 The structure of matter 78 D 8 Astrophysics D.1 Stellar quantities 202 D.2 Stellar characteristics and stellar evolution 205 D.3 Cosmology 210 D.4 Stellar processes (AHL) Energy production 8.1 Energy sources 84 8.2 Thermal energy transfer 88 9 Imaging D.5 214 Fur ther cosmology (AHL) 217 Wave phenomena (AHL) 9.1 Simple harmonic motion 92 9.2 Single-slit diraction 96 9.3 Interference 9.4 Resolution 101 9.5 The Doppler eect 102 97 Internal assessment 221 Practice exam papers 226 Index 241 Answers to questions and exam papers in this book can be found on your free suppor t website. Access the suppor t website here: w w w.ox f or dsecondar y.com / ib-pr epar ed-suppor t iii I N T R O D U CT I O N This book syllabus provides in full Physics coverage and offers of the support IB diploma Over view of the book structure to The students preparing for their examinations. the book will help you revise the study book material, essential terms and concepts, internal skills and strengthen improve your IB examinations. worked best All examples practices topics answers explain are to and A separate book exam warn marks section by from may is questions, is packed tips against illustrated questions why practical The and dedicated are errors. to or most The core topics, four SL and cover additional data-based options examination largest section structure and which missed. data-based the (AHL) the practice the student examinations, scored that (A–D) and and a practical complete set papers. with common past sections approach demonstrate annotated be which that several your of to into assessment, level questions, problem-solving divided learn higher the is The covers skills of all 9–12 nature of book, diploma statements. material topics the IB understandings assessment common of the are for SL for and topics, HL HL follows syllabus applications Topics and intended core physics 1–8 and contain students, students while only . The and science concepts are also mentioned where distinctive applicable. feature of how tackle and to the analyse syllabus. papers your and to a provide complete knowledge and monitor Full solutions data, experimental further and the set all and skills, IB-style boost of problems data-based (Chapter Practice examination to your your The suggest procedures. of show interpret opportunities progress to examples situations, experimental improvements problems Numerous unfamiliar check and appear section studies. and sample papers are a is A of paper 3. by followed of that Similar to worked by section analysis experiments illustrated scripts, questions detailed laboratory discussion often core topics, examples IB-style practice problems. and The examination in practical provides problems the confidence 13) and given online options section reviews the material assessed in at the second part of paper 3. Each of the four options www.oxfordsecondary.com/ib-prepared-support. is As any study guide, th is book is not presented The to replace your cou rs e ma teria ls , s uch laboratory m an ua ls, p a st internal pa pe r s notes. To the IB Ph ysics succeed in sy lla bu s th e and of need to use a broa d of of which are a va ilable this that this critical bo ok part preparation for of w ill y our online. th e na viga te studies, ex a m les s The you stres sf u l collect to more final this and higher and and section examination how process present level the to will select a the have to suitable experimental your marking at the These to test same for contains papers book. problems and outlines you report in data, a draw suitable criteria and achieve grade. opportunity (SL) explains satisfy highest The your and DP Physics assessment level that t h rough ma kin g for standard section a u t h or efficient. All subtopics. res ou rc e s , the hopes AHL you format many and y ou r ex amin a ti on, ra ng e SL investigation and conclusions will of assessment the out topic, own series a nd carry markschemes, a as nature textbooks, as in t e n d e d 1, IB-style and papers of practice written give before provide topic 3, will yourself time every 2 exclusively you the actual additional core and an exam practice options (HL) material. students must complete and three papers take assessment. day and external in the DP iv Papers Paper 3 a 1 table Physics at the and or of from internal part 2 assessment their external usually later. are page 1 of are two marks top grade, as day assessment the sat The V to give to 7 one internal combined (lowest) on as and shown your overall (highest). The answers and examination and solutions papers to are all practice given online problems at www.oxfordsecondary.com/ib-prepared-support. Blank answer available at sheets the for same examination address. papers are also Assessment over view SL Assessment Description HL Topics marks Internal Experimental work with a written repor t Paper 1 Multiple-choice questions — weight 24 Core: marks 20% 30 weight 24 20% 20% 40 20% 1–8 (SL) Paper 2 Shor t- and extended-response questions 50 40% 90 36% 1–12 (AHL) Section A : data-based and practical — 15 15 questions Paper 3 20% Section B: shor t- and extended-response 24% Option of your 20 questions The final points IB from diploma theory of 30 choice score is calculated knowledge and by combining extended essay grades for six subjects with up to three additional components. Command terms Command terms command For example, levels of are term the detail, (“discuss”), as pre-defined specifies the command from a shown and terms single in words type the state, word, next and phrases depth of outline, short the used explain sentence in all response or and IB Physics expected discuss numerical questions from require value you in answers (“state”) to and a problems. particular with Each question. increasingly comprehensive higher analysis table. Question Answer guidance mass of an object × its velocity State what is meant by momentum. momentum is a vector quantity Outline why the kinetic model of an ideal gas Empirical implies that the result follows from an experiment. The kinetic model is is an example of a theoretical model rather theoretical and follows from a series of assumptions. than an empirical model. Explain why there is a force acting on a garden The answer should consider the momentum of the system when water is leaving hose when water is moving through the hose. the hose. A boy throws a ball ver tically into the air and The answer will need to: later catches it. • state Newton’s third law of motion • consider the action-reaction pairs involved Discuss, with reference to Newton’s third law, the changes in velocity that occur to the ball and the Ear th at the instants when the ball is • released and caught. A list of commonly Understanding examination. questions in the used exact Therefore, this discuss the magnitude and direction of the velocities when the ball is thrown and when it is caught. command meaning you terms of should in Physics frequently explore this examination used command table and use it questions terms is is given essential regularly as a in for the your reference table below. success when in the answering book. Command term Denition Annotate Add brief notes to a diagram or graph Calculate Obtain a numerical answer showing your working Comment Give a judgment based on a given statement or result of a calculation Compare Give an account of the similarities between two or more items Compare and contrast Give an account of similarities and dierences between two or more items Construct Present information in a diagrammatic or logical form Deduce Reach a conclusion from the information given Describe Give a detailed account Determine Obtain the only possible answer Discuss Oer a considered and balanced review that includes a range of arguments, factors or hypotheses Distinguish Make clear the dierences between two or more items Represent by a labelled, accurate diagram or graph, drawn to scale, with plotted points (if Draw appropriate) joined in a straight line or smooth curve Continued on page VI v INTRODUCTION Command term Denition Estimate Obtain an approximate value Explain Give a detailed account including reasons or causes Formulate Express precisely and systematically a concept or argument Identify Provide an answer from a number of possibilities Justify Give valid reasons or evidence to suppor t an answer or conclusion Label Add labels to a diagram List Give a sequence of brief answers with no explanation Outline Give a brief account or summary Predict Give an expected result Represent by means of a diagram or graph (labelled as appropriate), giving a general idea of the Sketch required shape or relationship State Give a specic name, value or other brief answer without explanation Suggest Propose a solution, hypothesis or other possible answer There at is a fuller explanation of all the command terms and additional advice for answering each one online www.oxfordsecondary.com/ib-prepared-support. Preparation and exam strategies In addition study 1. and Get ready drink night’s exam the for sleep day , Organize and study place papers Create make them an by and and Make a list more for you A well, 5. by rules the a 6. of organized. Break studying complete tasks up book large time 7. as y o ur f ir st point Work your way understanding on the Check more topics your and preparation an d where s k il ls . the g ap s Sp en d a nd on lin e for notes the to for exams. Practice Physics use efficiently . papers as papers at you the the Solve can. end as a this worked looking at answering data constraint. booklet problems trial than reference. time many Take of a every before future to how solve problem rather definitions exam quickly from using past the book. Optimize terms. on your Keep possible. exam paying approach. extra your attention answers Double-check and quote significant all answers figures. as Read to short and numerical to Label sensible graph all questions command clear values as and numbers axes is and unit and annotate with diagrams. the Use topics in ex t r a imp rovemen t textbook information . a ny Make Try and of th rou g h iden tify words. practice points Learn tips from this book. you r t i me re qui re d . res ou rc e s f or 8. Do not panic. concentrate realistic them. goals to Take on Reflect mistakes vi your understanding under exam systematically on key questions carefully , and task Focus Recite own and ready quantity reference. during and parts. each Get and material. your your answer. units this follow exam-style of Use should actively. example time. 4. you learning. using the distractions. files Read rote good before Find manageable your can eat lighting, offline importance. agenda that simple score. Avoid computer online smaller, sure your adequate ventilation. studies. into some stress exercise. environment. with Plan tasks are sleep, your important improve useful arrange physical can and there enough reduce it Bookmark your and particularly your your Have water is as temperature 3. study. of suggestions, itself. thinking comfortable Keep above exam plenty positive 2. to the a positive things and on you work your improve attitude can and improve. systematically performance your future Set to and achieve learn results. from Key features of the book Each chapter typically covers one core or option topic, and starts with Theoretical concepts and key “You should know” and “You should be able to” checklists. These definitions are discussed at a outline the understandings and applications and skills sections of the IB level sufficient for answering diploma Physics syllabus. Some assessment statements have been typical examination questions. reworded or combined together to make them more accessible and Many concepts are illustrated simplify the navigation. These changes do not affect the coverage of by diagrams, tables or worked key syllabus material, which is always explained within the chapter. examples. Most definitions are Chapters contain the features outlined on this explained in a grey side box like page. this one. Example Examples common offer solutions to problem-solving typical problems and demonstrate techniques. This feature highlights essential terms and statements that have appeared in past markschemes, Nature of science relates a physics concept to the overarching principles warns against common errors of the scientific approach and the development of your own learning skills. and shows how to optimize your approach to par ticular questions. Sample student questions (most answers of show which are typical taken student from past responses to examination IB-style papers). In Links provide a reference to each response, the correct points are often highlighted in green while relevant material, within another incorrect or incomplete answers are highlighted in red. Positive or par t of this book or the IB Physics negative feedback on the student’s response is given in the green and data booklet, that relates to the text red pull-out boxes. An example is given below. in question. You will icon Number The may on marks have their the student earned marks of on see the question available from a an has past exam right been IB paper when the adapted paper based response Examination question SAMPLE STUDENT ANSWER An electric cable contains copper wires covered by an insulator. [3] Negative feedback An electric eld exists across the cable. Discuss, in terms of charge carriers, why there is a signicant current only in the copper. ▼ This answer could have achieved 1/3 No statement charge marks: carriers electrons. number Insulator is made from specic material, which the pass However, through. copper wires Hence can no carry electric large free-moving electrons, which have there is signicant electric Questions student’s not conductor between carriers are the mentioned in and the the number very good conductivity. Includes have high current in copper the idea numbers whereas that of conductors charge insulators do not. wires. Positive The charge field carriers, Hence, link the question. current. ▲ of of the electron electric cannot No that in feedback response taken from past IB examinations will not have the exam paper icon. Practice problems Practice problems are given at the end of each chapter. problem-solving skills. Some questions introduce These are IB-style questions that provide you with factual or theoretical material from the syllabus that can an oppor tunity to test yourself and improve your be studied independently. 1 M E A S U R E M E N TS 1 1 . 1 U N C E RTA I N T I E S M E A S U R E M E N T S I N You must know: ✔ the SI denitions of P H Y S I C S You should be able to: fundamental and derived ✔ use units SI what ✔ the is units meant by scientic meaning of metric you signicant levels of gures precision in use are used to indicate in what is meant by an format when calculated presenting raw expressing answers and and processed data scientic notation as multipliers concise a to in conjunction way express with as answers and data possible measurements ✔ ✔ are correct nal multipliers metric that a notation ✔ ✔ in measurements, when ✔ AND order of quote magnitude and and compare ratios, approximations to values, the nearest estimations order of magnitude ✔ what is meant by an estimation. ✔ estimate quantities signicant Scientists need a shared to an appropriate number of gures. language to communicate between themselves The change in definitions of the and with the wider public. Part of this language involves agreeing SI fundamental units in May 2019 the units used to specify data. For example, if you are told that your does not affect your IB Diploma journey to school has a value of 5000 then you need to know whether Programme (DP) learning as you are this is measured in metres (originally a European measure) or fet (an not required to know the definitions old Icelandic length measure). except as indicated in the subject guide. However, you should be The aware that textbooks written d’Unités agreed before this date may give the older fundamental definitions. these. You seventh in the The is IB six shown set of (almost units (base) are the units required unit Diploma of this rules is known abbreviated are to defined use six luminous Programme fundamental in and always units as and of the all will as the In this other seven intensity , physics you SI). the Système Internationale system, units seven are derived fundamental candela, that units; is not from the used course. use in the DP physics course are table. In physics, unless you are providing a final answer as a ratio Measure Unit A bbreviation or as a fractional difference, you mass kilogramme kg length metre m time second s quantity of matter mole mol temperature kelvin K current ampère A must always quote the correct unit with your answer. Marks can be lost in an examination when a unit is missing or is incorrect. You should always link your answer value to its unit (together with the prefix where appropriate). There are many expression when you derived 2 of meet units other these the in derived derived include units fundamental joule, unit volt, used units for the watt, in is the course usually first time. pascal. and given the in Examples this of book these 1 .1 Often, the use of a derived unit avoids a long string of 1 units at There the are end also of a number, some units so used 1 volt in the ≡ 1 J C course 1 kg m that are I N P H YS I C S fundamental 2 ≡ M E A S U R E M E N TS 3 s 1 A not SI. Many marks are lost through Examples 2 include in MeV c some Their parts , of meaning light the is year and subject parsec. and explained are when These used you by meet have special scientists them in in this careless use of units in every meaning those DP physics examination. When fields. a question begins ‘Calculate, in book. kg, the mass of…’, if you do not The SI also specifies how data in science should be written. Numbers in quote a unit for your answer then physics can be very large or very small. Expressing the diameter of an the examiner will assume that 10 atom as 0.000 000 000 12 m is unhelpful; 1.2 × 10 m is much better. This you meant kg. If you worked the n format of n.nn × 10 is known as scientific notation and should be used answer out in g and did not say so, whenever possible. It can also be combined with the SI prefixes that then you will lose marks. are SI permitted. prefixes canbe are written allowedis during added as in front 1.012 ks. included in of The the a unit full data list to of booklet modify its prefixes and you value, that can you refer so 1012 s are to it examinations. Prefix Symbol Factor Decimal number 1 deca da 10 hecto h 10 kilo k 10 10 2 100 3 1 000 6 mega M 10 giga G 10 tera T 1 000 000 9 1 000 000 000 12 10 1 000 000 000 000 15 peta P 10 deci d centi c milli m 1 000 000 000 000 000 1 0.1 10 2 0.01 10 3 0.001 10 6 micro µ nano n pico p femto 9 are • Only some one 2.5 µkg rules prefix for is 0.000 000 001 10 12 0.000 000 000 001 10 15 f There 0.000 001 10 0.000 000 000 000 001 10 here too. allowed per unit, so it would be incorrect to write 2.5 mg. 1 • You can put one prefix per fundamental unit, so 0.33 Mm ks would 1 be acceptable near as for 330 m s (the speed of sound in air) but nowhere meaningful. Significant figures distinguish (sf) between can lead significant to confusion. figures and It is important decimal places to (dp). For example: • 2.38 kg has • 911.2 kg The rule Specify with has for the the 3 sf 4 the and sf 2 and dp 1 number answer smallest to dp. of the number sf in same of a calculated number as answer the is quite quantity in clear. the question sf. 3 1 M E A S U R E M E N TS AND U N C E R TA I N T I E S Example 1.1.1 A snail travels a distance of 33.5 cm in 5.2 minutes. In Example 1.1.1, rounding up is needed. You should do this for Calculate the speed of the snail. every calculation– but only at State the answer to an appropriate number of significant figures. the very end of the calculation. Rounding answers mid-solution Solution leads to inaccuracies that may The answer, It is incorrect is only 3 to 7 sf, is 1.073718 × 10 1 m s take you out of the allowed to quote the answer to this precision as the time tolerance for the answer. Keep all quoted to 2 sf (the fact that 5.2 minutes is 312 s is not possible sf in your calculator until 3 important). The appropriate answer is 1.1 × 1 10 1 m s (or 1.1 mm s the end and only make a decision if you prefer). about the sf in the last line. In Example 1.1.1, an examiner would Sometimes estimations are required in physics. This is because either: be very happy to see … 3 = 1.073718 × 10 • an 1 m s educated guess is needed for all or some of the quantities in a so calculation, or the speed of the snail is 3 1.1 × 1 m s 10 (to 2 sf) … • there is an assumption involved in a calculation. as your working is then Often it will be appropriate to express your answer to an of ten. order of completely clear. magnitude, meaning rounded to the nearest power The best n way to express any order of magnitude answers is as 10 , where n is an integer. Example 1.1.2 You may see order of magnitude answers in Paper 1 (multiple Estimate the number of air molecules in a room. choice) written as a single integer. Solution When the response is, say, 7, this The calculation is left for you, but you should use the following 7 will mean 10 steps. It is also permissible to talk about • Estimate the volume of a room by making an educated guess at ‘a difference of two orders of its dimensions, in metres. magnitude’; this means a ratio 3 2 of 100 (10 ) between the two quantities. • The density numbers • The easy mass nitrogen of of is 1 air is about 1.3 kg m 3 —call it 1 kg m to make the later. mol of 28 g—call oxygen the molecules answer 30 g is for 32 g both and 1 gases mol of combined. 23 • Each If the command term ‘Estimate’ is used in the examination, it will always be clear what is required as you will lack some or all data for your calculation if an educated guess is needed. In estimation questions, such as Example 1.1.2, make it clear what numbers you are providing for each step and how they fit into the overall calculation. 4 The mole volume number of contains and 6 × density moles and 10 → molecules. mass of Avogadro’s gas in number room → and molar answer. mass → 1.2 1 . 2 U N C E R TA I N T I E S A N D You must know: ✔ what is meant by what random errors and systematic ✔ is meant by absolute, fractional and ✔ that error bars uncertainties are in used that gradients on graphs to an ✔ intercepts on random and that and include uncertainties uncertainty systematic errors can reduced absolute and go and/or on to state these range indicate data and data fractional determine with ✔ how identied collect uncertainties as ✔ ERRORS E R R O R S explain be percentage AND You should be able to: errors ✔ U N C E R TA I N T I E S graphs the overall uncertainties are uncertainty combined when in data calculations have involving addition, subtraction, multiplication, uncertainties. division ✔ and determine intercepts All measurement is prone to error. The Heisenberg raising the of to a power uncertainty in gradients and graphs. uncertainty Random errors are unpredictable principle (Topic 12) reminds us of the fundamental limits beyond changes in data collected in an which science cannot go. However, even when the data collected are experiment. Examples include well above data you this limit, then two basic types of error are implicit in the uctuations in a measuring collect: random error and systematic error instrument or changes in the environmental conditions where Random errors lead to an uncertainty in a value. One way to assess the experiment is being carried out. their impact times and on then a measurement use half the is to repeat range of the the measurement outlying values as several an estimate Systematic errors are often produced within measuring of the absolute uncertainty instruments. Suppose that an ammeter gives a reading of +0.1 A Uncer tainty in measurement is expressed in three ways. when there is no current between Absolute uncer tainty: the numerical uncer tainty associated with a quantity. For example, when a length of quoted value 5.00 m has an actual value somewhere between 4.95 m and 5.05 m, the absolute uncer tainty is ± 0.05 m. the meter terminals. This means that every reading made using the meter will read 0.1 A too high. The eect of a systematic error can The length will be expressed as (5.00 ± 0.05) m. produce a non-zero intercept on absolute uncertainty in quantity a graph where a line through the Fractional uncer tainty = . numerical value of quantity origin is expected. A fractional uncer tainty has no unit. Percentage uncer tainty = fractional uncertainty × 100 expressed as a percentage. There is no unit. Example 1.2.1 Five readings collected 0.972 m, of 0.975 m, a) Calculate the b) Estimate, for i) absolute ii) length of a small table are made. The data 0.979 m, average the 0.981 m, length length of of the 0.984 m the table. table, its: uncertainty fractional iii) the are: uncertainty percentage uncertainty . 5 1 M E A S U R E M E N TS AND U N C E R TA I N T I E S Solution a) The average (0.972 + length 0.975 + is: 0.979 + 0.981 + 0.984) = 0.978(2) m 5 b) i) The outliers are 0.972 and 0.984 which differ by 0.012 m. Half this value is 0.006 m and this is taken to be the absolute uncertainty . The length (This absolute standard case should is be error deviation 0.004 m. expressed is an of (0.978 estimate; the 0.006 m as is set of thus ± 0.006) m. another estimate measurements an is which the in this overestimate.) 0.006 ii) The fractional uncertainty is = 0.006(13) 0.006 . = 0.9782 This iii) You The will lengths, length. is a ratio percentage often both This need with to lengths combine quantity also 100 = 0.6%. mathematically: to have be added an to a pair give a of total uncertainty . Raising quantities to a power 2 When ± will need × uncer tainties are still added. added and subtracted a 0.006 quantities may unit. When the answer is found by division, the fractional The absolute uncer tainties are added when quantities are = is no 2 3) cm. What is the total perimeter of the table? y has The answer should be expressed as (1.08 ± 0.08) m The two sides of a table have lengths (180 ± 5) cm and When and uncertainty uncertainty , derived Combining uncer tainties (60 ± of b then ∆y = y = a , this is the same as a × a so using the ∆y ∆ a + ∆b algebraic rule above: ∆a = y In this case, the perimeter of the table is 2∆ a ∆a + . = a a a 180 + 180 + 60 + 60 = 480 m. The absolute uncer tainty is ∆y ∆a n In the general case, when 5 + 5 + 3 + 3 = 16 cm. y = a , = , where || n y The perimeter is (480 ± 16) cm or 4.8 ± a 0.2 m. means the absolute value or magnitude of the expression. Notice that when the quantities themselves are subtracted, When a quantity is raised to a power n, the fractional the uncer tainties are still added. uncer tainty is multiplied by n. What is the area of the table? The radius of a sphere is (0.20 ± ∆y ab When then y = c ∆a = y ∆b + a 0.01) m. What is the ∆c + b volume of the sphere? c 4 3 3 Volume of sphere is: The fractional uncer tainties are added when quantities are πr = 0.0335 m 3 multiplied or divided where r is the radius. 2 The area is 1.8 × 0.60 = 1.08 m . The two fractional 0.01 Fractional uncer tainty of radius = uncer tainties are = 0.05 0.20 0.05 0.03 = 0.050. = 0.028 and 1.8 So, the fractional uncer tainty of the radius cubed is 0.6 3 × 0.05 = 0.15. The sum is 0.078 and this is the fractional uncer tainty of The absolute uncer tainty is the answer. 3 0.335 × 0.15 = 0.0050 m The absolute uncer tainty in the 3 The volume of the sphere is (0.335 ± 0.005) m area = 0.078 × 1.08 = 0.084 It is on possible a graph. that data Therefore, points, there all are with an errors associated associated error, with the are presented gradient and There is more information about any intercept bars to on the graph. The way to treat these errors is to add error this topic in Chapter 13, which deals the graph. These are vertical or horizontal lines, centred with Paper 3, Section A. data 6 point, that are equal to the length of the absolute errors. on each 1.3 Maximum the true and minimum best-fit line. The best-fit lines gradients of can then these be drawn each side maximum–minimum of V E C TO R S AND S C AL ARS 20 lines 15 a range intercepts that For can the of of be values the that maximum–minimum associated graph corresponds in with Figure the 1.2.1, error the to the lines in the gradient error also true is in have the a gradient. range in m / ecnatsid give The values intercept. 1.6 with a range 10 5 between 0 1 2.1 and 1.1, so (1.6 ± 0 0.5) m s –5 time / s The intercept is 2.4 with a range of 1.0 to 5.8, so ( 2.4 ± 3.4) m. Figure 1.2.1. Maximum and minimum best-fit lines each side of a true best-fit line 1 . 3 V E C T O R S A N D S C A L A R S You must know: ✔ what ✔ that are meant You should be able to: by vector and scalar quantities ✔ solve vector problems graphically and algebraically . vectors (split into Quantities in can two DP be combined separate physics and resolved vectors). are either scalars or vectors. (There is a third Scalars are quantities that have type of physical quantity but this is not used at this level.) magnitude (size) but no direction. They generally have a unit A vector can be represented by a line with an arrow. When drawn to associated with them. scale, is as the length of the line represents the magnitude, and the direction Vectors are quantities that have drawn. both magnitude and a physical Both scalars and vectors can be added and subtracted. Scalar quantities direction. A unit is associated with add just as you need any other number in mathematics. With vectors, however, the number par t of the vector. to take the direction into account. For example, the scalar quantity Figure drawn 1.3.1 to shows the the same addition scale and of the two vectors. direction The angles vectors drawn must speed is written as v; the vector be accurately quantity velocity is written as v too. A further construction produces the parallelogram with the red (sometimes as v solid or v , but this notation is not used in this book). and dashed lines. Then the magnitude of the new vector v + 1 by the length of the blue vector with the direction as v is given 2 shown. v 2 v 2 v v 1 v 1 v 1 Figure 1.3.1. Adding vectors v and v 1 V ectors you can meet each also be the DP in other (Figure added 2 algebraically . physics course is The when most the common vectors are situation at 90° to 1.3.2). v 2 As v before, and 1 v addition by drawing . Algebraically , the use gives of the red vector trigonometry which gives the is the sum magnitude of of 2 θ 2 the resultant (added) vector as v + 1 1 2 and v 2 the direction θ as v 2 tan v Figure 1.3.2. Adding two vectors at 1 right angles 7 1 M E A S U R E M E N TS AND U N C E R TA I N T I E S Example 1.3.1 A girl her walks position 500 m due relative to north her and then starting 1200 m due east. Calculate point. Solution This has is a similar to the magnitude of situation 500 m and in Figure the 1.3.2 second a where 2 The magnitude is tan vector known skill as resolved shows = 1200 down the is vector 1200 m. 2 500 + 1200 1300 m. = 22.6° required into two resolving in the DP physics components the components the resultant first of Another 500 1 θ of the magnitude vector. will be at course right A right angle independent is angles is of that to chosen each of breaking each other because other. – the Figure a this 1.3.3 process. Fsinθ F The is vector the F points hypotenuse lengths F cos θ upwards of and the from the right-angled horizontal triangle. at The θ. This other length sides F have F sin θ Example 1.3.2 θ 1 An object angle moves N30°E. with a velocity Determine the 40 m s at component an of the velocity Figure 1.3.3. in the direction: Resolving a vector due east b) due north. 40 ms V y a) 1 °06 nis 04 = Fcosθ Solution a) The angle between the vector and east is 60° 60° 1 So the component due east = 40 cos 60° = 20 m s is 40 cos 30° = 40 sin 60° V = 40 cos 60° x 1 b) Due north, the component = 34.6 m s Example 1.3.3 A girl cycles 1500 m south-easterly due direction. north, 800 m Calculate due her east overall and 1000 m in a displacement. Solution A drawing of the of the journey displacement component is is shown. 800 + 1000 cos 45° is 1500 − 1000 cos 45° = displacement is 1700 m at tan = 790 horizontal 1510 m. 8 total 790 m 1 The The 1510 = 28° The component total is two vertical 1.3 You can now add or subtract any non-parallel vectors algebraically . Figure V = y1 V V y2 Figure 1.3.4. addition gives V gives V = 1 V 2 cos θ = V 1x 1 1 Algebraic method for adding or subtracting non-parallel vectors = V x addition 1 V = 2 2 2x the method. θ cos θ = V V V ertically the + V + which V 1x y is 1 which V 1y cos θ V 2 x is V V 1 sin θ 1 2 y + cos θ 2 V 1 2 . sin θ 2 These new vector lengths to give the new vector length V = V V V y 1 2 + x with an angle can be 2 2 added S C AL ARS θ nis θ nis 1 2 2 the shows AND 1 2 V V θ Horizontally 1.3.4 V E C TO R S to the horizontal tan of = y V x To subtract original two vectors, direction but simply leaving form the the length negative vector unchanged) and of the add one this being to the subtracted other (by reversing its vector. Practice problems for Topic 1 b) Number of free electrons in the charger lead to your Problem 1 computer. You will need to have covered the relevant topic before answering this question. c) Volume of a door. a) Express the following derived units in fundamental d) Number of atoms in a chicken’s egg (assume it is units: watt, newton, pascal, tesla. made of water). b) Give a suitable set of fundamental units for the e) Number of molecules of ink in a pen. following quantities: acceleration, gravitational field strength, electric field f) Energy stored in an AA cell. strength, energy. g) Number of seconds you have been alive. Problem 2 h) Thickness of tread worn off a car tyre when it Express the following physical constants (all in the travels 10 km. data booklet) to the specific number of significant Problem 5 figures. Determine, the following, with their absolute and Quantity Significant figures required percentage uncer tainties. Neutron rest mass 3 Planck’s constant 2 a) The kinetic energy of a mass (1.5 ± 0.2) kg moving at 1 1 Coulomb constant 2 (21.5 ± 0.3) m s 2 (use E = mv ). k 2 Permeability of free space 5 b) The force acting on a wire of length (3.5 ± 0.4) m Problem 3 carrying a current (2.5 ± 0.2) A in a magnetic field of Express the following numbers in scientific notation to strength (5.2 ± 0.3) mT (use F = BIL). three significant figures. c) The quantity of gas, in mol, in a gas of volume a) 4903.5 b) 0.005194 c) 39.782 d) 9273844.45 3 (1.25 ± 0.03) m 5 , pressure (2.3 ± temperature of (300 ± 0.1) × 10) K (use pV = 10 Pa at a nRT). e) 0.035163 Problem 6 1 Problem 4 A car is driven at 30 m s for 30 minutes due east and 1 Estimate these quantities. then at 25 m s a) Length of a DP physics course in seconds. Calculate the final displacement of the car from its for 45 minutes nor theast. star ting point. 9 M EC H A N I C S 2 2 . 1 M O T I O N You must know: ✔ the meaning of displacement, ✔ the ✔ difference the average acceleration that the uniform ✔ how uid ✔ what for ✔ for instantaneous velocity , use acceleration ✔ the kinematic uniform represent speed equations of motion apply ✔ resolve by affects terminal into motion and speed interpret and analyse ✔ describe and motion (suvat) of displacement–time, velocity and graphs displacement components motion experimental acceleration using acceleration–time horizontal projectile an of motion speed–time, and acceleration, vertical ✔ resistance. equations acceleration distance–time, acceleration resistance meant and velocity–time kinematic to is values distance, velocity between and only terms speed, and uid You should be able to: determination of the free-fall. Distance is a scalar quantity: it is the length of a path between two points. Vector and scalar quantities are discussed in Topic 1.3 Vectors Speed is a scalar quantity: and scalars. total distance travelled by an object average speed = total time taken Displacement is a vector quantity: it is the dierence between an initial position and a nal position. When a question asks for a vector Velocity is a vector quantity: quantity, such as velocity or change in displacement average velocity = displacement, you must give the time taken for change direction and magnitude of the change in velocity quantity to gain full marks. Acceleration is time taken for change The Y motion of displacement, N these are Look at The of distance Figure direct XY and line its X Motion graphs. can object and 2.1.1. from is and described acceleration. by the The vector scalar The X to length Y is of the relative the curved path displacement. to north are is Both required be represented quantities using can be distance–time shown as the distance. the to length specify and acceleration–time and the speed–time displacement–time, Figure 2.1.1. 10 of completely . V ector velocity–time quantities counterparts speed direction displacement 45° an velocity graphs. 2 .1 M OT I O N Example 2.1.1 There is more advice for A ball is a against time of the 25 plots ball v t Estimate the distance fallen the ball by determining the gradient of a graph 20 m / v llab eht fo speed graph rest, the deeps lacit rev vertical This from above s ground. a) released, distance 1− from constructing graphs and at a point and the area under the 15 graph in Chapter 13. 10 5 0 from 0 t = 0 to t = 2 4 6 8 8.0 s. Quantities can be derived from the time t / s motion graphs as follows: b) Determine the acceleration of the ball when t = 4.0 s. • Distance–time graph: speed is Solution the gradient of the graph a) The T o distance calculate the curve. fallen this Then by area, the ball count multiply is the this equal to number number the of by area under whole the the squares area of one curve. under square. • Speed–time graph: distance travelled is the area under the graph and acceleration is the In this example, there are approximately 28 squares under the gradient of the graph 1 curve. Each square has an area of 1 s × 5 m s , which is equivalent • Acceleration–time graph: speed to a distance of 1 × 5 = 5 m. change is the area under the So the distance travelled is 28 × 5 =140 m. graph. b) You can find drawing a the acceleration tangent to the of curve the at t ball = when 4.0 s. t The = 4.0 s by gradient of this The symbols used for motion in the tangent is the acceleration. Make sure you read off values as far DP physics course are: apart as possible (at 25 least half the length of the line). s displacement u initial speed v nal speed a acceleration t change in time 12 2 acceleration = = ≡ 2.03 2.0 m s 6.4 The suvat means equations ‘constant’ apply here. for The uniform speed–time acceleration graph only; must be a ‘uniform’ straight line. Example 2.1.2 These are connected by the four kinematic equations of motion for 1 A car changes its speed uniformly from 28 m s 1 to 12 m s in 8.0 s. uniform acceleration, often known as the suvat equations: a) Calculate the b) Determine acceleration of the car. v the distance travelled by the car during the = u + 2 of at 8.0 s acceleration. v 2 = u + 2 as 1 2 Solution s a) Use the suvat calculate by the writing equations acceleration. down what including symbols what want you to to find u initial speed v final speed a acceleration t change in time at 2 28 m s (v + u) t 1 12 m s s = 2 know, units, ut + 1 Start you and = 2 ? m s and 8.0 s out. The kinematic equations for The equation you need is v = u + at uniform acceleration can be Note that the units match; if they didn’t, you would need to used to determine the acceleration convert them. Take care to substitute correctly into the equation: of free fall g for objects released 2 12 = 28 + a × 8.0 and therefore a = near the Ear th’s surface; the −2.0 m s distance fallen from rest and time The car is slowing down so the acceleration is negative. to fall are required. 1 1 2 b) Use s = ut + at 2 2 = (28 × 8) + × − 2.0 × 8 = 224 − 64 = 160 m 2 11 2 M E C H A NI CS Solving • resolving Look for sentences such as ‘Assume air resistance is questions the involving initial projectile velocity into motion involves: horizontal and vertical components • recognizing negligible’; this makes your job vertical that the acceleration due to free-fall acts only on the component. easier. You will not have to carry As g is constant, the suvat equations can be used. The horizontal out calculations involving air component of velocity does not change (assuming negligible air resistance for projectile motion, resistance). but you may have to answer qualitative (non-mathematical) questions about it. Example 2.1.3 Idris kicks of32 m a) The a soccer away . ball ball Assume takes 1.6 s over air to a wall that resistance reach a is point is a horizontal distance negligible. vertically above the wall. The resolution of a vector into Calculate the horizontal component of the velocity of the ball. two components at right angles is considered in Topic 1.3. b) The ball State c) is the at Determine it was its maximum vertical the height component magnitude of of as its the it passes velocity initial at above the maximum velocity of the wall. height. ball when kicked. d) Deduce the angle to horizontal θ the at which the ball was kicked. Solution a) The ball travels The horizontal 32 m horizontally in a time of 1.6 s. 32 1 component of velocity = is 20.0 m s 1.6 b) At its maximum vertical c) You component need vertical so you height, to know is can of combine horizontal ball moves horizontally only , so the zero. the component the it initial the 1 u initial speed v final speed 0 t change in time 1.6 s ? m s velocity , with the component. Use acceleration a a suvat equation. (Note that component Use v = u + upwards at. is is negative because the initial positive.) Substituting gives 0 = u + −9.8 × 1.6 1 Rearranging Y ou now gives have the u = 15.68 m s information needed 2 velocity . Its magnitude is: 20 to calculate 2 + 15.68 the vector 1 = 25.4 m s 1 = 25 m s (2 s.f.). 15.68 1 d) The angle to the horizontal is tan = 38° 20 The effect of air resistance on an object’s motion is covered in more Resistive forces oppose the motion of objects detail in Topic 2.2. liquid. 12 They are considered on page 15. moving in a gas or in a 2.2 FOR C E S S AMPLE STUDENT ANS WER ▼ The not A tennis ball is hit with a racket from a point 1.5 m above the oor. The 1 ceiling is 8.0 m above the oor. The initial velocity of the ball is 15 m s suvat quoted equation and the appears directly . an here error it used is substitution Had would there been have been at 50° above the horizontal. Assume that air resistance is negligible. difcult Determine whether the ball will hit the ceiling. This answer could have achieved 2/3 [3] carried ▲ The component of sin50° = an error to be forward. speed ▲ The 15 × allow marks: −1 V ertical: to is 11.5 m s total height reached is correct. calculated correctly . 2 0 = (11.5) 132.25 = −2 × 9.8 × 5 ▼ The 19.65 total compared S = 6.75 m ceiling (8.0 m) conclusion 6.75 8.25 + > 1.5 8 = T he 8.25 ball won’ t hit the A could = Acosθ = 15 = 9.6 but (8.25 m) height the is of the wrong drawn. ceiling. ▼ The answer is the (m) component incorrect. This height with have achieved 0/3 This is of the speed is horizontal component. marks: v ▼ The × cos 50 answer calculated distance Y es, 9.6m > the 6.5m required to hit the ▲ There is a stated in real objects what ✔ Newton’s ✔ that meant × weight and dynamic are object the out a more vectors give equilibrium. force, the of be mass for has the to than using zero is no addition position of where the the 1.5 m ball begins. comparison. by points ✔ represent ✔ draw ✔ interpret a force as a vector object be But the a free-body diagram to its the Newton’s ✔ carry out ✔ identify forces rst acting on a system in terms law calculations using Newton’s second law or pulls. be a between and a contact through gravitational must force acts vector (net do action Newton’s and third reaction pairs using law. field field. an direction, Newton’s object so between interactions, and both a three force. must be uniform. object for static implies arise interaction force forces This also with to an using friction. magnitude rules the are of external resultant equal forces the occur of described interacts motion interpret fall pushes both is and equilibrium motion free can one When forces of an of describe acceleration to laws another. velocity unchanging When represented coefcients to and motion Uniform For be translational friction said example, of by acceleration solid laws the You should be able to: can three that Forces for is the mass one the the space ✔ ✔ to F O R C E S You must know: that that equal travelled. ▼ There ✔ is ceiling. from 2 . 2 assumes speed subject a the cancel to body , addition. force) not be on When body out a to is net external you in give need the to forces force. sum cancel translational zero resultant unbalanced 13 2 M E C H A NI CS Example 2.2.1 Physics often simplifies situations to make the A cyclist travels along a horizontal road at a constant velocity . calculations (mathematics) more Explain why the cycle is travelling at constant velocity . manageable. It represents real Solution objects using single points. For The translational motion, this is the system velocity . centre of mass—where all the mass net acts. In gravitational fields, this is force. equals These the weight acts. states external cycle also the centre of gravity—where all (cyclist N1 the forces and that, The cycle) forward frictional are equal is moving because in the force forces horizontally velocity of acting magnitude the in and is cyclist the in at constant constant, there pedalling opposite opposite is no the direction. directions. Example 2.2.2 2 A car of travels mass in a 9.0 × 10 straight 15 kg line This book uses N1, N2 and N3 to away from traffic lights after they turn green. and third laws of motion. It is best not to use your own abbreviations a) Calculate the in examinations unless you state acceleration what they are. between t = t = of the 2.0 s car s m / deeps 1– abbreviate Newton’s first, second 10 5 and 5.0 s. 0 b) Calculate the force used 0 to accelerate the 1 2 3 car. states that an object will remain 4 5 6 7 8 time / s Newton’s rst law of motion (N1) Solution at rest or continue with uniform 1 a) The change in speed is from 3.0 to 10.5 m s . It takes 3.0 s. velocity unless a net external force changeinspeed 7.5 –2 acts on it. acceleration = = = timetakenforchange b) The force that accelerates the car 2.5 m s 3.0 = mass of = 900 2.5 × car = × acceleration 2.25 kN. Newton’s second law of motion Since (N2) relates the resultant force F the acting on an object of mass m to the figures, acceleration of the object a with the well equation F = (= data you used should in the give calculation the answer were to 2 given to significant 2 significant figures as 2.3 kN). ma. Both force and acceleration are vectors, whereas mass is scalar. The direction of the net force and the direction of the acceleration it produces are identical. N2 denes the SI unit of When you apply a law in an answer, always state the law; that way, the examiner force, the newton. One newton (1 N) will be clear that you understand the physics underlying your answer. is the force that will give a mass of 2 1 kg an acceleration of 1 m s . In 2 fundamental units, this is kg m s N3 implies that • that action–reaction pairs exist. These are pairs of forces are: equal in magnitude and opposite in direction Newton’s third law (N3) states that • the same type of force (such as gravitational, electrostatic or action and reaction are equal and tension opposite. When a forces). ball is gravitational the an of 14 ball by the upwards the ball. released force acts Earth. from At the gravitational This is an rest above downwards same force on time, that is action–reaction the the the Earth’s ball ball equal pair. due in surface, to the attracts a attraction the magnitude Earth to the of with weight 2.2 When This the pair ball is sits rest electrostatic • the upwards • the equal These at forces force on in of table, origin; the downwards arise a because it table force the an force pair now acts. is: on of additional FOR C E S the the table ball ball and on the the ball table. are both deformed by Other interpretations of N2 and the other object. The electrostatic forces arise from the attempt by each N3 are considered in Topic 2.4. to return to its Example original shape. 2.2.3 2 A car a of mass horizontal 750 kg road. accelerates A resistive a) Calculate the resultant b) Calculate the force 0.30 m s force force that at the of that in 550 N acts engine a acts on straight on the car on the exerts the line along car. car. Solution a) The accelerating acting. b) The The resistive provide 550 + the 225 Note: = the The other. In is friction both normal of there called is cases is ≡ 750 550 N. So, the of the 0.3 2 The total = 225 N. engine engine is the must overcome must figures is no not is used be acting between surfaces are moving motion, the friction relative static to force the friction. When there is the In of the reaction frictional force the final µ is two surfaces to between relative each the movement, R force that acts F depends between R F = on the the surfaces. friction µ R d coefficient where µ of is the coefficient of d general, dynamic the magnitude friction, decreases Different until relative Dynamic friction. dynamic this and dynamic. s static force be: s where µ net figures. force friction ≤ This significant significant frictional or magnitude Static × to 780 N. (perpendicular) F is 230 N, answer. the said the is number whether When surfaces the on 775 N and magnitude depends force 230 N. full round-off force answer as the so the surfaces language is of size static of begin used for friction the to is friction. greater frictional force than that between for two surfaces slide. fluid friction (which occurs in liquids Fluid friction is dealt with in and gases). The terms air resistance and liquid drag are used to describe Option B of the DP physics course. the friction The it on amount moves The As object of the further the speed. reach in speed; of the drag so it equals net medium. are speed, the on In drag on drag falling fluid. object the the speed size at and the freely depending in on which shape when vehicle a size fluid speed the which between force. magnitude The two forces and there this of occurs the the is object its is no called and maximum at the force terminal gravitational and in zero travels planet’s their to the drag and force. falls automobile, force, the increases force and relative greater accelerating The an a the the resultant (N1). depends speed between frictional the the through depends relative velocity that terminal moves medium) this equals Objects a the other, engine it frictional accelerates, change terminal the as eventually each ‘stickiness’ of solid higher drag oppose of (called them. an a field also shape. 15 2 M E C H A NI CS Example 2.2.4 A small effects The a) steel in graph Draw a time is released from rest into a fluid. Ignore buoyancy question. shows how the speed for the v of the ball varies with time t v free-body diagram i) ball this ball at: t 1 ii) time t 2 b) When t = t , the gradient 1 0 of the graph is a 0 t t 1 Deduce an expression acceleration the a frictional and in terms acceleration force F acting of of on the free the mass fall ball at of g t Deduce, using frictional your force the for = ball M, the magnitude of t f c) t 2 1 answer acting on to the part ball a) ii), when t the = magnitude of the t 2 Solution You may be asked to draw a free- a) i) The ball is accelerating; there (i) body diagram to show the relative magnitudes and directions of all must be on There it. force the forces acting on a single body. a net is downward a smaller (ii) drag force force drag force drag upwards. The body is usually represented ii) The vector lengths should be as a dot. Take care that: equal The • the vector lengths represent the as net the velocity force must is be constant. zero (N1). weight force relative magnitudes b) The net force F = F total − weight F , f • the directions are correct F = Mg and F weight = Ma total weight force • every force is labelled So F = M (g – a) f unambiguously c) As the velocity is constant, N1 predicts that the upward and • only forces acting on the object downward forces must be equal. The magnitude of the frictional are shown. force (drag) must equal the magnitude of the weight force. In the sample student answer, drag force should be much shor ter S AMPLE STUDENT ANS WER in length. The vector sum of the weight added to the drag should be equal to the vector length of the An unpowered glider moves horizontally at constant speed. The wings of the glider provide a lift force. The diagram shows the lift force acting on the glider and the direction of motion of the glider. lift force. a) Draw the forces acting on the glider to complete the free-body diagram. The dotted lines show the horizontal and ver tical directions. This answer [2] could have achieved 2/2 marks: vertical direction of ▲ The diagram is well drawn motion lift force and everything opposes the is clear. forward The drag motion of drag the glider and the weight acts horizontal downwards, as expected. weight 16 2.3 This answer could have achieved 1/2 WORK , ENERGY AND POWER marks: vertical ▲ The drag force (here labelled direction of ‘friction force’) is correct. force motion lift force ▼ However, an upward force friction force horizontal is shown There is acting on certainly the a glider. force vertically weight upwards, is the lift but vertical it is already component there of – it the force. b) Explain, using appropriate laws of motion, how the forces acting on the glider maintain it in level ight. [2] ▲ The law (although This answer could have achieved 2/2 1st Law states that an object will move at a constant example clearly speed and vertical in a T he of straight lift’s the vertical glider , keeping line the or at rest providing component meaning glider in there level is is answer could have the no there same net is but no resultant opposite resultant to the vertical force. the weight equal to action to the force remains B no to stated Newton). here the is is referenced equality of correctly between the lift the and stated. force ight. achieved Newton’s clearly weight 1/2 quoted marks: glider According is component ▼ There This use attributed marks: The T he in not third exerting change law, on or the A. force And constant A also exerting the when rst no on B law external N3 is is is no laws link and situation. not between the The the physics of reference the to required. the force Do not simply repeat the question added, hence it maintain its level ight. back to the examiner. This does not gain credit. 2 . 3 W O R K , E N E R G Y You must know: ✔ work done transferred ✔ there are kinetic, ✔ energy ✔ power is a measure is be the of two forms gravitational can of the energy energy energy and ✔ stores store, of including ✔ elastic energy identify the specied calculate power conserved rate P O W E R You should be able to: between many A N D transfer as ✔ interpret ✔ include energy energy work force the ✔ efciency pathways for a as force × distance and speed force–distance effects energy-change and transfer done × stores of graphs resistive forces in calculations is: useful work out total work in useful power out total power in ✔ calculate changes in kinetic energy and changes or in ✔ Energy is transferred from one energy store to another via gravitational calculate an potential energy efciency . energy pathway. 17 2 M E C H A NI CS Energy stores include: Energy transfers and pathways appear throughout the DP physics • elastic and magnetic • chemical • kinetic • gravitational • electromagnetic • nuclear • thermal course. For example: • electrical in Topic 5 • heating in Topics 3 and Option B • waves in Topics 4 and 9. Conservation laws are impor tant Energy pathways (or transfer mechanisms) include: in science. They recur many times in physics and are a way • electric (a charge • mechanical • heating • waves moving through a potential difference) to learn (and later revise) the (a force acting through a distance) subject effectively. Always look (driven by a temperature difference) for the links between topics in this subject. Try to link ideas to check (such as electromagnetic radiation or sound waves). your understanding of the whole The rule of conservation of energy is never broken—but you will subject and to ensure that you can sometimes have to look hard to see where some energy goes. think in an unfamiliar context. In mechanical gravitational systems, to the kinetic) is energy transferred equivalent to the between work stores (such as done. work done = force (F ) × distance (s) The equation work done = force × distance applies when the force is When there is an angle θ between constant, However when the force is not constant with distance you the force direction and the need to calculate the area under the force distance graph. displacement then work done = Fs cos θ Example 2.3.1 The unit of energy is the joule (J). 1 J is equivalent to 1 newton metre A boy drags a box to the right across a rough, horizontal surface (Nm). In fundamental units, this is using 2 kg m a rope that pulls upwards at 25° to the horizontal. 2 s a) Once the load horizontal Change in kinetic energy ∆E is moving frictional at force a steady acting on speed, the the load is average 470 N. of an Calculate k the average value of F 1 2 object of mass m is m(v 2 − u ) b) The load is moved a horizontal distance of 250 m in 320 s. 2 Calculate the work done on the load by F when its speed changes from u to v Solution Change in gravitational potential energy ∆E a) of an object of mass m The frictional force acts to the left. The horizontal force to the p is mgh when it is raised through a right must ver tical height ∆h change in b) Conversion of ∆E k to ∆E for The the work 470 N for the resultant force to be zero with no velocity . Resolving as equal F cos 25 given done = 470 gives F = 518 N (= 520 N to 2 sf, data). = force × distance = 518 × 250 = 130 kJ. is a p convenient way to solve problems When an object is moving, energy has been transferred into kinetic when bodies fall in gravitational energy. Movement of an object within a gravitational field can also lead fields. You must use this approach to transfer when the acceleration is not is uniform: for example, when a skier slides down a slope with a changing gradient. 18 used. of energy . In this case, the term gravitational potential energy 2.3 WORK , ENERGY AND POWER Example 2.3.2 A diver The the a) climbs height diver of is a the diving diving board board and dives above the from floor it. is 4.0 m. The mass of 54 kg. Calculate diver to the climbs gain to in the gravitational diving potential energy when the board. 1 b) The diver kinetic c) energy Suggest from enters why the the of water the the at diver kinetic gravitational a as speed she energy 8.0 m s enters of potential of the the diver energy . Calculate the water. in part gained in b) part is different a). Solution a) = Δ gh = 54 × 9.81 × 4.0 = 2.1 kJ k 1 2 b) ΔE = m k 2 (v − ) u 2 Here u = 0 as the 1 diver starts rest. 1 2 ΔE from = 2 mv = × 54 × 8 = 1.7 kJ k 2 c) 2 A number • • Work The to is of factors done distance the water can against be air travelled is not discussed here. resistance. by equal the to centre the of mass distance of the gained in diver falling climbing the stairs. • The • Energy Energy diver can gains usually be gravitational goes transferred into at potential rotational different energy kinetic rates. in taking off. energy . Think of boy the two boys of Power is the rate of doing work . equal weight who run up a hill: the quicker is more powerful energy transferred of the two. power = time taken for transfer distance force × distance As work done = force × distance, power = or 1 force × Its unit is J s 1 . 1 J s = 1 watt (W). In time time 2 fundamental units, this is kg m Therefore power = force × speed. The area under a force–speed 3 s graph ul ener rre useful energy transferred gives the power developed during an energy transfer. eciency = total energy input As you saw in Example 2.3.2, not all energy is necessarily transferred power output from the original source into its final useful form—in most real cases, ≡ power input there is is a transfer efficiency. This via can friction. A measure be defined in terms of of the effectiveness energy or in of terms a of transfer power. Example 2.3.3 Water the in rate The a of water hydroelectric 12 000 kg takes 2.0 s at the a) Calculate b) A small electrical system. All to falls vertically to a river below at minute. fall this distance. It has zero velocity top. the Determine c) system every Outline the height the the through generator water goes electrical energy which of power in water efficiency through transfers the the output this 20% falls. is at the foot of the generator. of the generator. system. 19 2 M E C H A NI CS Solution a) The acceleration, g, 1 is uniform, = ut + From equations can be used. 2 at = 0 + × 2 b) suvat 1 2 s so 9.81 × 2.0 = 20 m to 2 significant figures 2 this point, it is best to work in seconds rather than minutes. 12 000 Mass of water flowing every second = = 200 kg 60 Gravitational mgh = As = the potential energy 200 × 9.81 × 19.8 generator is 20%, = transferred 38 847.6 the every second J power output is 38 847.6 = 7769.52 W (= 7800 W to 2 significant figures). 5 c) The the water has waterfall. energy . At As the energy has kinetic energy the gravitational water bottom occurred. energy of kinetic energy frictional stored the of the falls, the The and is is the enters a dynamo. energy , energy at transferred maximum turbine transferred the electrical energy waterfall, water water turbine into of potential into The into of kinetic the rotational thermal top transfer where dynamo wasted the of linear kinetic converts energy this and losses. S AMPLE STUDENT ANS WER An electric motor pulls a glider horizontally from rest to a constant speed 1 of 27.0 m s ▼ This is not the easiest way with a force of 1370 N in a time of 11.0 s. The motor has an to overall eciency of 23.0%. carry out energy the calculated used problem. gained to by and the then calculate First, glider the the the Determine the average power input to the motor. State your answer to an is efciency energy is appropriate number of signicant gures. This to by the motor. time to This give is the then answer are power gures in too many the answer the been data restricted which in the to 2 = or should 3 3/4 marks: F .S = 1367 .6 N × 148.5 m = 203094 J W = W × 23% = 883017 .3 9 J useful given W question. W used = used This ≈ = 80274 .3 w 11 s t solution achieved useful P ▲ The have input. signicant used have could divided W There [4] input answer could have achieved 4/4 marks: 148.5 uses -1 power = force × speed leading Average to speed = = 13.5 ms 11 the power of the glider directly . Power = Force × velocity = 1370 × 13.5 = 18495 W = 23% Then it uses a simple efciency 18495 calculation and correct rounding to × get the answer number 20 of with an signicant appropriate gures. 100 = 80413.043 W = 80.4 kW 23 2.4 2 . 4 M O M E N T U M You must know: ✔ that law MOMENTUM an of You should be able to: alternative motion is form force = for rate Newton’s of change second ✔ dene of and understand the meaning of momentum momentum ✔ ✔ ✔ that impulse and this is the energy can during an Momentum never lost is or is equal area be to the under change a transferred in force–time to kinetic apply momentum to graph along energy ✔ the law analyse a of straight identify conservation collisions and of momentum explosions for motion line collisions as elastic or inelastic. explosion. a conserved gained in momentum = mass × a quantity . collision The momentum unless external of a forces system act on is it. The fundamental unit of momentum velocity 1 is kg m s ; this is equivalent to the Velocity is a vector quantity; mass is a newton second (N s). scalar, so momentum is also a vector − always specify both its magnitude and direction. The act law on equal a of conservation system, to the the vector of momentum vector sum sum after of the states the that, when momenta no before external the forces collision is There are two par ts to this law. collision: • There is no change in m × u 1 + m 1 where × u 2 m , m 2 m 1 velocities + = × v 1 … are the + m 1 × v 2 masses of + momentum during a collision. 2 the objects u 2 , u 1 and v , v 1 … are the final velocities of … are the • Providing no ex ternal forces act, initial 2 the internal forces do not make any objects. 2 difference as they must This can also be expressed as ∑ mv = 0, meaning that the sum of the act equally (N3) on all par ts of mass × A girl she throws exerts The ball When still velocity the gains Earth, each ball the gains object high ball into consists upward force which before the causes momentum system downwards the a on for it the recoils of the the an Earth as the remains plus she opposite the the collision stationary and external However, ground in She after accelerate because momentum. on air. to and gain force girl girl throws has and also to but vertical giving direction must the on the force Quote both par ts when writing about this law. it. her momentum ball’s the object. ball increases the 0. speed. acted ball, be to velocity . Newton’s second law of motion These momenta are equal and opposite. (N2) can be written in terms of change in velocity: Example 2.4.1 ⎛ Δv ⎞ F = m× An apple Earth. is released Discuss how from the rest and falls conservation of towards the momentum surface applies of to , where ∆ means ⎝ Δt ⎠ the “change in”. Earth–apple system. This can be rewritten (provided mass is constant) as Solution The points to make are: ( m × Δv ) F = Δ(momentum) = Δt • the forces on the Earth and the apple are equal and Δt opposite which means • no external force acts on this isolated system change in momentum • changes and in the momentum of the Earth and the apple are equal opposite change in time This interpretation of N2 shows that force acting is the rate of change of • the momentum of the Earth–apple system stays the same and momentum. is conserved. 21 2 M E C H A NI CS Example 2.4.2 An air-rifle pellet is fired into a wooden block resting on a When carrying out calculations, rough table. keep track of the direction in wooden which objects move. Your work The needs to be presented so that it is then immediately clear what changes in during a collision. distance pellet and the block block a slide along straight the line table for a air-rie of 2.8 m before pellet coming to rest. One way is to define a direction as positive (perhaps even drawing it on a diagram); velocities in The speed of the immediately the opposite direction are then block after the 2.8 m 1 collision is 4.8 m s negative. When the answer The mass of the pellet The mass of the block is 2.0 g. is negative, indicate that the direction has changed. K inetic energy E a) Determine b) Compare the the is speed initial 56 g. of impact kinetic of the energy of pellet. the pellet with the kinetic is k energy of the pellet and block immediately after the collision. 2 1 1 1 2 mv = 2 2 m p 2 ( mv ) × , = Solution 2m a) The initial momentum of the system (pellet + block) is where p is the momentum of a 0.056 × 0 + 0.002 × u . The 1 initial velocity of the pellet is u movement to the and 1 mass m right is taken to be positive. 1 The final momentum The initial and final of the system momenta are is 0.058 × 4.8 kg m s equal. 2 × 5.8 × 10 4.8 1 This means = = that 1 × 2.0 2 significant 1 139 m s (= 140 m s to 3 10 figures). 1 3 b) The initial kinetic energy × is 2.0 × 10 2 × 139 = 19.3 J 2 The final kinetic energy of block and pellet is 1 3 × 58 × 10 2 × 4.8 = 0.67 J 2 About 18.6 J sound, deformation As the eventually Even though energy of Example the known be as in spring moving In (Example appears during a is the energy wood down thermal of the later, is transferred energy block all will and the from appear some kinetic as thermal energy will form. conserved is This in reduced all collisions, because of the the kinetic collision. this. kinetic energy energy collisions, 2.4.3) the the objects When other of collision. slows to illustrates which the of block transfer inelastic. elastic. during momentum 2.4.3 A collision 22 0.67 J) energy energy . to (19.3 J kinetic or is a removed conserved such when collision. is as the from in a release stationary gun the system collision, of is a it is is said compressed fired, kinetic energy 2.4 MOMENTUM Example 2.4.3 Two masses, connected Mass m Mass M m by and a M, on a frictionless compressed moves with spring velocity and horizontal table are released. v m moves with velocity v M a) State b) Deduce c) the change the Deduce the in the change change in in momentum momentum energy of of of the mass mass m M masses. Solution a) The initial change momentum in momentum of of the m system is m × (both masses) is zero; the v m b) initial momentum = final momentum mv m So 0 = mv + . Mv m Therefore, v = − M M M c) The initial energy is zero. The total 2 1 1 1 2 + m = This Impulse is the to a graph of in m M change 1 = 2 2 also links + m 2 m mv M 2 v 2 Mv energy is: 2 Mm 2 mv final m + M 2 2 v m M energy . the variation with time of the force acting on Impulse is the change in an object. The graph in Example 2.4.4 is typical of that often seen when momentum of an object and is one object collides with another. equal to force × time for which the force acts. Example 2.4.4 The unit of impulse is N s. In 1 A ball of vertical mass wall 0.075 kg with a strikes a fundamental units, this is kg m s horizontal The area under a graph of force F max 1 velocity of 2.2 m s impact is The time rebounds for against time is equal to the impulse ecrof horizontally . and the and can be used to estimate the 90 ms. momentum change when a force acts on an object. During the collision, 25% of the 0 ball’s initial transferred kinetic to energy other a) Determine the rebound b) Determine the impulse c) The graph varies shows with Estimate how 0 is energy 90 time / ms stores. speed given the of to force the the F ball ball from by exerted the the by wall. wall. the wall on the ball time. F max Solution 1 2 a) The initial kinetic energy of the ball × 0.075 × 2.2 is = 0.182 J 2 25% of this 0.182 × energy is transferred from kinetic energy . 3 So, = 0.136 J remains. 4 2 × 0.136 1 This gives a rebound speed v = of 1.90 m s 0.075 The direction is to the left. 23 2 M E C H A NI CS b) The in impulse momentum m(v u). speed the The to × of to the However, the change 0.075 c) given right of (1.9 impulse the wall ball care and v and is is momentum ( 2.2)) equals ≡ is the in needed the to 0.075 the by × area the 4.1 = under the speed left is equal same with final the ball the direction. sign to to the as u left. is change This the is initial Therefore, is 0.31 N s the to the left. force–time graph. 1 0.31 = × 0.090 × F so max F = 6.8 N. max 2 Practice problems for Topic 2 Problem 1 Problem 4 An object is thrown ver tically upwards at the edge of a A bus travels at constant speed of 6.2 m s ver tical sea cliff. The initial ver tical speed of the object inclined upwards at 6.0° to the horizontal. The mass of the 1 1 is 16 m s along a road 3 and it is released 95 m above the surface of bus is 8.5 × 10 kg. The total output power of the engine of the sea. the bus is 70 kW and the efficiency of the engine is 35%. Air resistance is negligible. a) Draw a labelled sketch to represent the forces acting on the bus. a) Calculate the maximum height reached by the object above the sea. b) Calculate the input power to the engine. b) Determine the time taken for the object to reach the surface of the sea. c) Determine the rate of increase of gravitational potential energy of the bus. Problem 2 d) Estimate the magnitude of the resistive forces acting An object is at rest at time t = 0. The object then on the bus. 2 accelerates for 12.0 s at 1.25 m s e) The engine of the bus stops working. Determine, for time t = 12 s: (i) Determine the magnitude of the net force a) the speed of the object opposing the motion of the bus at the instant at which the engine stops. b) the distance travelled by the object from its rest (ii) Discuss, with reference to the air resistance, the position. change in the net force as the bus slows down. Problem 3 An automobile of mass 950 kg accelerates uniformly Problem 5 1 from rest to 33 m s in 11 s. a) Calculate the resultant force exer ted by the A hammer drives a nail into a block of wood. The mass of the hammer is 0.75 kg and its velocity just 1 automobile to produce this acceleration. before it hits the nail is 15.0 m s ver tically downwards. After hitting the nail, the hammer remains in contact b) The manufacturer claims a maximum speed of with it for 0.10 s. After this time, both the hammer and 180 km per hour for the automobile. Explain why an the nail have stopped moving. automobile has a maximum speed. a) Deduce the change in momentum of the hammer during the time it is in contact with the nail. b) Calculate the force applied by the hammer to the nail. Problem 6 A magazine ar ticle suggests that wearing seat belts in vehicles can save lives in collisions. Explain, using the concept of momentum, why this is correct. 24 THERMAL 3 3 . 1 T E M P E R AT U R E A N D You must know: ✔ the molecular P H YS I C S E N E R G Y You should be able to: theory of solids, liquids and gases ✔ describe internal ✔ the meaning connection absolute of to internal the energy concepts of and what is meant by a phase ✔ what is meant by a temperature and ✔ describe change ✔ that ✔ describe solids, by 1 K the when specic capacity is temperature there latent change the is heat state of no is 1 a phase the of energy 1 kg change the kg of energy of a the change molecular liquids terms of in terms of molecular and differences between gases use Kelvin of a and Celsius temperature scales and required convert change in scale ✔ to change behaviour ✔ heat temperature energy its temperature temperature specic C H A N G E S between them substance state and required that ✔ sketch to and interpret temperature substance and with of a energy graphs substance showing varies transferred to or how with from the time the substance. ✔ how to calculate specic latent capacity energy heat changes differences changes and and apply involving specic these heat concepts experimentally . When energy is transferred to a substance, its temperature rises. When Physicists use macroscopic the substance cools, energy is transferred away from it. Temperature and microscopic models. In is sometimes described as the ‘degree of hotness’ of a body . At the Topic 3.1, the behaviour of bulk microscopic level, temperature is related to the motion of the atoms materials can be modelled using and molecules. Temperature can be identified as the mean kinetic latent heat and heat capacity. energy of particle in the ensemble. Alternatively, the kinetic theory of Temperature scientific is scale temperature defined is the scale. (temperatures using Kelvin temperature scale. It A temperature defined in terms is also scale of the scales. known requires The as present-day the absolute two fixed properties of a Topic 3.2 models the motion of atoms and molecules—imagined points as infinitesimally small par ticles. substance). Use the differences and The connection in substance a microscopic When than In the that between gives energy average of a a direct link and between the the motion of the macroscopic particles and assist your learning. kinetic energy group, terms, the the of one first kinetic group group energy of has of a the particles higher average is higher temperature. particle is 3 given by E Celsius temperature is always expressed in °C. The kelvin unit kT , = similarities of such models to descriptions. second mathematical temperature where k is the Boltzmann constant and T is the k 2 absolute (kelvin) never has a ° sign and is written temperature. K . Temperature differences are either written as K or deg (meaning ‘change of degrees’). When you are giving an answer that is a change in temperature, never use °C. 25 3 THERMAL P H YS I CS The potential-energy contribution to internal energy appears only in Temperature is a measure of liquids and solids where the substances are bound by intermolecular the average kinetic energy of a forces. Gases with two or more atoms per molecule can have rotational collection of moving atoms and and vibrational energies as well. molecules. Example 3.1.1 For the Kelvin scale, the two xed points are 0 K and 273.16 K . These Outline are absolute zero (the temperature internal the difference energy of an in internal ideal energy of a piece of metal and the gas. at which atoms and molecules Solution have no kinetic energy, equal to The internal energy of the metal equals the total kinetic energy of –273.15 °C), and the triple point of the water (equal to 0.16 °C). The triple atoms energy point is where all three phases of plus arises the potential from the energy bonds of the system. between the metal This potential atoms. water co-exist in a sealed container; The molecules of an ideal gas have only kinetic energy . One in not of the it occurs at a unique temperature assumptions of an ideal gas is that the particles it do interact and pressure. through molecular bonds and so there is no potential energy to Kelvin dened the degree to be consider. identical in both kelvin and Celsius. To conver t from °C to K , add 273. To conver t from K to °C, subtract 273. Example 3.1.2 Sketch a graph to show the relationship between the internal The internal energy of a substance energy of an ideal gas and its temperature measured in degrees is equal to the sum of the random Celsius. Explain the key features of this graph. kinetic energy of a collection of Solution par ticles plus the potential energy that arises from intermolecular The forces. is ygrene lanretni The concept of an ideal gas is graph some This the is gas because is absolute is covered in Topic 3.2. at the indicates internal 0 K the directly this is at in 0 °C, the internal is energy axis; to Absolute the of the zero intercept in there gas. proportional temperature. and temperature value that, energy on Celsius, this –273 °C. 0 0 The term par ticle describes atoms and molecules and There high temperature / °C are four known temperatures. states The of other matter. three One states of these, are solid, plasma, liquid only and exists at gas models them as small point-like All substances are made up of atoms and molecules and the differences objects without size or shape. between states Normally , a liquid The topic. 26 when and phase is due • melting then • boiling names and to a link used freezing nature energy changes changes and the thermal change Phase to in is the condensing These the DP (changes the bonding transferred gas. to of are latent Physics a solid, it called phase heat between (changes to between changes course solid between them. first changes later in are: and liquid becomes liquid) and gas). this 3 .1 Phase a changes solid. with The can graph energy be in demonstrated Figure 3.1.1 by transferring shows the thermal variation of energy T E M P E R AT U R E AND ENERGY C H A NGE S to temperature transferred. Gases have individual par ticles that are independent of each other molecules and move freely within a container, lling it completely. Pressure arises condenses D as the par ticles interact with the E erutarepmet gas container walls. boils Liquids can move around within the freezes bulk of the material. The par ticles B C only interchange with nearest liquid neighbours, which enables a liquid melts to have a denite volume but be able to ow. A Solids allow little, if any, movement solid between par ticles, which rarely energy transferred exchange positions with each other. Solids have a xed shape. Figure 3.1.1. The graph Variation of temperature with energy transferred can also be plotted as temperature against time (when the The specic heat capacity c of a energy is input at a constant rate). of a substance substance is the energy required When heat the state capacity change. The does internal not change, energy of the its temperature substance is rises—a increasing to change 1 kg of the substance by 1 deg (or 1 K). The unit of specic 1 while potential energy is largely unchanged (this is only approximately heat capacity is J kg 1 K . In 2 fundamental units, this is m true, especially when expansion or contraction 2 s 1 K occurs). Q While the state changes, the temperature is constant. Energy is c = m × being transferred constant—a into latent the heat potential form and the kinetic energy ΔT is where change. Q is the energy transferred (in J) m is the mass (in kg) Example 3.1.3 ΔT is the change in temperature The internal heating. Its energy of a subsequent piece of increase zinc in is increased temperature by is 1.5 kJ 11 deg. by (in K) The piece The specic latent heat L of a of zinc has mass 0.35 kg. substance is the energy required a) Explain the b) Calculate meaning of internal energy and to change the phase of 1 kg of the heating substance. The unit of specic the specific heat capacity of zinc. 1 latent heat is J kg 2 units, this is m . In fundamental 2 s Solution a) The internal energy is the sum of the potential energy and the Q L kinetic energy amount of of the energy zinc stored atoms. in the It can also be described as the = m zinc. To specify a specic latent heat change, state the type of phase Heating is the process of transferring energy using a non- change that is occurring. For mechanical or thermal pathway from an energy source to the example, “ The specic latent zinc. The zinc is acting as an energy sink. heat of freezing of ice (in other Q 1500 words, going from water to ice) is 1 b) c = = m ∆T The specific method of liquid, also are measured heat K 1 0.34 MJ kg capacity at A hot known The other can of a solid material of mixture When be can known temperature. resulting temperature. the 390 J kg 0.35 × 11 mixtures. known. known, = 1 The of one determined masses solid of be temperature the and of the liquid specific is using added solid heat a the to and reaches ”. a cold liquid final, capacities is determined. 27 3 THERMAL P H YS I CS S AMPLE STUDENT ANS WER In an experiment to determine the specic latent heat of fusion of ice, an ice cube is dropped into water contained in a well-insulated calorimeter Present any multi-step solution of negligible specic heat capacity. The following data is available. clearly with a clear description of Mass of ice cube = 25 g Mass of water = 350 g Initial temperature of ice cube = 0˚C Initial temperature of water = 18˚C Final temperature of water = 12˚C Specic heat capacity of water = 4200 J kg each step. An examiner can then give you par tial credit if you have made an error elsewhere. 1 1 K a) Using the data, estimate the specic latent heat of fusion ice. ▲ There are correct This of the water energy and energy the Then 7.56 kJ is by (heat gained melted. that lost by the ice is a once it Q = mcΔT melt the Q = 0.350 kg ▼ The equation heat achieved 2/4 marks: m = 0.350 kg c = 4200 J/kg.K ΔT = 6 ˚k × 4200 × 6 ˚k = 8.82 kJ = 0.025 kg Δ = 12 ˚C ice. 0.025 kg for have available m to could has recognition energy answer cooling capacity) the there the [4] calculations is capacity incorrect not latent (it 7 .56 kJ is = × 4200 0.375 × × 12 ˚k 4200 = × 1.26 kJ c = 4200 J/kg.K ? heat). 7.56kJ = 4 .8 ˚K + 273 = 277 .8 ˚K 1.575kg.K ▼ The the answer answer is fails correct, to but appreciate the b) The experiment is repeated using the same mass of ice. This time, the signicance of the command term ice is crushed. ‘suggest’. a This hypothesis explanation answer means and of needs it the to to propose requires proposal. go on to Suggest the eect of this, if any, on the time it takes the water to reach some The say its nal temperature. This the surface when ice area crushed, can of so interact the the more 3 . 2 ice answer could is molar ✔ the meant mass water by and equation of and the T he time it takes to reach A the of ✔ ✔ gas one the laws gas that gas pressure, the ideal Avogadro state and model laws is marks: the nal temperature will decrease. You should be able to: for an gas, mole, ✔ sketch constant ideal to gas ✔ an experimental between are an empirical ideal and and interpret interpret solve changes problems ideal p–V, of using P–T state the and of a V–T graphs gas equation of state for gas investigation law differences 0/1 G A S an ✔ achieved quickly . M O D E L L I N G what have increases You must know: ✔ [1] that and that a real the gas ✔ investigate ✔ understand model of model and at least one aspects an ideal gas law experimentally of the molecular gas: the assumptions kinetic of the kinetic how they lead to a theoretical model. theoretical. Pressure arises walls a of transfer gas with all momentum Quantity of A mole atoms, of three container matter a when is the mole phases depends they way of of on matter. the collide with scientists electrons, a rate The at the of the on gas the particles wall. compare mole pressure which numbers ions, of always objects. gets you 23 about 28 6.022 × 10 objects (atoms, electrons and ions respectively). 3.2 A mole of atoms of a chemical element has a mass equal to its MODE LLING A GA S atomic force mass number in grams. For example, one mole of the isotope of carbon-12 The ( area C 6 as ) behaviour a mass of 12 g. This is known as the molar mass 2 unit is N m of gases at extremes of temperature and pressure , which is the same as a pascal, Pa. In fundamental units, is 1 complicated. particles . Its Pressure is dened as 12 A simplified collide model elastically and of in an ideal which gas no is used in which intermolecular the forces 2 s this is kg m act. In solids, a normal force applies For a real gas, there are effects between molecules and with the walls. through a contact area between the Within to a make few the degrees gas of absolute zero, other effects become important non-ideal. solid and the surface on which it rests. Real gases have behaviour close This can be summarized in the to ideal only for low pressures, general gas equation for two states, There is more detail on pressure in a liquid in Option B.3. low densities and moderate 1 and 2, of a gas: temperatures. Treat gases as ideal p V 1 p 1 V 2 unless told otherwise. 2 = T T 1 2 The mole is the fundamental SI An equation of state describes a The equation can also be written as gas using three variables: pressure pV = and RT pV = Nk T B Equations of state are possible for unit for the quantity of matter of a substance. It corresponds to the N p, volume V and temperature T. mass of a substance that contains N A 23 6.022 × 10 real gases but need extra terms par ticles of the where substance. to account for high densities and R is the (ideal) gas constant 23 par ticle interactions. 1 (8.31 J mol The number 6.022 × 10 1 K is known ) as the Avogadro constant N The equation of state for an ideal A k is the Boltzmann constant B gas is 23 (1.38 × 10 pV = The number of moles 1 J K ) n of a substance = nRT N is number of molecules. number of molecules N where n is the number of moles. = N N A A Example 3.2.1 5 An ideal gas in a container of volume 1.2 × 10 3 m has a pressure of Notice that both sides of the 5 1.5 × 10 Pa at a temperature of 50 °C. equation of state have the units Calculate the number of molecules of gas in the container. of energy. You can think of R as being analogous to the specific Solution The temperature must be in kelvin: 50 + 273 = heat capacity of one mole of a 323 K. gas, with k 5 pV n = = = 8.31 × solution being analogous to B 5 × 1.2 × 10 4 RT The 1.5 × 10 requires 6.71 × 10 the specific heat capacity of one mol 323 the molecule. number of molecules = nN A 4 = 6.71 × 10 23 × 6.02 × 10 20 = 4.04 × 10 molecules When using the general gas You must law. This know the details of one experimental investigation of a gas equation, the units of pressure might be tested in Paper 1, 2 or 3. and volume must match on both Historically , involving equation of gas three state behaviour separate was gas identified laws that, through taken experiments together, reflect the sides of the equation. The only unit allowed for temperature in the equation is kelvin. 29 3 THERMAL P H YS I CS These graphs summarize the essential features of the gas laws. Charles's law Pressure law Boyle's law V ∝ T p ∝ T pV = constant pressure constant volume constant erusserp emulov erusserp temperature constant 0 0 0 0 0 1 0 absolute temperature absolute temperature volume The gas kinetic model particles and of a their gas is based on the following assumptions about behaviour. The gas laws were suggested by scientists in • A gas consists the 18th century following negligible experimental work . Such laws are say , said to be empirical. On the other individual the of particles; compared average with distance the the total total between volume of volume of particles is the the particles gas greater (or is you than could their size). hand, the kinetic model of a gas • Particles have the • Particles are Particles collide same mass. stems from a theoretical standpoint involving assumptions in constant, random motion. about the gas. These macroscopic • elastically with each other and the walls of the and microscopic approaches fit container. together confirming our view of gas • Interactions between particles can be ignored (so they do not exert proper ties. force • The on each time between • Gravity for other). a particle collision is negligible compared with the time collisions. can be ignored. Know the meaning of these assumptions and recognise how they affect the kinetic model. The Example 3.2.2 model leads to the relationship between the pressure of the gas and the mean square speed of A particle lengths x, of y mass and m moves with velocity u in a box with side z the par ticles. The steps in the y derivation of the model are given in Example 3.2.2. m z u x The particle repeated a) i) ii) Calculate State with b) strikes elastic The the the box the the time t shaded with N faces with at for right opposite between expression contains x-direction end collisions the angles collisions change and makes faces. in with the shaded momentum per face. collision face. identical speed u , particles, making all elastic moving parallel collisions at the to ends. x Determine c) The model the is average refined so force that F N on the particles 2 speed 30 c move randomly in shaded the box. of face. average the squared 3.2 The speed in direction y is v The speed in direction z is w Deduce an expression for u A GA S 2 2 i) MODE LLING in terms of . c 2 ii) Deduce an expression for F in terms c of 2 Nmc iii) Show pV that = , where p is the pressure of the gas and 3 V is its volume. Solution 2x a) i) The particle travels a distance 2x at a speed u, so t = u ii) The change face = Δp = in momentum for each collision force = rate the of change of momentum ∆p = 2 mu = ∆t 2x So, for N shaded 2mu b) at particles, force F = u ∆p , N ∆t which is: 2 u N Nmu × 2 mu × = 2x c) i) For one 2 c particle, 2 = x 2 u + the magnitude v + all ii) The the particles, average direction its velocity c is given by w 2 For of 2 as squared we do 2 c = 2 u + speeds not see v 2 + w must be the directional 2 same in differences each within gases. 2 c 1 Nmc 2 Therefore, u = and F = 3 3 x 2 F p iii) 1 = 2 Nmc 1 = × A 3 x 1 Nmc 3 xyz = but yz xyz 2 1 Therefore, p The equation speeds of average the of state particles. squared = 3 the pressure and temperature can volume also 3 mc , T = V of link a gas to directly the to the 2 2 T box speeds: Nmc Nk pV V The of Nmc or links volume 2 Nmc = 3 = 1 2 and = k therefore T = mc B B 2 3k 3 2 B The right-hand The left-hand molecule ( EK ) side side . is is The a the average measure units of k T of kinetic the are energy kinetic of a energy equivalent to gas of a particle. gas joules. B 31 3 THERMAL P H YS I CS Example 3.2.3 A cylinder pressure a) b) of of fixed volume 490 kPa Determine Calculate the the and a volume average contains 15 mol temperature of the kinetic of of an ideal gas at a 27 °C. cylinder energy of a gas molecule in the cylinder. Solution 15 × 8.31 × (27 + 273) 3 a) pV Use = nRT to give V = = 0.076 m 5 4.9 × 10 3 3 −23 b) Use E = k K . T E B = × 1.38 × 10 −21 × 300 = 6.2 × 10 J K 2 2 S AMPLE STUDENT ANS WER 0.46 mol of an ideal monatomic gas is trapped in a cylinder. The gas has 3 a volume of 21 m ▼ The kinetic assume answer model constant does not does velocity make it a) State how the internal energy of an ideal gas diers from that not and of a real gas. this or the it real is referring to [1] clear This whether and a pressure of 1.4 Pa. the answer could have achieved 0/1 marks: ideal case. Internal energy is constant as molecules move at constant velocity. ▲ The and distinction real gases is between clear even ideal though This the word ‘real’ the answer. The intermolecular potential does by appear deduction force energy reinforced not is the that implies correct no Ideal gas have achieved 1/1 marks: ignores intermolecular force between molecules is between collision. So there is no potential energy, and answer statement of begins the gas with a clear This energy only. be clear used. (because substituted in The the in substitution the the equation) numbers same and answer could have achieved 2/2 marks: equation pV to are order the = nR T is 1.4 × as answer 21= 0.46 × 8.31 × T 29.4 = is 3.8226 quoted to an appropriate number = of signicant 7 .7 Kelvin gures. T he 32 contains energy . b) Determine, in kelvin, the temperature of the gas in the cylinder. ▲ The in about kinetic kinetic could no and statement answer in temperature of the gas is 7 .7 Kelvin. [2] 3.2 MODE LLING A GA S Practice problems for Topic 3 Problem 1 Problem 5 a) X and Y are two solids with the same mass at the The pressure in a container is increased using a bicycle 3 pump. The volume of the container is 1.30 × 10 same initial temperature. Their temperatures are raised by the same amount; they both remain solid. 4 The pump contains 1.80 × 10 The specific heat capacity of X is greater than that of 3 m 3 m of air at a pressure of 100 kPa and a temperature of 300 K . Y. Assume that the air acts as an ideal gas. Explain which substance has the greater increase in Assume that all the air molecules from the pump are internal energy. transferred into the container when the pump is pushed in. b) Cold water, initially at a temperature of 14 °C, flows a) The air in the container is at an initial pressure of over an insulated heating element in a domestic 150 kPa and a temperature of 300 K . water heater. The heating element transfers energy at a rate of 7.2 kW. The water leaves the heater at a (i) Calculate, in mol, the initial quantity of gas in the container. temperature of 40 °C. 1 The specific heat capacity of water is 4.2 kJ kg 1 K (ii) Calculate, in mol, the quantity of gas transferred to the container every time air is pumped into it. (i) Estimate the rate of flow of the water. (iii) The temperature of the gas in the pump returns (ii) Suggest one reason why your answer to par t b) i) to 300 K after the pump has been used. is an estimate. Calculate the pressure in the container after the Problem 2 pump has transferred one pump-full of air into a) Distinguish between thermal energy and internal the container and its temperature has returned to energy. 300 K . b) Outline, with reference to the par ticles, the difference b) Explain, with reference to the kinetic model of an in internal energies of a metal and an ideal gas. ideal gas, why the gas in the container has pressure Problem 3 and why this pressure will increase when gas Use the kinetic model to explain why: molecules are transferred to the container. a) the pressure of an ideal gas increases when heated Problem 6 3 Air in a container has a density of 1.24 kg m at constant volume at a 5 pressure of 1.01 × 10 Pa and a temperature of 300 K . b) the volume of an ideal gas increases when heated at constant pressure. a) Calculate the mean kinetic energy of an air molecule in the container. Problem 4 A quantity of 0.25 mol of an ideal gas has a pressure of b) Calculate the mean square speed for the air 5 1.05 × 10 molecules. Pa at a temperature of 27 °C. a) Calculate the volume occupied by the gas. c) The temperature of the air in the container is increased to 320 K . 1 b) When the gas is compressed to of its original Explain why some of the molecules will have speeds 20 6 volume, the pressure rises to 7.0 × 10 Pa. much less than that calculated in par t (b). Calculate the temperature of the gas after the compression. 33 O S C I L L AT I O N S 4 4 . 1 WAV E S O S C I L L AT I O N S You must know: ✔ what is meant by You should be able to: an oscillation ✔ sketch and harmonic ✔ the AND denitions of time period, interpret motion displacement, for simple displacement–time, frequency , velocity–time, amplitude, graphs of phase acceleration–time and difference acceleration–displacement ✔ the conditions for ✔ the relationship simple harmonic motion. ✔ displacement in between simple acceleration harmonic describe and one the cycle energy of an changes that take place in oscillation. motion. A pendulum—a mass swinging at the end of a string—is an example Time period T is the time for one of an oscillating system. A cycle for this system is the movement of the cycle of the oscillation. Its unit is mass from the rest position at one end of the swing, through to the the second (s). opposite Frequency f is the number of cycles side to side the and other back is half to a the original rest position. Motion from one cycle. of the oscillation in one second. Its The 1 unit is the her tz (Hz or s rest position (in the middle of the swing for a pendulum) is also ). known as the equilibrium system will Simple harmonic position which is the position where the Time period and frequency are be when not oscillating. 1 connected by f = T Amplitude x motion (shm) is an oscillation for which is the maximum 0 displacement of the oscillating acceleration ∝ −displacement object from its equilibrium position. For simple harmonic motion: This can be expressed as an angle or a distance. Displacement x is the distance between the equilibrium and • • the acceleration the vector this is the of the direction meaning object of of is directly acceleration the negative is proportional opposite sign in the to to its the displacement displacement; equation. instantaneous positions and, as a vector requires a direction, can be Example 4.1.1 positive or negative. The equilibrium position is the position to which the system A mass, and with released an equilibrium from rest. Its position motion is at O, simple is displaced to point X harmonic. returns when it is not oscillating. Identify where the acceleration of the mass is greatest. Phase dierence is the dierence, Solution in degrees or radians, between two oscillations at the same instant in time. Because the So, acceleration distance from acceleration Graphs for shm, is O, is proportional the greater greatest showing at the the to –displacement, magnitude mathematics of shm in HL Topic 9.1. Concepts in Topic 4 are directly connected to the mechanics of Topic 2. 34 with time, are the the greater acceleration. X. variation of acceleration, There is more about the displacement of given in Figure 4.1.1. velocity and 4 .1 (a) 1.0 (b) 2.0 0.5 (c) 1.0 4.0 2.0 0 0 s m / 2− x 0 s m / v 1− mc / 0 0 0 a t / s t / s t / s –0.5 –1.0 –2.0 –1.0 –2.0 –4.0 Figure 4.1.1. x Instantaneous the t, v t and a velocity at displacement–time a O S C I L L AT I O N S t graphs for simple harmonic motion particular time is equal to the acceleration gradient of graph. 0 Instantaneous acceleration is equal to the gradient of the 0 velocity–time ∆s Mathematically , v ∆v = and a makes This is the a–t (Δ = ∆t This displacement graph. means ‘change in’). ∆t graph the inversion of the x–t graph. Figure 4.1.2. expected, since acceleration ∝ −displacement (Figure Acceleration– 4.1.2). displacement graph for simple • The v–t graph lags the x–t graph by 90° (or is 90° out of phase). harmonic motion • The a–t Energy graph transfers mass–spring from is elastic sustains 180° occur system, potential the out of throughout the kinetic energy oscillation phase in with the indefinitely x–t oscillator energy the the of the spring. when cycle. mass This no graph. For a transfers energy friction to and Although for standard level you only need to describe energy transfer changes in shm qualitatively, you acts. may find Figure 4.1.3 a good way Figure 4.1.3 shows the energy variations plotted with time. to remember the links between E total kinetic and potential energies. E k The energy graphs vary as 2 sin rather than the sine curves of E p Figure 4.1.1 (a–c). This is covered in more detail in Topic 9.1. 1 2 3 4 5 Phase is used extensively in Topic 9. displacement–time Figure 4.1.3. Energy variations against time for shm There are also links to Topic 6 where radian measure and degrees are Example 4.1.2 π rad used: 90° can be written as 2 A mass hanging on a spring is pulled vertically down 0.15 cm from and 180° written as π the equilibrium equilibrium position position and 0.75 s released. after The mass returns to rad. the release. The kinetic energy cycle: State: • has double the frequency of the a) the amplitude b) the time motion 2 • is never negative (E period for the oscillation. ∝ v ) k • has a dierent shape from the Solution sine curves in Figure 4.1.1. a) The amplitude is the distance from the equilibrium position to The total energy is constant with the maximum displacement. This is 0.15 cm. time when energy losses are zero. 1 b) The mass has travelled a cycle when it reaches the equilibrium The variation with time of the stored 4 elastic potential energy E has the p position for the first time after release. The time period is same frequency as the E –time k 4 × 0.75 s = 3.0 s. graph but is π out of phase. 35 4 O S C I L L AT I O N S AND WAV E S Example 4.1.3 Which of simple A The the following harmonic statements motion acceleration is about always The acceleration and velocity C The acceleration and the D The is true for an equilibrium away B same an from are object performing position O? O. always displacement in opposite from O are directions. always in the direction. graph of acceleration against displacement is a straight line. Solution The correct The acceleration velocity , means A and so that C answer B is is can D. be acceleration cannot in incorrect. be the The and correct. same or the negative opposite sign displacement Response D is in are the the direction definition always to of opposed alternative way shm so to Multiple-choice questions demand express this. care. The incorrect responses in Example 4.1.3 test the relationships between acceleration, S AMPLE STUDENT ANS WER velocity and displacement. The A mass oscillates horizontally at the end of a horizontal spring. The mass negative sign in a ∝ −x means moves through a total distance of 8.0 cm from one end of the oscillation that, because the acceleration and to the other. velocity are 90° out of phase, the a) State the amplitude of the oscillation. acceleration and velocity can be in the same direction or opposite. This 8.0 answer could have achieved 0/1 [1] marks: cm ▼ The student has not visualized the arrangement. from one passes The extreme through to the mass the is b) Outline the conditions that the system must obey for the motion to be moving other simple harmonic. and This position half distance. way The through amplitude is answer could that are would are the two required. be symbols better a 4.0 cm. x and a 4 . 2 dene for must be in what wave ✔ meant the opposite by wavelength, is overall disturbance transmitted in of a frequency the wave and ✔ without ✔ medium explain between transverse particle transverse sketch and of electromagnetic ✔ the nature of sound ✔ the of a medium distance–time graphs the of that solve problems for and transverse waves and involving describe an wave speed, wavelength experimental wavelengths electromagnetic for the spectrum. the speed method of leads waves waves magnitude regions in longitudinal interpret investigating 36 be waves ✔ principal and longitudinal frequency nature of must and ✔ the motion distance–displacement waves ✔ order a You should be able to: to distinction and W AV E S speed longitudinal to x clarity . energy the direction to x and ✔ marks: it T R AV E L L I N G is 2/2 the You must know: ✔ achieved conditions However to have this proportional ▲ These [2] equilibrium sound. for 4.2 Waves transfer energy without any overall change in the T R AV E LLING WAV E S medium Polarization of transverse waves through which they pass. is discussed in Topic 4.3 For transverse direction For of waves, energy longitudinal direction of both wave waves, energy Longitudinal particles is the medium oscillate at 90° to the particles oscillate in the same direction as the propagation. waves types of propagation. cannot be polarized, but otherwise the physics of similar. 1 cm equilibrium position with no wave position of par ticles with wave wavelength displacement / cm displacement 0.5 / cm 0.5 amplitude amplitude 0 0 2 4 dist ance / cm –0.5 −0.5 trough Figure 4.2.1. A graph of crest trough centre of centre of centre of compression rarefaction compression Displacement–distance graph for transverse and longitudinal waves displacement–distance provides a ‘snapshot’ of the shape of Wavelength λ is the distance a wave at one moment in time. Figure 4.2.1 shows the graph together between the two nearest points on with its interpretation for each type of wave. the wave with the same phase. A representation given same the by as time a of the motion displacement–time the graph period of in Figure the wave, of an graph 4.2.1, individual (Figure but rather this than point 4.2.2). graph its on This gives a the wave may look direct is Wave speed c is the speed at which the value the wave moves in the medium. for wavelength. displacement / cm 0.5 0 T time 3T T 2 Ensure that you read graph axes 2T 2 carefully both for the quantity and –0.5 the unit. Figure 4.2.2. Displacement–time graph for wave motion You need to be able to derive the equation for the speed of a wave: c = f λ Remember that a displacement– whereas a displacement–time graph gives the time T that one par ticle takes to go through one cycle. The wave speed is therefore distance graph shows that the wave λ 1 = moves forward by λ in one cycle, T f λ because f = T 37 4 O S C I L L AT I O N S AND WAV E S Example 4.2.1 The shortest distance between two points on a progressive π transverse wave which have a phase difference of rad is 0.050 m. 3 The frequency Determine of the the speed wave of is the 500 Hz. wave. Solution 1 π There is a discussion of radian rad is 60° which is of a cycle. 6 3 measure in Topic 6.1. This means that the wavelength of the wave is 6 × 0.050 = 0.30 m. –1 The speed Sound The is There The The the of are areas and c = f through gas the of are low high positions λ = 500 gases the × and medium of positions 0.30 = 150 m s liquids and as move waves of is is longitudinal as the wave waves. passes and maximum show almost (see 4.3). below above atmospheric atmospheric minimum pressure are at minimum displacement pressure) the points zero and atmospheric refraction, (rarefactions: (compressions: maximum reflection, Topic pressure pressure displacement pressure Sound a is them. where • wave transmitted pressure) • the molecules through • of (in all other the diffraction words, common and are where average). properties interference, but of waves: not polarization Example 4.2.2 a) Outline the transverse b) State i) c) a an difference a longitudinal wave and a wave. example transverse Sound between with a of: wave ii) frequency of 840 Hz a longitudinal travels through wave. steel with a −1 speed of 4.2 km s Calculate the wavelength of the sound wave. Solution a) In a transverse direction of wave, energy the vibrations are perpendicular to the propagation. Notice that the answer to Example 4.2.2 par t b) ii) specifies In ‘sound wave in a gas’. A solid a longitudinal same direction wave, of the vibrations of the particles are in the propagation. surface can transmit a sound b) i) An c) = electromagnetic wave. ii) A sound wave in a gas. wave as a transverse wave. c c f λ so λ = 4200 = f Electromagnetic • do not • travel need at the a = • have the 38 words medium same at and speed 3.0× decreased frequency . m waves: in 8 other 4.9 860 10 speeds can a travel vacuum through a vacuum irrespective of frequency , 1 m s in matter; this speed depends on in 4.2 T R AV E LLING WAV E S wavelength increases energy increases –5 10 –3 nm 10 3 1 nm nm Gamma rays 24 10 6 nm 10 3 nm 1 m 10 m X rays 22 Hz 10 10 20 Hz 10 18 Hz 10 16 Hz 12 10 10 Hz 10 8 Hz 10 6 Hz 10 4 Hz 2 10 Hz high frequency 10 Hz low frequency visible light 14 7 × 10 Figure 4.2.3. 14 Hz 4 × 10 Hz The principal regions of the electromagnetic spectrum and their wavelengths S AMPLE STUDENT ANS WER A longitudinal wave is travelling in a medium from left to right. The graph shows the variation with distance x of the displacement y of the par ticles in the medium. The solid line and the dotted line show the You should know the order of displacement at t = 0 and t = 0.882 ms, respectively. magnitude of the wavelengths for each region, a use for each wave and a disadvantage of using the 4 wave (which may be a medical disadvantage). 2 mm / y x 0 / m –2 –4 ▲ The sense student that this has conveyed direction to the displacement in the wave of is the the parallel particles The period of the wave is greater than 0.882 ms. A displacement to the right of the equilibrium position is positive. better to (although talk propagation a) State what is meant by a longitudinal travelling wave. This answer could have achieved 1/1 wave where the energy is moving in the same direction as the two graphs the b) Calculate the speed of this wave. could have achieved [2] 0/2 marks: by The time taken the solid the amplitude particle fλ for by the line to dashed λ = 1.6(m) f = to travel represented that line 0.882 ms. the speed m in 0.882 of time < 1.13 Hz the wave. There is no credit T 0.3 by represented is gives = motion displacement λ V the separated particles. from answer show particles maximum This of energy’). the 0.3 m. of be ‘direction marks: of motion of the would [1] ▼ The A about it in this 1 approach which tries to use 0.882 1 V < 1.81 ms as an the frequency . additional the but time used was Note that there is error as power of ten quoted in milliseconds here in seconds. 39 4 O S C I L L AT I O N S 4 . 3 AND W AV E WAV E S C H A R A C T E R I S T I C S You must know: ✔ wave shape is You should be able to: indicated by a wavefront ✔ sketch and wavefront ✔ the ✔ a meaning of amplitude and is locally at 90° to a wavefront and direction in which a wave solve is intensity of a wave is power per unit intensity of a wave is involving inverse square amplitude, intensity law sketch proportional and interpret the superposition interpret diagrams of pulses area and ✔ problems the moving ✔ ✔ involving shows and the diagrams rays intensity ✔ ray interpret and waves to ✔ sketch and that illustrate 2 amplitude polarized, ✔ superposition ✔ are added the meaning occurs when two or more only of transverse transmitted beams ✔ solve problems ✔ calculate using Malus’s law the resultant of two waves or pulses polarization waves can be of waves When in move motion. (Figure source algebra or graphs. polarized. As can change and rays both are their used to shape and visualize their these direction changes 4.3.1). waves space The they Wavefronts and wavelength rate of intensity. travelling time, or they in amplitude transfer A point of the same superpose. of wave source of the two energy a medium coincide Superposition at occurs the same whatever place the waves. is measured wave spreads is energy out using the through quantity space. The circular wavefronts plane wavefronts source Figure 4.3.1. and together using ✔ reected waves Plane and circular At radiates distance r power from the P (this source, the this spreads it transfers over a in sphere one of second). radius r wavefronts and rays P 2 which has an area equal to 4 πr .The intensity is given by I = 2 4 πr 2 The intensity I of a wave is related directly to its amplitude A: I ∝ up by A A wavefront shows the shape of When a wave has its amplitude doubled, the intensity goes the wave at one instant. Rays are 4×, so four times as much energy falls on a given area every second. at 90° to the wavefront locally and they show the direction of wave movement. A series of successive Example 4.3.1 wavefronts makes it possible to deduce the origin and history of the A bicycle lamp cone covers and a floodlight deliver their output energy into wave, assuming that wavefronts are that 0.20 of the area of a sphere. one wavelength apar t. The output power of the bicycle The output power of the floodlight a) Calculate the intensity of the lamp is is 1.5 W. 300 W. bicycle lamp when viewed Superposition is when two (or more) by waves add. This addition is vectorial an observer from 20 m away . and you must take account of the b) The observer has an eye with a pupil diameter of 5.0 mm. sign of the displacement. Two waves Determine the light power entering the eye from the bicycle that have the same displacement lamp. but opposite signs give zero displacement overall. c) The the bicycle observer. 40 lamp observer. and Deduce the the floodlight distance of are equally the bright floodlight to from the a 4.3 WAV E C H A R A C T E R I S T I CS Solution a) 1.5 W of light is delivered to 0.20 of a sphere of radius 20 m. Try to use technical language P 1.5 −3 I = = = 2 −2 1.5 × 10 4 × π × 20 × 0.2 superimpose for superpose but 3 b) The correctly. You may be allowed mW m 2 4 πr area of the pupil is π × (2.5 × 10 if there is doubt about what you 2 –5 = ) 1.96 × 10 2 m mean you will not receive credit. 3 So c) the The total power intensity entering en tering is the 1.5 eye × –5 10 × mu s t be = 1.96 × 10 the sa me 29 nW. f or bot h lamps. 2 P P b r f P b Intensity is the amount of energy b = and 2 r b P b = 2 r r b therefore, = 2 P r f r f f that a wave can transfer to an area P f f of one square metre in one second. As the power ratio is 300 : 1.5 = 200 : 1, the distance ratio is –2 14 : 1 The unit of intensity is W m . In –3 and the floodlight Polarization plane the is (Figure energy cannot displacement Polarization glare • stress • 3D has A polarized is = of 280 m the away . fundamental units, this is kg s oscillation waves the can at 90° to because the energy be of the wave polarized particle to one When the wavelengths or because frequencies superposing are displacement; propagation identical, other effects (described in and Topics 4.4 and 4.5) are observed. in uses, where two lenses wave typically: sunglasses modelling polarizing 14 Transverse many analysis × parallel. reduction movies and are 20 restriction 4.3.2). propagation longitudinal • the is of the films are behaviour with used intensity I of different to isolate will be engineering structures polarizations the two reduced in are shown images. intensity when it is 0 incident on polarizer. a polarizer Malus’s law with a polarizing predicts that the axis final at θ to that intensity I of of the the original polarized 2 beam will be I = I θ cos 0 Figure 4.3.2. Polarization of a transverse wave Example 4.3.2 a) State what b) Polarized is meant light of by polarized intensity I is light. incident on an analyser. The 0 transmission axis of the analyser makes an angle θ with the 2 direction of the electric field of the It is easy to forget the cos light. in Malus’s law. Remember that the Calculate, in terms of I , the intensity of light transmitted 0 component of wave amplitude through the analyser when θ = 60 . in the final beam is A cos θ and 0 2 Solution I ∝ A 2 , so I ∝ A 2 cos θ 0 a) Polarized in the light same has the direction of the electric field always plane. I 2 b) I = I cos 0 2 θ = I cos 0 60 = 0 4 41 4 O S C I L L AT I O N S AND WAV E S S AMPLE STUDENT ANS WER a) Radio waves are emitted by a straight conducting rod antenna (aerial). The plane of polarization of these waves is parallel to the transmitting antenna. polarized radio waves 56 km transmitting antenna receiving antenna An identical antenna is used for reception. Suggest why the receiving antenna needs to be parallel to the transmitting antenna. This ▲ The answer implies that radio is waves required will go in ‘all answer not receiving antenna through’. does issue must the be that the aligned polarization receiving with and achieved 1/2 marks: antenna needs to be parallel to the transmitting at an because angle of we 0˚ need or to 180˚ make to the sure the coming receiving radio antenna, so waves all the address radio the have the is ▼ The could a T he maximum answer [2] plane that waves will go through and won’ t be Polarized. antenna of this will b) The receiving antenna becomes misaligned by 30˚ to its original be parallel to the transmitting position. antenna. transmitting antenna original position position after misalignment 30° receiving antenna The power of the received signal in this new position is 12 μW. Calculate the power that was received in the original position. ▼ There the is answer depends a suggestion that on the the This answer 90˚ → could have in component angle rather 0 W than 1 the cosine of the angle. There is no 30˚ is of that 3 attempt no 42 to credit use can Malus’s be law gained. and so 12 μW × 3 = 36 μW achieved 0/2 marks: [2] 4.4 4 . 4 W AV E the meaning law, critical of You should be able to: reection, angle BE H AV IO U R B E H AV I O U R You must know: ✔ WAV E and refraction, total internal Snell’s ✔ solve reection problems Snell’s law, involving critical angle reection, and total refraction, internal reection ✔ an experimental refractive method for determining index ✔ interpret diagrams boundary ✔ the meaning with all of wave diffraction and that it the meaning of the ✔ path difference describe the waves at the diffraction are pattern normally formed incident on slit details of a double-slit interference describe the waves are incident on the interface intensity pattern produced changes in direction Reflection—the of wave the wave are continues to between two different interference. normal media, reflected ray observed. travel in the original medium. θ θ • Refraction—the wave travels in in pattern. incident ray • when a waves double-slit When showing media between ✔ ✔ the waves single two two occurs types plane ✔ of the new medium. 1 1 medium 1: refractive index = n 1 The angles of incidence, reflection and refraction are always measured medium 2: from the normal to the interface between the media. Incident, reflected refractive index = n 2 and refracted rays all lie in the same plane. θ n c 1 The 1 = ratio is n 2 the (relative) refractive index going from c 2 2 refracted ray medium 1 to medium 2 and is conveniently written as 1 speed of light in a . n The 2 Figure 4.4.1. Rays at a boundary vacuum between two media absolute refractive index is but, speed the speed refractive in air index is close going to that from in air of a to light in vacuum, the the in medium as medium practice, is the relative The reected angle is always equal to the incidence angle θ used. when 1 measured from the normal to the Media with large optical densities have large values of n and interface between the media. correspondingly small wave speeds. The refraction angle θ is 2 Rays same are reversible: path but in when the waves opposite are reversed, their rays trace out the direction. related to the incidence angle by sinθ n 1 c 1 = When waves travel between media from an optically more dense to a sinθ n 2 less dense incidence medium, (Figure the angle of refraction is greater than the angle of 1 , = c 2 2 where n is the absolute refractive index of the medium and c is the 4.4.2). speed of the wave in the medium. Light incident This is Snell’s law at any angle > θ Though not c is totally internally bent, par t of the reected. normal ray is reected. θ c When a wave goes from medium n 1 1 1 through medium 2 to medium 3, 2 high index 3 n then 4 1 critical angle 3 n = 1 2 n × 2 . Expressions 3 material 5 such as this can be useful when θ c the rays travel through more than light source one medium. Figure 4.4.2. Critical angle and total internal reflection 43 4 O S C I L L AT I O N S AND WAV E S As the angle of incidence increases from a small value, there comes a The critical angle c occurs when point incident ray θ where the angle of refraction grazes the interface—the reflected ray is incident gives rise to a 1 angle refracted ray with θ is the critical angle. A weak observed for these = 90°. So 2 angles. sin c n 2 larger incident angles, there is no refraction; only a strong 1 1 = For and sin c = total = internal reflection is seen. 1 n n sin90 1 2 1 2 since sin 90° = 1. Take care with the Example 4.4.1 n subscripts here. Total internal reection occurs when waves travel from a medium with high refractive index to one A plane a mirror reflector. enters the consists A ray glass of at of light an a is parallel-sided incident angle of 30° from to the glass air sheet above glass coated the glass with and surface. with a lower refractive index, and The refractive index of the glass is 1.5. the angle of incidence is greater than the critical angle. a) Determine b) Calculate the the angle of critical reflection angle for at the the reflector. glass–air interface. Solution a) The law angle shows of incidence at the air–glass interface is 60° and Snell’s that: sin 60 sin r . = Hence, r is 35°. Because the glass has parallel sides, 1.5 the angle of incidence at the glass–mirror 1 interface is also 35°. 2 1 b) Using sin c gives = c = sin 1 Waves incident ‘spreads out’ on as it and = n c = 42°. 3 2 a slit or an interacts obstacle with the undergo aperture diffraction. (Figure The wave 4.4.3). Take care drawing the single-slit diffraction pattern. Remember: • the central maximum is about nine times more intense than the first maximum • maxima fur ther from the central position have even smaller intensities • maxima are not symmetrical. Figure 4.4.3. The effect same is order Diffraction at a slit and around an edge most as or demonstrated obvious are with long-wavelength Take care not to confuse refraction reception in light radio what is when smaller but the than is waves aperture the observed are apparently a dimensions wavelength. for diffracted shadow all by waves. hills to area. and diffraction. It is easy to write one instead of the other when Example 4.4.2 under exam pressure so be careful to avoid this error. Monochromatic light is Solution intensity incident on Sketch graph the a variation intensity from the narrow of with slit. showing light distance centre diffraction 44 a of the pattern. are of Diffraction For is the easily example, allow radio 4.4 When is light of diffracted blue a by single a wavelength single slit, the (monochromatic red light is meaning diffracted more one WAV E BE H AV IO U R colour) than the light. When a wave interact. This distinctive travels through produces fringe two double-slit pattern (Figure (a) parallel slits, interference the and diffracted gives rise beams to a 4.4.4). diracted beam from top slit destructive interference (crest meets trough) S constructive interference (crest meets crest) lamp double slit diracted beam from bottom slit (b) The double-slit interference λD equation s gives the = d separation s of successive bright fringes when light of wavelength λ is incident on two slits separated by distance d and the screen is placed a distance D from the slits. For constructive interference two rays must have path dierence nλ For destructive interference two rays must have path dierence 1 n + Figure 4.4.4. , where n is an integer. 2 (a) Interference of two diffracted beams from double slits and (b) the appearance of the pattern The maxima (bright fringes) of this pattern are caused by two waves Take care with the phrase out of arriving at the screen in phase. The two waves then superpose as phase. This describes two waves described in Topic 4.3. The dark fringes (minima) occur when the that have any phase difference, waves arrive 180° (or π) out of phase. other than zero, and is ambiguous. Always quote the phase difference Example 4.4.3 as well, for example, Two identical loudspeakers are placed 0.75 m apart. π Each rad out of phase. loudspeaker between them. Y the is a emits sound of loudspeakers position Positions X intensity . There and Z of are are frequency and 5.0 m maximum equidistant further 2.0 kHz. away sound from maxima Y and Y from is the midway line that 2 joins intensity . and have minima zero sound beyond X and Z. −1 The a) speed Explain i) a ii) a of sound why the maximum minimum at at in air sound is 340 m s is: Y X and Z. 45 4 O S C I L L AT I O N S AND WAV E S b) Calculate: i) the wavelength ii) the distance of the sound XY . Solution a) i) Position travel are Y an in At X equidistant equal phase. intensity ii) is distance This and Z, the waves 1 is n + interfere the Y . loudspeakers, When the waves constructive so both waves superpose, interference and they an 2 arrive out of phase; the path λ destructively superposed 180° of to produces the maximum. difference They from to produce complete cancellation waves. 330 b) i) λ = = 0.165 m 3 2 × 10 0.165 λD ii) The separation between maxima = which is the separation between × = d minima. 5 = 1.10 m 0.75 XY = 0.55 m. S AMPLE STUDENT ANS WER A student investigates how light can be used to measure the speed of a toy train. direction of travel double slit toy train 1.5 mm laser light light sensor 5.0 m not to scale Light from a laser is incident on a double slit. The light from the slits is detected by a light sensor attached to the train. The graph shows the variation with time of the output voltage from the light sensor as the train moves parallel to the slits. The output voltage is propor tional to the intensity of light incident on the sensor. egatlov tuptuo 0 25 50 time / ms 46 75 100 4.5 S TA N D I N G WAV E S a) Explain, with reference to the light passing through the slits, why a series of voltage peaks occurs. This answer Light hits could the propagating have two achieved slits waves. [3] If and the 3/3 marks: diffracts path into difference two at radially the sensor is ▲ This a from multiple of the wavelength of the light then by superposition 1 a maximum occurs (constructive interference). n If a is question of a complete student thoroughly . superposition and the destructive interference occurs (the peaks are created because they are at multiples of links Because the voltage is proportional to described between and is destructive clear. intensity The of answer the fringe the pattern wavelength. well the process A also of read The 2 troughs). interference series is distinction constructive occur answer has + multiple who intensity which ofthe to is the also voltage a clear peaks requirement question. 2 which is proportional interference This answer form could to voltage have , (amplitude) the constructive the slits, peaks. achieved 0/3 marks: ▼ Although Because and as the light minimum passing will appear through as light sensor is in the different that there this really of position. T he direction of energy transfer in this case is the direction of particle movements in the the are is a recognition maxima just graph and restates in the the minima, details question. There parallel is to there maximum no attempt fringes medium. form variation voltage to or explain how relates shown to on how their changes the the intensity in graph. b) The slits are separated by 1.5 mm and the laser light has a 7 wavelength of 6.3 × 10 m. The slits are 5.0 m from the train track . Calculate the separation between two adjacent positions of the train when the output voltage is at a maximum. This answer could have achieved 1/1 [1] marks: 7 λD ( 6.3 × 10 )( 5 ) 3 S = = = 2.1 × 10 m = 2.1 mm ▲ A careful, 3 d ( 1.5 × 10 a 4 . 5 S TA N D I N G the nature of ✔ waves describe waves how nodes and ✔ how boundary substitution answer and with evaluation. You should be able to: standing ✔ clear W AV E S You must know: ✔ legible ) antinodes the in nature terms of and formation of standing superposition form ✔ conditions inuence distinguish between standing and travelling the waves node–antinode pattern in a standing wave ✔ ✔ the use of the term observe, sketch patterns in ✔ solve wave waves from superposition the are constant of in two space but waves vary with travelling in time. They opposite strings problems harmonic, Standing and interpret standing wave harmonic. and pipes involving length of the the frequency standing wave of a and the speed. arise directions. 47 4 O S C I L L AT I O N S AND WAV E S Permanent zero positions on a standing wave are nodes; the peak amplitude position is an antinode This table shows comparisons between standing and travelling waves: amplitude Standing wave Travelling wave Zero at nodes; maximum at antinodes Same for all par ticles in wave equal to 2x amplitude is x 0 energy 0 No energy transfer but there is energy Energy transfer associated with the par ticle motion frequency Same for all par ticles except those at Same for all par ticles node (at rest) phase Phase for all par ticles between adjacent All par ticles within a nodes is the same. Phase difference of wavelength have different π rad between one internodal segment phases (difference varying and the next from 0 to 2π rad) The formation of a standing wavelength Distance between nearest 2 × distance between any pair of wave is closely linked to resonance, par ticles with the same phase adjacent nodes or antinodes which is described in Option B.4. Standing waves superposes possible with when another string the a the when on a original wave string depends quantity , form with of the smaller wave wave. travelling greater mass the or of reflected along a density . string at 4.5.1 lesser the wave is Figure string per a boundary shows meets The unit the an and cases interface wave length: speed the a larger this speed. Pulse moving from low density Pulse moving from high density to high density string to low density string Reflected wave Transmitted wave in Transmitted wave Reflected wave Figure 4.5.1. When the When free, Pulses moving towards free and fixed ends of strings end of the string is fixed or free, Figure 4.5.1 still applies. The behaviour of the linked the reflected pulse is the original amplitude non-inverted; strings should remind you of the when fixed, the reflected pulse is inverted. There is no transmitted behaviour of light rays travelling pulse is these cases. between two media from Topic 4.4. There are also links to Topic 9.3, Notice that: where thin-film interference obeys • the phase of the reflected wave depends on the nature of the interface similar rules. Use these links to help • there is both a transmitted and reflected wave when the string is your understanding. linked The another. boundary Figure two to 4.5.2 fixed fixes shows points frequency f , a the standing-wave what and a happens wave is standing-wave shape when a superposed forms with for string a a is with specific its peak frequency . stretched between reflection. amplitude in At the 1 centre of harmonic the and instruments. harmonics 48 string. is an are important When are There there produced. are nodes at the oscillation further fixed mode nodes on ends. for the This many is the first stringed string, higher 4.5 S TA N D I N G WAV E S If the length of the string is L, you get the following harmonic series: λ c First harmonic, L = f N = 1 = 1 2L 2 c Second harmonic, L = λ , f N = 2 = 2 f = 2 1 L N = 3 … and so on. Figure 4.5.2. Formation of standing wave on a string fixed at both ends Example 4.5.1 A standing 0.78 m. wave The oscillates frequency a) Calculate the speed b) Calculate the first of with three oscillation of is transverse harmonic loops on a string of length 140 Hz. waves frequency along of the the wire. 0.52 . wire. Solution 2 × a) The wavelength of the wave 0.78 is = As the 3 1 frequency b) The is 140 Hz, wavelength the will wave now speed be 0.78 f λ is × 2 = = 140 × 1.56 m. 0.52 f is = 73 m s 47 Hz. 1 Example 4.5.2 A horizontal single glass frequency tube and a contains standing fine powder. wave forms A loudspeaker in the tube. at one Powder end heaps of a horizontal occur at glass tube emits a nodes. 1 Speed of sound a) Identify the b) Determine, waves type of in air wave = 340 m s formed in the tube. P i) its ii) its c) P for the sound in the Q tube: wavelength 0.27 m frequency . and Q are Compare points the in the frequency , tube. amplitude and phase of air particles at P with those at Q. Solution a) Longitudinal b) i) 0.270 λ = c) The Q P is because between are directions; a sound wave in a gas. 0.18 m f ii) so, the same. antinode adjacent the P and is midway node; node–node phase difference so, between the amplitude segments. is two This λ nodes of means P is 340 = are the in c = 3 Q is × 2 frequencies and this = = so must greater that 1.9 kHz 0.18 the be than wave an antinode. the amplitude displacements of are Q. in opposite π rad 49 4 O S C I L L AT I O N S AND WAV E S Standing waves harmonics also shown in form in Figure the air in pipes. These lead 1 the series of 1 λ λ In pipes, the nodes and antinodes to 4.5.3. 4 2 are formed at points of zero 1st harmonic 1st harmonic amplitude and peaks, respectively, as for strings. 3 λ 4 1λ A closed end is a displacement node; an open end is a 2nd harmonic 3rd harmonic displacement antinode 3 5 λ λ 2 4 3rd harmonic 5th harmonic Figure 4.5.3. The phase nature of of the the the of original the open is axis end, therefore a so an to Air at that pressure pressure is amplitude a of ends wave the must node. there the molecules incident ensure the atmospheric maximum reflections boundary . along an Standing waves in pipes open at one end and open at both ends at must Topic air a a 4.2 pipe be arises closed be molecules always end reflected remain that at the the cannot as an move inversion stationary . where maximum. molecules from atmospheric, showed displacement the of and the There open At this is pressure must be a end. Example 4.5.3 An a) organ Sketch pipe the frequency i) the of length L is displacement of pipe lowest closed of the at air one in end. the pipe when the emitted is: possible f 1 ii) the next harmonic above the lowest possible f 2 f 1 b) Deduce an expression for f 2 Solution a) i) and ii) see Figure 4.5.3 (left-hand column, harmonics). b) L is the length of the pipe. 4L So λ = 4L and λ 1 = 2 3 c Use c = f λ and f = f 1 1 = f 3 2 50 f . = 2 4L So 3c and 1 4L first and third 4.5 S TA N D I N G WAV E S S AMPLE STUDENT ANS WER In another experiment the student replaces the light sensor with a sound sensor. The train travels away from a loudspeaker that is emitting sound waves of constant amplitude and frequency towards a reecting barrier. The sound ref lecting barrier sensor gives toy train a graph of the sound sensor variation of loudspeaker output voltage with time along ▼ The the track that is key question similar in shape to the graph on page 46. Explain how this effect arises. standing This answer could have achieved 0/2 to is to the marks: wave barrier and A reference T he sound particle is emitting further apart, as the more barrier go particles moves to sound sensor , the higher amplitude on they move further , the lowest point appear , when the of sound a reect from reecting barrier , new maximum point this that is to to forms a the the loudspeaker. reection required. describe between This the at needs the to formation of appear , the when answering recognise [2] appear . standing nodes and number can be points wave of its antinodes. separate made will and and be any given sequence There points two full are that relevant credit. Practice problems for Topic 4 (ii) the variation of displacement with distance Problem 1 An object performs simple harmonic motion with displacement x along the wave. and time period T. 0 Label your graphs and axes clearly. a) Identify the phase difference between velocity and b) Label, where possible, on your graphs displacement for this motion. (i) the amplitude x of the wave 0 b) Sketch a graph to show the variation of kinetic energy with time for the object for time T. Explain (ii) the period T of the vibrations your answer. (iii) the wavelength λ of the wave Problem 2 (iv) points, P and Q, which have a phase difference A par ticle oscillates with simple harmonic motion π without any loss of energy. What is true about the acceleration of the par ticle? A It is always in the opposite direction to its velocity. of . 2 Problem 5 The frequency range that a girl can hear is from 30 Hz B It is least when the speed is greatest. −1 to 16 500 Hz. The speed of sound in air is 330 m s C It is propor tional to the frequency. . Calculate the shor test wavelength of sound in air that the girl can hear. D It decreases as the potential energy increases. Problem 6 Problem 3 Plane-polarized light is incident on a polarizing filter that For a sound wave travelling through air, explain what can rotate in a plane perpendicular to the direction of is meant by par ticle displacement, amplitude and the light beam. wavelength. −2 When the incident intensity is 32 W m , the transmitted Problem 4 2 intensity is 8.0 W m A transverse wave travels along a string. Each point in the string moves with simple harmonic motion. Calculate the transmitted intensity when the polarizer has been rotated through 90°. a) Sketch a graph showing (i) the variation of displacement for a point on the wave with time 51 E L ECT R I C I T Y 5 5 . 1 M A G N ET I S M E L E C T R I C F I E L D S You must know: ✔ the nature of You should be able to: electric charge ✔ identify of ✔ what is meant ✔ Coulomb’s by electric an charge ✔ what identify what of charge and the direction them the sign and the drift nature of charge carriers in electric current is movement metal of ✔ identify ✔ solve speed of charge carriers carriers is meant by direct problems is meant by involving electric elds and current Coulomb’s ✔ forms between law a that two forces eld ✔ ✔ AND potential difference. current, When there is an electric law, the potential field acting at drift-speed difference a point, a equation, and charge. charge placed at that The electric field concept is point will have a field acting on it. If this charge is free to move, then it developed in Topic 10.1. will be accelerated. Electric eld strength E is a vector. It acts in the direction Electric charge is dened in terms of the ampère. The of the force acting on a positive charge. current is 1 A when one coulomb of charge ows past a point in 1 s. Its unit is coulomb (C). In fundamental units, F force acting on a positive point test charge E = = this is 1 A s. q magnitude of the charge ∆q 1 The unit of E is newtons per coulomb (N C 3 fundamental units, this is kg m s I ). In = where I is the electric current, q is the charge that ∆t 1 A ows, and t is the time. The denition involves a ‘point test charge’, this is taken q When the current is constant, = It to be a charge so small that it does not disturb the eld. Electric current is the movement of charge. Its unit is the ampère (A)—an SI fundamental unit. It is dened in terms of magnetic eects. Electric currents can exist in solids, liquids and gases. Drift speed v of the charge carriers is related to the density of carriers Conduction n the charge on each carrier q the charge cross-sectional area of the carrier A contain and the electric current I ions in carrier. fixed release solids Solid is usually due conductors positive electrons ions to a that ‘sea’ to the (metals, make of free up movement and the of materials bulk electrons as of the part one such type as material. of the of carbon) These chemical I bonding. v = nqA The electrons are accelerated in the electric field. Kinetic energy is Density of the carriers is the transferred to them from the energy source (an electric cell, battery number of carriers in each cubic power metre of the conductor. 52 supply). Electrons then collide with the fixed positive ions. or 5. Because gain the and speed charge lose for the carriers kinetic make energy carriers as known repeated they as the collisions travel. drift This with leads to the an ions, 1 ELECTRIC FIELDS they average speed Assigning the direction of Overall, the field transfers energy to the fixed ions from the energy conventional current needs care. Early scientists labelled source. the charge carriers in metals as Coulomb’s law describes how the force between two point charges positive. We now know that charge varies with their separation r: carriers in metals are electrons kq q 1 F and are negative. Conventional 2 , = where q and q 1 2 are the magnitudes of the point charges 2 current is due to the flow of r (+ or the ). k units is the used constant for of distance proportionality and (its magnitude depends on charge). positive charges; direction rules rely on this. Unless you are told otherwise, always assume that ‘current’ Constant of propor tionality, k is q 2 N m 2 C Coulomb’s law (measuring F refers to a conventional current. q 1 9 8.99 × 10 2 = when the 2 4 πε r 0 charge in coulombs, force in newtons, charges are in a vacuum (or in air, and distance in metres). which has a permittivity almost equal Permittivity of free space, ε is 0 12 8.85 × to that of a vacuum). Electric charge exists in positive 1 10 F m (F is the farad; see When a medium other than a vacuum Topic 11.3). or air is in use, ε is replaced by ε 0 ε 0 r and negative forms. Like charges repel. Unlike charges To rationalise electric and magnetic where ε is the relative permittivity of r attract. A charged body can also 1 units in SI units, k the medium. = attract an uncharged body due to 4 πε 0 the charge movement within the neutral object. When both positive. charge charges The is forces negative, are positive, repel. the the When force is force one in charge negative. Coulomb’s is The positive forces law and is also the other attract. Example 5.1.1 Two identical separated, spheres the have attractive charges force of magnitude between them is F . q and The 2q. When spheres touch Potential dierence V is measured in volts (V). One volt is the energy W transferred between two points to share charge and are then returned to their original separation. when one coulomb of charge q Calculate the force between the charges after the separation. moves between them. W Solution 1 So V = and 1 V ≡ 1 J C 2 kq × Call the original separation d. F 2q = = 2 2 d This is an attractive force, so one q 2 kq of This means that electric eld strength d the charges must be positive and the When other they must be combine, negative. therefore, the total remaining charge is shared between the two spheres, q F each. The new can be written as is q. q This F = E force × distance energy 1 is = 2 q × q × distance distance 2 k F ′ q pd 2 2 kq 2 = F , = = which is repulsive and equal to distance 2 d 4d 8 F So E = q When charges move, energy is transferred to them from the V = d source. This alternative way to express E is This energy is called potential difference. It is measured in terms of the the subject of Topic 10.2. energy transferred per unit of charge. 53 5 ELECTRICIT Y AND M A GN E T I S M Example 5.1.2 4 A charged point sphere of mass 2.1 × 10 kg is suspended from an Example 5.1.2 par t a) is a ‘show insulating thread between two vertical parallel plates, 80 mm apart. that’ question. You must make every step clear including the A potential difference d of 5.6 kV is applied between the plates. final calculation. The a) thread Show then that makes the an angle electrostatic of force 8.0° F to on the the vertical. sphere is given by 0.29 mN. b) Calculate: i) the electric ii) the charge field on strength the between the plates sphere. Solution a) Horizontally , F T sin 8, = vertically , mg = T cos 8 4 F = mg tan 8 = 2.1 × 10 V × 9.81 × tan 8 = 0.29 mN 5600 1 b) i) E = gives = 70 kV m 3 d 80 × 10 4 F 2.9 × 10 9 ii) q = gives = 4.1 × 10 C 4 E 7.0 × 10 S AMPLE STUDENT ANS WER a) An electric cable contains copper wires covered by an insulator. An electric eld exists across the ends of the cable. Discuss, in terms of charge ▲ The answer includes the idea carriers, why there is a signicant current only in the copper. that of conductors charge have carriers, insulators do not high This whereas (by answer pass charge is no statement carriers electrons. in There the is no that number the electric question. as the of charge eld This conductor link question are carriers clearly same and states electric that and, The difference electrons eld and it is are to the the current ▲ All no material, electric which current. the electron However , copper cannot wires can number of free-moving electrons, which have very conductivity. Hence, there is signicant electric in copper wires. answer could have achieved 3/3 marks: the therefore, across motion wire is a conductor which means it has delocalized or free them. by that the electrons (the charge carriers) which can move. T he electric eld gives a force on them making them move which is the denition current. three marking ‘Accelerate’ points would and are have electric than ‘move’, but the current. without T he many insulators free moving not be electrons charge are not carriers free to move, (electrons) the been electric better specic both have of here. by hence electron exerts rise marks: and in accelerated this 1/3 required insulator eld large Copper potential made through carry This conductor achieved between mentioned is have the good the could implication). Insulator ▼ There [3] numbers sense current will signicant. is correct. b) A wire in the cable has a radius of 1.2 mm and the current in it is 3.5 A . The number of electron s per unit volume of the wire is 28 2.4 × ▼ The value for q 10 3 m . Show that the drift speed of the electrons in the wire is 4 (which 2.0 × 10 1 ms . [1] 19 should be 1.6 × 10 C) seems to This be replaced by N (the answer could have achieved 0/1 marks: Avogadro A number – see Topic 3.3) which is I incorrect. Furthermore, there is 3.5 no v = = 28 evidence actually 54 that this carried calculation out by the was student. nAq 2.4 × 10 3 ×( 1.20 × 10 2 ) 12 × Na 5.2 5 . 2 H E AT I N G E F F ECT OF You must know: ✔ Kirchhoff ’s ✔ what is current and by its AN EFFECT OF AN E L ECT R I C ELECTRIC CURRENT CURRENT You should be able to: circuit meant H E AT I N G laws the heating effect of an electric ✔ draw ✔ identify consequences and interpret ohmic considering V and I circuit diagrams non-ohmic characteristic components by graphs V ✔ that resistance is expressed as R ✔ = solve problems involving potential difference, I current, ✔ Ohm’s charge, resistance ✔ the denition of resistivity ✔ the denition of power ✔ what ✔ and ✔ is meant by ideal and uses including divider the circuit laws, power, non-ideal ammeters ✔ investigate over of potential divider a series simple circuit. electric current of a resistor causes the that resistivity the arrangements factors advantages and investigate dissipation. voltmeters practical An Kirchhoff ’s law combined of series resistance and experimentally affect the parallel and of resistors describe resistance of a the conductor. circuits, potential in controlling transfer of a thermal energy when charge Resistance flows in a component. An electric current can also cause chemical and potential difference magnetic effects. V across a component R Electrical power dissipated is the rate with which energy is = = transferred. current Remember that the potential difference V across the component is in the component I the Resistance is measured in ohms energy W transferred per unit charge. The power P dissipated in a (Ω). 1 Ω is the resistance of a W component in time t component that has a potential is dierence of 1 V across it when the t W But V = and q = It . So V and = q current through it is 1 A . W W IV = P = t It The resistance of a component depends on its size, shape and 2 V V 2 Using the definition of resistance R leads = to P = IV = I R = the material from which it is made. R I Experiments show that resistance These are important equations—you should try to memorise them. R is: The electrical resistance of a component indicates the difficulty that • propor tional to length l charges have when moving through the component—or, alternatively , • propor tional to the resistivity of the ease with which the component. energy can be transferred from charge carriers to the material ρ, which is a shape- independent quantity • inversely propor tional to area A Example 5.2.1 The ‘lead’ in a pencil is a conductor. It is made from material which 3 has a resistivity Determine diameter the of 4.0 × 10 resistance Ω m. of a pencil ‘lead’ of length 80 mm and RA 1.4 mm. Resistivity ρ . The unit of = I Solution resistivity is the Ω m. All samples −3 ρl R = 4.0 × 10 −3 × 80 × 10 = of the same pure material have the = 3 A π × (0.70 × 10 0.20 kΩ to 2 significant figures. 2 ) same resistivity. 55 5 ELECTRICIT Y AND M A GN E T I S M Example 5.2.2 A heating element wire is 4.5 m long with diameter 1.5 mm. 6 Its It resistivity is used is with Calculate the 9.6 a × 10 110 V power Ω m at its operating temperature. supply . rating of the heating element wire. Solution 3 Area of the wire π = × (0.75 × 10 2 6 ) = 1.77 × 10 2 m Ohm’s law states that the potential dierence across a conductor is Substituting into the resistivity equation gives directly propor tional to the current 6 ρl in the conductor provided that R 9.6 × 10 = × 4.5 = = 24.4 Ω 6 the physical conditions remain A 1.77 × 10 2 constant: V 2 V The power dissipated is P 110 = = ∝ I R 500 A component the = 496 W which rounds to 24.4 W. variation graphs. that of They V can obeys with be I Ohm’s for used a to law is an component show ohmic are whether a conductor. called V–I conductor Graphs of characteristic is ohmic or non-ohmic. The characteristics can be plotted Figure 5.2.1 shows the V–I characteristics for three different conductors: with V on the x- or y-axis. Make sure you read the axes labels I I I carefully. Be prepared to draw either version and remember to 0 0 construct all four quadrants of the 0 V V 0 V 0 0 graph even though, for an ohmic conductor, the line continues through the origin without change ohmic conductor lament lamp semiconducting diode in gradient. Figure 5.2.1. When resistances resistance 1 In V–I characteristics graphs of the are joined in combination series using the parallel, these series Add or 2 values In two you can calculate the rules: parallel Add the reciprocals of the values The parallel equation for two R = R + R 1 + R 2 + … 1 3 1 resistors can be written as 1 = + R R R 1 R + … R 2 3 R 1 R 1 + 2 = R 1 + R 2 When there parallel is part a of combination the network of series and parallel, calculate the first. Example 5.2.3 An electrical each of cable diameter consists of eight parallel strands of copper wire, 2.5 mm. 8 The resistivity The cable a) the ii) the a is current cross-sectional resistance Calculate strand 56 carries copper 1.6 of × 10 Ωm. 20 A. Calculate: i) b) of the cable. of a area of a 0.10 km potential single length difference strand of the of copper wire eight-strand between the ends of cable. the eight- 5.2 c) State core one advantage cable with of copper using of a the stranded same total cable rather than cross-sectional H E AT I N G a EFFECT OF AN ELECTRIC CURRENT solid area. Solution 3 a) i) The area of one strand π is × (1.25 × 10 2 6 ) = 4.91 × 2 10 m 8 1.6 × 10 ii) The resistance of one strand × 100 = is 0.326 Ω 6 4.91 × 10 1 The resistance of the cable of is this because they are in 8 parallel: b) c) V = IR Possible one the Some 20 × larger 5.2.2 electrical Ω. 0.0407 answers strand; rules Figure of = 0.0407 the will area from shows leads. 0.81 V include: cable surface follow = the the gives the conduct a better in currents a of of circuit a even heat conservation currents These flexibility still cable if one strand with breaks; dissipation. charge that represent compared and lead charge energy . into (and, a junction I therefore, 2 I 3 electrons) flowing to and from the junction. If charge builds up here, it I 1 prevents further flow. The sum of the currents at the junction must be I 4 zero – This is having taken into account their signs. ∑ I 0 I 5 Kirchhoff ’s first circuit law: = I (junction) 7 I 6 For by an electrical emf the sources loop This is by loop, is equal energy Kirchhoff ’s the energy to the per energy coulomb per transferred coulomb into transferred the out loop of sinks. second circuit law: ∑ V = 0 (loop) I 1 Sources of energy are represented by the emf of a cell (see Topic + I 2 + I + I 4 sinks by the ∑ sourcesofemf resistors. − + I 7 – I 3 – I 5 5.3) Figure 5.2.2. and 6 ∑ IR foreachresistor = Currents leading to 0 junction in a circuit, and the sum of the currents Example 5.2.4 Determine currents and I 1 Ω I I for this circuit. 2 1 2 Ω I 1 2 4 Ω Solution First, label I the junctions, 1 Ω 2 Ω X 1 currents and emfs. I 2 Voltmeters and ammeters are used to measure and record quantities in a circuit. In 4 Ω theoretical work , meters are often I 3 assumed to be ideal and to have Y no influence on the circuit. In At junction X: I + 1 I = 2 practice, voltmeters and ammeters I (Kirchhoff ’s first circuit law; Y is similar 3 are non-ideal. but with opposite signs). 57 5 ELECTRICIT Y AND M A GN E T I S M Voltmeters are placed in parallel For loop 5 A: = 1 × I + 4 × I 1 (Kirchhoff ’s second circuit law). 3 with a component to measure the For potential dierence across it. loop B: −10 = −2 × I + (− 4 × I 2 law again). The next ) (Kirchhoff ’s second circuit 3 Ideal voltmeters have innite resistance (otherwise the current in step is to I eliminate : 3 them would aect the circuit). 5 = 1 × I + 4 × (I 1 + I 1 ) and 10 = 2 × I 2 + 4 × (I 2 + 1 I ) 2 Ammeters are placed in series to measure the current in a Solving these two simultaneous equations gives component. I = −0.71 A and I 1 = 2.1 A 2 Ideal ammeters have zero resistance and do not dissipate any Potential-divider energy. to control circuits current. determine the V–I Figure have 5.2.3 advantages over shows arrangements characteristic for a both metal potential divider variable resistors used to wire. variable resistor Be prepared to answer both qualitative and quantitative questions about potential dividers. V You should be able to show that, for the circuit below, A A R 1 V V = 1 and supply R + R 1 2 V R 2 V Figure 5.2.3. V = 2 V–I characteristic for a metal wire supply R + R 1 2 S AMPLE STUDENT ANS WER V supply 0.7 0.6 0.5 T R A / I R R 2 1 0.4 0.3 V 0.2 V 2 1 0.1 0.0 0 1 2 3 4 5 6 7 V / V a) The graph shows how current I varies with potential difference V for a resistor R and a non-ohmic component T i) State how the resistance of T varies with the current going through T. ▼ It is easy to get this wrong. One This answer could have achieved 0/1 [1] marks: V way to avoid this is to work out I at both ends of the graph. For T he resistance of T increases at a decreasing rate with the low current. current, At high the resistance current, This would on calculator a it only is about about take to is a work 50 Ω 10 Ω. few seconds out. ii) Deduce, without a numerical calculation, whether R or T has the greater resistance as I = 0.40 A. This ▲ This time, a reference is . The fact that current could have achieved 2/2 marks: made V to answer [2] is R has greater resistant R has greater voltage at I = 0.40 A. As shown on the graph, the I same is for both greater the for answer components, one but component, V gives the formula, R T when when = V is I = 0.40 A. greater I directly . R, 58 then According to V hence R has greater resistance. it will have a greater 5.3 ELECTRIC C E LL S b) Components R and T are placed in a circuit. Both meters are ideal. Slider Z of the potentiometer is moved from Y to X. X Y Z A V ▼ The R focus current in potential should the be ammeter. divider on the This is a arrangement, T so moving the i) State what happens to the magnitude of the current in the ammeter. will This answer could have achieved 0/1 also of R and T’s to X reduces at the Z. current This in the marks: the sum slider difference reduce ammeter T he the potential [1] (because resistance of V R = IR, and T and does current. not change). ii) Estimate, with an explanation, the voltmeter reading when the ▼ This ammeter reads 0.20 A . approach difcult This answer could have achieved 0/2 marks: (though Answering this 1 = I × R I = 1 0.20 A 1 + 0.3 0.06 = remember and the T. 0.06 (T) R = 0.21 V = 0.21 × 0.2 = 0.042 = 0.14 (R ) R Kirchhoff ’s sum Because are Looking = = when is you of the laws. The currents they are in in R differences the parallel, across same. (v) 0.14 V the potential them = be R 0.7 0.3(Ω) 0.2 V to impossible). = 0.2 A is = going not question straightforward V is [2] 0.7(Ω) see that, at the when graph, V = you 2.0 V, can the two 0.2 currents are about respectively . and, 0.06 A These therefore, add 2.0 V is and to the 0.14 A 0.20 A answer. 0.4 0.3 A / I 0.2 0.1 0.0 0 1 2 3 4 5 V / V 5 . 3 E L E C T R I C C E L LS You must know: ✔ ✔ that electric cells are sources of the distinction You should be able to: are chemical electromotive between energy force primary stores and ✔ (emf) and charge secondary ✔ cells ✔ the distinction real cells between potential have the emf of a cell and its ✔ difference internal a the ✔ solve how to investigate practical electric cells emf problems describe and and potential ✔ the of charge ow required to cell and internal resistance of a experimentally resistance resistance direction secondary determine cell terminal ✔ identify involving other interpret difference the emf quantities the with of and a variation time as a internal cell of terminal cell (both discharges. primary and secondary). 59 5 ELECTRICIT Y AND M A GN E T I S M Alessandro V olta invented the first battery in 1800. This was a primary Primary cells are cells that operate cell constructed from alternating copper and zinc discs separated by until their initial chemical store brine-soaked cloths. Soon, other types of primary cell and the first is exhausted; they cannot be secondary cell — the lead–acid cell — were developed. Society’s need for recharged. Examples include zinc– portable carbon dry cells and alkaline cells. and innovative compact secondary power supplies continues to drive research into cells. Secondary cells can be recharged when their initial chemical store When is exhausted. The chemicals a across charged the cell cell is not terminals is transferring equivalent energy , to the the potential electromotive difference force (emf) ε of then rever t to their original form. the cell. Examples of this type include ε lithium, nickel–cadmium and represents coulomb nickel–hydride cells. a real of cell This loss maximum charge always terminals There is detail about charging the passing gives because appears energy a as an through smaller energy that is it. cell When potential required internal the to can deliver transferring difference move resistance r of each energy , reading V charge the for through however, across the cell its itself. cell. cells in Topic 11.2. A cell a with internal resistance V = ε − or IR resistance enclosed ε = I (R + in r ). can dotted These be represented lines (see equations Figure follow as the 5.3.1). from a ideal For cell this plus circuit, Kirchhoff loop. r ε The V A equation V both r for real is a (from used the cell. to ε = − IR provides negative Figure control of 5.3.1 the the graphical gradient) shows current. a As a circuit current method and in ε for (from which increases, a determining the V-intercept) variable terminal resistor potential I R difference Figure 5.3.1. A circuit to determine energy to across drive the cell charge drops because through the a higher current requires more cell. the internal resistance and emf of a practical cell Example 5.3.1 The circuit Figure to is determine variation 2 in 5.3.1 of the circuit. are given Voltmeter reading / V Ammeter reading / A 1.25 0.10 0.70 0.20 0.43 0.25 0.15 0.30 used the V with The I in results 1.8 in this table. 1.6 a) Plot a suitable graph 1.4 from these data. V / egatlov 1.2 b) Use your graph to determine: 1 i) the emf of the cell 0.8 ii) the internal resistance of the cell. to the y-axis 0.6 Solution 0.4 a) 0.2 b) See i) graph on the Extrapolating left. the line 0 gives 0.7 0 0.05 0.1 0.15 0.2 0.25 0.3 ii) r is the negative of the gradient = 60 – emf = – (1.8 V). 1.8 − 0.2 current / A the 0 5.5 Ω 5.4 M A GN E T IC E F F E C TS OF ELECTRIC C U R R E N TS S AMPLE STUDENT ANS WER A student adjusts the variable resistor and takes readings from the ammeter and 10 voltmeter. The graph shows the variation of 8 the voltmeter reading V with the ammeter ▼ You reading I clearly show what you are when command the doing 6 V / V Use the graph to determine must examiner the (ideally , explain is this ‘determine’ in words). 4 a) the electromotive force (emf) of In the cell. 2 [1] this dot example, drawn student at read the the the only point clue is where graph. The a the clue 0 This answer could have achieved 0/1 is marks: 0 1 2 3 4 that there are two lines for the 5 answer—if the examiner simply I / A required have 9 V b) the internal resistance of the cell. answer could have achieved 1/2 marks: The is is of the slope as v = ε − r, that of 5.6 5.6 = there would one. is, a perfect makes the a it answer. clear (negative) positive that the slope value. The Ir gradient 9 − not student negative Negative answer, been [2] ▼ This This the only calculation actually 9.0 0.81 Ω and 4.2 is account must be 4.2 taken 5 . 4 M A G N E T I C E F F E C T S O F of the negative gradient. E L E C T R I C C U R R E N T S You must know: ✔ the a magnetic long You should be able to: eld straight patterns conductor for and a a bar magnet, ✔ solenoid sketch long bar ✔ what ✔ how is meant by magnetic solve problems currents and straight involving magnetic determine solenoids patterns for and or the direction solenoid of k n ow ing a ma gn et i c the curren t e l d f or a di re c t i on charges. determine a charge the of conductor produce, and interact with, magnetic fields. This direction moving direction currents conductors, eld forces, ✔ Electric magnetic magnets wire elds, interpret force ✔ to and the in a in a force of the force magnetic acting magnetic on eld a acting and on the current-carrying eld. is Magnetic eld lines: the third effect of electric current. A visualization using field lines has • are in the direction that a nor th evolved to indicate both the direction and strength of a field. Field lines pole would travel if free to do so also give a visual meaning to gravitational and electrostatic fields. • cannot cross Although the lines are not real, properties can be assigned to them. • show the strength of the eld— the closer lines are, the stronger the eld • star t on nor th poles and end on south poles • act as though they are elastic strings, trying to be as shor t as possible. 61 5 ELECTRICIT Y AND M A GN E T I S M This a table bar provides magnet, Certain rules a information long can straight help you about (current remember the magnetic carrying) the field field wire and patterns a for solenoid. patterns. Bar magnet The field lines go from the nor th pole to the Notice the use of the phrase south pole—in the direction that a free nor th pole would travel. conventional current. Early scientists believed that current was due to the flow of positive, not negative, charge. It is impor tant to take this into account when N S answering questions that involve direction rules. Long straight (current-carrying) wire A pattern of circular field lines surrounding Use the right-hand rule to remember the field a wire. pattern. Imagine holding the wire with your right-hand thumb pointing in the direction of I the conventional current; your fingers curl in the direction of the field. B direction current direction of magnetic eld Solenoid Field lines outside the solenoid are the same Looking into the end of the solenoid, when as a bar magnet. Inside, the field lines go from the conventional current circulates anti- the south end to the nor th end to be continuous clockwise, that is the nor th pole. Clockwise inside and outside. circulation corresponds to the south. The right-hand rule can also be used. Field (Nor th) Current N S I 62 5.4 Magnetic that The fields charges occur moving interaction force acting of on in the the when a electric charge pre-existing moving charge moves. magnetic and the It field is not are magnetic M A GN E T IC E F F E C TS OF ELECTRIC C U R R E N TS surprising also field affected. lead to a The force between a moving charge and a magnetic field originates in a relativistic effect. charge. This is explored in more detail in The direction of the magnetic force is at right angles to the plane Option A. containing itself is the magnetic proportional • the velocity of • the magnitude • the strength of the of the field and the velocity of the charge. The force to: charge the v charge magnetic F B q field B The direction of the magnetic force acting on a moving charge is always at 90° to θ the plane that contains B and v. The equation is F = qv . When the velocity B sin V q of the charge and the magnetic field direction are at right angles (in other words, θ = 90° so sin θ Imagine a carriers, with = 1) the magnetic force acting on a moving charge is qvB conductor conductor is carriers the n A, of charge and length carriers the charge l that per on carries cubic each a current metre. carrier The is q. I of area charge of There the are nAl Magnetic eld strength B is in segment. The total charge is nAlq . The total force acting measured in tesla (T). Rearranging on the conductor when in a magnetic field of strength B is B( nAlq ) v , the F where v is the average drift speed of a charge = qvB equation shows that carrier. the fundamental units for tesla are 1 So F = ( nAlq)Bv current in a = ( nAvq)Bl = BIl because I = nAvq relates drift speed to conductor. N s m 1 A 1 s 3 ≡ kg s 1 A The tesla is a large unit. The magnetic eld strength of the Ear th The equation F = qvB sin θ is equivalent to F = BIl sin θ is about 50 µT. More details about the tesla are Rules can be used to help you with force directions. Fleming’s left-hand rule relates the directions of the (conventional) current and the field to the force. force (motion) given in Topic 11.4. There is a link between the force on a moving charge in a magnetic along thuMb eld along F irst nger F current-carrying conductor in a field. Remember that the charges B in a conductor move with an I left hand field and the force that acts on a current along seCond nger average drift speed. Drift speed is considered in Topic 5.1. You may have been taught other rules. Make sure that you understand how to use them – and don’t use the wrong hand! 63 5 ELECTRICIT Y AND M A GN E T I S M Example 5.4.1 A beta particle through angles the to moves same the at 15 times magnetic field. the speed Both of an particles alpha are particle moving at right field. magnetic force on the beta particle Calculate magnetic force on the alpha particle Solution F Bq β Use F = qvB. v β So = Bq F e β = α α v × 15 = 7.5 2e α S AMPLE STUDENT ANS WER The diagram shows a cross-sectional view of a wire. I = 3.5 A into page The wire, which carries a current of 3.5 A into the page, is placed in wire cross-section a region of uniform magnetic field of flux density 0.25 T. The field is directed at right angles to the wire. magnetic eld Determine the magnitude and direction of the magnetic force on one of the charge carriers in the wire. This answer could have [2] achieved 2/2 marks: 4 V ▲ The student has set out clearly and 2.0 × 10 19 I = 3.5 B the quoted = 0.25 = qvB sinθ = 1.60 × 10 64 for the force. = 1.60 = 8.0 × 10 N × 10 4 × a 24 direction q 19 F calculation = downwards 2.0 × 10 × 0.25 C sin90 = 1 5.4 M A GN E T IC E F F E C TS OF ELECTRIC C U R R E N TS Practice problems for Topic 5 Problem 4 Problem 1 A bicycle is powered by an electric motor that transfers energy from a rechargeable battery. When fully charged, The I V characteristic graph for two conductors A and B is shown. the 12 V battery can deliver a current of 14 A for half an 3.0 hour before full discharge. a) Determine the charge stored by the battery. 2.0 A A / I b) Calculate the energy available from the battery. B c) The cyclist uses the motor when going uphill and 1.0 1 maintains a constant speed of 7.5 m s Calculate the maximum height that the cyclist can 0 0 2 4 6 8 10 12 climb before the battery needs recharging. V / V Problem 2 A wire is 0.15 m long. A potential difference of 6.0 V is a) Explain which conductor is ohmic. applied between the ends of the wire. b) (i) Calculate the resistance of the conductor A when a) Calculate the acceleration of a free electron in the V = 1 V and V = 10 V. wire. (ii) A is a lamp filament. Explain why the values of b) Suggest why the average speed of a free electron resistance in par t i) are different. in the wire does not increase even though it is c) B is a wire of length 0.8 m with a uniform cross- accelerated. 8 sectional area of 6.8 × 10 2 m Problem 3 Determine the resistivity of B. A 6.0 V cell and a 30 Ω resistor are connected in series. Problem 5 The cell has negligible internal resistance. Wires X Y and PQ carry currents in the same direction, a) Calculate the current in the cell. as shown. b) An arrangement of a 30 Ω resistor and a 60 Ω resistor in parallel is connected in series with the original 30 Ω resistor. Calculate the current in the cell. State and explain the force on XY due to PQ. Problem 6 7 A wire has a cross-sectional area of 1.8 × 10 29 contains 3.0 × 10 2 m and 3 free electrons per m When the potential difference across the coil is 12 V, the current in the coil is 7.2 A . a) Calculate the mean drift velocity of electrons in the wire. b) The current is switched on for 12 minutes. Determine the energy transferred in the wire in this time. 65 CIRCULAR 6 6 . 1 C I R C U L A R what is meant velocity , ✔ move in by period centripetal M O T I O N a You should be able to: angular and force circle centripetal displacement, angular ✔ identify frequency acts and on an that acceleration object for centripetal act the examples towards it to ✔ force the describe and centre for qualitatively horizontal for the circular centripetal motion both be and the centripetal force for motion of circular motion quantitatively solve problems involving centripetal force, motion speed, ✔ of circular examples centripetal ✔ origin of of ✔ the AND G R AV I TAT I O N You must know: ✔ M OT I O N in a into force vertical gravitational taken motion and is circle at a constant constant at constant centripetal forces acceleration, angular displacement, angular velocity . period, linear frequency , speed and speed, need to account. Quantities involved in circular motion include period, frequency, angular Period T is the time taken to travel displacement and angular velocity once round a circle. The DP Physics course treats angular velocity as a scalar even Angular displacement θ is the angle through which an object moves; it is measured in degrees or though axis of it can be considered as a vector with its direction along the rotation. radians. Radian Angular velocity, with = r (rad) and is arc used = time taken for length θ angular displacement ω measure radius s angles. relate to Figure angle 6.1.1 shows how a sector θ 2π r For a complete circle, θ = 2π = t rad r where t is the time to travel θ rad. It The trigonometrical definitions for sine, cosine and tangent are in is a vector quantity. terms of Angular speed has the same the sides of a right-angled opposite sin θ triangle: adjacent = cos θ ; opposite = tan θ ; = denition as angular velocity, but it hypotenuse hypotenuse adjacent is a scalar quantity. When is small, ≈ sin θ ≈ tan θ because the curved arc length in the The angular displacement for a radian definition, and the opposite side in the triangle from which sin θ complete circle is 2π rad and the is defined, are similar in length and the hypotenuse and the radius are 2π periodic time is T, so ω = the same. T Linear speed The angular speed ω of an object is linked to its linear speed v v = rω s s The radian definition gives: So, = = to give v = = r r rearranged 1 × t s × r v = t , which can be r rω. s r s θ The linear speed is constant but its direction changes, so the linear (in rad) = r velocity changes: that is, the object is accelerated. N2 tells us that θ velocity Figure 6.1.1. 66 this change centripetal is associated force is with directed a force towards that the causes centre of circular the motion— circle. 6 .1 Example 6.1.1 CIR C UL AR Centripetal acceleration 2 An astronaut M OT I O N is rotated horizontally at a constant speed to simulate 2 4π v r 2π 2 a = ω the forces of r = = , where T = 2 take-off. r T ω 2 The centre of mass of the astronaut is 20.0 m from the rotation axis. mv 2 Centripetal force F = mrω a) Explain speed why is a horizontal force acts on the astronaut when = , r the where m is the mass of an object constant. moving at linear speed v in a circle b) The horizontal force acting on the astronaut is 4.5 times that of of radius r normal gravity . Determine the speed of the astronaut. Solution a) V elocity is a vector quantity . The velocity of the astronaut is Sometimes you will come across constantly changing because, although the speed is constant, the the term centrifugal force. This direction is changing. This means that there is an acceleration force can only arise when an and, therefore, a force acting on the astronaut. This is a observer is in a rotating frame centripetal force. of reference—you should not 2 mv use this term in an examination = b) 4.5 mg for the centripetal force. r answer; always use centripetal 1 So v = 4.5 gr = 4.5 × 9.81 × 20 = 30 m force. s to 2 significant figures. Circular motion arises in the Example 6.1.2 following common contexts. A particle light P string of of mass 3.0 kg length rotates in a vertical plane at the end of a Gravitational: The gravitational force between a planet (or the 0.75 m. Sun) and a satellite supplies the When the particle is at the bottom of the circle, it moves with speed centripetal force that keeps the 1 6.0 m s satellite in its circular orbit. When the particle has moved through 90°, calculate: Electrostatic: The mathematics of circular motion was used by Bohr to a) the speed of the b) the horizontal particle model the proton–electron system force on the in the hydrogen atom. particle. Magnetic: The force that acts on a Solution charge moving in a magnetic eld a) The object rises by 0.75 m with a gain of gravitational potential is at 90° energy of mgh = 3 × 9.81 × 0.75 = to the plane containing 22.1 J. the eld direction and the velocity. 1 2 The initial kinetic energy is mv 2 = 0.5 × 3.0 × 6 = This is the condition for a centripetal 54 J. 2 Kinetic energy 2 × 2E at 90° = 54 force. 22.1 = 32 J, = 4.6 m s 3 2 2 3.0 mv centripetal force is is used to make × 4.6 = = r Banking is 1 = The speed 32 k m b) so 85 N 0.75 cornering easier in a vehicle. centripetal force When the vehicle corners on a horizontal surface, the friction between horizontal the tyres and the road surface must be large enough to supply the centre of circular path centripetal and will force. attempt If the to go friction in a is not straight sufficient, line then (obeying the vehicle skids N1). normal force mg When the normal of the to surface the circle is banked, surface (Figure the provides 6.1.2). horizontal the component centripetal force of the towards force the centre Figure 6.1.2. Forces in cornering on a banked surface 67 6 CIR C UL AR M OT I O N AND G R A V I TAT I O N Aircraft also horizontal the centre bank and of in the the order lift to force make now a turn. The produces a aircraft tilts horizontal out of the component to circle. S AMPLE STUDENT ANS WER a) A cable is wound onto a cylinder of diameter 1.2 m. Calculate the angular velocity of the cylinder when the linear speed of the cable 1 . is 27 m s [2] State an appropriate unit for your answer. This ▲ The calculation is correct answer explained. The unit is ω ▼ Although there is a reasonable the a radius equation, correct and the identication quoting solution the × 2π 27 = = This × π 1.2 is when incorrect for T marks: answer could = 45 radians s 0.6 have achieved 0/2 marks: of correct D = 1.2 m V = ωr r = 0.6 m becomes 2π confused 2/2 27 = r with achieved 1 also correct. start have and v well could appears. circular The unit 2π × r = = T motion. 0.6 m × 11 1 = 6 . 2 N E W T O N ’ S 0.343 ms L A W O F G R AV I TAT I O N You must know: You should be able to: ✔ Newton’s ✔ ✔ that law of gravitation describe force Newton’s masses of ✔ but uniform that the point is law can be of gravitation extended to relates to spherical the masses ✔ by a apply eld point strength force test per mass at centripetal Newton’s motion gravitational experienced and relationship between gravitational force point density gravitational the of an law object point mass solve problems of in a gravitation circular to orbit the around a a unit mass placed at ✔ that gravitational point. orbital ✔ point Newton’s law of period determine gravitation due the to relates involving eld of a gravitational strength, the force, speed and satellite gravitational two orbital eld strength at a masses. gravitational force between two The constant of propor tionality G is objects to their masses force between and separation. It states that the gravitational 2 Fr given by F two point objects of mass m G = and 1 m m 1 distance 2 r between the objects m is related to the 2 by In SI units, G has the magnitude m 11 6.67 × 10 2 N m × 1 2 kg F (or, in ∝ m 2 − 2 3 fundamental units, m 2 s r 1 kg ). The can negative ignore sign this indicates for Newton’s law masses of uniform centres of the that the force is attractive (although you gravitation). refers to point masses but can be extended to spherical There is more detail about the density . gravitational interactions of objects at the planetary scale in Topic 10. 68 masses. In this case, r is the distance between the 6.2 Example 6.2.1 N E W TO N ’ S L AW OF G R A V I TAT I O N For an object on the Ear th’s surface, a mass m is gravitationally Knowledge of the acceleration of free-fall g at the Earth’s surface is attracted to the centre of Ear th, E one way to determine the mass of the Earth m mass m E . The law becomes E 2 g r E a) Show that m E , = where r E is the radius of the Gm × m Earth. E E F = G 2 r E b) The gravitational eld strength of the Earth at its surface is 6.1 where r is the radius of the Ear th. E times that of the Moon at the surface of the Moon. Calculate the This is the force of gravity; the mass of the Moon as a fraction of the mass of the Earth. weight of the object is written as mg 6 Radius of Moon = 1.7 × 6 10 m; radius of Earth = 6.4 × 10 m leading to Gm E Solution g = 2 r Gmm 2 E a) mg = . E Cancelling and rearranging gives r E g E 2 = Gm E E r E and hence the result. Gravitational eld strength b) The equation also applies to the force acting on small test object Moon. g = 2 2 g g r E m m and mass of test object r M E = M = M E F G G = 2 m 2 g m gives 1 r M M Dividing . The fraction × is 2 m g E 1.7 M = 6.1 r E E is the = 0.012. The symbol g here does not refer 6.4 specically to the Ear th’s surface but anywhere in a gravitational Gravitational field strength force per unit mass that acts on a eld. 1 mass in a gravitational field. Two units can be used for g: N kg (when dealing with gravitational 2 Example 6.2.2 eld strength) and m s (when dealing with acceleration). The gravitational field strength at the surface of planet P is Test objects are required because 1 17 N kg . Planet Q has twice the diameter of P . The masses of the the object itself contributes to planets are the same. the gravitational eld where it is Calculate the gravitational field strength at the surface of placed. Their magnitude must be P . much smaller than that of the mass Solution producing the eld. This is usual 1 when dening eld strengths. From Newton’s law of gravitation, g ∝ 2 r 2 2 g r Q r = and force the of 6.1 is Q provides force the between motion of the = P = 2 4.3 N kg 2 r Q linked 1 × g r P Topic = g 2 g 17 P P Therefore, 2 Q here to describe centripetal satellite satellite force. and For planet is a orbits, circular at right since orbit angles gravitational (assumed to the here), direction satellite. Example 6.2.3 A satellite point on orbits above the equator so that Earth. it stays over the same Much of the theory of satellite motion is still correct when a satellite has an Calculate: elliptical orbit (the usual case). The a) the angular b) the radius speed of the satellite assumption in DP physics is that satellites have circular orbits. of the orbit of the satellite. 69 6 CIR C UL AR M OT I O N AND Equating centripetal gravitational G R A V I TAT I O N Solution 2π a) and ω forces (using v = rω = The satellite has a periodic time of 24 hours ≡ 86 400 s. ) T 2π gives T 2 4π for the orbit is So the angular speed ω 2 mr mv GMm 2 ω r = = = 2 2π 2 T r −5 r = = 7.3 × 10 −1 rad s . T where M is the planet mass, m is 1 the satellite mass, r is the orbital GM radius that separates M and m b) ω GM = , r is the 6.7 3 = = 3 v is the linear orbital speed, ω 11 × 10 24 3 × 6.0 × 10 2 r = 42 Mm. 2 5 ω ( 7.3 × 10 ) orbital angular speed and T is the orbital period. 3 S AMPLE STUDENT ANS WER r GM Also, T = 2π , v = GM r The two arrows in the diagram show the gravitational eld strength vectors at the position of a planet due to each of two stars of equal mass M GM and ω = 3 planet r Each star has mass M = 30 You should be able to derive these 2.0× 10 kg. The planet is at 11 equations and use them. m from a distance of 6.0 × 10 each star. a) Show that the gravitational eld strength at the position 11 of the planet due to one of the m 6.8 × 10 4 star ▲ The equation is quoted substitution has an is clear. appropriate The number answer could have achieved 1/1 (one of more (6.67 × 10 30 ) × (2 sf in the question) and G correct unit for the ) = there = 3.71 × 10 −1 N kg 2 11 is r a 10 −4 = 2 2 × than g the [1] marks: 11 gures 1 N kg answer m signicant 3.7 × 10 and This the stars is g = star ( 6.0 × 10 ) answer. b) Draw two arrows to show the gravitational eld strength at the position of the planet due to each of the two stars. This answer could have achieved 1/2 [2] marks: planet ▼ The arrows are the right-hand the edge drawn arrow poorly; clearly misses 11 needed of the with star. Greater diagrams to care score 6.8 × 10 is full m star star marks. c) Calculate the magnitude and state the direction of the resultant gravitational eld strength at the position of the planet. ▼ The value in part a) has to This be combined, the angle dividing having between it by two answer could have achieved 1/3 [3] marks: calculated the stars and (calculating the GM gravitational eld strength again is F = 2 11 not needed). not the However, 4.94 × 10 r is 11 value G quoted. = 6.67 × 10 2 Nm 20 2.67 F × 2 /kg 30 M 10 11 = 4 .0 × r = towards the barycenter 4 .94 × 10 2 Nm 10 = 23 ▲ The direction ‘barycentre’ is is correct: another word 2.44 × 10 m for 27 F centre bodies stars. 70 of mass and is of two (or midway more) between the = 4 .48 × 10 Newtons of the two stars. 6.2 N E W TO N ’ S L AW OF G R A V I TAT I O N Practice problems for Topic 6 Problem 1 Problem 3 A toy train moves with a constant speed on a horizontal An Ear th satellite moves in a circular orbit. circular track of constant radius. Explain why its speed is constant, even though a force a) State and explain the direction of the horizontal force acts on it. that acts on the train. Problem 4 b) The mass of the train is 0.24 kg. It travels at a speed A beta par ticle is emitted from a nucleus, P . The 1 of 0.19 m s . The radius of the track is 1.7 m. subsequent path of the beta par ticle is par t of a circle of radius 0.045 m. Calculate the centripetal force acting on the 7 engine. Problem 2 Two charged objects, X and Y, are positioned so that the gravitational force between them is equal and opposite The speed of the electron is 4.2 ×10 1 m s a) Calculate the momentum of the electron. b) Calculate the magnitude of the force acting on the electron that makes it follow the curved path. to the electric force between them. c) Identify the direction of this force. a) State and explain what happens when the distance between the centres of X and Y is doubled. b) The mass of X is now doubled with no other changes. Deduce how the charge on Y must change so that the resultant force is unaltered. 71 ATO M I C , 7 7 . 1 PA RT I C L E D I S C R E T E E N E R G Y You must know: ✔ what is energy ✔ what NUCLEAR meant by is A N D discrete energy and discrete ✔ describe of meant by P H YS I C S a transition between two ✔ solve levels that occur during and the decay which nucleus is a random, radioactive of an atom emission that there are four electromagnetic, and ✔ the weak gravitational, of ✔ emitted complete changes beta describe the and changes and the strong to beta decay the nuclear ✔ decay decay describe particles and equations ✔ and what is meant by what half-life of investigate that of during atomic the spectra, wavelength atomic of transitions equations describe of decay for the alpha and absorption particles is a meant by nuclide half-life half from life a and decay experimentally determine curve (or by a simulation) nucleus ✔ ✔ spectrum forces: the alpha radiation absorption spontaneous emissions nuclear properties gamma fundamental involving calculations characteristics ✔ and gases problems including radioactive process the common photons ✔ R A D I O A C T I V I T Y You should be able to: levels energy AND background demonstrate radiation. the same numbers A neutral outside atom it. neutrons. The In electrons. consists nucleus neutral An ion of a of isotopes of the when an element but have different neutrons. positively number a of protons positively-charged contains atoms, forms that number neutral of nucleus with charged protons protons atom equals gains or electrons and the loses neutral number one or of more electrons. The energy The levels energy is levels lead to of an atom emission transferred to are discrete, spectra the atom that and having are is also a lowest discrete subsequently ground and arise state. when re-emitted. Example 7 .1.1 The energy levels of the hydrogen −0.378 atom are shown. −0.544 −0.850 The Planck constant is covered a) in more detail in Topic 12. a photon Identify of eV , the energy wavelength the transition of 658 nm. that is related gives to its frequency in rise to a photon of this f and wavelength λ wavelength. E = hf or = −1.51 hc hc by Ve / ygrene b) The energy of a photon E Determine, λ = c) λ Explain why the lines in an E emission spectrum involving where c is the speed of light and h only the ground state become is the Planck constant. −3.40 72 closer together of emitted the as the wavelength photons decreases. 7. 1 DIS CRE TE ENERGY AND RADIOACTI VIT Y Isotopes of an element have the Solution same number of protons but a 34 hc 8 6.63 × 10 × 3.00 × 10 dierent number of neutrons. 19 a) E = = = 3.02 × 10 J 9 λ Isotopes have the same chemistry 658 × 10 but dierent physics. To convert to eV: 19 3 02 × 10 1 60 × 10 = 1 89 eV 19 b) The 1.89 eV concerned. between excited c) At The −1.51 be only and higher energy The difference possibility levels, the second and the difference decreases on which the are between diagram the is ground the the state two levels change and the first excited have radioactive nucleus the energy differences in level the levels between and above wavelength of become the first become the closer excited more emitted level equal photons too. elements During energy 3.40, energy and Many the state. together. the must isotopes decay which ionizing undergo radiation is natural radioactive emitted and the decay. nature of changes. Radioactive decay and nuclear Radioactive we cannot predict radioactive pressure, Some of decay decay and the properties, so is random when cannot spontaneous. nucleus be will affected decay; by Random means spontaneous changing the that means fission can be confused. Decay that occurs naturally; most fission temperature, events are induced by the arrival on. of a neutron (Topic 7.2). emissions are a and listed in in radioactive this decay , together with some of their table: + Alpha α Nature Beta β Beta β Gamma γ Helium nucleus; Electron from the nucleus; Positron (anti-electron) from 2p + 2n β the nucleus; β Overall change in proton: Change of d quark to u quark Change of u quark to d quark Removal of excess energy neutron ratio as neutron changes to proton as neutron changes to proton from nucleus Ionizing power Strong Medium Penetration Few cm of gas; few mm of Few cm of aluminium + Origin ; e Weak paper Notes Electromagnetic radiation + ; e Quickly annihilated with Many metres of lead or electron concrete One or two energy states for Broad energy spectrum; Broad energy spectrum; Discrete energies tied to a decay an electron antineutrino is an electron neutrino is nuclear energy levels emitted in addition emitted in addition Decay equation A A 4 4 X → Z Y+ Z A α 0 A X → 2 2 Z Y+ A β +υ 1 Z+1 0 A X → e Z Y+ Z 1 A + β +υ +1 * X e Z A → X + energy Z A The notation used in the table X , is: where X is the chemical symbol Z for the element, A is the nucleon number (total number of protons + Equations for decay processes neutrons in nucleus), Z is the proton number (total number of protons must balance both for A and Z. in nucleus). Sometimes the symbol N is used for the neutron number. Use the correct notation for the A = Z + N neutrino and the antineutrino. Some of the table entries are covered in later topics (7.3, 12.1 and 12.2). 73 7 ATO M I C , NUCLE AR AND PA R T I C L E P H YS I CS Activity is the total rate at which a Activity and count rate are dierent; sample is decaying. it is dicult to count every emission from a decaying sample. Some You should be able to list the The unit of activity is the Becquerel sources of background radiation. emissions will be absorbed by the (abbreviated Bq). 1 Bq is the activity sample itself. The count rates from radioactive of one disintegration every second. sources used in laboratory Count rate is the measured number experiments need to be corrected of counts being detected in one second. for the background. Example 7.1.2 shows how this is done. Natural Each sources nucleus of of a radiation naturally give rise decaying to background element has radiation. an identical chance Half-life is the time taken for half of of decay per second. The total number of decays in one second, the the atoms initially present in a pure decay rate, is proportional to the number of nuclei of the element in a sample of a radioactive nuclide to sample. This original nuclei leads to behaviour where the time to halve the number of originally in decay. It can be determined from is constant, irrespective of how many were a graph of corrected count rate the sample. This time is known as the half-life against time Example 7 .1.2 The decay every Half-life is a value that relates activity 30 s. The of a pure background radioactive activity is sample was measured 0.4 Bq. Time t / s 0 30 60 90 120 Count rate / Bq 14.0 9.6 6.6 4.6 3.2 Determine the half-life of the sample. to a large sample of decaying Solution atoms. An individual nucleus Subtract the background activity (0.4 Bq) from each count rate does not have half-life. Topic 12.1 value. introduces the idea of a decay constant which is the probability Time, t / s 0 30 60 90 120 Count rate / Bq 14.0 9.6 6.6 4.6 3.2 13.6 9.2 6.2 4.2 2.8 that an individual nucleus will decay in the next second. Count rate minus background activity Plot on these a 16 values graph: least different intervals graph on (this shows lives at the from time the graph half- 14 to etar tnuoc for 12 / three the s half-life 1− Determine 14 10 8 6 4 2 7, 0 10 to 5 and 6 to 3 0 20 40 60 80 100 time counts The per (mean) average three 74 second). of these values is 52 s. 120 / s 7. 1 Nuclear between 1 is the properties particles strongest are in governed the force, Gravity by nucleus. these In the fundamental order of forces increasing that strength, DIS CRE TE ENERGY AND RADIOACTI VIT Y act There are examples of questions where using nuclear forces in Topic 7.3. are: • acts between masses • operates to infinity −38 • weakest known force; relative strength 10 Weak nuclear force • acts on quarks and leptons • governs decay of nucleons • force carriers are W and Z par ticles • operates within nucleus only (shor t range) −6 • relative strength 10 Electromagnetic force • acts between charged par ticles • force carrier is the photon • operates to infinity −2 • relative strength 10 Strong nuclear force • acts between quarks and gluons • force carrier is gluon • operates within shor t range • strongest known force; relative strength 1. S AMPLE STUDENT ANS WER The graph shows the 500 variation with time t of the 480 activity A of a sample containing phosphorus-32 400 32 ( 15 p ) 300 qB / A 200 100 0 0 10 20 t 30 40 / days ▲ The answer is correct. 32 Determine the half-life of P . [3] 15 This answer could have achieved 1/3 marks: ▼ This The full 12 is a ‘determine’ command details are verb question. implies required for that the days answer. shows count So A close that rate this only view is any of are 480 Bq answer from look there one the graph and probably at by at a 240 Bq. comes determination. supported quoted at markings the lack This of averages. 75 7 ATO M I C , 7 . 2 NUCLE AR AND PA R T I C L E N U C L E A R P H YS I CS R E A C T I O N S You must know: ✔ the denition of You should be able to: the unied atomic mass unit ✔ solve problems binding ✔ that mass ways to defect and describe nuclear energy binding stored in energy the what is meant by nuclear ssion and nucleus by mass involving the defect and are ✔ solve in ✔ involving energy problems radioactive decay , nuclear energy fusion released and nuclear nuclear ssion fusion ✔ sketch and interpret the general shape of the 2 ✔ that the equation ΔE = c Δm can be applied to curve nuclear of average binding energy per nucleon changes. against The unified atomic mass unit nucleon is used number. in nuclear physics. Mass can be The magnitude of the unied expressed in energy terms. atomic mass unit (abbreviated u) is very close to the mass of a proton Few elements undergo spontaneous nuclear fission. More commonly , or neutron (these dier slightly fission is induced when a moving neutron interacts with a nucleus. from each other). Nuclear fission occurs for a uranium (U) nuclide. A neutron collides 1 u is dened to be the mass of with a nucleus of U-235 and is absorbed. This produces a nucleus of 1 of the mass of a stationary the highly unstable U-236. The U-236 splits into two (or more) nuclear 12 fragments with the release of several fast-moving neutrons. carbon-12 atom. Fusion star, also such leads as the to energy Sun. The release. stellar This is the hydrogen origin of is at such atoms to leave a the energy high in a temperature Energy and mass are considered that electrons are and electrons in stripped away from single protons 2 interchangeable using ΔE = c Δm: a plasma mass can be expressed in energy units: a kilogramme is equivalent to 8 (3 × 10 2 ) 16 or 9 × 10 1 J 1 H 1 H 1 H H 19 1 eV ≡ 1.6 × 10 Figure 7.2.1 in which to produce shows two the protons cycle fuse J Therefore, one electron volt is ν ν (H-2) a deuterium nucleus (an isotope of 36 equivalent to 1.8 × 10 kg. hydrogen) 1 2 1 H H with the release of H positron (an anti-electron) and 2 way, the unit of mass is eV c 2 So 2 eV c an electron fusions 36 ≡ 3.6 × 10 γ γ 3 with forming and, 3 He protons Further occur 1 1 H H 4 He sustain and is used in nuclear proton power stations. Details of these neutron positron γ gamma ray ν neutrino Figure 7 .2.1. helium-3 He-4. (He-3) There is an He the stations are given in Topic 8.1. first finally , overall The fusion reaction that produces helium from hydrogen 76 neutrino. kg. The fission reaction can self- a 2 H When expressed using eV in this energy cycle. release during 7. 2 NUCLE AR RE ACTIONS Example 7 .2.1 Uranium-238 undergoes alpha decay to form thorium-234. 238 The table shows thorium-234 The nuclei following Mass of a) i) b) i) = the State are = particle 92 ) and (helium-4). 238.0002 u of number the Determine b) alpha U ( uranium-238 4.0015 u uranium-238 part an the available: = number the of of 233.9941 u Determine ii) and data particle State ii) masses uranium-238 Thorium-234 Alpha the neutrons of protons mass change, undergoes the increase in in in uranium-238 a in alpha a nucleus kg, when of a nucleus. thorium. nucleus of decay . kinetic energy of the products in i). Solution a) i) 238 92 = 146 4 ii) The helium nucleus 238 He, is so two of the protons U from 2 are removed There b) i) The are, mass when the therefore, change is 92 thorium 90 protons. 238.0002 − 233.9941 27 0.0046 × 1.661 × forms. − 4.0015 = 0.0046 u. 30 10 = 7.6 × 10 kg 2 ii) This 1 u can ≡ be 931.5 done via ΔE = c Δm or by remembering binding energy per nucleon that MeV . total binding energy for a nucleus = 0.0046 Precise × 931.5 = determinations number of protons + number of neutrons 4.3 MeV . of atomic mass show that the mass most stable 10 of a nucleus parts (proton known energy and is as Binding mass the terms then less known Protons electrostatically . the quarks distances, within has the neutrons the the its nucleus, This in positively it is the of difference described mass forces and and and inside and nuclear repulsive attractive be nuclides (e.g. Fe, Ni) constituent This also charged protons is its 9 in energy energy a strong up force of mass). can binding origins make mass equivalence However, strong total neutron defect. as are which the the plus mass through energy nucleus. than the repel force each acts between neutrons. At but, at binds larger the other small distances protons and together. VeM / noelcun rep ygrene gnidnib egareva is is 8 heavier nuclides 7 (e.g. U, Pu) 6 ssion 5 4 3 fusion lighter nuclides 2 (e.g. H, He) Energy must be added to a nucleus to separate it into its 1 component its parts. individual Similarly , nucleons to form together a nucleus from infinity by bringing requires 0 the 0 removal of energy . This is the source of the binding 50 100 150 200 250 300 energy nucleon number and mass defect. Figure 7 .2.2. Plotting a graph of binding energy per nucleon Graph of binding energy per nucleon against against nucleon number nucleon When a number large fragments larger are demonstrates nucleus, near magnitude the of such as highest binding nuclear U-236, part undergoes of energy stability . the per curve. nucleon fission, These (they the two smaller fragments are more have a strongly 77 7 ATO M I C , NUCLE AR AND PA R T I C L E bound) P H YS I CS and therefore this energy must be transferred out of the system. The change in binding The system moves to a more stable state. energy per nucleon for each In reaction is greater for fusion than a fission. On the face of it, fusion will fusion, helium Again, be better than fission for power generation. Unfor tunately, many hydrogen nucleus the nuclei nuclei (more lose (zero stable; binding larger energy binding per nucleon) energy per form nucleon). energy . S AMPLE STUDENT ANS WER of the engineering problems 32 raised by nuclear fusion have yet A nucleus of phosphorus-32 ( 15 P ) decays by beta minus (β ) decay to be solved. 32 into a nucleus of sulphur-32 is 8.398 MeV and for P mistake mass of the here is to include beta-minus the not no calculation. bound to binding way , as it particle anything energy . hydrogen-1 energy The In has consists else the it is 8.450 MeV. Determine the energy released in this decay. [2] so answer could have achieved 1/2 marks: 2/2 marks: is has 32 × 8.398 32 × 8.45 = 268.74 same zero of S particle This in . The binding energy per nucleon of ) 16 15 ▼ The S 32 32 the ( 16 binding one = 270.4 proton 2 in the M nucleus. = 0.511 MeVC = 2.171 MeV e ▲ Although not at all well This explained, correct this does answer and arrive scores at answer even though “determine” quality of this question, explanation 7 . 3 was 32)-(8.398) × a T H E quarks are used (32) high = S T R U C T U R E particles leptons have O F M AT T E R You should be able to: to nuclear uncharged strong ✔ the explain the patterns in ✔ use the lepton to are three quarks, the the forms that – charged conservation number neither two families and form six takes quarks present of that and view hadrons: mesons part in ✔ describe of six baryons are a ✔ matter there lepton and nuclear are mediate with the particle ✔ quark– and baryon to number, solve interactions neutrons interaction fundamental describe the particles laws of number range of charge, in terms forces mediation through strengths including of the exchange of the gravity fundamental particles and the ✔ sketch and and ✔ describe why ✔ describe the strong particles fundamental Feynman diagrams free quarks are not observed strangeness electromagnetic exchange the interpret baryon and details of the Rutherford–Geiger– nuclear, experiment that led to the discovery forces how the nucleus they ✔ what constitutes ✔ describe a Feynman diagram forces. quarks ✔ connement are not and explain why free observed explain that there is a Higgs for the mass of quarks 78 charge, quarks compare of ✔ involving protons Marsden weak of strangeness pair conservation nature uses our and and of Model describe there number, ✔ and Standard antiquark ✔ two interactions leptons ✔ – 32(8.450-8.398) 1.664 MeV problems ✔ = expected. You must know: ✔ achieved a where is have full (8.450× marks could the boson which accounts and charged leptons. 7. 3 In the and early 20th Marsden found that through a should small very Rutherford century , investigate number large to Rutherford angles propose positively-charged of that with a particle alpha thin the the alpha the by suggested atomic foil. is were This small, electrons Geiger scattering. particles gold nucleus that STRUCTURE OF M AT T E R most par ticles scattered par ticles are undeected They scattered result dense outside THE allowed and the nucleus. thin gold foil beam of par ticles Later in the positive century , (protons) Experiments interactions Model was then in it and became neutral began the to nucleus agreed and clear that there (neutron) reveal until, an in confirmed were particles increasing the by the nucleus. complexity mid-1970s, particle both in the in the Standard α par ticles Figure 7 .3.1. the Geiger through small and gold foil number angles more uorescent screen The Rutherford– Geiger–Marsden experiment Example 7 .3.1 In circular source of physicists. of Marsden with no particles than experiment, change in deviate most direction from their alpha or particles energy original loss. pass A direction by 90°. The Standard Model uses three groups of par ticles to describe Explain, with reference to Rutherford’s atomic model and the forces interactions within nucleons. acting in the nucleus: Quarks: These combine to form a) why some alpha particles are deflected through large angles hadrons which divide into the b) why most of the alpha particles are not sub-groups baryons (which contain deviated. three quarks) and mesons (which Solution contain two quarks: a quark– a) A few the alpha nucleus particles and the approach alpha very particle close are to a gold positively nucleus. charged Both and antiquark pair). so Leptons: These cannot combine there is a repulsive force between them. with each other. 1 The force varies as , where r is the distance between the Exchange par ticles: these convey 2 r information from quark to lepton centres of the alpha particle and the gold nucleus. and between members of each When r is particle b) very undergoes According nuclei of the is Quarks mass. and They generation charge This to the very alpha enough to are do gives in Every are the a with by and and the group. alpha has but other a between diameter. gold electric with an still The nucleus charge mass the gold majority closely have of and rest increasing antiparticle details First generation space force. their up (u) Quarks the their ‘generations’ particle large deviation. approach uncharged names Charge not very significant classified grouped is model, compared particles number. force significant experience leptons are a the Rutherford large (neutrinos table small, with with opposite antiparticles). the particles. Second generation charm (c) Third generation Lepton number top (t) 2 + Baryon number 1 e 3 3 down (d) strange (s) bottom (b) 1 1 e 3 Leptons 0 3 electron muon tau neutrino(ν ) +1 for leptons τ neutrino (ν ) neutrino(ν ) µ e 1 for antileptons e electron (e ) negative muon (µ tau(τ ) ) increasing mass 79 7 ATO M I C , NUCLE AR AND PA R T I C L E Quarks Baryons proton p uud neutron n udd has The P H YS I CS have strangeness strangeness number constitution of number = 0, except for the strange quark that –1. some baryons and mesons is given in this table. 0 Mesons neutral kaon K ds Conservation 0 reaction positive kaon negative kaon laws allow us to decide whether a proposed particle sd K is possible or not. + K us K su Example 7 .3.2 neutral pion 0 π uu or dd positive pion negative pion Reaction ① p + n → p + n Reaction ② p + n → p + µ + π π + p + + ud du Explain, in between a terms of proton a) reaction b) reaction ① ② − + µ baryon (p) and p conservation, a is possible is impossible. neutron (n), for the interactions why: The conservation rules for par ticle interactions involve: Solution a) The baryon number The right-hand on the left-hand side is 1 + 1 = 2. • charge • baryon number side is 1 + 1 + 1 − 1 = 2 and 1 • lepton number b) The baryon number for a meson + is so this is possible. 1 − = 3 0 3 • strangeness. The These must balance on both the baryon baryon number number on on the the left-hand right is 1 side so is this 1 + 1 = would 2 whereas violate sides of the equation for a par ticle conservation of baryon number. interaction to be possible. Confinement (also quarks strong are never known interaction. As quarks separate Contrast this as observed. but colour This energy the behaviour force with is is confinement) because supplied acting that of to a are that them free subject two-quark between the means quarks does to system, not electromagnetic the the decrease. field where 1 force is proportional to . For quarks, there comes a point where it is 2 r energetically than The to red, of separate blue mesons of ‘colour ’. anti-green 80 favourable interpretation flavour as more completely and green. colours. must this Each The of colourless. Quarks are when emerge in they effect create a a new therefore, in flavour Baryons consist to and, the can Standard have antiquarks consist of quark–antiquark observe three have three observed large-energy Model quarks pair. bound collisions. is colours, anti-red, colour–anti-colour always individual in via pair the quark described anti-blue making Thus rather quarks. all and white; hadrons colourless are groups 7. 3 As quarks move closer together under conditions of high THE STRUCTURE OF M AT T E R antiquark energy , (or antilepton) the force between them becomes small. This is known as asymptotic ver tex entering the ver tex freedom The existence of the Higgs boson was postulated in the 1960s. exchange par ticle In the Higgs theory , mass is a property conferred on particles when quark they interact with the Higgs field. (or lepton) One way to refractive explain index of a the Higgs field transparent is to use medium. an We analogy know the with entering the ver tex the medium is time there because speeds. We different could say wavelengths that blue travel photons through are it heavier at different than Figure 7 .3.2. red! A ver tex in a Feynman diagram Similarly , degrees. friction’ We one vertex think it The Photons Quarks and or are the the to either W from interact quark Higgs with mass to a left it and at direction right as of Higgs being vertices. least quark–quark to the field due to to a different ‘dynamic field. combinations arrowhead going bottom of and pointing represents time going particles diagrams arrow transition. have can between Feynman least different is one A vertex away transition crucial. (some or The diagrams from a has it. lepton–lepton diagrams are at The usually drawn with time top). and leptons Z are particles shown are as shown straight as wavy lines. lines Example 7 .3.3 Solution In electron minus capture decay), a (inverse proton and beta- an Feynman diagrams have electron interact to form a neutron mathematical signicance for and another ν n particle. par ticle physicists. In the DP Draw the Feynman diagram physics course, they are used to to illustrate the interactions between represent this interaction. W Identify the particles par ticles and exchange par ticles. involved. p e 81 7 ATO M I C , NUCLE AR AND PA R T I C L E P H YS I CS S AMPLE STUDENT ANS WER a) A par ticular K meson has a quark structure us. State the charge on ▼ The answer is that the a this meson. This 2 charge has u [1] answer could have achieved 0/1 marks: 1 e of and the s e. 3 The 3 negative overall charge “negative” is is –e. not Just writing enough to score. b) The Feynman diagram shows the changes that occur during beta minus (β ) decay. Label the diagram by inser ting the four missing par ticle symbols. This ▲ The quark change is of a d quark to a answer could have achieved 1/2 u u d d remaining proton. boxes This is include not p u d ▼ The marks: u correct. n emitted [2] an p correct β in a Feynman diagram even though ν e on a baryon given level the proton is out. c) Carbon-14 (C-14) is a radioactive isotope which undergoes beta minus (β ) decay to the stable isotope nitrogen-14 (N-14). Energy is released during this decay. Explain why the mass of a C-14 nucleus and the mass of a N-14 nucleus are slightly dierent even though they have the same nucleon number. This ▼ The have energy come mass to nucleus released from energy . is the The more must transfer During the which and the a has more smaller result of mass. and does focuses not on This not just exchange. The make the is However this 82 of a proton and decay, the a product a which in neutron. the same, clear difference proton. mass achieved 0/2 marks: neutron is turned into a proton, (N-14) has one more proton than causes their mass to be slightly different. energy the answer have and binding particle beta means Carbon -14 therefore could of nitrogen stable answer [2] nucleon because for number a beta (neutrons+protons) decay, a neutron stays turns into a 7. 3 THE STRUCTURE OF M AT T E R Practice problems for Topic 7 Problem 1 Problem 3 A student determined the half-life of a radioactive Plutonium-240 240 ( Pu 94 ) decays to form uranium (U) nuclide by placing it near a detector. He recorded the and an alpha-par ticle (α). number of counts in 30 seconds every 10 minutes from the star t of the experiment. The following data are available: The results given in the table were obtained. Mass of plutonium nucleus = 3.98626 × 10 −25 kg −25 t / minute 0 10 20 30 40 50 60 60 42 35 23 18 14 10 Mass of uranium nucleus = 3.91970 × 10 kg −27 Mass of alpha par ticle = 6.64251 × 10 Number of counts in kg Speed of electromagnetic radiation = 2.99792 × 30 s 8 10 −1 m s a) Explain what is meant by half-life. a) State the equation for this decay. 240 b) Determine the half-life for this nuclide. Pu → 94 Problem 2 b) Determine the energy released when one nucleus The energy levels of an atom are shown. decays. 0 _____________________________________________ Problem 4 a) (i) Identify, using the graph of binding energy per nucleon versus nucleon number (Figure 7.2.2), the _____________________________________________ −19 −2.42 × 10 level 2 J nucleon for the most stable nuclide. _____________________________________________ −19 −5.48 × 10 nucleon number and the nuclear binding energy per level 1 J (ii) Calculate the binding energy of the nuclide in par t a) (i). _____________________________________________ 18 −2.18 × 10 ground J b) Two protons fuse to form a deuterium nucleus as state described by: 18 An electron with a kinetic energy of 2.0 × 10 J makes 1 1 H + 2 H→ 1 an inelastic collision with an atom in the ground state. 1 0 H + 1 e + ν + 1.44 MeV 1 Identify: a) Calculate the speed of the electron just before the 0 e (i) collision. 1 b) (ii) ν (i) Deduce whether the electron can excite the atom to level 2. (iii) Show that this reaction satisfies the conservation laws for charge, baryon number and lepton number. (ii) Calculate the wavelength of the radiation that will result when an atom in level 2 falls to level 1. Problem 5 (iii) State the region of the spectrum in which the State the quark composition of radiation in par t b) ii) belongs. a) a proton + b) a positive pion, π Problem 6 State, with a reason, whether the following par ticle reactions are possible. − a) p + π − →K + + π + b) p+ ν → n+ e 83 E N E R GY 8 8 . 1 E N E R G Y S O U R C E S You must know: ✔ the meaning density ✔ what ✔ the a for of distinction energy You should be able to: specic energy Sankey P R O D U CT I O N energy and energy ✔ solve sources diagram between is primary and secondary ✔ sketch ✔ describe sources electricity is a energy and interpret the basic transformations the ✔ specic versatile form and energy density problems of transfer of Sankey features in, power energy diagrams of, and the stations from fossil energy based fuels, on nuclear secondary fuels, the wind, solar energy and water-based energy systems ✔ renewable and non-renewable ✔ a thermal nuclear a method of energy sources ✔ reactor control and requires a way to a discuss moderator, the issues of and risks nuclear associated with power exchange ✔ thermal safety production describe the differences between solar power energy . cells Much of the energy we and use solar photovoltaic originates in the cells. Sun. This includes: Fuels can be characterised as renewable or non- • fossil fuels that were once produced via photosynthesis renewable. Fossil fuels regenerate • the direct transfer of energy from radiation arriving at the Earth’s in timescales of millions of years surface and cannot be replenished as • the generation of weather and tidal systems that are harnessed for quickly as they are consumed. energy transfer. Some sources can be regenerated rapidly; wind, wave, solar energy and biomass are examples. These are include known nuclear as fuel primary and energy sources. geothermal Other primary sources energy . One possible definition of a Other types of energy source involve an intermediate transfer step renewable resource is that it can before the energy can be used; these are secondary sources. They include be regenerated at the same rate as electrical energy and some chemicals, such as petrol, where refinement that at which it is used up. of the Some crude energy within 84 fossil a fuel sources realistic is are time required before renewable. and are said use. Others to be cannot be regenerated non-renewable 8 .1 ENERGY SOUR CE S Example 8.1.1 A coal-fired ε. is It power burns a station mass of has coal M a power of output ρ density P . every Its Specic energy is the energy that efficiency second. can be transferred from 1 kg of the Derive 1 anexpression for the energy density of the ). fuel (unit: J kg coal. Energy density is the energy that Solution 3 can be transferred from 1 m Energy density is the energy available from unit volume of the of the fuel. 3 fuel (unit: J m ). M Use ρ The volume = V M V of coal consumed every second is ρ P The energy input to the station is , allowing for the inefficiency . E energy The energy density creating energy power steam, to dynamo the (a mechanical which dynamo’s The electrical grid) The to way coil, thermal to (kinetic) in a can and energy . can the be The field. kinetic is a energy to transfer turbine This is through to a boil of charge an water, thermal attached causes energy to a to flow electrical cable in form. network end-users. transferred and ε M M thermal transported geothermal represent the commercial energy E use magnetic be = turbine—this transferring energy nuclear, a Pρ × volume stations turns rotates domestic initial including One generating which ρ P = coal Thermal input is fossil quantitative from various fuels, fuels. energy transfers is to use a Sankey diagrams are a type of ow Sankey diagram. diagram that show energy transfers in a system or process. The width of an arrow in the diagram represents gear losses the relative size of its contribution (6%) to the total energy involved in the motive power (29%) transfer. transmitted energy in fuel energy (23%) (100%) engine losses (7 1%) noise exhaust (1%) friction (22%) (40%) radiated heat (8%) Figure 8.1.1. Nuclear power A Sankey diagram showing the energy losses in a car engine stations use nuclides that undergo induced fission. The physics of nuclear fission is Neutrons smaller, highly neutrons reaction absorbs The bombard further controlled nuclei fissions, using nuclei and leading more leading control rods to a to the neutrons. chain made from formation The described in Topic 7.2. emitted reaction. a of The material that neutrons. neutrons neutrons further uranium radioactive initiate is the released with low fissions. Moderator So atoms during speeds emitted have fission have the neutrons nuclei with are emitted highest are slowed which at high probability down neutrons speeds. of in But causing a moderator. collide elastically . 85 8 ENERGY PRODUCTION The thermal energy released during fission is used to boil water for the The physics of the energy turbine steam. transfer from generating station to user is discussed in Topic 11.2. In fossil-fuel stations, to boil In geothermal water surface to for the water is heated water is then The exchanger technique temperatures Solar power Solar water heat that storage to photovoltaic photon energy analternating in the at the releasing is high thermal energy forced hot local below the Earth’s temperatures. surface The energy where to transfer energy hot as steam water to can nearby geology water from stored water energy the in for a also turbine. be used in premises. has for use to an created domestic or high radiation. an the then antifreeze sunlight be used to via a a use. form. supply in can materials to solar with panel water electrical locally from (usually hot the semiconductor sunlight to to cycle. thermal energy use current pressure are circulating produce cells directly surface. contain from the methods The high rocks used transfers tank. exchanger Solar two panels added) to burned returned supply mainly close uses heating agent is and at the reused to is steam. water where The heat fuel turbine stations, regions The a the to transfer This the generates grid system. The flow equation does not T ake care with the distinction between solar heating panels and solar represent the maximum photovoltaic cells. The former do not have an electrical energy output; power that can be extracted from they store thermal energy in water. The photovoltaic cells are the type the turbine. The equation assumes that transfer the energy arriving with the photons into an electrical form. a final speed of zero for the fluid and the fluid must have some Example 8.1.2 kinetic energy remaining to move it away from the turbine. There Outline why most of the world’s energy consumption is provided have been several estimates for by fossil fuels. this residual energy; one of the earliest is due to Alber t Betz. He Solution suggested that no turbine can Fossil fuels are widely available and large quantities still remain in 16 the (about 60%) transfer more than 27 built of the theoretical maximum power. a) b) ground. so can blade for is rotor flow be the the can fuel small through used both. ρ, to relatively Fluid rotor diameter rotor close They with When power be source. volumes a is wind speed available of Fossil used or the from easily fuels produce turbine either the transported have power a significant to water. fluid the or is fluid generate The v high of renewable the arriving energy amounts same and stations flow a be density , energy . energy . equation density at can of turbine is the of It used fluid cross- rotor 1 diameter 3 height sectional blade area A Aρ v is 2 rotor Wind blade turbines can be in a horizontal or vertical format (see Figure 8.1.2). tower Water turbines can generate electrical energy in several ways. These include: horizontal axis Figure 8.1.2. ver tical axis • hydroelectric • pumped ver tical wind turbines • tidal and • 86 systems Horizontal and storage techniques Sun wave hydroelectric (using the systems movement of water bodies due to Moon tides) techniques (using the movement of waves onto a shore). 8 .1 ENERGY SOUR CE S Example 8.1.3 2 A wind turbine’s blades have a total area 65 m Analyse and describe energy 1 The turbine is used The density of in a wind of speed transfers in a generating system 15 m s carefully. Topics 2, 5 and 11 (HL) 3 air is 1.3 kg m have direct relevance here. It may a) Determine b) Calculate the mass of air incident on the turbine every second. be wor th reviewing the language of energy transfer in Topic 2.3. turbine c) Only the total every 40% of kinetic energy of the air that arrives at the second. the total kinetic energy of the wind can be converted Pumped storage hydroelectric into electrical energy . systems use electrical energy to Calculate the electrical power output of the wind pump water to a higher reservoir. turbine. The water stores gravitational Solution potential energy. Then, at times of a) The air volume equivalent to incident the on volume of 2 65 m the a turbine cylinder in of one length 15 m: 65 m is cross-sectional 2 and second peak demand for electrical energy, area the water runs back through the 3 × 15 m = 975 m pump. The pump motor is now in reverse, so works as a dynamo, To calculate the mass of air arriving in one second: generating electricity. The energy 3 mass = density × volume = 1.3 kg m 3 × 975 m = transfer is from gravitational 1270 kg. potential to kinetic to electrical. 1 2 b) The kinetic energy of this air is × 1270 × 15 = 140 kJ 2 c) 0.4 (40%) output = of this energy can be transferred to the electrical 57 kW. Scientists continue to research storage-battery technology. You S AMPLE STUDENT ANS WER should be aware of new types of a) The hydroelectric system has four 250 MW generators. The specic energy storage and generation 1 . Determine the maximum energy available from the water is 2.7 kJkg techniques as they develop. time for which the hydroelectric system can maintain full output when a 10 mass of 1.5 × This answer 10 kg of water passes through the turbines. could have achieved 2/2 [2] marks: ▲ A well-presented, 10 1.5 × 4 250 3 10 × 2.7 × 10 1 × 6 × × 10 4 .05 × 10 J answer, answer 9 = correct 13 = 10 including given to 2 a clearly-shown signicant Watts gures. 13 4.05 × 10 = 40500 ≈ 41000 seconds 9 1 × 10 ▼ There appears to be the b) Not all the stored energy can be retrieved because of energy losses in suggestion the system. Explain one such loss. [1] lead to energy certainly This answer could have achieved 0/1 temperature will cause high loss. frictional temperatures There losses are in the marks: turbine High that energy loss in thermal energy. losses this bearings in the answer and resistive electrical is too cables. vague for But credit. 87 8 ENERGY 8 . 2 PRODUCTION T H E R M A L E N E R G Y You must know: ✔ the three You should be able to: methods convection, T R A N S F E R of thermal conduction, energy transfer: ✔ sketch radiation and intensity thermal ✔ the solar arriving constant at the is the Earth’s amount orbit of from interpret with the nature ✔ what of black-body the Sun ✔ solve radiation meant by emissivity and the albedo problems describe of the Earth’s surface varies the mean of emitting temperatures involving law and the use of Wien’s the Stefan–Boltzmann the Earth’s law effects surface of atmosphere on temperature solve problems involving albedo, emissivity , seasonally solar ✔ variation daily ✔ and various the bodies albedo the ✔ at of for radiation ✔ is graph energy displacement ✔ a wavelength what is meant by the greenhouse effect. constant and the Earth’s average temperature. Thermal energy is transferred by three processes. Conduction occurs through collisions between electrons and atoms and Conduction: The principal mechanism in solids. through intermolecular interaction. Convection: The principal mechanism in fluids. Convection occurs in uids. The hot areas of a uid are less dense than Radiation: The only mechanism for transfer in a vacuum. the cold areas, so the par ticles rise All atoms An atom possess kinetic energy at temperatures above absolute zero. from the hot areas to the cold areas. moving about its fixed position can transfer energy to other The denser, cold areas then fall into atoms close by when it has a greater energy (temperature) than its the hot areas, creating a convection neighbours. In this way , energy can be transferred from current. high-temperature regions of a solid to low-temperature regions. Thermal radiation is emitted as electromagnetic waves by all Example 8.2.1 objects at temperatures greater than 0 K (absolute zero). Thermal Energy loss from a house can be reduced by placing foam between energy passes through a vacuum. the inner and Explain how exterior can outer heat be house transfer reduced walls from (called inside cavity a warm wall insulation). house to the cold by: a) a cavity wall with no foam b) a cavity wall with foam. Solution a) Air is wall b) a is not A foam within parts is insulator, conducted a the of energy 88 good mixture solid the transfer. are energy through of and foam so gas and cannot poor transferred the through the inner cavity . solid. convect. The gas Both conductors the and do is fixed solid not in and allow position gas effective 8. 2 All objects radiate electromagnetic radiation. They also THERMAL ENERGY TRANSFER absorb Black-body radiation is a continuous electromagnetic radiation incident on them. A black body absorbs all spectrum that depends only on the the radiation incident on it. A black body also emits radiation with temperature of the radiator. a pattern characterised by its temperature; this is called black-body energy radiated radiation. The extent to which an emitter is imperfect compared with a from the surface black body is described by its of an object emissivity Emissivity e = energy radiated Wien’s displacement law predicts the from a black The wavelength peak shifts to shorter body at the same characteristic peak in the graph of wavelengths as the temperature of the relative intensity against wavelength: black body increases. temperature and viewing conditions 3 The area under the curve is given 2.90 × 10 For a black body, e = 1. λ (in m) = by the Stefan-Boltzmann law. This max T (in K) predicts that the total power P output For a perfectly reecting and non- radiating object, e = 0. by a black body radiator is 4 P = e σ The fourth-power dependence of the AT ytisnetni evitaler total emitted power on temperature is where Visible shown in the graph above. Doubling 6000 K σ is the Stefan–Boltzmann constant 8 5000 K (5.67 × 10 4000 K 2 W m the temperature from 3000 K to 4 K ) 6000 K means that the emitted power is increased by a factor of 16. e is the emissivity (included for 3000 K bodies that are not perfect black bodies). 100 500 1000 1500 2000 wavelength / nm The Earth’s and emitted climate atmosphere. important Radiation is affected radiation Global areas of arrives at the by many surface, warming and factors, and the climate including the composition change incident of the modelling are research. at the Earth’s surface from the Sun. The spectrum of Temperature of the Ear th if it this black-body radiation is determined by the temperature of the Sun 278 K were a black body 2 (about 5700 K). A power of about 1400 W m is delivered to the top of Temperature of the Ear th with the atmosphere. the peak The atmosphere is effectively transparent to most of 255 K an albedo of 0.3 wavelength although there is some scattering of the shorter wavelengths. Actual temperature of the Ear th, including albedo and the Roughly 70% of this energy is absorbed by the Earth itself – the rest 287 K is greenhouse effect reflected. The The Earth proportion then re-emits reflected radiation is the albedo because it is at a temperature greater The greenhouse gases are water than 0 K. For the average temperature of the planet to be constant, vapour (H O), carbon dioxide (CO 2 there must be a dynamic equilibrium between the incident and ), 2 emitted methane (CH ) and dinitrogen 4 radiation powers. monoxide (sometimes called However, the Earth’s temperature is much lower than that of the Sun nitrous oxide) (N 0). 2 and the peak radiation wavelength with a longer greenhouse gases) is inside trapped are emitted wavelength. opaque the by to Earth is Certain infrared Earth–atmosphere in the infrared atmospheric wavelengths system. and region— gases so (called energy This contributes This overall to The greenhouse effect is vital to an increased average temperature for the Earth. warming ensure the average temperature of of the system through atmospheric absorption is known as the the Ear th is high enough to sustain greenhouse effect life. However, the enhanced When the increased remains the concentrations amounts in the of of these outgoing system. temperature of the The gases infrared overall planet, increase are albedo which in the absorbed, becomes affects climate atmosphere, and more smaller. and sea This energy increases level. greenhouse effect increases the average temperature of the Ear th and causes significant climate change. Be clear about the distinction. The as effect the of the enhanced increased greenhouse concentration of greenhouse gases is known effect 89 8 ENERGY PRODUCTION The value for the albedo shows both Climate modelling includes total scattered power Albedo is daily and seasonal variations for the factors discussed here total incident power any one location. The latitude of the and many more. The atmosphere– The global annual mean albedo for location is also impor tant as this ocean system is extremely Ear th is 0.3. This means that 70% of determines the angle of the Sun in the complex, and many national and the incident power from the Sun is international research groups are sky. Cloud coverage also affects the absorbed by the planet. value of albedo. focusing their effor ts on the implications of atmospheric change for the Ear th’s climate. S AMPLE STUDENT ANS WER The following data are available for a natural gas power station that has a high eciency. 1 Rate of consumption of natural gas = 14.6 kgs Specic energy of natural gas = 55.5 MJkg Eciency of electrical power generation = 59.0 % Mass of CO = 2.75 kg 1 ▼ It is important calculation carefully The rst rate at to in this present with step a is full to type your of work explanation. calculate the generated per kg of natural gas 2 which energy is converted. 7 This is the specic rate energy of of gas the consumption gas = 55.5 × One year = 3.16 × 10 s ×14.6. a) Calculate, with a suitable unit, the electrical power output of the 59% of this energy is eventually power station. transferred This into answer multiplies an electrical divides by the rather specic [1] form. than This answer could have achieved 0/1 marks: energy . 55.5 The presence of the unit (kg) is also = 3.8 MJkg/s at 59% confusing. 14.6 = ▲ Every kilogram of gas burned 2.2 MJkg/s b) Calculate the mass of CO generated in a year assuming the power 2 in one second gives rise to 2.75 kg station operates continuously. of CO . This answer [1] evaluates 2 This this the and converts mass of CO (in one step) produced in answer could have achieved 7 (14 .6 There signicant three been are, however, gures signicant in the gures too would have for (3.16 × 10 ) = 1268740000 kg decrease their dependence on fossil fuels. answer should international the × c) Explain, using your answer to ii), why countries are being asked to answer reduction could have achieved 0/2 [2] marks: focus With an 2.75) answer; best. ▼ The × many This on marks: one 2 year. 1/1 to the huge amount of carbon emissions that these power perspective of CO stations release produce, the issue of disposal of the emissions 2 in the world. of ‘disposal The consideration arise no meaning. have to of the The considered reduce the emissions’ answer the should global emissions without every affordable or environmentally efcient has options. need that arise d) Describe, in terms of energy transfers, how thermal energy of the from CO release. 2 burning gas becomes electrical energy. This ▼ The question discussion place (the in key a of electron 90 in power the transfers’). microscopic view energy a transfers gas-red phrase is‘energy requires the is of This could causes have achieved molecules to 0/2 marks: increase motion. Increased take station question the not Heat that answer [2] incorrect transfer of appropriate. of electrons causes increase of electrical energy. motion 8. 2 THERMAL ENERGY TRANSFER Practice problems for Topic 8 b) The maximum power output of a coal-fired power Problem 1 A 250 MW generating station is to provide energy for station is 2.3 GW. a large and isolated town. Residents are to choose 1 The energy density of nuclear fuel is 82 TJ kg between a nuclear fission station and a coal-fired station. Both stations have lifetimes of about 25 years. Determine the minimum mass of fuel that would be required by a nuclear power station to provide the Compare the relative costs and the environmental same maximum annual energy output as the coalimpact of both types of generating station. fired station. Problem 2 Problem 5 Water falls through a height of 4.8 m in a hydroelectric Solar cells are to provide the electrical energy for a small power station to provide electricity for a village. village with 29 houses. Each house uses an average a) Calculate the change in potential energy of a 1.0 kg power over one year of 800 W. mass of water falling through a ver tical height The intensity of solar radiation at the surface of the of 4.8 m. Ear th is 650 W. b) Discuss factors that affect the usefulness of The efficiency of the conversion of solar energy to hydroelectric power stations for electricity electrical energy is 15%. production. a) Estimate the total area of solar cells needed to Problem 3 provide the power for the village. When the concentration of carbon dioxide in the atmosphere doubles, the albedo of the Ear th increases b) State one reason why the area covered by solar cells will need to be greater than your estimate. by 0.01. Average intensity received at Ear th from the 2 c) Suggest fur ther problems that may occur when using only solar cells to provide the energy for the Sun = 340 W m village. Average albedo = 0.30 Problem 6 a) Determine the change in the intensity of the The solar intensity arriving from the Sun at the radius of radiation being reflected into space by the Ear th. 2 the Ear th’s orbit is 1400 W m b) State one reason why the answer to par t a) is an Mean radius of the Ear th’s orbit around the estimate. 11 Sun = 1.5 × 10 m Problem 4 8 Radius of the Sun = 7.0 × 10 m a) Coal-fired power stations emit greenhouse gases. a) Estimate the total output power of the Sun. Outline what is meant by a greenhouse gas b) Use your estimate in par t a) to deduce the temperature of the Sun. 91 WAV E 9 9 . 1 PHENOMENA (AHL) S I M P L E H A R M O N I C You should be able to: You must know: ✔ the dening motion ✔ how M O T I O N equation for simple harmonic ✔ identify ✔ describe simple context of harmonic the simple motion arises pendulum in and mass–spring system how problems and the in solve about energy how energy potential solve forms problems velocity to changes in shm moves in between kinetic shm the ✔ ✔ energy (shm) graphs transfer and and involving acceleration displacement, during shm using algebra. shm. 2 equation a The In Topic 4.1 you were introduced Its to the equation that defines simple solution is = x −ω = x can sin ωt x be if solved. the object 0 begins harmonic motion: 2 velocity) × displacement of object x tnemecalpsid acceleration of object = −(angular time rest t its = at motion 0. If one the at the object extreme, centre is then when released x = from cos ωt. x 0 T Figure time t 9.1.1 t. For cosine the It shows against is a the sine graph (or result, of cosine) the x wave. graph begins Angular velocity (the constant ω) at the maximum amplitude (x ) and 0 is related to the time period T by continues as a cosine graph. 2π = and to the frequency of the Figure 9.1.1. Displacement– 2 The T equation v = (x ±ω oscillation f by ω = 2πf variation time. the It has a motion, sign ± the 2 − ) x 0 time graph for shm because, object can at be any with single moving displacement position either rather between towards or gives the away the than ends from of the centre. Example 9.1.1 Notice that the displacement and acceleration equations satisfy the original shm definition because x = x A particle line the with of mass m amplitude executes A and simple frequency harmonic f. motion Calculate the in total a straight energy of particle. sin (ω t ) and 0 Solution 2 a = −x ω sin(ω t) so that 0 This can be approached in a number of ways. 2 a = −ω x, as required. The maximum value of v occurs at the centre of the motion, x = 0. 1 2 So = v Aω. As ω = 2πf, v max = 2Aπf and the kinetic energy = mv max 2 1 2 or m 2 92 4A 2 π f 2 2 or 2mA 2 π f 2 9 .1 Knowing how displacement acceleration–time gradients in the of the graphs to respective varies be with drawn. time These displacement and allows are velocity–time connected velocity as H A R MONIC M OT I O N and through graphs, S IM PL E the shown table. Displacement Object star ts in Object star ts at centre of motion ex tremes of motion sin ω t x = x x tnemecalpsid T 3T 2 2 cos ω t x = x 0 0 t T 2T Velocity v = ω x cos ω t v = −ω x 0 x v yticolev v = ±ω sin ω t 0 2 ( x0 2 2 − x ) v = ±ω ( x0 2 − x ) t T 3T T 2T 2 2 2 Acceleration a = ω x 0 2 sin ωt a = a noitarelecca x T 2 x a = −ω x t 3T T cos ωt 0 2 a = −ω ω x 2T 2 2 Example 9.1.2 P the with graph displacement time a) moves This x simple shows of P in harmonic the the variation medium of with t Calculate the of b) its Calculate of the 1 maximum mc / x acceleration magnitude 2 thgir eht ot motion. tnemecalpsid Particle P . speed at t = 0 0 0.12 s. t / s its direction of motion at t = 0.12 s. Solution a) T = tfel eht ot State tnemecalpsid c) –1 –2 0.20 s. 2 a = max b) The c) To T 2 = x T the 2 x 0 = 19.7 ≈ 20 m s 2 x −2 2.0 × 10 displacement = −2 × 31.4 0 2π v 2π at t = 0.12 cm is 2 1.62 cm. 2 −2 = 31.4 ( 2.0 × 10 −2 ) − ( 1.62 × 10 −1 ) = 0.37 m s right 93 9 WAV E PHENOMEN A (AHL) 1 2 The equation for kinetic energy E is . mv k Combining 2 In Topic 4.1 it was noted that the this with both the displacement and time variants gives shapes of the energy–time graphs 1 1 2 are not sine or cosine curves and E mω = 2 2 (x k − x 0 2 ) mω = 2 2 x (ω t) cos 0 2 2 that one cycle of energy variation For the potential energy in the system, the key is to recognise that, has half the time of one cycle of for true shm, no energy loss occurs and the total energy is constant shm. 1 2 and E The graph of E k at E against mω = T 2 . x This must be the sum of E 0 p and E k , so E p is always p 2 displacement is parabolic: 1 1 2 total energy mω 1 2 2 x − E 0 1.000 mω = 2 x 2 2 mω = 2 x k (ω t) sin 0 J / ygrene 2 2 2 potential energy Two kinetic and contexts the for simple mass–spring harmonic motion are the simple pendulum system. energy 0.20 0.00 0.20 displacement / m Simple The pendulum simple shm. This pendulum is because mathematics, harmonic Mass–spring is meaning for small an there example is that a of approximate simplification the motion is in only the Providing extension simply swings. the force spring system that is the spring always acting system on is is directly it, the exactly elastic and its proportional motion simple of the to mass– harmonic. restoring force relaxed spring θ mass extended spring mass ℓ F t m base base x mg cos θ mg sin θ initial position initial position position of left of left edge of left edge edge when θ spring extended The force acting on the pendulum bob to return it to the The centre is mg sin θ and this is equal to force F acting proportional So, gsin θ = a and, for small angles < 10° sin θ rad). x However, θ = leading to g the return the = −a So, = ma and sign a to its spring mass −kx because x is measured away from positive extension is to the (the = − g is towards the l but the force Comparing this with a = −ω is shows the l T = 2π m g ω = 2π . k 94 kx force the acts acts to a = − k because the in left). m x which shm equation, giving to position. right and l a sign to x matches = gives 2 x. g that = vertical). F equilibrium negative direction Rearranging a x: extended, Thus directly the diagram vertical is l the negative spring x l (the a ≈ θ When (in on ma. the 9 .1 S IM PL E H A R MONIC M OT I O N Example 9.1.3 The for graph a shows the mass–spring maximum kinetic variation system. energy of The in total mass the potential of the system is energy spring is with 0.32 kg time and the 20 mJ. ygrene laitnetop latot 0 0.2 0.4 0.6 0.8 1.0 1.2 time / s a) State, the b) c) with a reason, the time period of oscillation of the mass on spring. Calculate the Determine spring the constant amplitude of k of the the spring used. oscillation. Solution a) The potential system. The energy time cycles period of twice the for mass one on time the period spring is of the 0.80 s. 2 0.32 4π × 0.32 1 b) T = and 2π so k ; = k = 20 N m 2 k ( 0.80 ) 2 1 1 2 2 2π 2 c) 2.0 × 10 mω = x 0.32 = 0 2 × 2 0.80 which gives x 0 = 0.045 m. S AMPLE STUDENT ANS WER A small ball of mass m is moving in a horizontal circle on the inside surface of a frictionless hemispherical bowl. a) The ball is now placed through a small distance x from the bottom of the bowl and is released from rest. The magnitude of the force on the ball towards the R equilibrium position is R given ball relessed here x by mgx R equilibrium position where R is the radius of the bowl. Outline why the ball will perform simple harmonic oscillations about the equilibrium position. This answer could [1] have achieved 1/1 marks: ▲ This T he acceleration of the ball is proportional to displacement and answer essential directed towards the equilibrium position. captures both is of shm: points the about the magnitude denition and the direction. 95 9 WAV E PHENOMEN A (AHL) b) The radius of the bowl is 8.0 m. Show that the period of oscillation of the ball is about 6 s. This ▲ This is a good answer. answer could have achieved 2/2 [2] marks: The x g R R mgx 2 acceleration deduced thus of (and the is identifying sphere shown the as is F a = R ∝x) constant of g 2 proportionality . This is ω and leads 1 ω to a correct ω = = = − = calculation. 1.1 s R 2π 2π = T = 5.7 5 ≈ 6 s = 1 ω 9 . 2 1.1 s S I N G L E -S L I T D I F F R A C T I O N You should be able to: You must know: ✔ the nature of single-slit ✔ diffraction describe the diffraction ✔ how to determine the position of the in a diffraction of a produced single-slit by monochromatic rst light minimum appearance pattern and by white light pattern. ✔ describe the appearance effect of a of changing diffraction slit width on the pattern. The rst minimum position of a wavelength and b is the distance from single-slit diraction pattern is the single slit to the observing screen. λ θ must be in radians for the given by θ = , where θ is the angle The proof of this equation uses the b approximation sinθ equation θ . Ensure that = ≈θ and so the between the central maximum equation is only true for small angles. b and the rst minimum, λ is the your calculator is set correctly. This advice applies to all angle calculations in DP physics. Example 9.2.1 Sound of waves width of wavelength 35 cm are incident on a gap in a fence 2.7 m. Topic 9.3 provides advice about precision in drawing the intensity– a) The θ position graph for diffraction. first from minimum the Calculate b) The central in the intensity of the sound is at an angle of maximum. θ frequency of the sound is reduced without changing its amplitude. All wave types demonstrate State and explain how this will affect the position of the first diffraction; you could be asked minimum about diffraction in sound waves or microwaves, and so on. The Solution λ basic physics is unaltered. a) θ = 0.35 b b) be leading to 7.4°. 2.7 A reduced will 96 rad, = frequency diffracted means through a a greater larger wavelength, angle. so the sound 9.3 Monochromatic light, diffracted by a slit, At HL, gives the INTERFERENCE instantly Some situations you meet in recognizable diffraction pattern. you should be able to sketch the course have circular aper tures the diffraction patterns with precision and confidence. You should also rather than single slits. There is an be able to carry out straightforward calculations of the position of the additional factor in the diffraction first minimum position. 1.22 equation which becomes = b S AMPLE STUDENT ANS WER when the aper ture is circular. Microwaves of wavelength 32 mm leave a transmitter through an aper ture of width 64 cm. Estimate the angle, in degrees, between the central maximum and the first minimum of the diffraction pattern. [2] Some problems ask for the pattern This answer could have achieved 1/2 differences between red and blue marks: in the context of a white-light λ 32 ▼ The θ = = = answer converted b has not source. The best way to answer been 0.05 from radians into these is from first principles: 640 degrees. The answer should be 2.8°. λ = b Another way to remember the effect is that in refraction, red light is refracted less than blue light. In diffraction red is diffracted more than blue. 9 . 3 I N T E R F E R E N C E You should be able to: You must know: ✔ what is meant by Y oung’s double-slit ✔ experiment describe two-slit including ✔ how to investigate the Young’s double-slit the interference modulation patterns, caused by single-slit (twodiffraction slit interference) the appearance arrangement experimentally ✔ ✔ of the two-slit sketch of pattern and diffraction how of the it is modied waves at and interpret intensity–position graphs interference (modulated) each of the two-slit interference patterns by ✔ slits solve problems involving the diffraction-grating equation ✔ the interference patterns ✔ the interference pattern from to multiple slits ✔ from a describe the conditions required for diffraction constructive and destructive occurring in thin lms including the interference grating ✔ what how is to meant solve by thin-lm problems interference including and thin-lm interface and phase the and at changes effect of their that interfaces, occur refractive at the index. interference. Here’s how to investigate the optical behaviour of a pair of slits. The double-slit interference • Take your could ruler use to pair a do of slits and microscope measure the simultaneously distance d imaging between the slits them. and a You metal pattern was introduced in Topic 4.4. The basic equation developed there λD this. was s = . In Topic 9.3, we look d • Project the light from a laser pointer (taking care not to look at the at how the slits themselves modify direct beam or a reflection of it) through your double slit and onto the basic double-slit pattern. a screen at least screen — this is 3 m D. away . Measure the distance from the slits to the Topic 9.3 provides advice about precision in drawing the intensity– position graph for diffraction. 97 9 WAV E PHENOMEN A (AHL) • You should see a fringe pattern on the screen. Use a ruler to measure The Young’s slit interference pattern the distance between a known number of fringes (10–20 of them (each fringe of equal intensity) is about right). Use this measurement to determine s, the distance is modulated by a single-slit between adjacent fringes. diraction pattern (see Figure 9.3.1). The diraction pattern acts • Use the data to calculate the wavelength of the laser. as an envelope for the interference fringe pattern, setting a maximum So far interference limit on the intensity at any infinitely position. both thin. This interference Assume width that a by patterns interference fringe is and been a of a poor model occurring for must distance. are two as experiment small that these treated diffraction double-slit separated diffraction has offset slits interference occur has the at two Both by diffracted at slits slit beams the by that multiple slits as slits. slits of give the rise separation. that are gives same to It rise (finite) identical is to the the final pattern. white light source diraction grating Figure 9.3.2. The diffraction grating arrangement Figure 9.3.1. As slit the A single-slit diffraction pattern modulates the interference fringes number width and of slits increases spacing to three unchanged, then or more, other with effects the individual appear. The The order of a spectrum is n. The fringes become sharper. Subsidiary maxima appear between the fringes wavelength of a line in the spectrum but, as the number of slits increases, these become relatively less is λ. The distance between adjacent intense than the main peaks between them. slits, the slit separation, is d. For a diraction grating, the angle θ between the straight-on beam and a line is given by n = d sin When that the of a becomes lines number of diffraction a with bright slits becomes grating. central darkness The very pattern maximum between them. high for the surrounded These arrangement monochromatic lines by intense are orders becomes light sharp and are The number of slits per metre is how diraction gratings are often 1 given integer labels many wavelengths, counting the out from individual the lines centre become which is zero. spectra—lines With or a specied = continuous red to violet band (Figure 9.3.2). d Example 9.3.1 5 A diffraction 486 nm order the is grating normally diffracted diffraction has 4.5 incident image that × 10 on can 1 lines m the be . Light grating. of wavelength Determine produced for this the highest wavelength by grating. Solution There are a number of times in The highest order occurs at θ = 90°. Using this value in the equation the DP physics course where an 6 d sin θ answer can only be an integer; gives n = 2.22 × 10 × sin(90°) = = 4.6 7 λ 4.86 × 10 this is one of them. Take care and round up or down to the nearest integer as appropriate. 98 This means observed. that the fourth order is the highest order that can be 9.3 When a wave boundary strings when pulse travels determine are easy light to the to shape imagine. travels the between end of Less two the of a string, reflected easy are conditions pulse. phase Such changes at the effects that INTERFERENCE Boundary conditions were in covered in Topic 4.5 for fixed and occur free boundaries, and for those media. where two strings of different mass per unit lengths are joined. The the reflected and associated reflected and refracted phase waves changes of play light their interfere part in with each determining other what and is transmitted. X π phase change The phase changes at boundaries depend on the refractive indices n Z 1 of n 0 phase change the two < n 1 2 n Y When n 2 media. the electromagnetic wave is reflected at an optically Figure 9.3.3. medium (higher refractive index), there is a phase change 1 denser Phase changes at of π rad interfaces (≡ 180°). When no the phase wave is reflected at an optically less dense medium, there is change. In these diagrams the incident, The phase changes are summarised in Figure 9.3.3. reflected and refracted rays are There are two cases that illustrate the effect. drawn well away from the normal to make things clear. In reality, the rays ① A camera lens can be made anti-reflective at one wavelength by lie almost on top of one another. coating the front refractive index There phase is a surface with midway change a transparent between of π at that both X of substance the and air Y; and the that that has of refractive a the glass. index λ increases at both interfaces. When the thickness of the coating is , the You may find it odd that 4 λ light from Y has travelled an extra cancellation of the reflected light distance (there and back). 2 In phase change terms, there is π (≡ 180°) means that more light enters the difference between the ray at lens. Remember that light is an X and the ray at Z. electromagnetic wave and our When d × n, the real where thickness n is the of the refractive coating index. is d, The the effective light, thickness therefore, travels model is a simplistic one. As usual, is a total XZ distance of 2nd, and when this is equal to (or odd multiples of it) then, allowing for energy loss in 2 the light from Z and the light from X are π (≡ energy must be conserved. When none is reflected from the surface λ 180°) out of phase. the coating and glass, overall more The must be able to pass through the 1 condition for destructive interference 2 nd is = m + , 2 where m is an lens system. integer. θ All the reflection from the front surface of the lens is eliminated at one n air specific θ n oil ② a Patterns wet X road of coloured (Figure fringes can sometimes by seen in the oil layer on d 1 Y n water 9.3.4). Figure 9.3.4. The Z wavelength. relationship between the refractive indices of the layers is Interference at an oil different film on water from the the lens water oil) but coating beneath not at the it. case. Now bottom The oil there is has a reflection a larger phase refractive shift at (oil–water). the index top When than surface the film (air– has a λ thickness of then the total path difference in the oil is λ and this will The data booklet only gives one 2 set of equations for thin-film lead to destructive interference between the ray from X and the ray interference and there are two from Z. This time, the destructive interference case is 2 nd = mλ for light possible sets depending on the at normal incidence. layer refractive indices. It is best The light look see a at appears the film coloured different multi-coloured because colours fringe are pattern. one wavelength removed at is missing. different As angles so you you to learn how to solve individual cases rather than rely on the equations—just use these as a memory aid. 99 9 WAV E PHENOMEN A (AHL) S AMPLE STUDENT ANS WER direction of travel A student investigates how light can be used to measure the speed of a toy train. Light from a laser is incident on a double slit. The light from double slit the slits is detected by a light toy train sensor attached to the train. laser light 1.5 mm The graph shows the variation light sensor with time of the output voltage from the light sensor as the train 5.0 m moves parallel to the slits. The output voltage is not to scale propor tional to the intensity of light incident on the sensor. egatlov tuptuo 0 25 50 75 100 time / ms a) Explain, with reference to the light passing through the slits, why a series of voltage peaks occurs. This ▲ An answer that understanding of shows the a answer could have achieved [3] 3/3 marks: When light wave pass through the slit, it creates two wave source good unusual with same frequency. In some position along the train track, context. The through the an train is moving interference question asks for an pattern and the path length difference between the two source is integer value explanation of the wavelength λ, which creates constructive interference, of the fringes differences, in terms and a link of path between the therefore peaks are created. At some other places path length voltage minima and the intensity λ minima. account This of answer both of gives difference is n an these. , so the wave interfere destructively, creating a 2 place with no intensity, therefore no voltage. b) i) The slits are separated by 1.5 mm and the laser light has a 7 wavelength of 6.3 × 10 m. The slits are 5.0 m from the train track . Calculate the separation between two adjacent positions of the train when the output voltage is at a maximum. This ▲ The question routine continues calculation as involving answer could have achieved 1/1 [1] marks: a the 7 double slit equation. The answer ( 6.3 × is S 10 )( 5 ) = = 0.0021 m 3 correct. 1.5 × 10 (ii) Estimate the speed of the train. ▲ This problem involves This the rate at which the fringes (from the train the graph) answer could have achieved passes to the 0.0021 1 distance between fringes in (b)(i). = 3 The 100 solution is correct. [2] linking 25 × 10 0.084 ms 2/2 marks: 9.4 9 . 4 R E S O L U T I O N You must know: ✔ the meaning of You should be able to: diffraction-aperture size ✔ solve problems resolvance ✔ what is meant single-slit by resolution diffraction and in the context diffraction the denition of the resolvance of a of involving diffraction resolution and gratings of gratings ✔ state and explain resolution ✔ RE S OLUTION of an the Rayleigh image of two criterion slits for the diffracted by diffraction a single slit or circular aperture. grating. You see a whether light this was asked that two one As on is by one the images diffraction Figure a total the Rayleigh patterns are lamp can or But from a two be in at of this the a and distance close Lord resolved Figure and that apart long lamps coincides shows, criterion well night scientist, just pattern intensity . distinguished at British 9.4.1(b) the hill the the summed single there the is a there intensity In This is a pronounced idea of the was in than cannot 9.4.1(c) of other. dip closer pattern (a) question small are Figure decide maximum minimum patterns pattern. you Rayleigh’s central first separation, 9.4.1(a) Can together? Rayleigh. when with away . be the images unresolved dip. Example 9.4.1 (b) Two An vertical observer distance of straight views 2.5 m. of wavelength a) Determine, in the the b) slit and Deduce the filaments the The of filaments filaments lamps are through are just a separated narrow resolved by slit with 5.0 mm. from a yellow light 580 nm. degrees, blue slit the minimum angle of resolution using light. width. images just resolved c) State and viewed explain at the whether same the position filaments using the will same be resolved slit and a when red filter. (c) Solution a) Geometrically , the minimum resolved angle is 3 distance between filaments 5 × 10 3 θ = = distance to 2 × 10 = filaments rad. 2.5 3 This b) answer Using the needs to Rayleigh be converted criterion and to degrees; 2 × rad 10 ≡ 0.11° rearranging, 9 λ 580 × 10 3 b = = = 0.29 × 10 m. images fully resolved 3 θ 2.0 × 10 Figure 9.4.1. The slit width is Resolution 0.29 mm. (a) not resolved (b) just c) Red light has a longer wavelength than yellow light. This will resolved (c) well resolved mean a wider will overlap and the diffraction more. images The will angle ‘dip’ not be in and the so the total diffraction intensity will patterns disappear resolved. When writing about resolution, remember that it is the image that is There is a example, a coarse question when grating of resolution viewing with spectral slits well when lines using within separated a diffraction the means same that grating; order. two for Using similar resolved, not the object. Resolution is a property of an optical or other image-forming system. 101 9 WAV E PHENOMEN A (AHL) wavelengths spread of a out will merge. the grating in grating to spread With a fine spectrum spectral grating, enough lines in this the to two be way wavelengths distinguished. is called will The be ability its resolvance λ Resolvance R = , where = m ∆λ Example 9.4.2 m is the order of the spectrum and N is the total number of slits Two violet lines in the hydrogen spectrum have wavelengths illuminated. 434 nm and 410 nm. These lines are just resolved in the second- For the examination you can 1 order use either of the λ spectrum by a diffraction grating with 30 lines mm . values or the Estimate mean λ The resolvance of a diraction the width of the light that is incident on the grating. Solution grating is the reciprocal of the λ Δ λ 24 nm λ and = 410 so nm N 410 = = = 8.5 fractional dierence in wavelengths mΔ ∆λ 2 × 24 between two wavelengths that can It is best to round up to 9 at this point; this means that the light just be resolved; in other words . incident on 9 = ∆λ must be illuminating nine slits and is, 3 therefore, = = 0.3 mm wide. ∆ 30 grating λ 1 the 10 λ Notice in that terms of using an 434 nm estimate, changes this is the not an number of important slits to 9(.05). So, difference. S AMPLE STUDENT ANS WER A diraction grating is used to resolve two lines in the spectrum of sodium in the second order. The two lines have wavelengths 588.995 nm and 589.592 nm. Determine the minimum number of slits in the grating that will enable ▼ There is calculation an of error Δλ; in this the the two lines to be resolved. should This 0.597 nm. detail However, for the there examiner to is error answer through. is correct 9 . 5 The and answer could that of there an ✔ is in = 1.597 nm; N = 589.29 D O P P L E R mean λ ÷ 1.597 there effect as sound are motion are observed between a ✔ when source 589.29 nm ÷ 2 = 184 describe is and useful Doppler differences observed in between light waves the and ✔ Doppler sketch effect in situations where and problems solve the Doppler using effect the effect diagrams showing how the Doppler occurs waves. ✔ interpret the When rest a source relative frequency when the sound 102 = E F F E C T observer that marks: You should be able to: frequency relative 1/2 the scores1. T H E shifts achieved follow rest You must know: ✔ have enough Δλ this [2] be to of an a source and Doppler sound wave observer, emitted by and the wave both or diagrams observer to you to explain effect. electromagnetic source and source. A Doppler electromagnetic given move shift relative radiation are radiation observer to occurs each treated agree in other. is at about the The differently . the frequency cases for 9.5 When the (taken to be imagine. (the speed. were once here) When but cross are spheres—centred wavefronts This line theory this data moves on a their line the booklet that be the the for a are in with the observer is its effect at a own medium more EFFECT medium the emitted D O P PL E R is waves. at easy constant the quickly to rate These wave than they detected. distances between case and waves the are observer detects emitted. by a stationary the detected source When and this is or the are more emitted concentric crosses slowly the (when again. observer not and observer approaching) is source, movement However, effect the to source true, are the moving speed along the components used. gives moving stationary , stationary two separate one transmission up’ move of (when centres. must to relative Doppler is source ‘catches source. quickly so the the that the frequency away , distorted assumes between along The more and higher with to observer source than not the from position a moving observer relative continue compared wavefronts receding) the so frequency the and the and source lower moving emitted, They the is wavefronts emitted ‘stretched’ a air The frequency) waves, For source THE f source = ′ equations for the Doppler effect: v f and another for a moving v ± u s v f observer ′ = f ± of the o emitted u (source) observer . Here frequency . is u f ’ is the observed with The the speed source of the speed u o know which sign frequency and f is the v sound . is v and However, you the speed need to s to use. Example 9.5.1 Work from first principles each A whistle emits sound of frequency of 1000 Hz and is attached to time. Imagine you are standing by a string. The whistle is rotated in a horizontal circle at a speed of the roadside and a police car goes 1 30 m s 1 . The speed of sound in air is 330 m s along the road at speed sounding An observer horizontal is standing a long way from the whistle on the same plane. its siren. The sound goes from a high to a low frequency, changing as the car draws level with you. In a) Explain why the sound heard by the observer changes the moving source equation, the regularly . negative sign must be used for b) Determine the maximum frequency of the sound heard by the observer. the approaching police car and the positive used for the receding police car. Solution a) The whistle away . wavefronts the (source) When are observer frequency the is moves whistle is compressed increases first towards moving and the compared increased. When the towards rate with source a at observer the which they stationary and and observer, observer then the cross whistle. are The An alternative way to remember this is to memorise the full expression but only for the approaching case, this is moving v + u 0 apart, the wavefronts are further apart and the frequency f ′ = f v − u s decreases. As the frequency heard relative speed is constantly changing, the When either u changes regularly too. s or u is zero, the 0 term simply disappears from the b) The maximum frequency occurs when the source and observer equation. are approaching. When the objects are moving 1000 × 330 apar t, reverse the signs. f = f = = ( 330 1100 Hz 30 ) 103 9 WAV E PHENOMEN A (AHL) The Doppler effect with electromagnetic waves needs care and so DP Redshifts occur in Options Physics only treats cases where the observer and source speeds are A and D. much less than the speed Option A of ∆f With this ∆λ assumption, higher speeds, the ≈ λ f (At v = An object moving relative to another experiences time dilation which light. c equation is more complicated.) affects the perceived rate at which it When the source and observer are moving away from each other, the emits pulses; this is the relativistic observed wavelength is longer than that emitted, so the frequency is Doppler shift. reduced. Radiation originating from a region closer This is together, a redshift. there is a When the source and observer are moving blueshift. of a large star or black hole has a The Doppler effect has many applications. It can be used with radar gravitational redshift as energy is to provide In radar the speeds of moving objects, and in weather forecasting. required to move it to infinity. This astronomy , microwaves are reflected from objects energy is gained at the expense of Moon. frequency (E = such as the v For a reflection, the frequency shift Δ f ≈ is 2 f hf). c Option D S AMPLE STUDENT ANS WER Light takes a long time to reach us from distant galaxies. The Universe Police use radar to detect speeding cars. A police ocer stands at the side of the road and points a radar device at an approaching car. The has expanded during this time and device emits microwaves which reect o the car and return to the so the wavelengths have stretched device. A change in frequency between the emitted and received as a consequence. This is the microwaves is measured at the radar device. cosmological or galactic redshift The frequency change Δ f is given by 2 fv Δ f = c Where f is the transmitter frequency, v is the speed of the car and c is the wave speed. The following data are available. Transmitter frequency f = 40 GHz Δ f = 9.5 kHz = 28 ms 1 Maximum speed allowed a) Explain the reason for the frequency change. This As answer the have ▼ The a answer discussion However, of begins well relative towards the of with end there an the motion. car is f ’ = moves to the statement the device that faster the “waves than they at the not The waves speed change of are light in physics perceived police ± U ofcer , frequency the due to device’s the waves relative motion ofcer . s would the reected microwaves and this on are reected on the car , they travel does back This disqualies major the as if they had the additional speed of the car . As a result, the waves reach the device faster than they would have if they error whole were reected off a non -moving answer. frequency 104 police travelling depending speed. the marks: V the observer ’s towards 0/3 f reach As have”. achieved V a have increased car could [3] change. source, and therefore there is a 9.5 This answer could have achieved 2/3 ▲ There marks: the T his case is a Doppler effect where there is a moving source the waves and a stationary obser ver is is reduced, source which is moving, causes the also in the distance same waves to that be the waves ‘squeezed’ the smaller distance. T he remains the wavelength same. = f that (this stating is then decreased T herefore, frequency if and is is not always not). into while the distance the is the idea waves that travel incorrect – the is better to increases say (c recognition increases of shift. travel reduced wavespeed frequency question ▼ However, a identication the the EFFECT (receiver). worth the D O P PL E R clear for frequency given Because is reason There of THE λ). that the wavefronts because the source since the emission has of are closer moved the previous wavefront. b) Suggest why there is a factor of 2 in the frequency-change equation This answer could have achieved 0/1 [1] marks: ▼ The Because the distance travelled by the wave is two times which distance between the source and the radar factor reection the of from acts rst 2 is the as because moving observer of the object and (car). then idea as source. here is The echo distance wrong. Practice problems for Topic 9 b) Calculate the angular width of the first-order Problem 1 3 A crane moves a load of mass 1.5 × 10 kg that is spectrum. suppor ted on a ver tical cable. The mass of the cable is c) A detector is positioned a distance of 2.0 m from the negligible compared with the mass of the load. When grating to detect the maxima. Calculate the distance the crane stops, the load and cable behave like a simple between the extreme ends of the first-order spectrum pendulum. The oscillation of the load has an amplitude in this position. of 5.0 m and period of 8.0 s. Problem 4 a) Calculate the length of the cable that suppor ts the When white light is reflected from a thin oil film floating load. on water, a series of coloured fringes is seen. b) (i) Determine the maximum acceleration of the load a) Outline how the fringes arise. as it swings. b) State and explain the changes to the appearance of (ii) Calculate the force on the load that produces the the oil film when it is viewed from different angles of acceleration in b) ii). incidence. Problem 2 Problem 5 7 Laser light with wavelength 6.2 × 10 m is incident on a a) Outline what is meant by the Rayleigh criterion for single slit of width 0.15 mm. resolution. a) Calculate, in degrees, the angle between the central b) A radio telescope with a dish diameter of 120 m maximum and the first minimum in the diffraction 2 receives radio signals of wavelength 6.0 × 10 m. pattern. Calculate the minimum angular separation b) Describe and explain the change in the appearance between radio images which this telescope can of the pattern when the monochromatic laser light is just resolve. replaced by white light. Problem 6 Problem 3 A train horn emits a frequency f. An observer moves 1 A diffraction grating with 10 000 lines m is used to towards the stationary train at constant speed and analyse light emitted by a source. measures the frequency of the sound to be f′. The 1 The source emits a range of wavelengths from 500 nm speed of sound in air is 330 m s to 700 nm. a) Explain, using a diagram, the difference between f ′ a) Calculate the angle from the central maximum and f at which the first-order maximum for the 500 nm b) The frequency f is 300 Hz. The speed of the observer wavelength is formed. 1 is 15.0 m s Calculate f ′. 105 FIELDS 10 1 0 . 1 D E S C R I B I N G (AHL) F I E L D S You must know: ✔ what is meant electrostatic by You should be able to: a gravitational eld and an ✔ represent eld suitable ✔ what is electric meant by masses gravitational gravitational potential potential that a eld and that line ✔ indicates the direction strength of density a of eld lines map elds of a ✔ what that is indicates an object by ✔ an with equipotential mass can move describe equipotential surface using that a a and eld describe line and the an surface the electric charge, on two eld between around two point a single charges, describe the charged and plates a gravitational eld around a point without around a spherical mass, and close to the work surface ✔ potential between between mass, doing and using surface ✔ gravitational electric lines) the eld meant and (eld eld point ✔ charges, force symbols equipotential the of and relationship ✔ and lines charged equipotential object surface can move without on an doing of a large massive object. electric work. A field is a region (gravitation) or a in which charge a force acts ‘at a distance’ on a mass (electrostatics). Electrostatic fields and force Field strength is defined as acting on a test object the gravitational fields involve very size of test object similar concepts and vocabulary. Use your knowledge of one of Electric field strength Gravitational field strength these fields to reinforce your Definition F F understanding of the other. E g = = q Nature of test object small positive charge 1 Unit In N C both cases, strength Because in the work a not 106 of force it test is acts object is small mass 1 ≡ 1 V m small N kg and does not change 2 ≡ m s the field placed. on an object, work must be done to move the object field: done the the where Potential of m = is force a object × distance quantity on which individual independent the objects force of acts. moving in the magnitude Potential it. is a of mass property of or charge the field, 10 . 1 Electric potential at a point is the Gravitational potential at a point is work done W in moving unit positive the work done W in moving unit mass DE S CRIBING FIELDS This topic picks up from Topics 5.1 and 6.2 which covered charge from innity to the point. from innity to the point. Electric potential is given by Gravitational potential is given by Coulomb’s law and Newton’s law of gravitation. You should understand W = q ΔV W the principles of these laws. = m Δ V e g where q is the charge and ΔV where m is the mass and Δ V is the is the gravitational potential dierence. electric potential dierence. topics include: 1 1 Unit of electric potential: J C Equations introduced in the earlier g e ≡ q Unit of gravitational potential: J kg V q 1 Coulomb’s law F = 2 k ; 2 r Newton’s law of gravitation The concept of innity produces a At very large distances (innity) the ‘standard’ place where potential is force must be zero. The potential at m m 1 F = 2 G 2 zero. In both electric and gravitational r innity is zero. 1 elds, force varies as . 2 distance An is important that distinction gravitational potential can be between potential either is positive gravitational always or and negative electric whereas To help your understanding, link fields these new ways that describe electric energy change in a field to the negative. meaning of potential difference in To see this, first think about gravity . Gravitational force is always current electricity. This is explored attractive. To move a test object to infinity from somewhere close to a in more detail in Topic 10.2. mass to At means be work overcome). infinity , been that the negative must Energy potential before is done (this transferred of the be the system transfer to positive test to is is because the system zero. infinity an to attraction move Therefore, took it the must has mass. have place. Q + In electric charge, fields, the However, has to stored they be situation a positive done to positive will when fly is a the same charge keep and them potential as a charge for energy , attracted gravity; positive together. is test When simply the release a potential charge you by repel, want the to two negative is negative. and work regain charges the and apart. Q Point have charges, radial point fields masses, charged spheres and spheres with associated 0 V +6.0 V +4.5 V –1.5 V with them Figure (Figure 10.1.2 uniform 10.1.1). shows field and + + + + + + –3.0 V the +3.0 V Figure 10.1.1. the – – – – parallel between charged two +1.5 V –6.0 V the positive and negative point charges 0 V plates. Figure 10.1.2. Between Radial fields for – –4.5 V equipotentials mass plates the Electric field line pattern for field parallel plates lines are parallel and equally You can ignore the effect of edge spaced; the field is uniform. Outside the plates, the field strength must effects in the DP Physics course fall away to the magnitude outside. The curved edge effects are the but not their existence. Always way in which the system makes this transition. draw them when you have to Although the Earth has a radial gravitational field, we live close to the represent the field between two surface so the separation of the lines is not apparent to us—broadly charged plates. speaking, we live in a uniform gravitational field. 107 10 FIELDS (AHL) Figure Points of the the 10.1.3 on shows the sphere green surface, green and which no a gravitational surface so surface on the have are the overall charge at the same work or field same done. can to a spherical distance potential. is mass due When This move from a gives planet. the mass centre moves on an equipotential without work being transferred. Because work is done when a charge or mass moves along a field line, –80 V equipotentials must always meet field lines at 90° –90 V Example 10.1.1 –100 V A precipitation consists of maintained a) b) Figure 10.1.3. system two at large collects parallel potentials Explain what A small dust is of meant particle dust +25 kV by particles vertical an moves plates, and in a chimney . separated by It 4.0 m, 25 kV . equipotential vertically surface up the centre of the Field lines and chimney , midway between the plates.The charge on the dust equipotentials around a planet particle i) is+ Show 5.5 nC. that there is an electrostatic force on the particle of about 0.07 mN. 4 ii) The up mass the of the centre of dust the particle chimney is 1.2 at a × 10 kg constant and it vertical moves speed of 1 0.80 m s Calculate strikes the one of minimum them. Air length of the resistance is plates so that the particle negligible. Solution a) An equipotential means that no surface work is is a done surface in of moving constant charge potential. around on This the surface. Vq b) i) The force on particle = qE where = d is the distance between d the plates. The potential difference 4 5.0 × 10 is 50 kV . 9 × 5.5 × 10 5 So force = = 6.875 × 10 N 4.0 5 force 6.875 × 10 2 Example 10.1.1 b) i) is a ‘show ii) The horizontal acceleration = = = 0.573 m s 4 mass that’ question. You must convince The the examiner that you have particle is in the centre of the 1.2 plates, × 10 so has to move 1 2 completed all of the steps to carry 2.0 m horizontally to reach a plate. Using s = ut + at and 2 out the calculation. The way to do knowing that the particle has no initial horizontal component this is to quote the final answer to 1 2 × 2.0 2 at least one more significant figure of speed gives 2.0 = 0 × t + 0.573t so t Here it is quoted to sf—and in therefore, the length must be = = 2 (sf) than the question quoted. 2.63 s and, 0.573 2.63 × 0.8 = 2.1 m. this situation this is fine. S AMPLE STUDENT ANS WER Explain what is meant by the gravitational potential at the surface of a planet. ▲ There question are two and two marks for points to This this answer done per taking be 108 unit the ‘small’ from has mass, mass in a innity them (it both: and does potential to the could have achieved 2/2 marks: is the work done per unit mass to bring a small test mass work the not idea have of from a point of innity (zero to denition) surface. answer make— It this [2] (in the gravitational eld). PE) to the surface of that planet 10 . 2 1 0 . 2 F I E L D S AT the distinction potential ✔ the eld ✔ the ✔ what between and potential and ✔ determine and between potential gradient the the potential potential ✔ solve problems ✔ determine ✔ solve energy energy of involving a of a point point mass charge potential energy strength meaning is WORK You should be able to: energy relationship and AT W O R K You must know: ✔ FIELDS of meant orbital escape by the potential inside a charged sphere speed orbital motion, orbital speed problems satellite energy of an involving orbiting object a planet escaping the the orbital and the speed escape gravitational of a speed eld of aplanet ✔ the magnitude the variation charged of of electric electric eld strength potential and inside a ✔ solve sphere problems charged masses ✔ aboutforces and inverse-square law particles in solve on problems charges uniform a radial field that obeys in circular the circular orbital orbital orbital energy motion of and motion behaviour. ✔ In involving inverse-square law behaviour, and involving the masses radial in forces acting and elds. potential This topic builds on Topics 5.1, 1 depends , on where r is the distance from the origin of the field to the 6.1, 6.2 and 10.1. r object In (for Topic work example, 10.1 done to a in moving point potential in moving particular a a point. mass or was unit charge defined mass Therefore charge or or to be unit there from point is one mass). zero at positive a infinity and charge from potential difference (non-infinity) point is the infinity involved in a field to another. Electric potential in a radial field Gravitational potential in a The value of the constant k radial field kq V = 1 − in the data booklet is e r V g q is the for a GM charge that produces πε = 0 − r the vacuum (or air, approximately), as M radial is the mass of the object field. 1 producing the radial field. in Topic 5, and when in a πε ε r The potential energy (with zero potential at infinity) in each case 0 is: dielectric material (see Topic 11.3 for more details on the meaning kq 1 E p = qV q 2 GMm = e E r p = mV = of ε ). r − g r Electric potential difference links to a term used in Topic 5 – simply called potential difference (pd) there. 109 10 FIELDS (AHL) An important relationship change field strength in graphical against terms, distance Electric field field strength and potential is potential = change In between field graph in distance strength at point at x). point This is x is –(gradient known as Gravitational strength, the field of the potential potential gradient. strength, ∆V ∆V g e E = g − = − ∆r ∆r For a uniform electric field electric strength field, For a uniform gravitational field is ∆V G close ( final potential initial planet surface, g = , − h ; in a ) potential Understanding the links change to where is ΔV the change in potential G distance between both types of field over and the connections between when the change is from +V to distance gravitational field strength and potential are mΔV impor tant for you to be able to h. The change in , which potential is mgh, energy when an is object G V apply your knowledge in this area. a 0, (a change V), E of = − V of = d mass and m moves through d vertical distance h htgnerts dle cirtcele Example 10.2.1 14 A charged ΔV two oil parallel drop of weight oppositely 3.0 charged × 10 N metal is held stationary between plates. E a) The electric field between the plates is uniform. distance i) Δr Explain ii) Sketch laitnetop cirtcele b) The what field plates this lines are means. to show separated by the electric 4.0 mm field and the between potential the plates. difference ΔV eld E = applied − between them is 380 V . Δr ΔV Calculate the magnitude of the charge on the oil drop. Solution distance a) Δr i) Field htgnerts dle lanoitativarg The strength field is strength the is force the per same coulomb acting everywhere on the between oil the drop. plates. distance 1 g ii) The field lines should be at 90° to the plates and parallel to 2 r each other; effects the should field be line separations should be constant. Edge shown. ΔV V 1 b) The field strength E = = 95 000 N C . The weight of the d drop = Eq 14 So 3.0 × 10 4 = 9.5 × 10 19 q and q = 3.2 × 10 C. Δr distance laitnetop lanoitativarg 1 r ΔV Figure g = Figure 10.2.1. shows, for both electric and gravitational fields, the − Δr The relationships between potential and field for both electric and gravitational fields 110 10.2.1 ΔV relationships graph is graph will between shown also for a field strength positive apply to and charge attraction in (a potential. repulsive the electric The electric effect). case. The field gravity 10 . 2 You should potential understand the distinction between potential electric potential at a point is the work per unit gravitational WORK and from charge infinity to the mass AT energy . moving FIELDS in point. electric potential energy is the work required to move a gravitational charge mass between two points at different potentials. Example 10.2.2 2 GM m p m s The equation v s = 2 A and Point B P are is separated 120 mm by from a distance A and of 160 mm 200 mm. from r GM B. p 2 ⇒ v = r 160 mm Point A has a charge of +2.0 nC. r P B This shows that the orbital speed Point B has a charge of 3.0 nC. −3.0nC GM p a) Explain why there is a point X 120 mm v 200 mm = orbit on the line electric AB at which potential is r the Notice that v zero. depends only on orbit +2.0nC A the orbital radius and the mass of b) Calculate the distance of the point the planet not on the mass of the X from A. satellite. Solution a) The is A geosynchronous orbit is one with potential always the due to negative. potentials sum A is always There to must, positive; the therefore, be potential a point due (X) at to B which zero. an orbital period that matches the Ear th’s rotation on its axis. A special case of this type of orbit is the geostationary orbit which b) Taking x as the distance from A to the zero potential point, is geosynchronous and also −19 k −19 × 2 × 10 k × (−3 × 10 ) positioned above the equator. + = x (0.20 0 . This solves to give x = 8.0 cm. A satellite in geostationary orbit x) remains apparently xed in position when viewed from Ear th. This type of orbit is used for many communications satellites. Rockets energy and and satellites stored that orbit a gravitational planet or potential Sun have energy GM when gravitational potential energy is E = in kinetic orbit. m p The both s where − r is the p r orbital radius (from the centre of the planet and not the surface) M is p the mass of the planet and m is the mass of the satellite. 1 (radius decreases) is often 2 The kinetic energy of the satellite m is The point that the kinetic energy increases when an orbit decays s v s misunderstood by students. 2 Gravity provides the centripetal force required to keep the satellite in GM Remember that v 2 GM m p its circular orbit. m s v GM 1 = ⇒ m 2 is term half on the the r left-hand magnitude of r s = s r The v m p 2 s So, side the is = orbit (M is the planet mass). 2 the 2r kinetic gravitational energy potential of the satellite energy . and This tells you that if r v is smaller, must be larger. orbit GM m p The total energy of the satellite at orbital radius r s is 2r 111 10 FIELDS (AHL) What happens to a satellite when it moves to a lower orbit? The Escape speed is the minimum equations predict that, at a lower radius, the gravitational potential speed required for an object to energy is more negative energy is higher. (has a larger magnitude) and the kinetic leave the Ear th’s surface and (just) The total energy is also more negative. The speed reach innity. of the satellite, therefore, increases even though, overall, it has lost The initial kinetic energy must energy . provide energy equal in magnitude to the gravitational potential energy GM The energy sign of follows a in charged a similar particle way . orbiting Equating another charge Coulomb’s law of opposite with the p at the surface; this is 2 kq r So m v s mv 2 force: = m p 2 q 1 centripetal GM 1 E ⇒ , 2E as before. K r r = = P 2 s esc 2 r Rockets 2GM bound can in leave orbits. Earth’s The gravity escape speed completely is the speed as well as required remaining to do this. p ⇒ v = esc r Example 10.2.3 2 This is × the orbital speed at the surface. 24 The mass of a) Deduce b) Calculate 750 kg of the the Earth radius the when is of a change raised 6.0 10 7 kg and its geosynchronous in to × a potential radius orbit energy geosynchronous of a is for 6.4 the from 10 m. Earth. satellite orbit × of the mass surface Earth. Equating the gravitational Solution force with the centripetal a) The time period for an Earth geosynchronous orbit is 24 hours. force leads to other results too. 2π Using the angular velocity ω, −5 T = (24 × 60 × 60) = GMm Equating 2 = . Therefore, ω = 7.3 × 10 −1 rad s ω rather than the linear speed, gives centripetal force with gravitational force leads to mrω 2 1 r r = 2π GM 2 And, because T . ω = 3 ω 7 Substituting gives an answer for the orbital radius of 4.2 × 10 m. 2 mr 4 π GMm , which gives = 2 2 r b) The change in potential = −GM 2 radius and r is the orbital 1 1 − T r R where R is the Earth radius. 4π 2 T 3 = r 11 This GM is 6.67 × 10 24 × 6.0 × 10 1 1 6 7 6.4 × 10 4.2 × 10 The relationship between T and r 7 = is due to German mathematician 5.3 × 10 1 J kg Johannes Kepler; it is known (Note the manipulation of the negative signs here – never ignore as Kelper ’s third law. Kepler the signs; always carry them through.) deduced this empirically from 7 the astronomical observations of The potential energy change is × ΔV g m 5.3 × 10 = 4.0 × 10 × 750 10 Danish astronomer Tycho Brahe. Moving charged magnetic the 112 = satellite fields. particles. particles The are fields affected give rise to by the presence different types of of J electric motion and for 10 . 2 Charged particles in an electric field Charged particles in a FIELDS magnetic AT WORK field +V force parabolic path current V d magnetic eld out = E d of plane of paper X 0 V force conventional current An electron enters perpendicular the opposite to a uniform the field direction to electric lines. these field A force An acts in lines. electron to the field to the velocity lines. centripetal The force on the charge is Ee and enters uniform The of force a the and magnetic magnetic electron. the force This electron is field acts the moves perpendicular always at 90° condition in a for circular a path. the 2 m acceleration a v e Ee The . = In terms of the pd force on the electron is Bev which must equal , V r m e where between the charged plates that produce r is the radius of the orbit. the eV field and their separation d: a = . m d e Example 10.2.4 A proton and a 7 of 1.5 The × 10 magnetic motion of positive pion travel along the same path at a speed 1 m s the into field a uniform direction proton and magnetic is the at 90° pion. to The field of flux the initial rest mass density direction of a pion 0.16 T. of is 28 2.5 × 10 kg. a) Calculate b) Comment of c) the The the the radius on how of the curvature path of of the the pion path will of the differ proton. from that proton. magnetic paths of field the strength is decreased. Suggest how this affects particles. Solution 2 m v m p a) Rearranging and = Bev gives r = r The data booklet provides m ; v p simplifying eB substituting gives p 27 1.67 r 7 × 10 × 1.5 × 10 = = 0.98 m. 19 1.6 × 10 b) A pion has charge so is c) less only smaller for Again, from and the so × 0.16 mass the the the path than mass pion is and radius of a both proton. different. so its is when is particles From path equation, particles The less the more B is have radius the same equation, r curved. decreased, r increases curved. 113 10 FIELDS (AHL) S AMPLE STUDENT ANS WER The gravitational potential due to the Sun at its surface is 11 1 J kg −1.9 × 10 . The following data are available. 2 Mass of Ear th = 6.0 × 10 kg 11 Distance from Ear th to Sun = 1.5 × 10 m 8 Radius of Sun = 7.0 × 10 m a) Outline why the gravitational potential is negative. ▼ While there the answer, far. This a is between some does answer question dened it is about not really the in us This answer answering It is works force and ▲ There are made. (i) This mass The has is to away zero be This must scores at in answer the Since to achieved 0/2 marks: the opposite direction to the gravitational both: move Sun the been to 0 could of have achieved gravitational gravity attracts, 2/2 marks: potential work is always dened has to be at to be have zero gained at on the object to get closer to 0 closer to innity, which is the same thing as (ii) potential, so potential is negative, since work need a innity . mass to be added to move away from the Sun’s gravitational eld. was negative potential innity. done for b) The gravitational potential due to the Sun at a distance r it force. be innity , to before have and need done from potential moved marks points answer potential work two two have the direction. Because therefore could very relationship gravitational positive truth get [2] and from its still centre is V innity . . [1] s Show that ▼ The rst sentence is not rV = constant s very the clear. other It way Fortunately answer should be round the makes expressed for up for This clarity . remainder of answer could have achieved 0/1 marks: the this. GM V rV = s s r Mass ▼ You maybe answer the scored command answer zero. term: correctly equation and surprised for shows constant is The that key show that GM, rearranges the but value that is that. gravitational the Sun in The the potential of is of this the This may seem picky to you but it is the essence of a show that question. You not must not expect the examiner to finish the work for you (which is what this enough. It should have gone on to answer requires)—write every step down. say that G M × 11 G is must constant. be Therefore, constant too. 10 . 2 FIELDS AT WORK Practice problems for Topic 10 Problem 1 Problem 4 Two parallel metal plates A and B are fixed ver tically a) Define electric potential at a point. 20 mm apar t and have a potential difference of 1.5 kV b) A metal sphere of radius 0.060 m is charged to a between them. potential of 50 V. a) Sketch a graph showing the potential at different (i) Deduce the magnitude of the electric charge on points in the space between the plates. the sphere. b) A plastic ball of mass 0.5 g is suspended midway (ii) Determine the magnitude of the electric field between the plates by a long insulating thread. The strength at a distance of 0.12 m from the centre of ball has a conducting surface and carries a charge of the sphere. State an appropriate unit for your 3.0 nC. answer. The ball is released from rest. Deduce the subsequent (iii) Identify the magnitude of the gradient of electric motion of the ball. potential at a distance of 0.12 m from the centre of Problem 2 the sphere. a) Outline what is meant by gravitational field strength. Problem 5 4 b) The radius of Jupiter is 7.1 × 10 km and the gravitational field strength at the surface of Jupiter is A satellite orbiting a planet has an orbital period of 60 minutes and an orbital radius of 9. Mm. 1 25 N kg . Estimate the mass of Jupiter. a) The satellite orbits with uniform circular motion. Problem 3 Outline how this motion arises. A proton from the Sun passes above the Nor th magnetic b) Show that the orbital speed of the satellite is pole moving parallel to the Ear th’s surface at a speed 1 6 of 1.2 × 10 about 2 km s 1 m s . At this point the magnetic field of the Ear th is ver tically downwards with a magnetic flux c) Deduce the mass of the planet. 5 density of 5.8 × 10 T. a) Calculate the radius of the path of the proton when it is above the pole. b) A helium nucleus moving with the same initial velocity of the proton is also above the pole. Compare the paths of the proton and the helium nucleus. 115 E L ECT R O M A G N ET I C 11 1 1 . 1 I N D U CT I O N E L E C T R O M A G N E T I C You must know: ✔ that there is a is a ✔ what is meant ✔ that I N D U C T I O N You should be able to: magnetic there magnetic ux in a region where ✔ describe eld magnetic ux ux density is equivalent eld relative solve that an force (emf) is induced in or when when that it moves there links relative to a Lenz’s law a when a conductor magnetic eld include and the magnetic use of ux Faraday’s and the are changes in the and explain ✔ law of linkage changes how conservation of energy leads to law magnetic conductor Faraday's ux magnetic explain that depends ✔ uniform that induction Lenz’s ux arises when a ✔ eld a emf and strength electromotive conductor to problems of magnetic ✔ induced changes to law magnetic the ux linkage ✔ magnetic how magnetic moves by (AHL) on the the magnitude change in of the induced magnetic ux emf linkage. induction. Scientists in the 1820s realised that relative motion of a conductor in Topic 5.4 showed that electric a magnetic field leads to the production of an induced emf and to currents in conductors in magnetic an induced current in the conductor. This introduces the concepts of fields lead to motion. Magnetic magnetic flux, magnetic flux density and magnetic flux linkage fields were visualized using field The units of magnetic ux density lines. The closer these were drawn, Magnetic u x ϕ the stronger the field represented. number of lines cut by a conductor is related to the 2 are Wb m or enclosed by a one-turn coil while The concept of magnetic field 2 1 T ≡ 1 Wb m in the magnetic eld. The unit of strength (symbol B was used in When a coil has more than one magnetic ux is the weber (Wb). Topic 5.4) is developed fur ther. turn (N turns), then each coil turn The density of eld lines indicates the links one set of lines, and the total strength of the eld, so the magnetic magnetic u x linkage is (ux for one This idea links to Topic 5.4. eld strength, measured in tesla (T), turn) × (number of turns); in other There, charge in a wire was moving can also be called the magnetic ux words, NBA or Nϕ in a magnetic field and this gave density (in symbols, ϕ = BA). Density rise to a force acting on a wire. Here in this case means the number of eld the wire is moving its electrons lines per square metre. along with it – but the physics is the same. An emf moves field the is is at generated right into plane the of between angles plane the to of a the ends magnetic the page, of a field. and the straight In Figure conductor 11.1.1 conducting the rod is when it magnetic moving in page. magnetic eld into page × × × × × × × × × × + × × × (a) × × + × × × × × × × × × + (b) conventional electron current ow (c) Figure 11.1.1. 116 An emf is induced in a conductor moving in a magnetic field 11 . 1 Individual free electrons are moved upwards by the moving rod. E L E C T R OM A GN E T IC INDUCTION This Lenz’s law states that the induced is a conventional current moving downwards and a motor effect acts emf gives rise to an induced current on each electron. on each free Fleming’s left-hand rule (or other rule) gives the force that opposes the change producing electron to the right. This produces a negative charge at it. It provides a statement about the R and an electron deficit (positive) at L. Eventually , the electron flow direction of the electromagnetic stops as electric electromotive When will there flow. predicts forces force is a (emf) has complete Kinetic that prevent energy this been motion generated circuit, is current further as shown transferred will produce to a along the in the in Figure An induced induction eect. rod. electrical motor rod. 11.1.1(c), charge energy . Lenz’s effect to the left Faraday’s law states that the law emf ε induced is given by to ∆ ε oppose the change made to the = , where Δϕ is the −N system. ∆t change in ux and Δt is the Work is done to overcome this opposing force and is the origin of change in time. N is the number the transferred energy in the system. Lenz’s law is equivalent to of turns of the coil for cases where conservation of energy . If the forces produced acted upwards, energy this applies. When the velocity would be created in the system. This is not possible. and eld are not perpendicular, Faraday’s the law makes electromagnetic a quantitative effect. In this statement case, about when the the magnitude conductor of of ∆ written as ε l moves at speed v in a uniform magnetic field of flux . This can also be = BA cos length density B, = ( BAcosθ ) −N , the ∆t where B and A are the magnetic lv change in the flux every second is Bvl because ε = − NB × and N = 1. ux density and the area swept 1 out respectively, and θ is the When there are N turns, ε = BvlN angle between the eld and the Faraday and others Figure11.1.1 can be was extended developed ‘cutting’ to other the field idea lines; conductor that this is shapes the a conductor simple idea conductor. in to grasp. It too. See Topic 5.1 for a discussion When the magnetic field moves and the conductor remains fixed, more about the meaning of or when the magnetic field changes magnitude and the conductor conventional current. remains the fixed, magnetic the effects are the same as when ΔB over the conductor moves in field. Example 11.1.1 The magnetic flux density changes by a time Δt. As a result, A coil with five turns has an ∆B 2 Faraday’s law predicts that the emf generated will be NA , where N area of 0.25 m . The magnetic ∆t flux density in the coil ∆B is the number of turns, A is the fixed area of the coil and is the rate changes from 60 mT to 30 mT ∆t is at which the magnetic flux density changes with a time of 0.50 s. time. Calculate the emf the magnitude induced in the of coil. S AMPLE STUDENT ANS WER Solution A ver tical metal rod of length 0.25 m moves in a horizontal circle about ∆B a ver tical axis in a uniform horizontal ver tical axis ε = − NA magnetic eld. ∆t metal road The The metal rod completes one circle magnitude is 3 of radius 0.060 m in 0.020 s in the ( 60 − 30 ) × 10 uniform horizontal magnetic eld of strength 61 mT. −5 × 0.25 × 0.50 magnetic eld Determine the maximum emf induced between the ends of the metal rod. This answer could have achieved = 75 mV [3] 2/3 ▼ This marks: attempt calculation of has the one error. rotation 1 S peed of rotation = 2π × 0.06 ÷ 0.02 = 0.6π m s The speed is 1 incorrect The nal (it should answer be 6.0π m s would have ). been 3 maximum emf is Bvl = 61 × 10 × 0.25 × 0.6π better = to two signicant gures. 9.15mV ▲ The and solution the steps is are well explained clear. 117 11 E L E C T R OM A GN E T IC 1 1 . 2 INDUCTION POWER (AHL) G E N E R AT I O N You must know: ✔ what is meant by AND TRANSMISSION You should be able to: an alternating current ✔ explain generator (ac) the operation generator of an including alternative changes of current generator frequency ✔ the meaning square of values average for power current and and root mean voltage ✔ solve problems problems ✔ what is meant by a step-up and a involving involving root average mean power square and (rms) and step-down peak currents and voltages in ac circuits transformer ✔ ✔ what is meant by a diode describe the use transmission ✔ what is meant by of transformers in the bridge full-wave and of ac electrical energy half-wave ✔ investigate diode bridge rectication circuits rectication experimentally ✔ how to describe qualitatively the effect of adding ✔ capacitance to a diode-bridge rectication solve problems down The involving step-up and step- circuit. simplest ac generator transformers. is a one-turn coil of wire rotating in a The principles of uniform magnetic field (Figure 11.2.1(a)). the is Zero flux links the coil electromagnetic induction (Figure 11.2.1(b)) when coil parallel to the field lines. When the discussed in Topic 11.1 apply to coil has turned through 90° it is perpendicular to the field lines and coils rotating in a magnetic field that the flux has changed, inducing an emf. When the coil turns through lead to the generation of alternating another 90° it is in its original orientation except that the sides of the currents (ac). coil (a) have reversed. through a final The in The 180° to flux linkage return to its has changed original again. The coil rotates orientation. ac generator emf negative the alternating Slip slip rings to rings the coil trough varies (Figure current (Figure load. The will continuously 11.2.2). be induced 11.2.1(a)) rings are With are in used attached between suitable a to to load positive coil energy ends and peak and connections, connected transfer the a electrical to the from coil. the rotate a an coil with it. V Stationary rings (b) and brushes, are often connected made to the of conductive carbon, press onto the load. Example 11.2.1 A rectangular dimensions coil of 20 mm 650 × turns 35 mm with rotates about a θ horizontal Figure 11.2.1. axis. The axis is at right angles N S Simple ac generator to a uniform 2.5 mT. makes a) b) c) At an magnetic one instant θ angle Identify the value the coil Calculate the Determine the with through of to field θ the of density the coil vertical. for a magnetic flux plane the be of the magnitude of the magnetic the coil ux minimum. ux maximum passing ux through linkage through the when θ coil as it is 30°. rotates. Solution a) The is b) c) coil the The The must be minimum magnetic (θ = 0) to the θ magnitude). ux maximum eld parallel ux which is ϕ = BA linkage eld is cos θ is either = when corresponds to for a the 90° ux or to be zero (this 270°. 1.5 μWb. the coil is perpendicular maximum ux of to 1.75 μWb. 6 The 118 maximum ux linkage is: 650 × 1.75 × 10 = 1.1 mWb turns. the 11 . 2 Figure an ac 11.2.2 shows generator the that is output constant speed; the G E N E R AT I O N AND T R A N S MI SS ION of rotating ux at θ or time linkage a POWER waveform is When the rotation speed is sinusoidal. Quantities associated doubled: rate of with the varying output are its change frequency and the peak emf • the frequency doubles because of ux ε 0 linkage and peak current the time for one rotation halves time I and 0 induced Alternating current and pd values • the rate of change of flux linkage emf doubles and so the peak emf can be compared with direct Figure 11.2.2. current (dc) values. doubles too. Variation of emf with A suitable time for one cycle of a coil in a uniform average is needed for ac. When magnetic field the current over the one cycle, power Figure and P pd are their averaged mean dissipated 11.2.3 shows in that value the the is load wave zero is and cannot always shape of be used. However, positive. the power dissipated is 2 always positive (sin ) and has a non-zero V V × 0 . This can be written that is equal to I 0 I 0 average 0 as × 2 The resistance of a device 2 2 in an alternating circuit can These separate quantities are known as the root-mean-square (rms) be calculated using either V I 0 values of the current and pd, with V = 0 and I rms = V rms V 0 2 2 R rms = or . However, you I I 0 rms tnerruc must use either rms or peak rotsiser ni detapissid rewop 0 time (a) (b) values – do not mix them. The peak power delivered by the mean value =V supply is P max I 0 0 of power This is only delivered twice in each cycle at the instant of each maximum and minimum of the waveform. 0 time The average power is Figure 11.2.3. rms Current, power and mean power values for ac rms 1 which is V I 0 The power delivered resistor of effective values resistance over R is one V 6 V direct 6 V rms current an . I rms of cycle Root an alternating mean square supply values to are a 0 2 the rms alternating supply by has power the same supply . A lamp brightness connected when to connected a to a The input voltage V alternating supply . and output s emf ε are related to the ratio of the s V n p Example 11.2.2 p numbers of turns: ε n s An ac power supply of emf V and negligible internal resistance where = s is is the number of turns on the n 0 p connected to two identical resistors of resistance R in parallel. primary coil and n is the number of s Calculate the average power dissipated in the turns on the secondary. circuit. This is written in the data booklet as Solution R The total resistance in the circuit is ε as the resistors are in N I p parallel. p = p = 2 ε N s I s s 2 2V rms The power dissipated in the circuit Energy is conserved in the is R transformer (this is approximately 2 V However, V true in practical transformers where V 0 0 , = so the power dissipated is rms eciencies are 90% or higher) so R 2 V p I = p ε I s , where I s p and I are the s primary and secondary currents. In Alternating current can easily be converted from one voltage to other words, the power input to the another using a transformer. connect low-voltage This is useful in domestic situations (to device is equal to the power output. energy over long devices to the mains) or to transmit electrical distances. 119 11 E L E C T R OM A GN E T IC INDUCTION (AHL) Transformers consist of: Energy losses in the transformer • a primary coil connected • a secondary coil • a core from to the input terminals Electrical resistance in the coils 2 leads to I R losses that can be connected to the output terminals reduced by using thick wires for made a soft iron material that is easily magnetised in the coils and connecting wires. presence of a magnetic field but loses this magnetism quickly when Core losses include the following. the field is turned off. • Eddy current losses: The Both coils are wound on the core so that the fields through both are as changing magnetic eld produces an induced emf in the conducting similar as possible. core as well as in the secondary. The operation of the transformer This loss is reduced by laminating the core – manufacturing it • in insulated strips so that the Alternating alternating electrical resistance is large but the the same current is magnetic rate as the supplied field in to the alternating the primary core. This and field establishes reverses an direction at current. magnetic proper ties are unaected. • As both coils are wound on the same core, this field is linked to the • Hysteresis losses: Energy is secondary . needed to re-magnetize the core in opposite directions. • There and is an a constantly emf is changing induced in magnetic field inside the secondary it. Less than 100% of the core ux will be linked to the secondary leading • When there is a resistive load attached to the secondary , there is an to ineciency. induced current in the load. Example 11.2.3 An ideal transformer with 800 turns produces an rms output of 3 2.0 × 10 V when a) Calculate b) Outline the it is connected number how eddy of to turns currents are the 230 V required rms on minimized mains the in a supply . secondary coil. transformer. Solution V n p p a) ; = n = 6960 turns s ε n s b) s There is an magnetic core the is eld for core has energy a of emf it. This and high one in the the leads therefore ux ac transmission loss in electrical magnetic conversion energy in metallic allowing The induced to to a due an long the induced resistance to to conductor. penetrate voltage over core current By but changing is using still because the laminations, effective in it. another is distances; it particularly allows a important reduction in cables. Example 11.2.4 A cable station is to Compare a) to a transmit factory . the 100 kW The power total losses of electrical resistance for a in power the transmission 250 V b) from cable is voltage a power 0.40 Ω of 25 kV . Solution a) 250 V ac: P = VI so the current in the cable is 400 A. The power 2 loss in total b) ac: 6.4 W. higher 120 cable power 25 kV is the I R = 64 000 W—a substantial fraction of the transmitted. The current There pd. is will in be the cable signicant is now 4.0 A and savings in the power power loss with loss the 11 . 2 Some devices cannot use alternating current—they require POWER G E N E R AT I O N AND T R A N S MI SS ION direct current. X The half-wave rectifier produces a pulse of current for half of every ac load 10 V ac cycle. resistor Y The diode allows conduction only in one direction (Figure 11.2.4) so it (a) conducts are for shown. peak only Note half that of each the cycle. average The for the input and output output current waveforms is not half the values. A full-wave output rectified output with two pulses per ac cycle can be pd obtained in two time ways: (b) input Figure 11.2.4. Half-wave rectifier D 1 X (a) C R pd time Y A rectier can be used to conver t ac to dc. D 2 Z A diode bridge is one kind of rectier. Two types are common: the half-wave rectier and the full-wave X (b) rectier. X D D 4 1 D A 2 D 3 Y + secondary of pd B transformer Figure 11.2.5. Figure then Full-wave rectification 11.2.5(a) output is shows positive diode time Z D at a the conducts. two-diode top of the When arrangement. diagram the and transformer When the negative output at has transformer the The basic proper ties of diodes bottom, reversed and are covered in Topic 5.2 where the 1 the positive output is at the bottom of the secondary coil, then diode D V I characteristic is given. When 2 conducts. Whichever resistor always is in diode the conducts, same however, the current in the load the diode is forward biased, charge can flow through it. But if the pd direction. across the device is reversed, it The alternative bridge arrangement (Figure 11.2.5(b)) uses four diodes. cannot conduct. This is an ideal When X is positive with Z negative, diodes D and D 1 point A positive and point B negative. When conduct making 3 point Z way to strip off the negative par t of is positive, diodes an ac waveform and just leave the D and D 2 One with way the When As conduct giving the same current directions as before. 4 but at smooth load the the positive-going par t. to resistor output output a slow out rate waveform (Figure goes drops the 11.2.5(a) positive towards because the for the the zero value of is to and first place capacitor in parallel (b)). time, value, RC a is the the capacitor capacitor chosen to be charges. discharges long The charging action of capacitors compared with the time for half a cycle. The voltage, therefore, drops is discussed in Topic 11.3. only slightly before the capacitor is charged for the second time. 121 11 E L E C T R OM A GN E T IC INDUCTION (AHL) S AMPLE STUDENT ANS WER In an alternating current (ac) generator, a square coil ABCD rotates in a magnetic eld. B slip ring C brush The ends of the coil are connected to slip rings and brushes. The plane of the coil is shown at the instant when it is parallel to the magnetic eld. Only one coil is shown for clarity. Explain, with reference to the diagram, how the rotation of the generator produces an electromotive force (emf) between the brushes. This T he ▼ A good answer to this answer rotation references the • Faraday’s concept of ux linkage law and the the rate upwards, the The direction not rotation eld of this forces required 1 1 . 3 in by answer and the to on currents of what is Faraday’s change of experiencing the the between coil marks: changes its ux over time and law, an ux. In emf will be produced as emf a the change diagram, in ux in as the AB rotate magnetic eld two ends direction is of the from magnet, N→S the force is (rightwards). upwards, Using is Fleming’s question. left right rule, the current will ow from A to B. C A PA C I TA N C E You must know: ✔ square 1/3 of coil emphasis the achieved to: is • of have question according requires could [3] meant by You should be able to: capacitance ✔ solve problems involving parallel-plate capacitors ✔ what ✔ how ✔ the is meant by a dielectric material ✔ capacitors combine in series and describe the effect of dielectric materials on parallel capacitance properties of a resistor–capacitor (RC) ✔ investigate combinations of series or parallel circuit capacitor ✔ what is meant by the time constant of an circuits RC ✔ describe the charging process for a capacitor in circuit series ✔ how to solve problems involving the ✔ of a capacitor through a xed with a resistor discharge determine the energy stored in a charged resistor capacitor ✔ how to solve problems for charge, current and ✔ voltage involving the time constant of an describe a circuit. 122 the nature of exponential RC capacitor through a resistor. discharge of 11 . 3 A simple air or an When the capacitor insulator the (Figure (Figure capacitor plates potential consists (Figure is of 11.3.1(c)). connected is to This between There parallel metal plates, separated by (a) 11.3.1(a)). 11.3.1(b)). difference two C A PA C I TA N C E the now a power flow eventually plates no supply , equals longer charge ceases that of sufficient flows when the the power energy to onto supply before transfer charging more fully electrons to and from the plates and the capacitor is said to be charged Figure stored 11.3.2 on shows one of how the pd between the plates varies with charge them. E This linear graph leads to the definition of capacitance q (b) Capacitance: C , = where q is the charge stored and V is the potential + V + difference at which it is stored. + −1 The unit is the farad (F); 1 F ≡ 1 C V . In fundamental units, this is electron + −2 −1 m 2 kg A 4 movement s + 1 The energy stored at pd V by a capacitor of capacitance C qV is (this 2 is the area under the graph) where q is the charge V < E during stored. charging 2 q 1 1 q 1 2 The definition of C as gives two qV variants: V = CV 2 = 2 2 C E Energy is transferred increases. The area to a charging under the capacitor graph (Figure as the 11.3.2) pd across gives the it energy (c) transferred during charging. + + + + + Example 11.3.1 + + A charge of 20 μC is stored at a potential difference of + 10 V . + Calculate the energy charging stored. nished E Solution 6 1 20 × 10 × 10 4 E = qV = = 2 1.0 × 10 J 2 E Figure 11.3.1. Charging a capacitor 1 qV of energy is transferred from a power supply but qV only of 2 energy is stored power power has supply supply . energy is been and When radiated capacitor lost in the capacitor the away during resistance and resistances from the charging. in the are of wires internal very system the Where as small, is the that resistance the bulk electromagnetic q connect of of the the radiation. gradient = roticapac no the It the derots egrahc remainder? by capacitance energy stored = area under the graph Example 11.3.2 V potential dierence A capacitor Calculate stores the 20 μC of capacitance charge of the at a potential difference of 10 V . across capacitor capacitor. Figure 11.3.2. Charge–pd graph for a 470 μF capacitor Solution 6 q 20 × 10 6 C = = V = 2.0 × 10 = 2.0 µF 10 123 11 E L E C T R OM A GN E T IC INDUCTION (AHL) Example 11.3.3 Capacitors consist of two plates of area A separated by Two capacitor plates are separated by a rubber sheet of thickness distance d. Capacitance is given A by C = 3.0 mm A ε ε = fills the space are r d 0.35 m × 0.45 m. ε them. The plate dimensions for the rubber is 6.0. r d the constant ε between , where ε 0 that is the relative Calculate r the capacitance of the arrangement. permittivity of the material between Solution the plates. ε for air is very close to r 2 The area of the plates is 0.158 m 3 . The distance apart is 3 × 10 m. 1; however, for other dielectrics,ε r A 0.158 12 takes values that range from 1 to C = ε gives ε 0 C = 8.85 × 10 × 6 × = r 6 10 9 2.8 × 10 F ≡ 3 nF 3 d . A dielectric between the plates 3 × 10 increases the capacitance so that a greater charge can be stored at a par ticular pd. Dielectric has one materials end are positive polar, and the meaning other that the negative, insulator either molecule permanently or E cap temporarily The (in molecules an electric line up in field). the field with the positive ends towards the E dielectric negative plate. This reduces the total field: E = net E . − E cap Because dielectric E net V the total field E = , V is also reduced. The dielectric increases net d + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + − + capacitance. + + Capacitors can be combined in series or parallel networks. + plate + + + + dielectric Figure 11.3.3. How a dielectric increases capacitance Capacitors in parallel Capacitors C V 1 +Q 1 Q 1 in series V 2 V 1 +Q Q +Q Q +Q Q V +Q Q V +Q Q 2 2 C C 1 2 C C 1 C = 2 C + C 1 The same pd V must be across total stored (and A charge each supplied in power supply) is Q = Q Q ows series, this onto charge be written pds are given by C V 1 + C 2 V and So V as C = V + V 1 = parallel same for as V in = both = Q + cases. C 1 + C 2 Q , = C where C is the C 2 + capacitor equivalent 1 1 = C series 1 + C 1 to + C 2 1 and 2 in they both. 2 Thus, 124 the – 2 1 = capacitor 2 C C CV C Q Q can is + Q 1 This each by The the C 1 are charge 1 + C 2 capacitor. The 1 = C series. 11 . 3 C A PA C I TA N C E 7 Example 11.3.4 6 A 20 μF and a arrangement The is are potential capacitor then fully are connected connected in parallel in series. with This capacitor series X. 5 The Cm / egrahc capacitors 60 μF charged. difference across X is 12 V and the total charge stored 4 is 4.0 × 10 C. 4 3 2 a) Calculate the total capacitance b) Calculate the capacitance of the three capacitors. 1 of X 0 0 c) Calculate the potential difference between the terminals of 20 40 60 80 100 120 140 the time / s 60 μF capacitor. Figure 11.3.4. Charge versus time for Solution capacitor discharge 4 q C a) 4.0 × 10 = C = = V 33 µF 12 The discharge equations are 1 b) For the series 1 1 = capacitors, Parallel capacitances add, . C is 15 μF. τ −6 20 × 10 therefore t + −6 C q 60 × 10 X must = q and 15) = = V a capacitance of τ e , 0 where q and V 0 (33.3 V 0 have t e are the initial 0 18 μF. charge and pd, q and V are the q c) The same charge is stored on both series capacitors and V charge and pd remaining at time t = C So, the ratio pd is must 1 : 3 so be the split pd in inverse across the ratio 60 μF to the capacitances. capacitor is and τ is the time constant which determines the discharge rate. The 3.0 V . τ = RC Also, since current When a capacitor loses its charge through a fixed resistor, the change I is the rate of is t exponential change of charge, I = I τ e 0 Figure 11.3.4 discharging shows the variation in charge Q with time t for a Q capacitor. 0 1 0.8Q 0 The decay λ constant in radioactive decay is analogous to . There 0.37Q at time RC 0 is, however, though no there is formal still a concept constant of half-life time for in the capacitance initial charge theory to even egrahc τ 0.6Q 0 2 (0.37) Q = 0.14Q 0 0.4Q at 2RC 0 0 halve. 0.2Q 0 Instead, remember that the charge reduces to 37% of its original value 0 in time τ (Figure 11.3.5). 0 RC Because C , = when C is constant, q ∝ V, so just re-scale the axes from a Figure 11.3.5. Q t to a V t t current 4RC during discharge is I Discharge graph in units of RC graph. The 3RC to V move 2RC time q = I τ e 0 Example 11.3.5 There are close similarities between the way the stored A 470 μF capacitor discharges through a resistor of resistance R. charge changes with time and The initial pd at time t = 0 is 12 V and at t = 1.0 s the pd across the the way the number of nuclei resistor is 8.0 V . decreases with time in radioactive Calculate decay. Use these similarities to aid R your understanding. Solution 1 RC pd across the capacitor at t = 1.0 s Rearranging and taking logs: 3 = 4.0 V so = 2e 1 8) − 1 and ln ( 3) = 1.10 The mathematics of radioactive = decay is covered in Topic 12.2; the RC RC physics of half-life is discussed in = = 4 4.7 (12 = 1 1.10 × 1 ln R is × 10 1940 ≡ 19 kΩ Topic 7.1. 125 11 E L E C T R OM A GN E T IC INDUCTION When curve takes time (AHL) a is an capacitor charges asymptotic infinite to time. the In through final q a fixed and practice, the V q resistance, values. and V So, in values the charging principle, reach 99% this by a 5RC S AMPLE STUDENT ANS WER An uncharged capacitor in a vacuum is connected to a cell of emf 12 V and negligible internal resistance. A resistor of A B resistance R is also connected. At t = 0 the switch is placed at position A. The graph shows the variation with time t of the voltage V across the capacitor. The capacitor has 12V capacitance 4.5 μF in a vacuum. R a) On the axes, draw a graph to show the variation with time of the voltage across the resistor. This answer could have achieved 0/2 [2] marks: 14 ▼ There about are the misunderstandings functions components. The of 12 the capacitor and 10 R are in them series; both the must total pd always across be 12 V . 8 is 2 V , 10 V the the and pd pd so across across on. the R The v / v When capacitor must two 6 be curves 4 will cross at the resistor the x-axis 6 V . will The be voltage for asymptotic to 2 as asymptotic the to capacitor pd is 0 12 V . 0 10 20 30 40 50 60 70 80 90 100 t / s b) i) The time constant of this circuit is 22 s. State what is meant by the ▼ The it has time the constant units of s is RC when time constant. and R is This ohms the and time C is in taken farads. for the It is a 37% discharging initial of its initial capacitor answer could have achieved 0/1 marks: also charge T he on [1] in to fall time constant is the time it takes to charge up half of the to capacitor . value. ii) Calculate the resistance R. This ▼ The answer ignores the answer could have [1] achieved 0/1 marks: factor 6 of 10 only from one the mark 4.5 μF for and the with question, T all T credit is = R.C = R T = time constant C = capacitance lost. c 22 s R = resistance = 4 .9Ω 4.5 µF ▼ The pd across the RC combination is determined by the c) A dielectric material is now inser ted between the plates of the fully power supply while the switch is at charged capacitor. State the eect, if any, on position A (we moved). Even movement, are not though the pd told that there stays is the it is charge i) the potential dierence across the capacitor. [1] same. This T he answer could potential have achieved difference will 0/1 marks: increase. ▲ This is correct although it is rather is too brief. A dielectric introduced increasing between the the material plates, capacitance. The pd ii) the charge on one of the capacitor plates. is This constant supply), have 126 (maintained so the increased by charge too. the answer could have achieved 1/1 marks: power stored must T he charge on one of the plates will increase. [1] 11 . 3 C A PA C I TA N C E Practice problems for Topic 11 Problem 1 Problem 5 The graph shows the variation of magnetic flux ϕ a) A parallel plate capacitor is made from overlapping metal plates with an air gap in between. through a coil with time t ϕ State t wo ways of increasing the capacitance of the capacitor. b) An RC circuit is constructed as shown. S 0 1 S t 0 S 2 a) Sketch the variation of the magnitude of the emf in R the coil with time over the same time period. 9.0 V b) Explain your answer to par t (a). Problem 2 C A square coil with sides 8.0 cm is made from copper wire of radius 1.5 mm. A magnetic field perpendicular to 1 the coil changes at a rate of 5.0 mT s . The resistivity of When the switch is moved from position S 8 copper is 1.7 × 10 to 1 Ω m. position S the capacitor begins to discharge. 2 Determine the current in the loop. (i) The capacitance of the capacitor C is 470 μF. It Problem 3 takes a time of 60 s for the pd across the capacitor to A vehicle with a radio antenna of length 0.85 m is fall to 5.0 V. 1 travelling horizontally with a speed of 95 km h . The horizontal magnetic field due to the Ear th where the Calculate the resistance of the resistor R. vehicle is located is 0.055 mT. (ii) Sketch a graph to show the variation with time of a) Deduce the maximum possible emf that can be induced in the antenna due to the Ear th’s magnetic the voltage across the capacitor, from the time when the switch goes to position S 2 field. (iii) State and explain how you would modify the b) Identify the relationship between the velocity of circuit so that it takes 90 s for the pd across the the vehicle and the magnetic field direction for this capacitor to reach 5.0 V rather than 60 s. maximum to be attained. Problem 6 Problem 4 A20 mF capacitor is connected to a 20 V supply. The Two identical circular coils, X and Y, are arranged parallel capacitor can be discharged through a small motor. The to each other and wound in the same sense. When energy of the capacitor can be used to lift a mass of direct current is switched on in X the current direction is 0.15 kg through a height of 0.80 m. clockwise in X. a) Calculate the initial energy stored by the Predict the direction of any induced current in Y when capacitor. a) the current in X is switched on. b) Determine the efficiency of the energy b) the current in X is switched off. conversion. 127 QUANTUM 12 1 2 . 1 T H E AND NUCLEAR P H YS I C S (AHL) I N T E R A C T I O N O F M AT T E R W I T H R A D I AT I O N You must know: ✔ what is meant by You should be able to: the term photon ✔ discuss the problems ✔ what that of ✔ ✔ it meant cannot by be the photoelectric explained by the effect wave theory ✔ explain cannot matter has a demonstrated the Bohr model quantization of wave for of nature and that this can ✔ pair hydrogen angular atom momentum and in the ✔ that the ✔ that there what is can have a wave function associated ✔ tunnelling a of probability a particle associated through a with the state ✔ and solve effect and photoelectricity classical by pair wave theory production an the evidence the for experiment wave nature order-of-magnitude solve potential meant including uncertainty is of by experimental waves, it features explained demonstrates matter with experiment photoelectric and annihilation discuss model ✔ which be explain electrons the the and light that be is photoelectric involving matter that of electrons estimates from the principle uncertainty principle for energy–time position–momentum. barrier. When a zinc incident with on sheet it, the radiation charged. The Einstein’s of is negatively sheet loses longer sheet is wavelengths losing explanation charged negative of charge or by and ultraviolet charge. This when the electron photoelectricity radiation does sheet not is is occur positively emission. included the following points. • Light • h • One • No is of the frequency Planck photon consists of photons, each with energy E = h f constant incident electrons minimum f are value on the emitted metal when can the corresponding to interact photon a with energy threshold one is electron. below a frequency f 0 • The minimum energy , known as the work function Φ (= h f ), is 0 associated that • Any with oppose excess the the energy removal energy (Φ – required of h f the ) is to overcome electron from transferred to the the the attractive forces metal. kinetic energy of the 0 emitted • An electron. increase incident electrons 128 the photons Therefore, kinetic in an incident but does increase emitted energy. per in light not intensity change intensity second—but the increases energy increases does not the of the the number photons. number change of of their maximum 12 . 1 These led to the Einstein photoelectric equation E = THE − Φ , hf INTERACTION OF M AT T E R WITH R A D I AT I O N which max anode 1 2 can also be written as m v e = − Φ , hf where v max is the maximum max 2 speed of the electrons. vacuum Figure 12.1.1 shows the experiment Millikan devised to test Einstein’s electrons V s result. + When monochromatic incident on the cell, light the with cathode large emits enough photon electrons. energy Increasing is the reverse pA pA pd between electrons pd V at cathode no longer this point and anode reach is the known (anode negative) anode—the as the gives current stopping falls potential a to and pd at zero. eV = which The hf cathode − Φ . s s potential of anode is made negative so 1 Figure 12.1.2 shows a graph of the variation of with V frequency electrons cannot quite reach it and the s λ The two different metals have the same gradient (which depends picoammeter reading becomes zero on h, Figure 12.1.1. c and e) but are offset depending on their work Photocell for function. measuring h Wave theory cannot instantaneous. The account wave for photoelectricity theory suggests that, as no the effect matter is how weak the metal 1 V metal 2 s light, given enough time, energy accumulates and releases an electron gradient for all wavelengths. The photoelectric effect predicts that, when the hc = photon energy what observed is is too in small, no electron will ever be released and this e is practice. –1 λ 1 λ 0 Example 12.1.1 hc eλ Electromagnetic radiation of 0 frequency 20 f is incident on a metal. The maximum Figure 12.1.2. energy E of k The graph 15 photoelectrons J kinetic max obtained in the Planck constant 91 the surface is measured. The graph 01 from experiment for two different metals 10 / the variation with f of E k Calculate, using the graph, max the 5 E a) xam k shows Planck 0 constant h. 10 20 30 –5 b) Determine required the the to metal minimum remove an 14 energy electron f / 10 Hz from surface. Solution a) The equation for the graph is E = hf − max Comparing the Planck this with y = mx + c shows that the 14 Reading gradient (m) is h constant. off the coordinates (30 × 10 , 14.5 19 × 10 ) and 14 (7.5 × 10 , 0), the gradient is 19 ( 16 × 10 − 0 19 ) 16 × 10 34 = 14 ( 30 × 10 = J s In Example 12.1.1 par t a), the 22.5 × 10 ) − 7.5 × 10 6.4 × 10 14 14 question begins ‘Calculate, using b) The minimum graph to the energy is the energy found by extrapolating the the graph, the …’. In an exam, you must follow such instructions. The y-axis. examiner will require the solution 19 This value is 5.0 × 10 J. to be based on a graphical analysis. (Alternatively , Φ = where hf 0 14 7.5 × 10 J.) f is the intercept on the x-axis: Few, if any, marks will be available 0 for alternative approaches. 129 QUANTUM AND NUCLE A R P H YSICS (AHL) 12 Light has some wave-like properties and some particle-like properties. The de Broglie wavelength λ is This leads to the idea of wave-particle duality. In 1924, French physicist h given by λ Louis , where p is the = p other momentum of the par ticle. de Broglie words, demonstrate He suggested under certain wave-like suggested that that this duality circumstances, properties particles have as a matter matter de applied also could to be matter. In observed to waves Broglie wavelength Example 12.1.2 This a) Calculate the wavelength 9.1 × of an 9.0 10 × crystal nickel. kg 10 of Explain and dark why to speed electrons can investigate The fringes The wavelength be a on of electrons, the an potential American diffracted physicists, a beam of Clinton electrons after diffraction, form a series of by a bright front of the tube. electron difference has been travelling accelerated at from non-relativistic rest speed, of be calculated by equating the kinetic energy to the electrical h 1 crystal 2 m v = eV and m e using that V, used energy: structures the They 1 m s can this by Germer. and through b) verified Lester electron 6 speed was and Broglie 31 mass hypothesis Davisson v = p = eV . 2m e λ So = . e 2m 2 electron eV e In diffraction. 1913, account Danish for the physicist Niels observations Bohr of proposed Geiger and a model Marsden of and the to atom to address the Solution problem that the concept of an electron orbiting a nucleus contradicted h a) Use λ = classical m physics. Bohr proposed that: v e • 34 electrons exist radiation (an in certain stationary states without emitting any 6.6 × 10 = 31 9.1 × 10 orbiting electron does this according to classical 6 × 9.0 × 10 physics) 11 = 8.1 × 10 m • b) The separation of atoms electrons emit or absorb radiation of frequency f only when moving is between stationary states (the energy E transferred is equal to hf) 10 of is the order roughly of the wavelength 10 m, same of the as so the • the angular in part of an electron in a stationary state is nh electron quantized found momentum with values , a). where n is an integer. 2π The third assumption wavelengths The basics of the Rutherford– Geiger– Marsden experiment are Austrian in as a is equivalent standing physicist understanding. Erwin He wave to onto Schrödinger described the fitting the an integral electron made the quantum number of orbit. next states breakthrough of particles using 2 a covered in Topic 7.3. wave function proportional to ψ that the gives an probability amplitude density for P(r). each position. ψ is Mathematically , 2 P( r ) The Bohr theory leads to the = ΔV, ψ distance r where from an P(r) origin is the and probability ΔV is the size of finding of the a particle volume a considered. nh suggestion that mvr and = One way to align the wave and particle approaches is to consider the 2π electron standing waves that exist in the negative potential well for that the energy, in eV, of a par ticular the electron–proton the probability system of the hydrogen atom. Figure 12.1.3 shows 13.6 orbital is given by E = − 2 functions for four states of the atom in a simple one- n dimensional version. V = V This prediction fits extremely 0 E n = 4 4 well with the energy level changes outlined in Topic 7.1. However, the E n = 3 3 theory was a poor fit for atoms more complex than hydrogen and helium. n = 2 E 2 E n = 1 1 V = 0 Figure 12.1.3. dimensional atom 130 The probability functions for four energy states of a simple one- 12 . 1 All the wave the probability positions of Probability in functions of the which can a be finding standing functions measurement physicist and Werner a the node introduce Heisenberg, of Heisenberg at the electron waves give the observation. measurement made. have it two atom there the is of a limit quantities on The the in as INTERACTION OF M AT T E R WITH R A D I AT I O N that antinode probability the (known that suggesting Heisenberg’s introduced places suggested zero. maximum concept First edge THE 1927 uncertainty by German precision conjugate uncertainty of density . with variables) position Δx h and uncertainty of momentum Δp are given by ΔxΔp ≥ . There is a 4π h corresponding relationship between energy E and time t: ΔEΔt ≥ 4π Where electric fields are strong (near an atomic nucleus, for example) The pair production must a photon with sufficient energy can spontaneously convert into conserve charge, lepton number, a particle–antiparticle pair. This is known as pair production and baryon number and strangeness. typically leads to the creation of an electron together with a positron These conservation laws were or a proton–antiproton pair. The incoming photon must have enough considered in Topic 7.3. energy to create both particles. 2 Energy is the requirements mass of one electron is must 1.02 MeV . The be total of meet from of Under some possible pair being the and emitted energy is must E = The 2mc rest energy likely annihilate, photons is created. minimum production to be forming equal of where the short two the , energy an photon lived. It photons. total m of will The mass–energy particles. circumstances, with photon particles electron both annihilating that Therefore, the another energy the the 0.511 MeV . positron soon of mean much less production than the of an 1.02 MeV of electron–positron pair energy—but for only is a h short time. The uncertainty principle ΔEΔt ≥ predicts that a low- 4π energy short The photon near wave function probability of probability may phenomenon The to an atomic nucleus can create two particles with lifetimes. alpha grouping of be a at particle any very (2 small protons nucleons suggests place (see, in but alpha-particle particle of of being the it is that the Universe non-zero. particle at one This has a time. finite This accounts for the decay . + for 2 neutrons) example, is the a particularly position of stable helium on the potential binding energy per nucleon chart in Topic 7.2). There is a tendency well of for the four particles to group within the nucleus. nucleus Figure 12.1.4 particle energy within barrier nucleons tunnel the The through function the for the is there not is nucleus. energy much happen next. enough an to average Alpha barrier, The escape a in the the force emitted reducing reduced—though of over attractive particles with energy Coulomb on the all probability decay amplitude finite—amplitude Coulomb barrier alpha inside of the outside. this wave escape. the relative can nucleus the a of particles constant by and existence alpha the what because within barrier wave shows alpha barrier function Quantities decay of heights an and beyond such as the the individual alpha barrier means radioactive nuclide energies. are that decay determined Figure 12.1.4. Quantum tunnelling of alpha par ticles 131 QUANTUM AND NUCLE A R P H YSICS (AHL) 12 Example 12.1.3 An electron is confined within a region of length Ψ 10 2.0 × The 10 m. variation electron at a a) Outline b) Estimate with distance particular what is x time meant of is by the wave Ψ function of the given. wave function –10 the momentum of the electron. x / × 10 0 1 c) Deduce the uncertainty in the momentum of the m 2 electron. Solution a) The wave function proportional particle b) There at are a to is the the specified three property square root of of a the particle. It probability is of finding the position. wavelengths of the electron wave function in 10 2.0 × 10 m. 10 2 0 × 10 11 Therefore = wavelength = 6.7 × 10 m 3 34 h 6.6 × 10 23 So p = = = 1.0 × 10 N s 11 λ 6.7 × 10 10 c) The uncertainty in the position of the electron is 2.0 × 10 m. 34 6.6 × 10 25 Heisenberg uncertainty = gives 2.6 × 10 N s 10 4π × 2.0 × 10 S AMPLE STUDENT ANS WER An apparatus is used to investigate the photoelectric effect. A caesium λ = 400 nm cathode C is illuminated by a variable light source. A variable power supply is connected between C and the collecting anode A. C A e I A A current is observed on the ammeter when violet light illuminates C. With V held constant the current becomes zero when the violet light is replaced by red light of the same intensity. Explain this observation. This answer could According up V of lower question comparison. to be clearer between and to. It can violet red and these be is and awarded—but of a the candidate. the work violet is to needs that There not is frequency, to remove than the no There A → is enough A → no the the energy and the fact 132 the work electrons. of the light where provided light. plate, T herefore, current → 0A. Violet a that function. to liberate the When the by light the red is is light light, energy electrons light electrons a current. to term no is is a statement that link red this that the light arrives each photon as light is and that less an than on for photons photon shines of energy frequency job link is of the T here’s indicates between the was work function—the mentioned. f), model made is which not has enough don’ t ow from C to has higher frequency, which lead ▲ There not the h the mark do (E = marks: light greater it examiner quantum 2/3 differences differences this on achieved about the inferred energy energy all answer about red what The is the photons dependent ▼ This to have [3] energy E = hf has off the plate C to travel to plate 12 . 2 This answer When is a violet used, ϕ where than the is lower violet could electrons is while have than shone the on is shone wavelength energy light, when have the of violet greater light is sufcient repulsion cathode. by electrons have red than given Since the cathode forces on lower caesium. red electrons electrostatic light of Electrons light kinetic kinetic the light, hf-ϕ, KE = has emitted shone from red on when energy the cathode. energy anode that are to when ▼ This emitted terms when red light is shone on the cathode do not have answer of the explain energy to do so. T his is why electrons cannot anode when subsequently by red violet why light is replaced current is zero with when red violet light, light N U C L E A R in energy attempt is of the made to photoelectric reach and is it is in but terms again of this is wrong as electrons. replaced P H Y S I C S You must know: that entirely light. 1 2 . 2 ✔ An Einstein’s equation, the is kinetic sufcient electrons. kinetic P H YS I CS marks: kinetic the emitted 0/3 has function electron overcome achieved which wavelength the violet light, work light Hence, have NUCLE AR Rutherford You should be able to: scattering leads to an estimate ✔ estimate the radius of a nucleus using the rule 1 of nuclear radius 3 that ✔ that a nucleus has nuclear energy the properties existence from was beta of the neutrino deduced using the law ✔ that the unit and how a minimum its Broglie observations scattering intensity experiment location that based on uses the the de wavelength decay of radioactive probability time A describe ✔ ✔ ∝ levels ✔ ✔ R is the of decay explain high decay decay of a nucleus ✔ per solve time constant. ✔ deviations from Rutherford scattering at energies problems intervals explain involving that methods are for not radioactive integer measuring decay for half-lives short and long half-lives. Head-on scattering means that the alpha particle returns along the Topic 7.3 discussed Rutherford same path by which it arrived. The point at which the alpha particle scattering and showed that the changes direction is where all its initial kinetic energy is transferred to following deductions can be made electric potential energy stored in the nucleus–alpha system. from the observations of the 1 k (2 e )Ze Geiger–Marsden experiment. 2 m v α , = α where r is the distance of clos est ap proa ch, 2 2 r • Most of the atom is empty space. is m the mass of the a lp h a pa rticle, v α is the initial speed of th e α • There are small dense regions of alpha particle, Z is the proton number of the n ucleu s and k is the positive charge in the atom. Coulomb This constan t. equation assumes that the nucleus does not recoil. The larger v , the α closer the alpha particle will be to the nuclear centre before deflection. 133 QUANTUM AND NUCLE A R P H YSICS (AHL) 12 The Coulomb constant is k in the kq F Example 12.2.1 q 1 formula 2 ; see Topic 5.1. = 2 a) r Determine the 7 2.00 b) × kinetic energy of an alpha particle travelling at 1 10 m s Calculate . the Ignore closest relativistic distance of effects. approach for a head-on collision 197 between The radius R of a nucleus with the alpha particle and a gold Au. nucleus, Assume 79 1 3 nucleon number A: R = R , A that the gold nucleus does not recoil. 0 where R is a constant with value 0 Solution 15 1.20 fm (1.20 × 10 m). 27 a) The mass of the alpha particle = 4.0026 × 1.66 × 10 27 = 6.64 × 10 kg 2 27 Example 12.2.2 6.64 × 1 7 10 × ( 2.00 × ) 10 2 Its kinetic energy = 12 mv = = 1.33 2 × 10 J 2 12 The radius of the C nucleus b) 6 The gain in electric potential at closest energy when the nucleus and alpha 15 is 3.0 × 10 m. Calculate the particle are their separation r must equal the loss in 20 radius of the Ne kinetic nucleus. energy . The alpha particle charge is +2e and the gold 10 nucleus charge is +Ze Solution 1 kq 3 R = R 12 ⇒ A Gain 0 in potential energy = 1.33 × q 1 10 k × (2 e ) × Ze 2 = = r r 1 A 3 Ne R = R Ne 2 19 × C k A Substituting C ( 1.6 × 2 × 79 × 15 3.0 × 10 20 so 3.6 × 10 r = 2.7 × 10 m r 3 12 Nuclear 15 = 14 gives 1 = ) × 10 radius measurements show that nuclear density is roughly m constant whatever the nucleon number A 17 is very only electron intensity high, by that of three-billion The approximately a neutron 2 × star — a the nuclide. This density 3 kg m 10 of . This matchbox-full value of is approached which has a mass of tonnes. expression for the radius of a nucleus is derived from the fact that AM nuclear ρ density is constant. Since ρ , = where M is the mass of a 3 4 πR 3 1 3M 3M 3 θ nucleon, min 3 R = A and R = A 3 4 πρ 4 πρ angle of diraction Figure 12.2.1. Intensity– Beams of protons diffraction. diffraction angle graph for intensity electron scattering or Figure with pronounced electrons 12.2.1 scattered minimum can shows a be (diffracted) at used typical to probe variation angle for a the of nucleus scattered metal. There is using electron a θ min λ This angle is related to the diameter D of the atom by sin θ , = where D In wave diffraction in Topic 9.4, the λ is the de Broglie wavelength of the scattered particle. diffraction equation was given as Deviations from Rutherford scattering are seen when high-energy λ θ (θ = in radian). The reason electrons (> 420 MeV or so) are scattered in experiments. These b deviations for the inclusion of sin θ occur because: is that the minima for scattering generally • the collisions occur at large angles >> 10°. The energy approximation θ emitted ≈ sin θ only that become is from inelastic; transferred the into the incident mass as electrons mesons are lose created kinetic and nucleus applies for small angles and the • deep inelastic scattering occurs as the electrons penetrate further into full sine relationship must be used the nucleus; they then scatter off quarks inside nucleons (this for scattering. scattering 134 has provided evidence for the Standard Model). type of 12 . 2 NUCLE AR P H YS I CS Example 12.2.3 a) Explain, with 400 MeV effects b) are in Electron nuclear your in diffraction the nuclear detail, why determining electrons nuclear size. of energy Ignore relativistic estimate. density Outline i) quantitative used experiments and main the average conclusions lead to information separation drawn of about nucleons. for: density ii) average separation of particles. Solution 34 hc λ a) 6.6 = 8 × 10 × 3.0 × 10 = ≈ 6 400 × 10 E The de Broglie 3 fm 19 × 1.6 × 10 wavelength for the electrons of this energy is 15 about 10 nuclear m which diameter. diffracting is of This the same means wavelength order that the comparable of magnitude 400 MeV to the size as electrons of their a have a target. 1 3 b) i) Nuclear density is constant; the relationship R ∝ A can be 3 written as R ∝ A volume This leads to = constant mass ii) As a consequence separation Nuclei, like atoms that are The gamma Some observed of the nucleus the with in of medical diagnosis. The decay full technetium 99 99 m Mo → Alpha emitted nuclei the and unstable states, a formed for moving of the nuclear number a the technetium, molybdenum in of energy time discrete (Mo) that The emitting state. nuclide decay . energies. state. before ground a levels radioactive metastable short to average constant. emitted in a be have are small state decay density , also nuclei at are excited energy must energy 0 + 99 m e 43 metastable constant This used in involves is * Tc 42 is in from (T c) the daughter this its involved of nucleons their when daughter remains is the radiation remainder release of and +υ 1 * 99 Tc e → 43 Tc + γ (99m and * mean a 43 state). decay and beta (β ) decay are very different in energy terms. energy spectrum of beta The 2 10 decay electrons from In particles alpha and the zero, emission, alpha the final daughter v emitted show only one one active nucleus particle. When the momentum nucleus move in must or decays initial also opposite two be energy into the momentum zero. The directions values. daughter before alpha with decay particle speeds nucleus in is and the Bi ytisnetni alpha the ratio m α d = 0 v α d In 0.2 0.4 0.6 0.8 1.0 1.2 m kinetic energy / MeV beta-minus decay , emitted electrons are observed to have a complete Figure 12.2.2. range of energies from zero to a maximum that is slightly less than The energy the spectrum from the decay of maximum energy believed to be available (Figure 12.2.2). Bismuth-210 135 QUANTUM AND NUCLE A R P H YSICS (AHL) 12 Italian are an physicist produced unknown in many In 1933, Enrico in the decay: particle. ways, Fermi The leading to interpreted the daughter three a energy to mean nucleus, particles beta this can the share that beta the three objects particle available and energy spectrum. Topic 7.1 describes radioactive decay as a random and spontaneous process in which an one”. individual nucleus decays into a been Fermi named Experiments indirectly this particle confirmed this the neutrino, prediction meaning and the “little neutrino neutral has since observed. daughter nucleus with the emission Neutrinos are difficult to observe. Large numbers emitted in fusion of par ticles. reactions can only special The in be the pass observed through when a few the Earth interact every second. indirectly with The particles nuclei under circumstances. properties zero Sun of the neutrino include neutral charge and effectively mass. Example 12.2.4 a) Calculate, from b) All in Figure the joules, the maximum beta-minus particle energy 12.2.2. beta particles emitted by Bismuth-210 arise from identical anti-electron neutrino The rate of change of nuclei with energy dN changes in the bismuth nucleus. time is the activity A of the dt Explain must how also be Figure 12.2.2 suggests that an emitted. sample. Solution This equation can be written as a) The maximum energy is 1.2 MeV . dN dN A = = −λ N = or − λd t , 6 This N dt corresponds to 1.2 × 10 19 × 1.6 × 10 13 J ≡ 1.9 × 10 J. where A is the activity (decays per b) The second) and this second equation total energy existence of a available beta energy from the decay spectrum is must constant. mean that The there is no λt leads to the solution N = N e unique , way to distribute the energy between the beta particle 0 where t is the time and N and is the the daughter product. 0 initial number of undecayed nuclei There are an infinite number of ways to distribute the energy at t = 0. three The activity A of the sample is ways. The energy and the This is is an shared indication between that the three particles daughter are nucleus, involved. the electron λt A = λ N e antineutrino. 0 Each identical decay per provides nucleus unit a time. of a This fundamental particular probability relationship nuclide is of has known as the same the decay radioactive probability constant λ; it decay: The negative sign in dN = −λ N rate of loss of nuclei The relationship = −λ × number of nuclei remaining arises because dt between λ and half-life t 1 the number of undecayed nuclei is an important one. After 2 N decreases with time. 0 one half-life, the number of atoms has halved. So = exp( − λ t 1 ). N 0 2 2 Eliminating N and taking logarithms gives ln 0.5 = −λt 1 0 2 ln 2 This can be written as t 1 2 136 = λ of 12 . 2 NUCLE AR P H YS I CS 1000 daughter growth N 100 mother decay 0 0 5000 10000 15000 20000 25000 30000 time Figure 12.2.3. Figure are 12.2.3 further The Then the the total an and Reasonably A measuring • taking the with nuclei nuclide is t both that for are stable the nuclei forming. and does The not that graph undergo a A pure to of using this nuclides up sample the the the to with and number sample counts the long of half-lives then be over N in the measured collected counts measuring atoms can involves a by a (this detector complete the mass sample. involves of finite sphere sample). λ N = short • of the equation in activity factoring for leads estimate surrounding λ of This varies decay . nuclide sample. N daughter daughter radioactive making The the how the determination the size and that obtaining of shows decaying assumes Decay and growth curves for stable daughter the is used half-lives of count calculate can background readings to be measured count rate the rate against in decay by: the time constant. laboratory until the value equals that background • subtracting • assuming that • plotting graph • finding a the the background the corrected of ln A gradient of from count against this each reading ∝ rate activity time graph, which –λ gives Example 12.2.5 Radioactive iodine (I-131) has a half-life of 8.04 days. 1 a) b) Calculate, Calculate sample c) Deduce in the with the decrease seconds to number an , the of activity time taken decay atoms of for constant of I-131 of I-131. required to produce a 60 kBq. the activity of the sample in part b) to 15 kBq. Solution ln 2 −6 a) λ = = 8.04 × −1 1.0 × 10 s 24 × 60 × 60 4 A 6.0 × 10 10 b) The number of atoms is given by N = = = 6.0 × 10 6 λ 1.0 × 10 1 c) A decrease from 60 to 15 kBq is ; in other words, two half-lives, 4 which is 16 days. 137 QUANTUM AND NUCLE A R P H YSICS (AHL) 12 S AMPLE STUDENT ANS WER 106 Rhodium-106 Rh ( 45 ) decays 106 3.54 Rh 45 106 by beta minus (β Pd ( 46 ) VeM / E into palladium-106 ) decay. β The diagram shows some of the nuclear energy levels of 0.48 106 Pd 46 0 rhodium-106 and palladium-106. decay. The arrow represents the β a) Explain what may be deduced about the energy of the electron in the β ▲ The answer subtraction maximum particle that is this a also – for 0.48 the a leads energy (3.54 There shows = decay. to This the the answer T he 3.06 MeV). recognition beta clear energy energy for the fails to achieved 2/3 marks: make minimum of the β particle is 3.54-0.48 MeV = 3.06 MeV . T he is smaller than 3.54 MeV because it shares its energy the with answer that have that particles. ▼ The it maximum could beta energy is [3] correct an antineutrino. b) Suggest why the β T hus energy < 3.06 MeV . decay is followed by the emission of a gamma ray beta photon. energyis This ▲ The answer recognizes in an that excited excess [1] 0. could T he palladium the gamma have achieved nucleus is 1/1 not marks: in its ground state and it releases correctly the state energy answer via nucleus and the is loses left ray to go to its ground state. the emission of a c) Calculate the wavelength of the gamma ray photon in par t b). gamma photon. This answer could have achieved 1/1 [1] marks: 6 ▲ The calculation is clear and 1.24 × 10 hc eVm -12 accurate. λ = = E 138 = 0.48 MeV 2.583 × 10 -17 m ≈ 2.6 × 10 m 12 . 2 NUCLE AR P H YS I CS Practice problems for Topic 12 Problem 1 Problem 4 Electromagnetic radiation incident on a metal causes a Electrons are emitted instantaneously from a metal photoelectron to be emitted from the surface. surface when monochromatic light of wavelength 420 nm is incident on the surface. a) State and explain one aspect of the photoelectric effect that suggests the existence of photons. a) Explain why the energy of the emitted electrons does not depend on the intensity of the incident light. b) The work function of sodium is 2.3 eV. b) Suggest why the electron emission is (i) Outline what is meant by work function. instantaneous. (ii) Electromagnetic radiation of wavelength 320 nm c) The work function of the metal is 2.3 eV and one is incident on sodium. electron is emitted for every 4500 photons incident Determine the maximum kinetic energy of the electrons emitted from the sodium. Problem 2 on the surface. The surface area of the metal is 6 4.5 × 10 2 m (i) Determine, in J, the maximum kinetic energy of an Electrons are emitted from a heated cathode and emitted electron. accelerated in a vacuum through a potential difference (ii) Determine the initial electric current from the as a narrow beam. This beam is fired at a polycrystalline surface when the intensity of the incident light is graphite target in a chamber. The inside surface of the 6 3.8 × 10 2 W m chamber is coated with fluorescent material that emits light when the electrons release their energy to it. Problem 5 137 Caesium-137 ( 55 Cs ) decays by negative beta decay to a) The electrons reach the inside surface travelling at a 7 speed of 4.0 × 10 form a nuclide of barium (Ba). 1 m s a) Write down the nuclear reaction for this decay. Calculate the de Broglie wavelength of the b) The half-life of caesium-137 is 30 years. Determine electrons. the fraction of the original caesium that remains after (i) Sketch the pattern of light you would expect to 200 years. see emitted by the fluorescent material. c) Caesium is a waste product in a nuclear reactor. (ii) Explain why the pattern suggests that electrons Suggest why the fuel rods in the reactor are removed have wave-like proper ties. well before the uranium is completely conver ted. b) Explain one aspect of the experiment that suggests that electrons have par ticle-like proper ties. Problem 6 12 The radius of a carbon-12 ( 15 C 6 ) nucleus is 3.1 × 10 m. Problem 3 22 Radioactive sodium ( 11 Na ) has a half-life of 2.6 years. a) Determine the radius of a magnesium-24 24 A sample of this nuclide has an initial activity of ( Mg 12 ) nucleus. 5 5.5 × 10 Bq. b) Sketch a graph to show the variation of nuclear radius a) Explain what is meant by the random nature of with nucleon number. radioactive decay. Annotate your graph with both the C-12 and Mg-24 b) Sketch a graph of the activity of the sodium sample nuclei. for a time period of 6 years. c) Calculate: 22 1 (i) the decay constant, in s , of Na 11 22 (ii) the Na in the sample number of atoms of 11 initially (iii) the time taken, in s, for the activity of the sample to fall from 100 kBq to 75 kBq. 139 D ATA - B A S E D 13 P R A CT I CA L ( S ECT I O N You must know: ✔ what in is meant by AND Q U E ST I O N S A) You should be able to: random and systematic errors ✔ plot points ✔ construct ✔ sketch ✔ determine accurately on a graph measurements. best-t gradients ✔ Paper to 3, Section answer these demonstrate This chapter practical A contains the provides work for and DP on graph graphs interpret errors closely in the that on meaning gradients to the and those you will of graphs questions. linked assessment physics a intercepts the guidelines the on data-analysis are internal bars lines and determine questions in error and The intercepts. skills practical described find in helpful you skills need you Topic in all 1. your course. There is a table in the Internal Communication using a graph is essential in physics; graphs allow Assessment chapter that lists many relationships between data points to be grasped visually and quickly . of the relationships you meet in It helps if you know the best graph to plot to display your data. First, the course and how you should consider how best to render your data as a straight line (in the form manipulate the data to obtain a y = mx + c, where y and x are the data, m is the gradient of the line and c straight-line trend. is the intercept on the y-axis). This may involve algebraic manipulation. Example 13.1 A simple D above ceiling. as a pendulum the The floor. period function pendulum Suggest a of is suspended There T the of is no from access oscillation vertical to of distance the the the h ceiling of pendulum from height suspension the is floor at the measured to the bob. suitable graph to display the data. Solution The length of The equation the pendulum is D h l for the period is T = 2π D = g A plot of T However, against h squaring 2 will both not be sides a straight and h 2π g line. rearranging gives straight of 2 4π 4π 2 T = − h + g This is D g now in the form y = mx + c 4π 2 So a plot of T against h will give a line gradient − g 4π with an intercept on the y-axis of D g 140 Once can be Here • you have collected data and identified a suitable graph, the graph plotted. are some Plotted guidelines graphs need for drawing labels and graphs. units on both axes; for example, 1 speed / m s . The / does not mean divide, it means measured in. You 1 should • Scales 1 : 10. 1 : 4 • write on the The fill • Data • All points can thick as for origin on a of rather than straightforward 1 : 7 the your or 1 : 9 be printed plots that of should graph thickest a lines be as the not m/s. using never 1 : 2, be the grid 1 : 5 used. in or Only use or the This have may a mean +. black when grid, to (0 , 0)). × sharp, Aim area. at using guideline, on possible. begin clearly should As grid half does marked mistakes). the be m s necessary . (one should erase form 1 : 6, much range false markings you as a as the should 1 : 3, absolutely minimum using in axes ratios when Always units pencil your pencil (so pencil is not that line is sharp A common question is ‘Draw the enough. best-fit line’. This does not mean draw the best-fit straight line, as • Draw • Use all straight lines with a ruler (preferably transparent). the best fit line could be curved freehand for curves. Practise the curve several times first, with the data points distributed drawing with your hand inside the curve, and then draw the line in evenly about it. one • continuous Try • to get the line, the line. Do not very the movement. same number minimizing force good your reason the line to of total to go distort points on distance through your of the line both in all sides points origin this of from unless you have a way . Draw the lines from which you Modern slope can and do (Figure is 13.1). these Once then line gradient have intercept this, straight plot calculators is your use to easily that, line this the points. your is and the many being into and from a data Another x them plotted, besides the data separate Join programs given facility . obtained graph Remember a divide Find two of internal quick two more of is If way groups: averages that way can points. two y that of your mean data determine your to upper values are and line. as possible so they stretch from best axis to axis. Then use at least half lower of this line for the determination. and The longer the line, the smaller The the fractional uncer tainty in your too. gradient result. available visualizing the groups straight will calculate a gradient as long calculator find both the your to you. data, y upper half a • graph the is an averaging gradient of y-coordinates a ÷ technique straight change line, in from by which you calculating can obtain: change in x-coordinates (x • the at gradient the new point and a point concerned straight mirror at line. align To it on a and draw so that curve, by drawing calculating this the tangent, when a into small it the 2 tangent gradient use looking the of this average curve , y 1 its reflection appear continuous. The mirror is ) 2 upper plane (x and , y then ) 1 at lower average 90° to the curve, the tangent line draw along directly the from mirror—a protractor gives this. lower half y – y – x 2 • The area under a graph can be found either algebraically 1 gradient = x 2 or can be made into a series of triangles and 1 rectangles x when found of the by each line is straight. counting square. When squares and it is a curve, determining the the area area can be value Figure 13.1. Dividing plots into two groups to find an approximate best-fit line 141 D ATA - B A S E D 13 QUE STIONS AND PRACTIC AL (SECTION A) • The it intercept off is a or by false on either using origin. axis can trigonometry This will be to almost found either calculate certainly the by directly intercept involve reading when there extrapolation of Always look closely at the origin your plotted line to the intercept. of graphs printed in examination booklets. When there is a false Every data origin, take this into account. When the When the origin is shown as grid (0 , 0), this may be impor tant in a is multiple-choice or other question. consider you point on absolute need roughly the the do of of the error graph in nothing size use your error the will have point further. smallest is an error much However, square or less associated than when larger, the the with size of absolute then you it. the error need to bars When no marking is shown where When the x every point on a graph has an error bar, it is possible to construct and y-axes cross, this is a the maximum and minimum gradients. The construction is shown in signal that you should only pay Figure 13.2. The two lines must lie on the outer edges of the relevant attention to the shape of the curve. error bars. calculated maximum Once in they the have usual gradient been way . The minimum drawn, the absolute two error gradients in the can gradient be is gradient An error bar is an I-shape or 2 H-shape centred on the datum point. The width or height of the bar indicates the error in the This is range similar of data to for the a calculation particular of an datum absolute error from the overall point. measurement. The nature of the The absolute error in the gradient can then be used in further error experiment will determine whether each point on the graph has the calculations. same or dierent absolute errors. Similar ideas will apply to the intercepts on the axes. In Figure 13.2, the 1 intercept on the y-axis is (108 ± 5) m s Topic 1.2 describes the treatment of errors in single data maximum 150 best t line points whether absolute, fractional line or percentage. 140 minimum s m / deeps 1– In Paper 3, Section A, error bars may be used and you will normally line 130 120 be told when the errors in one measurement are negligible. 110 100 0 1 2 3 4 5 time / s Figure 13.2. Minimum and maximum gradients on a graph Spreadsheet software, such as Excel, has graph-plotting tools that can quickly and accurately draw points and error bars, and carry out graphical calculations. It is wor thwhile teaching yourself how to use this software early If you are studying at Higher Level you will need to be able to in your DP physics course; it will save time and effor t throughout your studies and beyond. plot and analyse log–log and Remember that no graphing software is available to you during written log–linear graphs (eg in examinations, so make sure that you can draw a graph neatly by hand and you radioactive decay the variation of have the skills to calculate gradients and intercepts quickly and accurately. ln(activity / Bq) with time). Notice how the unit for activity is inside the bracket associating it with the quantity, not with its logarithm 142 Where there is a false origin, never extend a grid by drawing to obtain an intercept. Calculate the intercept using similar triangles or by substitution into y = mx + c S AMPLE STUDENT ANS WER A student suggests that the relationship between I and x is given by: 1 = Kx + KC I where K and C are constants. 1 Data for I and x are used to plot a graph of the variation of with x I 45 40 35 30 m 2 1 25 W / I√ 1 20 15 10 05 –5 0 5 10 15 20 25 x /cm a) Estimate C. This answer [2] could have achieved 0/2 marks: ▼ The y-intercept However, y intercept C = is (0, the is correct. candidate was 5) asked 1 for the value of C. When = 0, I 5m x = C the and this is the intercept on x-axis. b) Determine P to the correct number of signicant gures including its 4π unit, where P . = [4] 2 K This answer could have achieved 4/4 marks: 30 − From (b)(i), the gradient = 20 − considering (0, 5), (20, 1 5 K = 2 1.3W by ▲ There 0 points 30) and π From (a), K 2 K = π are about well makes a number this presented. the of answer: source of The the good it is clear candidate data for = so the P 2 gradient clear and the data P points concerned on line. are far apart and 2 K π 4π = ⇒ P The calculations are = 2 P 4 the correct and the unit and number of K signicant gures are correct too. 4π 2 T herefore, P = = 8.0W 2 m cm 2 K 4 = 8.0 × 10 = 8.0 × 10 2 W m 2 m 4 W 1 c) Explain the disadvantages that a graph of I versues has for the 2 x analysis in par t (a) and par t (b). [2] ▼ There This answer could have achieved 1/2 is some credit given marks: for the point However, about this is the not gradient. the whole 1 Since slope will be = , it will be too small to measure story: the equation becomes 2 K 1 I from graph to right no. of signicant = so, gures. K Y intercept will be too small → hard to measure c is (x known, plotted unless C 2 2 to + C) the graph give a cannot straight be line. 143 D ATA - B A S E D 13 QUE STIONS AND PRACTIC AL (SECTION A) S AMPLE STUDENT ANS WER A student measures the refractive index of a glass microscope slide. He uses a travelling microscope to determine the position of x of a mark 1 on a sheet of paper. He then places the slide over the mark and nds of the image on the mark when viewed through the slide. the position x 2 Finally he uses the microscope to determine the position of x of the top 3 of the slide. image of mark slide X 3 X 2 sheet of paper X 1 mark The table shows the average results of a large number of repeated measurements. Average position of mark / mm x 0.20 ± 0.02 0.59 ± 0.02 1.35 ± 0.02 1 x 2 x 3 a) The refractive index of the glass from which the slide is made is given by: x − x 3 x 1 − x 3 2 Determine: i) the refractive index of the glass to the correct number of signicant gures, ignoring any uncer tainty. This ▲ After an initial error, answer 1.35 the − could 0.2 have of the working is marks: = 1.51 carried 1.35 through 0/1 1.15 = remainder achieved [1] − 0.59 0.76 correctly . ii) Determine the uncer tainty of the value calculated in par t i. This ▼ There nd the are subtractions differences in the answer need to but be the this is obtained absolute be in a correctly . uncertainties added. x − 3 0.02 the + subtraction, x The and 1 0.02 x − 3 not solution 0.02 x 2/3 marks: Δx Δx = n 0.02 + x − x = x 1 − 3 x 0.02 + 1.15 0.76 2 the need absolute achieved image uncertainties 3 As have to Δh position could [3] Δn = 1.5 (0.0437) Δn ≈ ±0.0655 where n is the refractive index to errors = ±0.1 = Δn are 2 alone indicates. as b) After the experiment, the student nds that the travelling microscope is badly adjusted so that the n measurement if each position is too large by 0.05 mm. 144 Outline the eect that this error will have on the calculated value of the ▲ The refractive index of the glass. [2] index point is is that calculated the by refractive differences. x x 1 3 This answer could have achieved 2/2 marks: Each measurement in is x x 3 changed It will have no effect because this error will be cancelled x − x − x 3 n is 3 overall same amount and ratio does not change. have been expressed 1 = x the out the once by 2 This calculated. could more clearly in the answer. 2 Practice problems for Section A 20 Problem 1 Data are obtained for the variation in the extension of a 18 thin filament when a force is applied to the filament. A 16 graph of the results is shown. 14 9.0 lament 8.0 12 breaks 7 .0 10 5.0 01 / F 2– S / T N 6.0 8 4.0 3.0 6 2.0 1.0 4 0.0 2 0.0 1.0 2.0 3.0 4.0 5.0 6.0 –2 x / 10 m 0 0 05 10 15 20 25 30 35 40 45 a) Draw a best-fit line for this graph. m / kg b) The stress in the filament is defined as a) Draw the best-fit line for these data. force acting on the thread b) Determine the percentage uncer tainty in cross-sectional area of the thread T when m = 10 kg. The radius of the filament is (4.5 ± 0.1) µm. c) The student suggests that T ∝ m and plots a graph (i) Determine the absolute uncer tainty in the crossof T against m sectional area of the filament. 25 (ii) The filament breaks when the force acting on it is 2 20 85 mN. The stress at which steel breaks is 1.0 GN m 15 stress than steel. c) Determine the energy required to stretch the filament T Deduce whether the filament can withstand more 10 5 from an extension of 20 mm to 50 mm. 0 2 3 4 5 6 7 Problem 2 — √ m A student collects data for a system in which a mass oscillates about a given position. The graph of variation (i) Calculate the gradient of the graph. of time period T with mass m is shown. (ii) Determine the intercept when m = 0. The uncer tainty in T is shown and the uncer tainty in m d) Suggest the graph that the student should construct is negligible. n to confirm that n = 2 in the equation T ∝ m 145 R E L AT I V I T Y A A . 1 B E G I N N I N G S O F R E L AT I V I T Y You must know: ✔ the denition ✔ what is of meant a by You should be able to: reference Galilean frame relativity ✔ use the Galilean ✔ determine charged ✔ Newton’s ✔ how postulates regarding space and description of forces acting on charge or a moving charge ✔ the reference Maxwell’s the speed force on or a moving magnetic for a reference frame determine the nature of an electromagnetic eld depend observed from different reference frames. frame postulate of the electric a as on is equations time ✔ stationary whether particle specied the transformation light in regarding a the constancy of vacuum. A reference frame defines the motion of an object relative to others. An iner tial reference frame is Reference frames consist of an origin together with a set of axes. one in which Newton’s rst law of motion holds. No external force An inertial reference frame is an extension of the reference frame idea. must act on an object in the frame. Albert Einstein was not the first scientist to discuss reference frames; By implication, the frame is not this accelerating. was moving probably ship observed motion of Galileo’s y to and fly the Galileo suggested at that random. An ship ideas Galilei. by described butterflies observer watching suggest He the in in a the the windowless cabin cabin cabin would could not in a always deduce be the butterflies. that: y′ • direction is • position • stationary is relative relative X x′ is not an absolute condition. x′ x These lead to a set of equations (quoted for the x-direction only). x When y of y′ the origin another of a reference reference frame A frame (Figure A ’ is a distance A.1.1), and the X from position the of origin an v t = 0 vt object is frame A ’ is x relative When the to the origin in frame A, the position of the object in x′ x ’ = x − X x origin of an inertial frame A ’ is moving relative to the that the x′ x original of inertial frame A with speed v (and assuming x = vt + x′ Figure A .1.1. Two frames of reference connected by the Galilean transformation origins coincide related to the Combining coincidence at time t speed u ’ of distances of the = 0), the and the speeds origins. speed same These object leads u of in to x' equations an object frame = x are in frame A ’ by u ’ = vt at known time as u t − A is v. after the the Galilean transformations Newton took postulates 146 Galileo’s ideas and expressed them as the two Newtonian A .1 Example A .1.1 identical OF R E L AT I V I T Y Newtonian postulates ① Two BE GIN NING S spaceships, A and B, move along the same Space and time are absolute: a straight time interval in frame A is the same line. as a time interval in frame A’. (This A is moving away from Calculate away A at the from speed the Moon 0.30c velocity of B at relative relative speed to to 0.40c and B is moving disagrees with one of Einstein’s postulates.) A. the Moon, using the Galilean ② Two observers in separate iner tial reference frames must transformation. make the same observations of the Solution u′ = u + v physical world and therefore agree = 0.30c + 0.40c = 0.70c on the physical laws. (This agrees with one of Einstein’s postulates.) All observers in inertial frames make the same deductions about (a) physical 5.1) law. and James Clerk Maxwell electromagnetism (Topics connected 11.1 and electrostatics 11.2) by (Topic incorporating X the v speed of inertial is u’ = u in frames observe this light a are identical not − v vacuum. what to agree values the suggests It about for the Galilean that, if was clear that, physical speed of observer observers laws, light transformation one if then in a they at different must vacuum. predicts. moves in a The all However, + q equation constant velocity + q (u – u’) relative Transfer to between another, inertial then their frames observations leads to of c unexpected will differ changes in by v the (b) stationary, Y descriptions of moving obser ves a charge. magnetic eld Two identical speed (a) is (b) v. Figure the is positive view the A.1.2 from view charges gives the from two inertial the +q are moving views frame inertial of the parallel same associated frame of an at the same due to event: with an observer Y . both charges moving observer X + q X moves + q with the charges electrostatic describes current in the and origin; effects associated Parallel describes currents they the effect repelled. differently , with in are the between Y , them moving observing that at the a as purely different charges have speed, a Figure A .1.2. moving in parallel them. same direction have a magnetic attraction (a) between (a repulsion as the v. the by consider speed However, of predicted Now in them. two both a the two charges the charge Two point charges observations with the Newtonian +q In the wire, the wire are stationary . moving electrons X is at same and rest force Einstein parallel are lead also to a for metal to the same both X physics observers) postulates. moving relative to at the wire at a speed v. protons constant Protons and the v + q wire. Y is moving with the positive charge: the moving positive v is in the inertial frame of sees the moving force on electrons the in positive the wire charge produce as a magnetic magnetic in origin. centred experiences on a the wire. repulsive The charge magnetic is moving force at field outwards 90 that + + + − − + + v is from this the + + + circular (b) to + The o and − + Y . + X + charge Y and wire. v Y does the Y is not agree electrons so concerned. contracted for The attractive force this. Y the protons The proton Y electrons. with and have positive of the moves in the with wire are separations a higher protons electrons. feel in linear a the moving the wire density greater Once positive again, to charge the are left as with far as relativistically than repulsive the and that of force positive the than charge v the + q is v repelled in from origin. the Once wire again, but X Y and interprets Y agree the on force the acting physical as + electrostatic + + + + + − − − + + + v + + + effects. Figure A .1.3. Moving charges in iner tial reference frames. 147 A R E L AT I V I T Y S AMPLE STUDENT ANS WER You should be able to interpret Two protons are moving with the same velocity in a par ticle accelerator. this result with any combination of positive or negative moving protons charges and electron motion. Consider each possible case Observer X is at rest relative to the accelerator. Observer Y is at rest and ensure that you can explain it relative to the protons. Explain the nature of the force between the properly. protons as observed by observer X and observer Y. This ▲ The answer correctly by Y that will protons demonstrate as only (electrostatic) ▼ There Y is force no is Obser ver viewed no the mention repulsive. by discussion repulsive X; the clearly X force agged than a Y additional of the force need measures achieved 1/4 marks: for as a up to that size because moving relative to the reference frame. T herefore protons would experience not only an electric force, but a the magnetic force. protons, they a only Since would obser ver only Y is experience at rest an relative electric force, proton becomes magnetic once it is moving. If of observed the smaller attractive is There comparison in X force. since is have an also the could identies the the electric answer [4] v = 0, no magnetic force will be experienced as seen by the is equation question. F = qvB. repulsive there is an magnetic force. A . 2 L O R E N T Z T R A N S F O R M AT I O N S You must know: ✔ the two ✔ how ✔ the You should be able to: postulates two clocks Lorentz of can special be relativity synchronised ✔ solve ✔ use transformations problems the ✔ ✔ how to ✔ what ✔ the add velocities under special Lorentz describe in meant meaning by of an invariant proper time how proper frame length ✔ of the meaning of rest ✔ the meaning of spacetime ✔ the details determine measurements observers can be into measurements in either the position of various and time events mass ✔ show that two events that happen at interval different and two to: reference coordinates ✔ different by addition relativity quantity and velocity transformations spacetime converted is involving explanation of the muon positions in space and observed decay to be simultaneous by one observer are not experiment simultaneous ✔ the meaning of time dilation and in a different reference frame length ✔ solve problems involving the time-dilation and contraction. length-contraction ✔ solve problems equations involving the muon-decay experiment. In 1887, American demonstrated measurement of their direction of the Albert variations experiment. independent 148 scientists that were They in so Michelson light small confirmed reference speed frame that that in and due they the which Edward to changes fell below Morley in the of it measured. was light precision speed was A .2 This • result The laws (The • The of Einstein physics to as Newton’s speed of light of Lorentz inertial in reference. absolute the same first free the Newton in all postulates inertial as reference T R A N S F O R M AT I O N S follows. frames. postulate.) space (Replacing (a vacuum) Newton’s is the same concept of in all inertial absolute space time.) introduced frame modify are same frames and led LO R E N T Z with a gamma γ factor that compared the speed v of an c: 1 = 2 v 1 2 c He then is modified . Δt' γΔ ∆t the x refers difference the to Galilean the Δx between x of a and point x 1 two but Δx inertial not in reference according one frame to is – x 1 frame moving relative ), . so This that In to the relativistic also a important on the factor the changes speed of then depending time t’ in the the first v becomes t' = γ t − equivalent Lorentz the object in difference transformation, when Δx in a the difference second with γ is Δx = γ (Δx the inertial time also in relative appears inertial in reference vΔt). 1 difference frame – to the length the now first, equation. frame relative depends and that Similarly , to the first c position values an same x 2 Sometimes, on that second second of the 2 the are the is Galilean terms, 2 The ( x − vt ) and γ length difference 2 to = the 2 according Lorentz. (x 1 frames to x' transformations position in and frame transformations time S are are are known needed: in in frame S’ this case, the and the inverse required. Lorentz transformations and inverse Lorentz transformations (for Frame A is moving relative to frame B transformation in the x-direction). with a constant velocity v An object in A is moving at u Galalilean Lorentz Inverse Lorentz A relative to A; an object in frame B is x' = x − vt x' = γ ( x − vt ) Δx' = γ ( Δx relative to B moving at u x = γ ( x' B + vt' ) vΔt ) The speed of the object in A as measured by an observer in B is t' = t vx = γ t' u t − 2 c = γ ⎜ u t' + 2 c Δt 1 + 2 c The speed of the object in B as ⎟ 2 ⎝ v A vΔx ⎞ ⎛ v = A = γ u ' vx' t Δt' + A ⎠ c u − v B measured in B is u ' = B u v B 1 − 2 c These are given in the data booklet The transformations for length and time lead to expressions for the u − relative velocities when one or both objects are moving at speeds close as u′ v = uv to that of light. The equations follow from the usual definition of speed 1 − 2 c from Topic 2.1 but using the Lorentz transformations. 149 A R E L AT I V I T Y Example A .2.1 Two rockets, A and B, move along the same straight line when One way to get the signs correct is viewed from Earth. to note that: A is travelling away from Earth at speed 0.80c relative to Earth. • the signs are the same in the numerator and denominator B is travelling away from A at speed 0.60c relative to A. • the signs must be the same as a) Calculate the velocity of B relative to the Earth according to of B relative to the Earth according to the Galilean transformation, so Galilean relativity . when the frames are receding as b) Calculate the velocity the described by the observer, the theory of special relativity . signs are positive. c) Comment on your answers. Solution a) Using u ’ = b) Using the u The inverse Lorentz u' u − with v equation appropriate as 0.60 c + v = given transformations are not given 2 part the speed postulates. approximation The need that all to use The invariant quantities are: Proper time inter val Δt 1.40c booklet: 0.80 c are at exceeds small in this v < is is not not inertial (unchanged) possible valid under (except the as an c). transformations Rest mass m (often and relativity speeds different invariant c Galilean appropriate quantities quantities = c a), Einstein Lorentz equations. 0.80c 2 c In + 1 + in the data booklet, but you will c) data 0.60c 0.95 c = 0.60 c × 1 derived from the standard the gives 0.80 c = uv notice that they can easily be in signs does reference between not, frames however, differ. mean Some frames. is the mass of an object in the frame at 0 0 which it is at rest. shor tened to proper time) is the time interval between two events occurring at the same place in an iner tial reference 2 2 Space time inter val Δs 2 = (cΔt) – (Δx) 2 2 – (Δy) – (Δz) . frame. It is the shor test possible time that can be observed Note: in the DP physics course, we assume onebetween two events. In any other reference frame, the time dimensional motion (in the x-direction). between two events is dilated (longer) and the time in this So, O Proper length L 2 2 new frame is given by Δt = γΔt s 2 ( ct ) = − ( x) ignoring the Δ symbols. is the length of an object as measured 2 2 0 For any iner tial frame, Δ s 2 2 ( ct ) = − ( x) = ( ct' ) 2 − ( x' ) by an observer at rest relative to the object. Any other observation of the object will result in a length Electric charge (see Topic A.4, HL only) measurement L that is shor ter (contracted) than the L 0 proper length: L = γ The muon decay experiment This observation involves time dilation (or length contraction) for its explanation. Muons cosmic are produced rays probability of appreciable fraction muons of 150 the in a this frame of a speed will than frame frame the of atmosphere in is The and a light. reach are the It which is relative dilated and speed In to at the the rest. muon appears a known that to fact, because are high-energy have possible ground. they when muons known predicted moving rate muons. of molecules. formation that decay top air the ground in the the after reference muon of at with muons the observers) 0.98c, the decay fraction reach decaying (the interact is an predict many the more muons are However, frame much at longer to us about than in A .2 The muon observation contraction. us is to be a The few contracted W e know travel in muon as it our as of they is are but the from be the the explained top fast are the as Fewer atmosphere to in to terms the to our the are is measured the reference for by T R A N S F O R M AT I O N S length muons, decay measured muons of atmosphere to relative expected distance shorter. of However, moving muons much from also kilometres. many frame, frame travels distance tens how can LO R E N T Z every frame. metre observers removed by distance from in they the the beam surface. Example A .2.2 Muons 0.98c. are particles Explain, with with a proper calculations, lifetime why of 2.2 µs . this They experiment move downwards provides evidence to for the surface at a speed of relativity . Solution One of the conclusions of special relativity is that, to a moving observer, time is dilated. 1 Assuming Galilean relativity , muons must reach the ground in 6.7 µs. So, only of them should arrive 8 undecayed. 1 Special relativity predicts that γ for this event 1 = is = 2 1 To an 5 × observer 2.2 the 11 µs . = observed Measurements frame issue each unless of in the frame Therefore, outcome made we in apply measuring only and one the time of it reference a small reference in two the fraction confirms Lorentz of the Earth, should special frame are inertial frames that mean have in at lifetime decayed of before the muons they reach should the be ground. This is theory . another There are 98 the invalid 0 c relativity transformations. 5 2 1 v is also rest the relative to other. When 10 m, two an clocks are observer in near the to same one inertial clock will frame notice but that are the separated other by clock –7 reads a difference moving the clock of accelerating longer in another very is the inertial to slowly) do so and × 10 closer means an 3 s. This (assuming clock. frame. way slowly one not that During The infinitely transfer is clock a this to or simple they are motion, move to use reading a a to problem not the clock third the in the clock from clock second because same is place) no one frame (also to moving clock. S AMPLE STUDENT ANS WER An electron is emitted from a nucleus with a speed of 0.975 c as observed in a laboratory. The electron is detected at a distance of 0.800 m from the emitting nucleus as measured in the laboratory. a) For the reference frame of the electron, calculate the distance travelled by the detector. This answer could have [2] achieved 2/2 marks: 1 γ = = 4.5 2 ( 0.95 c ) 1 − 2 c l 0.800 m 0 l= = 0.178 m l= γ 4.5 151 A R E L AT I V I T Y b) For the reference frame of the laboratory, calculate the time taken for the electron to reach the detector after its emission from the nucleus. This [2] answer d v could t = t achieved 2/2 marks: 0.800 m d = have 9 = 2.74 = v s × 10 0.975 c c) For the reference frame of the electron, calculate the time between its emission at the nucleus and its detection. This answer could have achieved 2/2 [2] marks: 9 Δt 2.74 s × 10 10 Δt γΔt = Δt 0 = = γ ▲ This a)–d), sequence scores solutions in full of 4.50 answers, marks. a)–c) s = 6.08 × 10 0 are The clear and easy d) Outline why the answer to c) represents a proper time interval. to follow. to d) is an The written acceptable alternative This denition the one also the be for given proper on shortest possible to time time page acceptable to interval 150. say It that interval would it that answer could have Because it is the it is solve meant by problems kinematics point space. time in what inter val is using meant represent a spacetime involving spacetime diagram and simultaneity how ✔ represent and a by a moving worldline object by and its how represent same may be a straight an event on a spacetime diagram as to using line a time dilation spacetime and length contraction diagram (constant determine, for a specic speed, the angle velocity) a worldline on a spacetime diagram curve what and the worldline is meant by the twin how frame of to represent reference on more a than one spacetime requires you to use spacetime diagrams for one-dimensional time axis a describe inertial and spacetime resolve the twin paradox using a diagram. diagram. A spacetime The DP physics course only the paradox ✔ ✔ in apoint and ✔ obser vers diagrams between a two You should be able to: ✔ which or between D I A G R A M S ✔ ✔ marks: is You must know: to 1/1 is S PA C E T I M E what achieved to observe. A . 3 ✔ [1] response simple diagram (also visualisation Explanations diagrams of of known events relativistic as in a Minkowski one or phenomena more are diagram) reference facilitated allows frames. using the too. motion. Figure A.3.1 Particle from is the called frame, 152 A is shows the spacetime stationary origin the a of the worldline worldline in this ct-x diagram. of is this diagram reference The single particle. parallel to (with time ct frame line Because the a x-axis). and parallel it axis. is is situated to the stationary –1.0 m time in axis this A .3 Particle B is moving at a constant velocity of 0.5c. At t = 0 it was at S PA C E T I M E DI A GR A M S the A spacetime diagram consists origin of the reference frame of the spacetime diagram. The worldline of two axes: one for distance of this particle is a straight line moving through the spacetime (x-axis) and one for time (drawn diagram. The speed is 0.5c because, when the distance from 8.0 is 4 km, the time is ct = 8.0 and therefore t 4 and v = the 0 origin = = 8 c 0 on a conventional y-axis). The units for the x-axis can be in 4 c metres, kilometres, light years, etc. 8 However, the y-axis can be in time c units or ct Particle C is worldline accelerating for from acceleration is a rest. It began at the point (–2 km, 0); measured in distance units (light seconds or light years). the curve. A worldline is a sequence of spacetime events – it is the line The θ angle between the B worldline and the ct-axis gives the speed in spacetime that joins all the opposite X v 1 of tan θ B: = = = adjacent For Figure θ A.3.1, cT is so angle θ = tan v measured to be 27°and positions of an object throughout c c tan 27 = its existence. 0.5 so v = 0.5c. –5 ct / km Particle B from Figure A.3.1 can also be displayed on a t / 10 s spacetime B 10 A diagram drawn for the particle A reference frame (Figure A.3.2). 9 B is moving at 0.5c relative to A and is at the A origin (coincident 3.0 x with 8 the B the worldline origin) at t = 0 in both frames. Therefore, as far as A is concerned, 7 for B is the same as the time axis for B because B is 6 stationary in its own 2.0 frame. C 5 The B time and distance axes both swing around in the A diagram, as 4 shown in Figure particularly A.3.2. helpful The way B to axes compare makes paradoxes in A final element Figure A.3.2 from the event of E simultaneity that are occurs at is a labelled two easier a light as ct’ inertial to x’. reference This is frames = as has and been a 1 drawn = As 5.0 km. This shows nothing can exceed the A ct-axis The and intersection the of the light the path the of speed light of in light, the all reference ct’-axes x / km frame must lie –3 –2 –1 1 2 3 4 of A. Figure A .3.1. between Spacetime diagram in the reference frame of A cone. two θ time –4 ct 1.0 3 it 2 which 1.0 km cT a resolve. cone position x and sets of axes give us important ct / km information about the way A and B perceive the timing of E in ct′ their • W reference is the frames. time (ct = 5) at which E happens in the reference frame light light 8 of A. Y come come 7 • X (the an intersection observer on the of the light worldline cone with through ct) the is the origin time in when A sees X E 6 Z happen. 5 W • Y (the intersection of the light cone with ct’) is the time E when 4 an observer on the world line of ct’ sees E happen. 3 • Z (the intersection parallel to x’) is of E when with E a line happens constructed in frame through E and B. 2 x′ 1 –4 –3 –2 Figure A .3.2. –1 1 2 3 4 x / km Events and a light cone on a spacetime diagram 153 A R E L AT I V I T Y Observers in the two frames do not agree about the timing of events. This Problems and paradoxes featuring is a consequence of the constancy of the speed of light to all observers. timing discrepancies involve the concept of simultaneity Relativistic A classic speed simultaneity example towards a is that is often of a explored train through moving at the use of paradox. high tunnel. Example A .3.1 A train doors of at Explain proper each why , reference, the length end. in The the 100 m speed tunnel tunnel is too is of moving the train reference short for to is frame, this to the right towards a tunnel of proper length 80 m that has 0.8c the doors can be shut on the train but, in the train frame of happen. Solution γ for It the will the train is 1.67, obviously centre of the fit so in into tunnel the the at a time of frame tunnel (with 20 m in of the space the tunnel tunnel in the length of reference frame. The front of the train doors is 60 m. are shut when the train is in it). 20 20 So, rest = = 83 ns after the doors shut, the train hits the doors. 8 0.8 c 0.8 × 3 × 10 From ct the train frame of reference, the tunnel ct′ is contracted and approaches the train train in (moving to the left). The contracted length of train frame the tunnel is 40 m and so the train is too long right end of train when for train centred in the tunnel. left end of train enters tunnel frame tunnel in tunnel frame However, instant. both This is doors can because still the shut clocks for an (observers) left end of train in in the tunnel and on the train that measure the train frame when timing of the doors shutting cannot ever agree. right end of train emerges Suppose there is a pair of the tunnel.These in the tunnel of clocks clocks frame. In are the at each end synchronized train frame, right end of train the tunnel clocks the right-hand disagree. The one at enters tunnel of x′ proper length of tunnel the vL end left-hand 0 8 × of clock the by tunnel an is ahead amount of 80 = = 2 210 ns 8 c 3 In the × 10 time interval between the right-hand tunnel in clock and the left-hand clock reading zero tunnel frame according dnah thgir lennut fo dne dnah tfel of train lennut fo dne proper length x door of the left-hand This the train tunnel end problem diagram Figure A .3.3. to of is will the have the moved left-hand past the train. made (Figure observers, clear A.3.3). in The a spacetime tunnel frame is Spacetime diagram for the train–tunnel ct-x, the train frame is ct’-x’ simultaneity problem The t win paradox Mark, travel On a arrival find that 2T by γ 154 one to of a pa ir distant at his the of s ta r, twin, twins , s ta r at he a u ses tra vels Ma ria , a cons ta nt ha s sp aces h i p s p ee d ba ck a ged to by Earth 2T to wit h but l e a ve Ea r t h L ore n t z at the Mark sa me has a nd f ac t or γ s pee d only to a g ed A .3 This is what paradox from we arises him w ou ld exp ect beca u s e in from Ma rk’ s t i me di l a t i on. frame , Ma r i a However, has mov e d S PA C E T I M E DI A GR A M S a a wa y γ at Study the annotations on the So, why is Initially , Maria the two symmetrical. In not th e y oung er reference fact, they frames are twin in not. the Maria on Mar k’ s paradox has diagram carefully. re t u r n ? appear remained in ct an inertial hand, and he (non-accelerated) frame. accelerated beginning was decelerated also frame This needed breaks can (Figure be at to the the end. To accelerate. symmetry seen A.3.4) at the on a turn and in on of out explains the the around Moving spacetime drawn Mark, journey at of the the the Jay worldline other star Mark worldline inertial paradox. diagram Maria’s frame. S Maria remains on the worldline that is coincident Mark arrives Q ct′ with her ct axis. But Mark moves on worldline ct’ at star R out to the Maria thinks thinks that longer To we avoid clock Mark Maria on the imagine Jay’s that agree original a star. observes his simultaneous at the star arrival speed Jay comes towards Jay worldline as from Earth they at time Q. as Mark R—they no events. x′ acceleration–deceleration that with arrives the and issue star that at at the star, Mark’s Mark x synchronizes pass. returning to Maria is shown. Jay thinks Maria worldline that (so Maria far as is at Jay S as and he leaves. Mark are Maria’s ageing concerned) happens between Q and Figure A .3.4. S. The twin paradox in a spacetime diagram Example A .3.2 Four The light beacons events P , S, R are used to guide a rocket A to a docking station. and ct Q are four flashes from beacons shown in and the ct′ the are spacetime diagram. The diagram shows S the reference R frames Q of Earth (ct-x) and A(ct’-x’). x′ Deduce the order in which: P a) the beacons flash in the x b) reference frame an observer Earth of A sees the beacon flashing. Solution a) Construct P b) and Q Construct (and are lines occur light seen through at the cones on the same from Earth) in events time the the parallel and that events. order P , R to These R, x’. then Q These flashes cross and S show that before S. the ct-axis simultaneously . 155 A R E L AT I V I T Y S AMPLE STUDENT ANS WER An observer on Ear th watches a ct / km 3 rocket A. The spacetime diagram shows par t of the motion of A in the reference frame of the Ear th ct′ 2 A observer. Three flashing light Z beacons, X, Y and Z, are used to Y guide rocket A. The flash events 1 are shown on the spacetime x′ Parallel diagram. The diagram shows the X axes for the reference frames of x / km 0 1 Ear th and of rocket A. The Ear th 2 3 observer is at the origin. a) Using the graph above, deduce the order in which the beacons ash in the reference frame of the rocket. This ▲ This that is a shows complete a good answer the and the The of has is well taken because done time (with essentials of the the 2/2 marks: X and Y would happen at the same time before Z , word the worldline is parallel to the x’ axis, meaning that in to the indicate achieved spacetime diagram student have understanding implications diagrams. could answer Beacons of [2] parallel) reference frame of A, they would happen at the same time. Z the construction. would ash later , since it is further up the ct’ axis than X and Y . b) Using the graph above, deduce the order in which the Ear th observer sees the beacons ash. This ▲ This is correct and gains answer Implying full could all have [2] achieved beacons ash 2/2 marks: light going at the same velocity, X marks… will arrive ▼ …however, have made the this student clearer on Z ct Y and = Y Z cones intersect 2 km and Light the meaning are seen A . 4 is the obser ver because worldline the travel 0.4 km, hence they, Y and 0.4 km Z would before, but has and Z , arrive at the same time M E C H A N I C S meant by rest energy meant by relativistic ✔ what is meant by particle and total energy ✔ describe the the laws context of of energy special and relativity determine the potential difference necessary relativistic a particle to a given speed or energy mechanics ✔ solve problems involving momentum and photons relativistic ✔ that electric ✔ that mass charge and is of invariant momentum –2 units an MeV c can be quantity expressed energy in collisions MeV c respectively . and particle decays in ✔ calculate the wavelength of –1 and to acceleration accelerate of conservation in momentum ✔ implications ( A H L ) You should be able to: is 156 ashes Y simultaneously . what for Earth. afterX. at ✔ the Y the events momentum ✔ on from R E L AT I V I S T I C what by simultaneously, You must know: ✔ rst could drawn that seen the to diagram. be during a relativistic decay . photons emitted A .4 Einstein needed to modify for conservation physical ideas of energy and R E L AT I V I S T I C M E C H A NI CS (AHL) momentum The total energy of a par ticle is their laws to hold in all inertial frames. The rest mass E = E + E 0 m of a particle—the mass as observed in a frame in which the particle (rest mass + kinetic K energy E 0 , ignoring potential K 2 is at rest—is invariant. Einstein showed that there was an equivalence energy) and E = γ c m 0 2 between energy and mass expressed as ∆E = ∆m , c and for the rest 2 So, E energy E is E this = m 0 0 (1– γ)c = m K 2 0 c 0 Momentum p is conserved in special relativity provided γ 2 The equation E 2 = p 2 2 c + m 4 2 c can be rewritten as (m 0 c 2 ) 2 = is 2 E – (pc) 0 incorporated into the equations. 2 The quantity on the left-hand side of the equation is (rest mass) which 2 So p = γ m v and E 2 = p 2 c you know is invariant. The right-hand 2 for any two Placing The a inertial charged kinetic frames, particle energy ΔE E in side 2 – (pc) an E’ electric transferred must 2 = to be invariant c 0 and, 2 – (p’c) field a also 4 2 + m 0 causes particle of it to accelerate. charge q when ke accelerated through a potential difference V is ΔE =qV. Electric charge ke is an invariant When quantity . dealing with energy –1 MeV c mass kilogramme and and equation E MeV c joule). 2 the transfers involving particles, the units 2 for This 2 = for is energy the are same as often used setting c to (instead be 1 of and using 2 p + m . 0 Example A .4.1 Two protons each a) proton is Calculate as the accelerated of speed one order other by of an head-on. observer approach as The in a speed of laboratory . measured in the proton. potential in each measured relative frame Determine be approaching 0.50c the reference b) are difference to reach a through speed of which a proton must 0.70c Solution u + a) v c = u' 4 = = = 0.80 c 2 uv 5 0.25 c 1 + 1 + 2 2 c b) The c equation for the kinetic energy gain of the proton is 2 E = m ke (1 – γ)c 0 –27 γ = 1.67 and E = 1.67 × 16 10 × 0.67 × 9 × 10 –10 = 1.0 × 10 J k –19 =1.6 The pd required is × 10 × V 630 MV . Example A .4.2 Topics 7.1 and 12.1 described a) Calculate the potential difference through which a proton, starting photons as par ticles with energy from rest, must be accelerated for its mass–energy to be equal to but no mass. This implies that the four times its rest mass–energy . equation for their energy is b) Calculate the momentum of the proton after acceleration. 2 E 2 = p 2 c Solution 2 2 a) . Momentum can be assigned to a photon as Kinetic energy gain = 3m = 3 × 938c = 2810 MV E h E p p = and therefore = p = , 2 2 b) Using E 2 = p = p 2 c 2 + m + m c 4 c : c 0 where h is Plank’s constant. 2 16m 4 c 2 2 c 2 p 2 So p 4 c p 2 c 2 = 15m 4 c p 2 So p 2 = 15 m p 2 c 2 ⇒ p = 15 m 2 c −1 or 3630 MeV c p 157 A R E L AT I V I T Y S AMPLE STUDENT ANS WER 0 A lambda Λ par ticle at rest decays into a proton p and a pion π according to the reaction: ▲ The answer begins well. 0 Λ Although not conservation mentioned, of → p π + the momentum has where the rest energy of p = been used to recognize that = π particles (and solution the after the opposite) is decay completely calculation have momenta. of the equal The speed of the pion after the decay is 0.579c. For this speed The correct proton 140 MeV. up to γ = 1.2265. Calculate the speed of the proton. this incorrect: does not pion as of point, the speed equal they value the have momentum The the for is the speed equal but γ solution of be have achieved 3/4 marks: for π momentum of = 1.2265 × 140 × 0.579 = 99.42009 = of the p magnitude proton masses. 1.2265 incorporated could momentum = 99.42009 ÷ 938 = 0.105992 = γv and ∴ must answer is proton different not [4] speed. This ▼ At 938 MeV and the rest energy of the v for proton = 0.105992 ÷ 1.2265 = 0.086418c = ans as v = 0.105992c. The nal 2 v 1 2 c correct answer is A . 5 0.105c. G E N E R A L R E L AT I V I T Y ( A H L ) You must know: You should be able to: ✔ ✔ the equivalence principle use the equivalence explain ✔ why light paths are bent in the presence of the the denition ✔ details ✔ the denition of Schwarzschild ✔ the denition of an of the of gravitational why time dilates ✔ applications near a black ✔ calculate ✔ describe holes ✔ the general the Universe as a deduce and masses and dilation frequency shifts experiment the to observe and measure redshift Schwarzschild radius of a hole hole theory of apply the equation for gravitational time relativity dilation to an calculate ✔ of time gravitational gravitational horizon black to near experiment black ✔ light redshift Pound–Rebka–Snider event of mass gravitational ✔ principle bending near the event horizon of a black hole. whole. The Einstein cannot be outside using the Situation Observer of distinguished illustrated watching principle a equivalence from thought same events inertial states effects. experiment as one of that gravitational The involving them releases principle two an effects can be observers object with mass. ① X the is in an elevator elevator and not (lift) that has connected to no it; windows. Y can see Observer what Y is happens inside. The experiment gravitational X releases observers 158 is carried fields. the The object. agree It about out well elevator stays this. away is where from moving it is, as at no any masses constant forces or velocity act on it. when Both A .5 X repeats elevator X the describes does not the but, accelerate motion to Y , around Situation and to agree; accelerate When experiment begins of the this in the time, the object object as the direction as stays object of the accelerating stationary is released, elevator and R E L AT I V I T Y (AHL) the ceiling. downwards. the GE N E R A L elevator Y and X it. ② the elevator interprets downwards). the Y , is at rest motion on the on of the the other Earth’s object hand, surface, exactly will as X releases before interpret it as a the object (acceleration gravitational effect. X cannot distinguish between situation ➀ and situation ➁. This is the X principle The hold theory motion. in have the the principle same side. of status theory . status of and elevator line and elevator. the light is Both the also no leads under to on the some this the another agree motion, reference only does general not theory , conclusions Imagine surface, through of laws. light. through absolute frame same on Earth’s observers is inertial observers, fields shone passes there an All equivalence above of that of obey gravitational A pulse straight suggests The general same influence the equivalence. general relative The of (Figure time hole. hole with The on a hole light the about A.5.1) in travels opposite the the one in side a of this. Y Another time accelerate as the downwards elevator and the because hole the gravitational Observers the be as X hole and travels in the observer and both Y the must So in enters Earth’s the light is in curved path nearby mass. through when a in draw observer leave reconciled it pulse the gravity elevator and elevator, the field. still sees observer the The the are elevator light light enters exit subject starts to to the through the Figure A .5.1. Principle of equivalence for light same field. experiment. one light Y the Y the gravitational the field. that has physical also other. describes spacetime same must see The only light Light, been the as like conclusions light way arrive these moving any warped in a object, by the about through views can curved moves presence path in of a a Example A .5.1 A boy a stands direction downwards Explain on the initially curved whether floor of parallel path the boy an to to elevator. the the can floor. He The throws ball a ball follows in a floor. deduce that the elevator is at rest. Solution The boy states The cannot that alternative accelerating Gravity , affects is F, motion is this that upwards; according the A spaceship source make gravitational to of the the elevator these the inertial will general The equivalence effects and give the cannot boy identical theory and the be could principle distinguished. be outcomes. equivalence principle, light. contains at deduction. and an front observer of the ship and two near the light sources. observer and One the light other light 159 A R E L AT I V I T Y source to the At R, is at the Universe the instant sources begin rear. with when to The the the emit ship is velocity moving vector spaceship light of the in begins same at constant the to velocity direction accelerate, frequency . The of relative RF. both light light from R ∆h takes a time ∆t to = arrive at the observer. The light from F arrives c almost instantaneously , During the time the because light takes F to and the reach observer the are observer close. from R, the speed ∆h of the spaceship and observer will have by ∆v changed = a ∆t = a , c where light a is the from there is R acceleration. as now a c, as usual, difference The but observer measures experiences between the a source ∆f speed. The observed frequency shift speed ∆v = is the Doppler speed shift and of the because the observer ∆h a = 2 f The same resting The frequency shift is given observer to ∆h as problem on reach the from the can Earth’s R must observer. be considered surface have This with R in the loss c context below transferred energy c F. of the A photon energy to the corresponds spaceship reaching the gravitational to a reduction field in in the data booklet g 2 the frequency (because E = hf) and so there must be a redshift as the c wavelength increases. because the context is often that of the Ear th’s gravitational field. However, questions can be In developing tests, his including general the theory of gravitational relativity , redshift of Einstein proposed several light. asked for any gravitational field or This effect was confirmed by They a American physicists Pound, Rebka and acceleration. Snider 22 m in 1959. above the fired gamma gamma source and ray beam repeated upwards the to a experiment detector firing downwards. Gravitational time dilation: a The gamma photons should experience a fractional change in energy redshift is equivalent to a clock at ∆E ∆h R appearing to tick more slowly = . 2g The values that Pound and his co-workers measured 2 to the observer near F. Similarly, E c an observer on top of a mountain for thinks that time runs more slowly at change, sea level. 160 the fractional which energy confirms changes this test compared of general well with relativity . the theoretical A .5 GE N E R A L R E L AT I V I T Y (AHL) Example A .5.2 An observer Laser is close A directs a to light the Earth’s beam surface. horizontally towards the observer who 14 measures Another beam a) the identical vertically Calculate B, b) as laser State the B, be by the laser × 10 Hz. below towards the difference the observer, fires a light observer. between the two lasers, A and observer. has assumption 4.8 150 m frequency which one to upwards measured Explain c) frequency the made higher in frequency your to calculation the in observer. part a). Solution Δh a) Δf = 150 g f = 14 9.81 × × Laser A has energy as One German first that His solution R is of 7.9 Hz He a frequency B. = that This because loss of the energy photon is loses translated into a h ∆f the value astronomer was of the able spherical field ) of g does not change over the 150 m. and leads gravitational ∆E solutions theory . surrounds ∆λ from as change exact general shift physicist × 10 higher rises assumption vertical the the it frequency c) = 8 (3 b) 4.8 × 10 2 2 c Karl field to equations derive to the redshift the mass in M. that equations non-rotating of Schwarzschild constitute for uncharged wavelength This the one of Einstein’s gravity field mass. for fractional provided a photon in wavelength the shift GM s is = , = where r is the distance from the centre of the mass 2r 2GM to Schwar zschild radius 2 λ R = . s rc 2 c the point where the photon is emitted and is R the Schwarzschild radius This has the dimensions of length. s For values of r > R , gravity applies as normal, but inside the sphere of s radius R , the normal structure of spacetime does not apply . When s r = R , there is a transition between the two regimes. This distance from s the centre Near the strong of the event mass is horizon, gravitational known as the spacetime field. Mass is event horizon extremely collapses warped towards the due centre to the of the Events inside the event horizon black hole. dened by R cannot inuence s observers outside it. It represents Clocks in the region of a strong gravitational field run more slowly the surface where gravitational pull than in the absence of gravity . This is true near an event horizon. As a is so large that nothing can escape, clock moves towards the transition, external observers see it tick more not even light itself. The event and more The light slowly with the clock never quite crossing the event horizon. horizon is the surface at which the emitted by the clock is gravitationally redshifted as the clock speed needed to escape from the approaches R . The clock (and any observer unfortunate enough to mass is equal to the speed of light s be travelling horizon in a with finite it) will observe amount of its proper own time. passage through the event – this is the origin of the term black hole. 161 A R E L AT I V I T Y Example A .5.3 For a non-rotating mass with a Schwarzschild radius R a) Explain, , the the with reference Schwarzschild to the motion the event of light, what is meant by radius. s proper time interval Δt is related 0 b) Deduce the distance from horizon to the centre of a to the time interval Δt measured 30 star of mass 2 × 10 kg. by a distant observer at distance r from the centre of the mass by Solution Δt 0 Δt a) The Schwarzschild of mass radius is the largest distance from the centre = R s a to the point outside the mass at which photons of light 1 − cannot r This is escape because shortest path Photons When you write about spacetime its mass curves between cannot spacetime gravitational two escape prevents in field. spacetime points conditions photons and through from photons curved where follow the spacetime. extreme warping of escaping. in the vicinity of a black hole or 11 2GM event horizon, always give the b) R = 30 2 × 6.7 × 10 × 2 × 10 = = s 2 3.0 km 2 8 c impression of extreme warping. (3 × 10 ) Remember that spacetime is always warped by the present of mass. In the unusual conditions around a black hole, the warping is S AMPLE STUDENT ANS WER much greater than normal. It is believed that a non-rotating supermassive black hole is likely to exist near the centre of our galaxy. This black hole has a mass equivalent to 3.6 million times that of the Sun. ▲ The crucial cannot escape idea is that light here. a) Outline what is meant by the event horizon of a black hole. This ▼ However, it would have to have spacetime seen a warping reference rather could too that achieved 1/1 marks: event horizon is an imaginary surface strong light of a sphere where the to than gravitational theidea have been T he better answer [1] gravitational force pull is so that cannot escape. is strong. Star S-2 is in an elliptical orbit around a black hole. The distance of S-2 from the centre of the black hole varies between a few light-hours and ▼ The not terms in referenced used; 5.0 s proper The is time in not for phrases ‘dilated the equation the the event ‘dilated less’ answer mentioned beg the several light-days. A periodic event on S-2 occurs every 5.0 s. or as on more’ are a b) Discuss how the time for the periodic event as measured by an S-2. observer on the Ear th changes with the orbital position of S-2. and This ‘More points or to less be than made what?’ here answer the proper time for in is a 5.0 s, so different any the event times frame relativity, achieved the distance between is dilated less must, always when S-2 is = , and dilated 1 to the black periodic hole, will is at be its black 162 the observed dilated more extreme hole. time the black hole decreases, 0 5.0 s closer and Δt by be r • S-2 as s than marks: observed R greater 0/2 on Δt general have are: time S-2 could The When • [2] question, period than distance when from it the event changes. more when the distance increases, to the A .5 GE N E R A L R E L AT I V I T Y (AHL) Practice problems for Option A Problem 1 Problem 4 A space station is at rest relative to Ear th and carries An electron is observed in a laboratory to have a total clocks synchronized with clocks on Ear th. energy of 2.30 A spaceship passes Ear th travelling at a constant a) Show that the speed of the electron is about velocity with γ MeV. 0.98c = 1.25. a) Calculate, in terms of c, the speed of the spaceship b) The electron is detected at a distance of 0.800 m from its source in the laboratory frame. relative to Ear th. b) As the spaceship passes Ear th, a radio signal is emitted from the Ear th that is reflected by the (i) Calculate the distance travelled by the detector in the electron frame. spacestation and later observed on the spaceship. (ii) Calculate the time taken for the electron to reach Construct and annotate a spacetime diagram to show the detector from the source in the laboratory frame. these events. (iii) Calculate the time taken by the electron to move Problem 2 between its source and the detector in the electron a) Outline what is meant by proper length. frame. b) A pion decays in a proper time of 46 ns. It is moving (iv) Suggest which of your answers to (ii) and (iii) is with a velocity of 0.95c relative to an observer. a proper time interval. Calculate the decay time of the pion as measured by Problem 5 the observer. A spaceship leaves Ear th with a speed 0.80c. Problem 3 a) Draw a spacetime diagram for the Ear th’s frame An unstable par ticle A decays into a par ticle B and its including the motion of the spaceship. antipar ticle B b) Label your diagram with the angle between the A is at rest relative to the laboratory when it decays. The momentum of B relative to the laboratory is worldline of the spaceship and that of the Ear th. Problem 6 −1 7.4 GeV c An electron and proton with equal and opposite velocities annihilate to produce two photons of identical −2 The rest mass of B and B is 1.8 GeV c energies. The initial kinetic energy of the electron is Deduce the rest mass of A . 2.5 MeV. a) Determine the speed of the electron. b) Calculate the energy and momentum of one of the photons. 163 ENGINEERING B B . 1 R I G I D B O D I E S A N D P H YS I C S R O TAT I O N A L You must know: You should be able to: ✔ the denition of torque ✔ ✔ the denition of moment ✔ the denition of rotational calculate torque for single D Y N A M I C S forces and for couples of inertia ✔ equilibrium sketch and variation linear that equations of rotational time graphs of that angular show the displacement, under conditions of velocity and torque kinematics ✔ apply with equilibrium angular ✔ interpret and constant solve problems involving the rotational angular equivalent of Newton’s second law acceleration ✔ ✔ that angular momentum ✔ that Newton’s is solve problems rotational second law can be applied in form to angular the distinction and one that is between in both and translational equilibrium solve problems involving the rotational motion quantities ✔ objects a ✔ modied involving conserved an object that is rolling angular slipping. moment of acceleration inertia, as torque analogies to and linear quantities ✔ solve problems that involve rolling without slipping. Topic 2 covers linear mechanics When an object rotates about an axis When the initial angular speed ω with no translational motion and is changes to a nal angular speed ω and the interaction of objects that in f displaced through an angle θ in a time t, its angular acceleration are treated as points; this par t of time t, it has an angular velocity Option B deals with objects that (ω −ω ) f θ have shape and size. Many of the ω = mechanics. t t quantities in rotational mechanics have direct analogies in linear α = Remember from Topic 9.1 that ω = Here 2 πf is the correspondence between quantities motion: Linear quantities Rotational quantities You should be confident using the equations from Topic 2.1. Learn v = u + at ω = ω f + α t i the links between linear quantities 2 and rotational quantities and the v 2 = u 2 + 2 as ω 2 = ω f + 2αθ i way in which they are used. This 1 s = ut + 1 2 topic emphasizes these links. at θ = ω 2 α t t + i 2 2 You should also be familiar with the quantities described in (v + (ω u) + ω f s θ = 2t 164 ) i = Topic 6.1. 2t in linear and rotational B .1 R IGID BODIE S AND R OTAT I O N A L D Y N A MI CS Example B.1.1 A laboratory centrifuge reaches its working angular speed −1 of 1100 rad s from rest in −4 centrifuge Calculate is 7.6 the × 10 4.2 s. The moment of inertia of the 2 g m angular acceleration of the system. Solution (ω gular α acceleration ω i f ) ( 1100 = next step is to 2 = t The 0) = examine the 260 rad s 4.2 way in which the rotational equivalent 2 of force The connects rotational inertia with through the angular equivalent depend spheres, to on the mass centre of of mass particular m, rod, I is moment axes connected the Moment of iner tia acceleration. by is of a inertia rotation. light given of rod I. For Moments two length l very about = ∑ mr , where m is the mass of an object and r is of the distance from the axis for each small an I par t of the object. axis by The unit of moment of iner tia 2 is kg m 2 l I = m 2 + m 2 l 2 = However, 1 2 when ml 2 the rotational axis is changed to the centre of one of 2 the spheres, The I moment about an becomes of axis inertia of 0 + ml for rotation a bicycle through wheel the of centre mass m of the and radius wheel is r You will not be required to simply calculate moments of iner tia. 2 mr (assuming compared that with the the spokes rim and and the centre bearing are very light The value of I or the equation to tyre). compute it for a par ticular shape will be provided. The rotational a force a turning this is F acts equivalent at a effect. defined of distance Torque by a is r a linear from force an vector right-hand axis and, rule is torque. of A torque rotation therefore, (rather like and has a exists when produces direction; Fleming’s rule in F applied force electromagnetism was defined by a left-hand rule). The relationship is radius shown in Figure B.1.1. from r The link between change F in Newton’s second F = ma, axis and momentum also = time angular law, applies to rotational motion when taken torque momentum has been defined: direction τ change Γ in angular time Angular acts of momentum = on the the the of centre their they must rise is this of arms external give system. system example taken momentum speed to produced an by an system has ice body . their torque a modified is their into of This up acts to They conserved purely alter This torque. The a their of transfer of torque inertia Figure B.1.1. The right-hand rule to describe torque A common moment movement external speed momentum. energy an moment vertical rotational angular the the means. about their reduces because unless when internal spinning conserve here internal the with skater body . is consequences of axis by inertia Notice the through pulling arms involved and that can no only must be skater. Newton’s laws of motion can be expressed in rotational contexts. Newton’s rst law: Every rotating body continues to rotate at constant angular velocity unless an external torque acts on it. Newton’s second law: F = ma. Newton’s second law in a rotational Rotational kinetic energy links to linear kinetic energy . energy added As usual, context is changes the can changes be in inelastic, elastic momentum of or the have system or systems depending involved. = I on Newton’s third law: Action torque and reaction torque acting on a body are equal and opposite. 165 B E NGIN E E R ING P H YS I CS Angular momentum, L is dened as Example B.1.2 Iω . The unit of angular momentum 2 is kg m A flywheel 1 is accelerated from rest. The flywheel has a moment of s 2 inertia of 250 kg m −1 and takes 8.0 s to accelerate a) Calculate the angular acceleration of the b) Calculate the average accelerating torque c) Calculate the rotational to 90 rev min ywheel. The rotational kinetic energy of a body with moment of E acting on the ywheel. K rot iner tia I and angular velocity ω is 1 the 2 = end of its kinetic energy stored in the ywheel at acceleration. ω K rot 2 Solution The change in rotational kinetic 90 × 2π 1 a) energy E Angular speed of the ywheel = = 9.4 rad s of a body with moment 60 Krot 9.4 of iner tia I changing angular 2 and α velocity between ω and ω I = is = 1.18 rad s 8.0 f 1 2 I (ω f 2 −ω b) ) Γ Iα = = 250 × 1.18 = 295 N m 2 1 1 2 2 The unit of E is kg m 2 s c) which is Rotational kinetic energy = I (ω 2 −ω f 2 = ) i × 250 × 9.4 Krot 2 2 the same as a joule. 4 1.1 × 10 = When a cylinder between the When a (slips) along the cylinder at v the the of rω . horizontal ground on point used radius to is r the in is is at ground, the point the point of as to right v − moving at contact rest the at rω in rolling, the kinetic right v + = with 0 rω at and (point linear linear the at rest) speed speed bottom energy of a energy is gained by rolling object the object is then v. The must v = change of h, m gh = Iω 2 rω 166 mv 2 of the moving and the top 2 + down 2 + top be mv . When 2 1 2 height static 1 Iω rolling 1 vertical of 2v 2 kinetic moves coefficient 2 total contact calculations. 1 The of rest. ground, any rolls the However, cylinder and slides be moves along ground. contact should A cylinder of cylinder cylinder Because friction rolls J a slope with a this B .1 R IGID BODIE S AND R OTAT I O N A L D Y N A MI CS S AMPLE STUDENT ANS WER A satellite approaches a rotating space probe at a negligibly small speed in order satellite to link to it. The satellite does not rotate initially, but after the link they rotate at the same angular speed. The initial angular speed of the probe is probe 1 16 rad s The moment of iner tia of the probe about 4 the common axis is 1.44 × 10 2 kg m common axis The moment of iner tia of the satellite 3 2 about the common axis is 4.80 × 10 kg m a) Determine the nal angular speed of the probe−satellite system. This answer Angular could have achieved momentum is 2/2 [2] marks: conser ved, L = I.ω ωs = ωp 2 ▲ A well-presented ωp Ip 1 = Is.ωs + 1 ωp Ip 1 2 ωp = 1 (Is + makes 2 Ip 2 × obvious that examiner. The physics are stated and the are clear. to the principles ) 4 10 everything 2 4 1.44 solution ωp Ip × 16 = ωp (1.44 × 10 substitutions 3 + 4 .80 × 10 ) 2 230400 ωp = × 19200 2 ωp = 12 rad/s 2 b) Calculate the loss of rotational kinetic energy due to the linking of the probe with the satellite. This answer could have [3] achieved 3/3 1 1 4 2 E marks: = ω .I krot1 p = × 1.44 × 10 2 × 16 = 1843200 ▲ Again, out 1 1 the 2 = (I krot2 + p I ) s × ωp − krot1 E and full accurate. credit It here. is is well easy laid This is to a award model 2 = × 19200 × 12 = 1382400 of 2 2 E answer 2 2 E the p how to answer a question. 2 = 460800 ≈ 460000 J lost krot2 460000 J is lost. 167 B E NGIN E E R ING B . 2 P H YS I CS T H E R M O D Y N A M I C S You must know: You should be able to: ✔ ✔ the rst law of thermodynamics describe the statement ✔ the second law ✔ the denition of the denition of of thermodynamics of as a energy explain the sign conventions used in the rst entropy law ✔ law conservation thermodynamics ✔ of rst of isothermal and of thermodynamics, Q = W + ΔU adiabatic ✔ solve problems involving the rst law of processes thermodynamics ✔ that an isovolumetric process is carried out at ✔ constant describe using ✔ that an isobaric constant process is carried out at the can denition be of cyclic visualized sketch the processes using pV and that diagrams, describe they and terms what ✔ the is Clausius and interpret the of, of thermodynamics interpretation and the Kelvin interpretation meant by a second and as law of thermodynamics consequence of, in entropy how solve problems involving entropy changes and them describe ✔ law pressure ✔ to second (Joule–Kelvin) ✔ ✔ the volume Carnot processes in terms of entropy change cycle ✔ solve problems involving adiabatic changes for 5 denition of thermal efciency and how to 3 monatomic solve problems involving thermal gases where pV efciency . A thermodynamic system denes boundary. The system together with the items of interest in a par ticular the surroundings constitutes the context. The system is separated Universe As with Topic 3, Option B.2 covers the behaviour of gases, but takes a broader view. Here, we from its surroundings by a consider the general proper ties of systems in terms of changes they A pressure–volume (pV) diagram shows the changes in the pressure undergo and how these have an and volume of a gas as it moves between two or more states or around impact on the rest of the Universe. a closed The first energy by cycle. its law as it of thermodynamics applies to surroundings. systems; The law is is an expression specifically , written as a Q of the system = W + of ΔU . conservation a gas The acted terms on in the Some changes of state of a gas equation when positive are: have special names that you should recognize. • Q — the energy • W — the • ΔU — the transferred into the system from the surroundings Isobaric changes occur at constant work done by the system on the surroundings pressure. change in the internal energy of the system. Isovolumetric changes occur at constant volume. This equation Isothermal changes occur at Figure constant temperature. is B.2.1. trapped the can For inside boundary) be all a applied four cylinder with the to the changes, with four gas imagine a piston surroundings changes that at being an one shown ideal end gas in (system) (making everything else up in the Adiabatic changes occur without energy being transferred into or out Universe. of the gas. Isobaric change The work pressure. through W = The 168 done The by the energy distance x system transfer is pAx . on W But, the Ax is pΔV first law becomes Q = ΔU + surroundings when pΔV the the is at constant piston of area change in volume A is moved ΔV, of B.2 The work done the pV the graph graph. or by or This by on can the be system is evaluated the by equivalent either of the counting area squares T H E R M O D Y N A MI CS under under integration. When sketching pV graphs, make sure that the relative gradients Isothermal change of the isothermal and adiabatic There is no change to ΔU as this is the internal energy of the gas: Q = W. changes are correct. Remember that the area Any thermal energy transferred into the system must appear as underneath a pV graph is the external work done by the system on the surroundings. energy transferred. This can be Isovolumetric change work done on the gas, or work done by the gas depending on The term W is zero because no work is done by or on the system (gas): the direction of the state change. Q = ΔU When the gas expands, it is doing work; when it is compressed, work Adiabatic change is done on it. No energy is transferred into or out of the system, and so Q = 0 p × V has the units of energy. implying at the the that ΔU expense of = W. the surroundings, Any internal the external energy . temperature of work done by the Put simply , when the system must system work is must done be on fall. Example B.2.1 0.064 mol of an ideal gas is enclosed in a cylinder by a frictionless piston. Two isotherms are shown on the pV diagram for 300 K and 500 K. aP 01 / p 4 B C 8.00 A 500 K 300 K 0 0 2.00 –3 V / 10 a) Explain state b) of how the i) A to B ii) A to C Calculate the gas at at rst is law of changed constant constant the 3 m heat thermodynamics applies when the from: volume pressure. energy absorbed by the gas in the change from: i) A to B ii) A to C. 169 B E NGIN E E R ING P H YS I CS Solution a) The rst where W is is the The (W) = entering by the B done to to the is the system written gas as from ΔU and Q the is = W ∆U, + surroundings, the change in the gas. is at by reect A to move W of A to is change must Q done change increases The thermodynamics energy energy work ii) of the work internal i) law Q C the is allow constant the system change at volume, on the so gas. Q = ∆U The as no temperature ΔU constant pressure and so the piston this. ΔU + In this in expansion case, the temperature increases and work is done 3 b) i) ΔU = nR(500 300) = 1.5 × 0.064 × 10 × 8.31 × 200 = 160 J 2 T 3 2 ii) V = V 2 = 3.3 × 10 3 m 1 T 1 4 pΔV So = ΔQ 8.0 = × −3 10 ΔU+ × (3.3 pΔV = 2.0) = 104 J 264 J Example B.2.2 When the state of an ideal monatomic gas changes from An (p , V 1 ) to (p 1 , V 2 ideal monatomic gas is in an expansion pump at an initial pressure ) in an adiabatic of 2 100 k Pa and a temperature of 313 K. adiabatically to 1.7 When the pump is operated, 5 3 change, pV 5 p 1 V gas expands times its original volume. 5 3 So the = constant. a) Calculate the pressure of the b) Calculate the temperature air in the cavity after the expansion. 3 = p 1 V 2 2 of the air after the expansion. (The exponent is dierent when the gas has more than one atom in the Solution molecule.) a) γ 5 pV = constant, where γ = 3 This is why the gradient of an 5 adiabatic change on a pV diagram p = 100 × 10 1 3 × is steeper for the same gas than 1.7 3 = 41.3 kPa when it undergoes an isothermal pV b) change (for which pV = constant). = constant ⇒ T = 220 K T Carnot that Carnot this gave the operates cycle. case) at first Energy a high surroundings at does the work Energy Q on is description through a a is cycle of of transferred temperature, lower a gas theoretical state into and the working energy temperature. heat engine—a changes—known The is fluid (the transferred remainder of as gas out the device the to in the energy system. supplied to the gas trapped in a cylinder by a piston from 1 a hot reservoir that is at a high temperature T . The gas expands, hot and the piston atmospheric However, original will move pressure. the state. gas is This to can until The gas work only in the pressure has a done cycle, happen of work so it when the must an gas under is the these now amount go of same as conditions. back to its energy Q 2 rejected to a cold reservoir at a low temperature T cold 170 is B.2 The thermal efficiency of useful the work Carnot heat done Q engine = given by: Q 1 η is T H E R M O D Y N A MI CS 2 ≡ Carnot energy input Q The thermal eciency for the This assumes that all the energy Q − Q 1 is transferred into useful work 2 Carnot cycle can also be written as and there are no losses to friction, and so on (which is why Carnot’s T − T hot engine is only theoretical). T cold = cold . =1− Carnot T hot The cycle consists (Figure B.2.1). W The to X: gas is of at two T isothermal and and expands two adiabatic isothermally X to that Y: no The energy gas hot changes absorbing energy Q hot Note T . 1 goes expands into the gas because adiabatically and the the change is isothermal. temperature falls to T . cold The gas loses internal atmosphere — the Y to the Z Z: The gas to W: done gas is internal The on energy piston now continues as the compressed energy . The gas is gas increases the and moves work compressed, its gas to do isothermally done again internal work on the expands. on the to gas Z is adiabatically , energy to with no rejected and return all it change as the to energy . work to T hot The area a net work enclosed previous infinitely done by by the energy the gas curve. This state — this slowly — another on the surroundings cycle means reason is reversible that why the the cycle Carnot in one and it must cycle cycle can be is is the return to operated theoretical. W isothermal Q 1 adiabatic X isothermal T hot adiabatic Z Y T cold volume Figure B.2.1. The Carnot cycle Example B.2.3 This table shows measurements experimental a) Calculate possible some made heat the on an engine. temperature of heat source 830 °C temperature of cooling system 17 °C heat energy supplied per second 78 J power output of heat engine 1.5 W maximum efciency of the engine. b) Suggest whether approaches your the actual answer to efciency part of the heat engine a). 171 B E NGIN E E R ING P H YS I CS Solution T T hot a) η 1103 290 cold = = = T 74% 1103 hot P 15 out = b) = = P 19% 78 in The engine The first The second law is of significantly less thermodynamics efficient equates than work the and theoretical energy value. transfer. The Clausius statement of the law of thermodynamics lays out the situations in which second law: energy can be transferred into work. Energy cannot ow spontaneously from an object at a low temperature There are a number to an object at a higher temperature required in the DP without external work being done statement and Entropy defined the of ways physics entropy to state course: the the second law. Clausius formulation Three of statement, these the are Kelvin statement. on the system. is in terms of the energy ΔQ absorbed by a system and The Kelvin statement of the second the kelvin temperature T at which the energy transfer occurs. Entropy law: is Energy cannot be extracted from a a measure increase of the disorder in a system. Any real process tends to disorder. reservoir and transferred entirely Consider a crystal of common salt (sodium chloride). When solid, the into work. salt atoms are arranged regularly in a highly ordered way . Drop the The entropy formulation (due to crystal into water for solution. and it dissolves with many possible arrangements Boltzmann) statement: the The entropy of the system has increased. To restore For any real process the entropy of the order to the crystal, the water must be evaporated either naturally the Universe must not decrease. or The change in entropy ΔS of the ΔQ system is ΔS = by salt heating. to that Universe; of This the the process solid total of of again these decreasing will cause changes the other always entropy entropy of the dissolved increases in increases. . T 1 The unit of entropy is J K Example B.2.4 Calculate the entropy change when 1 kg of ice melts. 1 The specific latent heat of fusion of ice is 330 kJ kg Solution The melting occurs at The entropy change a temperature of 273 K. 3 330 × 10 1 is = 1.2 kJ K 273 S AMPLE STUDENT ANS WER A heat engine operates on the cycle shown in the pressure–volume diagram. The cycle consists of an isothermal expansion AB, an isovolumetric change BC and an adiabatic compression CA. The volume at B is double the volume at A . The gas is an ideal monatomic gas. 6 P = 4 × 10 A P a T = 612 K –4 3 V = 1.50 × 10 M Δs = + 0.6797 isothermal erusserp Δu = 0 W = 416 J P = ! Q = 0 B V = 2V B W + Δu = 0 A T = 612 K A560 Δs = 0 W = 0 C 2V V volume 172 ΔV = 0 because the B.2 T H E R M O D Y N A MI CS 6 At A the pressure of the gas is 4.00× 10 4 and the volume is 1.50 × 10 Pa, the temperature is 612 K 3 m . The work done by the gas during the isothermal expansion is 416 J. a) i) Justify why the thermal energy supplied during the expansion AB is 416 J. This [1] answer could have achieved 1/1 marks: 3 ▲ The Because it is isothermal, ΔT = 0, and Δu = answer correctly nRΔT identies that there is no change in 2 ∴ Δu = In this 0. Q = Δu + W , and if Δu = 0, Q = U W . and also case W = 416 J, so Q = therefore Q = W so that Q is 416 J. 416 J. The temperature of the gas at C is 386 K . ii) Show that the thermal energy removed from the gas for the change BC is approximately 330 J. This answer could have [2] achieved 2/2 marks: 3 PV ΔV = 0, W = 0 so Q = Δu nR = = 0.980 Δu = nRΔT ▲ BC no ΔT = 386 − 612 = is at constant volume so 2 T change in W for this part of the −226 K cycle; Q = ΔU and a calculation 3 3 Δu × = 0.98 × (−226)= −332.227 = −330 J using 2 Q = −330 J, so 330 J is taken out of the gas. clear iii) Determine the eciency of the heat engine. This nRT conrms the result. A 2 answer could have achieved 2/2 answer. [2] marks: W e = Q =416 J in Q ▲ Once useful work done = 416 − Δu Δu AC ∴ W = 416 − 330 = = −Δu AC = + presented aim 86 J to your W e again, clear and well- 330 BC work. achieve It this examination should sort of be your quality in answers. 86 = = = Q 0.207 416 0.207 or 20.7% b) State and explain at which point in the cycle ABCA the entropy of the gas is the largest. This B answer would [3] could have have the achieved highest 3/3 marks: entropy. Entropy difference is ∆Q calculated ∆S wrong. and for the change AB ∆T ΔS = +0.680. T his means that the system gains entropy ▲ The AB from as A→B. the A change and is C have the adiabatic. same entropy T herefore, if the because ΔQ entropy at B = 0, there than T herefore B at has A, it the will also largest be higher than that at is a that supports is stated chain the of clearly argument answer. Again, a is model higher answer and answer. C. entropy. 173 B E NGIN E E R ING B . 3 P H YS I CS F L U I D S A N D F L U I D You must know: ✔ the denitions of D Y N A M I C S You should be able to: density and pressure ✔ determine using ✔ Archimedes, principle buoyant liquid in a and why objects what is meant by ideal solve problems the meaning ✔ Pascal’s ✔ the of ✔ explain denitions of ✔ the continuity of streamlines ✔ Stokes’s the equation effects describe laminar and turbulent pressure, Bernoulli density , equation, and Stokes’s the law in uid ow using the Bernoulli principle, Bernoulli’s the frictional spheres during drag that laminar is exerted on ow ow ✔ and objects effect small the buoyant equilibrium principle ✔ on uid hydrostatic signicance acting principle involving principle, continuity ✔ force are ✔ an the Archimedes’ Pascal’s ✔ ( A H L ) dene and determine the Reynolds number. equation law. Figure filled B.3.1 with shows a hydraulic incompressible oil lift of used to constant raise a car. volume. The When lift a is force F X F F X X is applied at X, the pressure p in the liquid is . This pressure is A X transmitted F through the fluid so that, at Y , where the area is larger, Y A X A Y F A Y p = Y . The force exerted upwards on the car is much larger (by A ) A Y than X that exerted at X. However, the volume of fluid moved at X is pressure throughout fluid = p X Figure B.3.1. The hydraulic jack the same as at Y so the car moves upwards by a small amount (by the A X ratio ). The device is energy neutral assuming no frictional losses. A Y Buoyancy mass Density, ρ m = described forces by acting upwards Archimedes’ on an object submerged in a fluid are principle. = volume V Density is a scalar and has units Example B.3.1 3 kg m Roughly 90% of the volume of an iceberg is below the water Pressure surface. Estimate Density of the density of sea ice. force normal to a surface p = 3 water = 1000 kg m area of the surface Pressure is a scalar and has units of Solution 2 N m or pascal (Pa). Consider upthrust 2 1 N m a 1000 kg on this slab ice is of ice. 900g as By Archimedes’ this is the principle, weight of water the displaced. = 1 Pa. The fundamental unit 3 1 of pressure is kg m 2 However, because the volume of the ice is 1000 m , the density of s 3 the ice is 0.9g≈ 900 kg m An impor tant idea in uid dynamics is Pascal’s principle. Pascal stated that a pressure change that occurs anywhere in a conned An ideal, This non-viscous means that the incompressible motion of a fluid particle at undergoes a point in streamline the fluid is flow . the same incompressible uid is transmitted as the motion of particles that preceded it at the same point. A collection of through the whole uid. The key streamlines constitute a stream tube. The stream tube changes dimensions, words here are incompressible and but the amount of fluid inside it does not as fluid cannot enter conned. surface 174 of a stream tube. This leads to the continuity equation or leave the B.3 FLUIDS AND FLUID D Y N A MI CS (AHL) Archimedes’ principle states that for an object wholly or par tly in a A y uid, an upward buoyancy force v acts on the object that is equal to y the weight of uid displaced by v x A the object. x The buoyancy force B is usually Y known as the upthrust. X Upthrust is given by B = ρ V f Figure B.3.2. g f Streamlines, stream tube and the continuity equation where ρ is the density of the f uid and V is the volume of uid f The continuity equation describes the changes to the motion of fluid displaced by the object. inside and a stream leaving tube. the As stream entering and leaving constant and the is time fluid tube cannot must AvρΔt. is the be The same enter fluid for or leave, constant. is both So, the for mass incompressible, ends of entering time Δt, the the ρ so stream mass is tube. An ideal uid has no resistive force either to the uid itself moving, or The Bernoulli equation can be modified by multiplying each term by to solids moving through it. Such a 1 2 volume V . mv So + mgz + pV = constant × V . When this is done, the 2 equation kinetic The are an energy + gravitational Pitot–static one opening the of viscosity is considered later). becomes used: that the tube 90° to tubes potential system parallel at two Pitot uid is non-viscous (the meaning to the are tube — when a is the used to streamline same far energy flow is work measure (the streamline apart — so + static (the done fluid tube) Pitot on speeds. and tube). that the static steady , the pressure fluid tube = constant Two another tubes Continuity equation: Av = constant with and Assuming cannot is v x = A x v y y Av is known as the volume ow affect difference A 3 rate and has the units m 1 s a The Bernoulli equation, measure of the kinetic energy of the fluid: 1 2 ρv 1 p ρv + + ρgz + p = constant, is 2 x = p x ⇒ v y = 2 g( h x h y ) 2 x 2 a statement of the conservation Pitot tubes can be mounted on aircraft wings to measure the aircraft of energy in the context of uid speed relative to the air speed. dynamics. Each term in the equation has the unit of pressure, 2 Example B.3.2 N m or Pa. This leads to A tank two B. liquids, A is p = p + 0 contains ρ gd for f a uid at rest when comparing A and pressure at a point in the uid and vertically the pressure a height d above it. above B. 60 cm Calculate the liquid A speed with liquid B from a density = 1000 kg m emerges hole bottom −3 which of at the the tank. 40 cm liquid B −3 density = 13 600 kg m 175 B E NGIN E E R ING P H YS I CS Solution The pressure The = B pressure 53 310 + on at 5880 B due the = to base ρgh A is of the 59 250 Pa. = tank This is 1000 is × 9.81 13 600 × equivalent × 0.6 9.81 to × = 0.4 0.44 cm 5886 Pa + 5886 of liquid alone. 1 2 ρv Rearranging ρgz + + p = constant and recognizing that p is the 2 same on both sides of the equation gives 1 v = = 2 gh 2.9 m s Example B.3.3 Water flows into a long gardening hose of diameter 1.5 cm with a 1 speed of 5.0 m s Calculate the . The spray velocity of nozzle water has leaving a diameter the of 8 mm. nozzle. Solution The continuity A So v in = equation A in is Av = constant as density is constant. v out out 2 2 0.75 × 10 1 v and = A out × v in × 5.0 A in out 3 4.0 Viscosity η × 10 = 18 m s is the resistance of a uid to stress, where one plane of Real fluids have viscosity which varies according to temperature and the the liquid slides relative to another chemistry of the fluid. Some fluids are more viscous than others (compare parallel plane. the flow of honey with water at the same temperature, for example). The unit of viscosity is the Pa s or George Stokes analysed the resistive (drag) force F 2 N m kg m of 1 acting on a sphere D s. In fundamental units, this is 1 radius r due to a viscous fluid while the sphere falls through the s fluid under resistive conditions medium, it of will streamline reach a flow. terminal When an object falls in a speed. Stokes’s law: when the speed of the sphere is v,the resistive force F = Example B.3.4 6πrvη D A sphere The is falling following at data terminal are speed through a fluid. available. You saw terminal speed in Diameter of sphere = 3.0 mm Topic 2.2. 3 Density of sphere = 2500 kg m = 875 kg = 160 mm s 3 Density of fluid m Including the eects of buoyancy, 1 Terminal speed of sphere when the density of the medium is ρ and the density of the Determine the viscosity of the fluid. sphere is σ, the terminal speed Solution 2 2r v g (σ − ρ ) Rearranging = the Stokes’s law expression gives t 9η 2 3 2 2r η g (σ ρ 2 × ) = ( 1.5 × 10 × ) 9.81 × ( 2500 − 875 ) = = 9v 9 × 50 mPa s 0.16 t Laminar ow is steady, predictable Steady (streamline) flow is observed at low fluid speeds. As the speed streamline ow where the increases, the flow between layers of fluid sliding past each other becomes stream tubes remain intact and unstable; particles from different streamlines and stream tubes begin to material does not cross between interact and mix. The flow is no longer laminar but is now turbulent streamlines. Turbulent ow is unpredictable; it is To see an example characterized by the appearance of the eddies and vor tices in the uid. smoke begins source of 176 smoke rising the of laminar from with an smoke a flow piece orderly the flow of turning into smouldering flow but becomes a few turbulent wood or centimetres turbulent. flow, paper. watch The above the B.3 This transition between the two flow states and the transition speed FLUIDS AND FLUID D Y N A MI CS are The Reynolds number, R difficult to predict. There is a ‘rule of thumb’ for flow in a pipe (AHL) is a e using dimensionless quantity given by the Reynolds number R e vr ρ R = e When R is less than 1000, flow can be taken to be laminar; when R e greater is η e than 2000, the flow will be turbulent. These values are different where r is the radius of the pipe, v because there is a complex transition between the flow states — it is not is the speed of ow, ρ is the density possible to be precise about the nature of the flow in the transition. of the uid, η is its viscosity. Example B.3.5 3 A syrup flows at a rate of 4.0 m 1 s in a circular pipe of diameter 3 6.0 cm. The syrup has a density of 1300 kg m and a viscosity of 17 Pa s. Deduce whether the flow is laminar. Solution flow rate 4.0 3 Speed of flow = = = 1.4 × 10 1 m s 2 area of pipe π × 0.03 3 ( 1.4 vrρ R = ) × 10 × 0.03 × 1300 = = 3200 e η This is 17 greater than 2000, so the flow will be turbulent. S AMPLE STUDENT ANS WER A ball is moving in still air, spinning clockwise about a horizontal axis through its centre. The diagram shows streamlines around the ball. a) The surface area of the ball −2 is 2.50 × 10 2 m . The speed −1 of air is 28.4 m s under the −1 above ball and 16.6 m s the ball. The density of air is F −3 1.20 kg m Estimate the magnitude of the force on the ball, ignoring gravity. This [2] answer could have achieved 1/2 marks: ▲ The −2 A = 2.5 × 10 basic Bernoulli equation excluding z because including thickness height of the ball is so Pv 1 P + of the (reducing ignore the the Bernoulli it to two terms). 2 1 used 2 P in 2 ▼ However, − ball Pv 2 1 1 correct to 2 = 1 2 P is decision 1 2 + the negligible. equation 1 P method 2 m = × 1.2 × (28.4) in the the wrong equation area and so is the 2 − × 1.2 × (16.6) = 318.6 Pa = Δp answer 2 is incorrect. The cross- 2 2 1 sectional F area of the ball is of the 4 P F = = P A ΔF = ΔP .A surface area. A −2 ΔF = = 318.6 7 .965 × ≈ 2.5 × 10 8 N 177 B E NGIN E E R ING ▼ This is comment a repeat made in of P H YS I CS the the b) State one assumption you made in your estimate in par t a). This part but answer. is not The a good essential answer the the Bernoulli ow must point equation be is laminar. to will is turbulent, break across the down ball B . 4 the and will When what the height of the ball is negligible, so the difference in would be 0. streamlines the pressure equalize. is meant by You should be able to: the natural frequency of ✔ describe is examples critically-damped meant by damping, critically-damped and under-, over- ✔ describe, oscillations what is meant by the ✔ what is meant by a Q using vibration object ✔ close periodic stimulus and describe the and oscillations graph, with its over- how the driving natural amplitude frequency frequency of for of an vibration phase relationship between of a periodic stimulus and the forced frequency oscillations what to under-, by frequency driving a varies of factor ✔ ✔ marks: the vibration and 0/1 FORCED VIBRATIONS AND RESONANCE (AHL) what ✔ achieved that, You must know: ✔ have hold, values ow could enough T hat for [1] previous meant periodic by resonance in response to that result from it a ✔ solve problems ✔ describe involving Q stimulus. both useful and destructive effects of resonance. Simple harmonic was implicit motion has a constant amplitude and energy; this Simple harmonic motion is an assumption of earlier topics. In real cases, a freely covered in Topics 4.1 and 9.1. oscillating frictional system losses gradually are known loses as energy through resistance. The damping Under-damped oscillators lose energy gradually and come to rest taking many oscillations to do so. In under-damping, exponentially . This three-quarters of the amplitude implies the that energy) is of the a the time oscillation to constant lose for decreases half the the amplitude (and system. Critically damped oscillators stop moving (and therefore lose their A child on a swing is given a push to begin the oscillation. Without total kinetic energy) in the shor test intervention, the swing will eventually stop moving. Add energy to time possible. the Over-damped oscillators stop system in indefinitely . a systematic However, way when and the the oscillation driving frequency can (the be maintained frequency moving in a longer time than the applied to the swing) does not match the natural frequency of the swing, critically damped case. (See Figure B.4.1.) the swing the driving said to Figure amplitude will frequency undergo B.4.2 forced shows is be to larger the or smaller natural depending frequency . The on how driven close system is oscillations how the amplitude of the driven system varies with over-damped frequency edutilpma An critically damped of the important amplitude for driver. feature each of these level of curves damping; is that this there occurs is a maximum at resonance time Figure and B.4.3 the driven, shows driven then the system. the driver phase relationship When is 180° the out driver of between frequency phase with the is the driver greater driven system than system; under-damped when the driver Figure B.4.1. in frequency phase. At is smaller resonance, the Amplitude against 90° time graph for oscillators with various degrees of damping 178 is driver apart with the driver leading. than the driver natural and the frequency , driven the system are B.4 FOR C E D V I B R AT I O N S AND RE S ON ANCE (AHL) driver leads by half a period noitarbiv decrof eht fo edutilpma π lighter damping dar / gal esahp heavier damping π/2 driver leads by quar ter of a period driver and driven in phase 0 zero damping light damping heavy damping natural frequency driver frequency f 0 driving frequency Figure B.4.3. Phase relationships between driver frequency and natural frequency Figure B.4.2. Resonance curves for different degrees of damping Resonance tides are attraction large of and when is an caused an the important by phenomenon. resonances Sun and potentially oscillating in Moon. is It is oceans used amplitudes driven close to driven Disadvantages destructive system the to of in tune by the radios, resonance mechanical its natural and gravitational include systems frequency by a Figure B.4.2 shows four amounts varying driving system. of damping and the variation The in or width the of a system. “quality” system resonance This factor. takes to curve sharpness Q is decay the to is depends defined approximate zero when the on the using amount a quantity number driver of of has damping called oscillations been of the maximum amplitude. As the Q that a removed. damping increases, the point of maximum amplitude drifts to lower frequencies. Also, the curves are not symmetrical about the maximum amplitude. energy stored per cycle Q = 2π × energy dissipated per cycle where f energy stored = 2× f is the resonant frequency. 0 × 0 Q has no unit. power loss A high take Q many values implies cycles indicate damping and value system 30 000 for to that on the that stop a moving damping a system door is as lightly the power heavy . closure, vibrating is crystal Typical 100 in a for a damped loss Q is quartz small. values simple and are will Low Q 0.5 pendulum for in the air, watch. Example B4.1 An electric simple of the motor maintains pendulum motor motion of is the that 48 mW has a and pendulum the the bob The motor is switched off. oscillation time period of maximum is of a demonstration 10.0 s. kinetic The power energy of output the 2.3 J. Deduce the time the system will take to stop. Solution The pendulum being The supplied resonant is at storing the energy rate frequency is of at the rate of 2.3 W with power 48 mW. 0.1 Hz. 2.3 So Q = 2π × 0.1 × = 30 2 4.8 × 10 The system other is words, going in to about oscillate 5 about 30 times before stopping; in minutes. 179 B E NGIN E E R ING P H YS I CS S AMPLE STUDENT ANS WER ▲ (The upper provided.) curve The reasonably was lower drawn A driven system is lightly damped. The graph shows the variation with curve and is driving frequency f of the amplitude A of oscillation. incorporates a) On the graph, sketch a curve to show the variation with driving the features looking the that for. The original greater; examiners peak because also the is lower the peak is were than damping shifted is frequency of the amplitude when the damping of the system increases. This lower also frequencies, which [2] to answer could have achieved 2/2marks: is correct. A ▼ The the endpoints left, Try to was the avoid no are curves this penalty , not clear; appear to ambiguity . on cross. There however. 0 f 0 b) A mass on a spring is forced to osciliate by connecting it to a sine wave vibrator. The graph shows the variation with time t of the resulting displacement y of the mass. The sine wave vibrator has the same frequency as the natural frequency of the spring–mass system. [2] 30 20 10 mm 0 / 0 y 5 10 15 20 t /s –10 –20 –30 i) State and explain the displacement of the sine wave vibrator at t = 8.0 s. ▲ The answer correctly begins quoting relationship at the [2] by phase resonance. This answer could have achieved 2/2 marks: It π is important resonance it is not and to on to in use the condition identied. make the word answer mentioned the be to in of The correct Because it is resonating, is at therefore, zero a question resonance answer the vibrator difference will be 2 T herefore, the sine at wave vibrator at 0.80 seconds will be has goes deductions: maximum phase because the at one of its zeroes, which means the displacement s = 0. the Displacement = spring the 0. 8.0 s; must be at ii) The vibrator is switched o and the spring continues to oscillate. displacement. The Q factor is 25. energy stored Calculate the ratio for the oscillations of the spring–mass power loss system. This answer could have [2] achieved 2/2 marks: 1 E Q = stored 2π.f T = 8 seconds f = = 0.125 Hz = a f a T ▲ The answer provided from uses including earlier to all the provide the Powerless data Energy frequency a correct 25 = 2π × 0.125 × r r = = Power solution 180 to the ratio. stored 31.8 loss 31.8 B.4 FOR C E D V I B R AT I O N S AND RE S ON ANCE (AHL) Practice problems for Option B Problem 5 Problem 1 A potter ’s wheel is rotating at an angular speed of A flywheel consists of a solid cylinder of mass 1.22 kg 1 5.0 rad s and radius 240 mm. with no torque acting. The potter throws a lump of clay onto the wheel so that A mass M is connected to the flywheel by a string the wheel and the clay have a common axis of rotation. wrapped around the circumference of the cylinder. 2 The moment of iner tia of the wheel is 1.6 kg m and the The mass falls from rest and exer ts a torque on the 2 moment of iner tia of the clay is 0.25 kg m , both about flywheel which accelerates uniformly. the common axis. −12 After a time of 4.85 s, the velocity of M is 2.36 m s The angular speed of the wheel changes suddenly 1 2 Moment of iner tia for the flywheel = mass × radius when the clay lands on it and no net angular impulse is 2 added to the system. a) Calculate the angular acceleration of the a) Calculate the angular speed of the wheel flywheel. immediately after the clay has been added. b) Deduce the torque acting on the flywheel. The potter now applies a tangential force to the rim of c) Determine M. the wheel over 0.25 of a revolution to return the angular Problem 6 1 speed to 5.0 rad s An ideal gas is compressed quickly by a piston in a The wheel has a diameter of 0.62 m. cylinder. The gas is initially at a pressure of 110 kPa b) Calculate the angular acceleration of the wheel. and a temperature of 290 K . The volume of the gas is 4 6.0 × 10 3 m c) Deduce the average tangential force applied by a) Calculate the quantity of gas in the cylinder. State an the potter. appropriate unit for your answer. Problem 2 b) The gas is compressed to a pressure of 190 kPa and Explain why a piece of wood floats. 4 a volume of 4.0 × 10 3 m Problem 3 A Pitot–static tube is used in an aircraft that is travelling 1 at 75 m s (i) Suggest, with a calculation, whether the gas is 1 into a headwind of 15 m s . The density of compressed isothermally. 3 the air is 1.3 kg m . (ii) Explain why the compression may be Determine the pressure difference measured by the adiabatic. instrument. c) 15 J of energy is used to compress the gas. Problem 4 a) A mass of 24 kg is attached to the end of a spring −1 of spring constant 60 N m . The mass is displaced Determine the change in the internal energy of the air in the cylinder. 0.035 m ver tically from its equilibrium position d) The compression is repeated very slowly. and released. Discuss the entropy change that takes place in Determine the maximum kinetic energy of the the cylinder and its surroundings as the air is mass. compressed. b) The mass–spring system is damped and its amplitude halves by the end of each complete cycle. Sketch a graph to show how the kinetic energy of the mass on the spring varies with time over a single period. You should include suitable values on each of your scales. 181 IMAGING C C . 1 I N T R O D U C T I O N T O You must know: I M A G I N G You should be able to: ✔ the denition of a real ✔ the denition of linear image and virtual image ✔ construct for virtual magnication converging the magnication optical ✔ describe properties of converging the denition of of focal point, focal length the curvature optical for curved the properties thin denition properties the curved of a thin surfaces lens modify arise wave on them of axis the principal axis focal point and focal of a converging or diverging lens on a converging diagram and construct scaled ray diagrams to solve problems lenses of focal point, focal length for converging and converging and diverging lenses and ✔ principal mirrors mirrors for ✔ the way incident identify ✔ diverging problems and scaled ✔ solve and length radius diverging to mirrors ✔ ✔ and diagrams how the fronts diverging ray and from ✔ scaled solve problems involving the thin lens equation diverging and linear and angular magnication lenses. ✔ explain spherical aberration be Reflecting incident (a them converging shows concave mirror mirrors on the with to of describe and how the chromatic ir effects can reduced. spherical form mirror) features and aberration or images. convex both types surfaces These (a of modify surfaces diverging wave can be mirror). fronts either Figure of light concave C.1.1 mirror. convex mirror Q X principal axis F C C P F Y Q’ Z f focal length focal length R radius of curvature (a) (b) Figure C.1.1. The focal (c) Concave and convex mirrors length is half the radius of curvature for both mirrors. You must know the meanings of principal axis PC, focal point F, focal length f, centre of cur vature C and radius of cur vature R, as defined on the diagrams. 182 Figure C.1.1(a) shows mirror that parallel These focus) rays at are are the said focal to the to path the come point. of three principal from rays axis infinity incident (one and ray they on the converging coincident converge with (come to it). a C .1 Figure time C.1.1(b) they come for move from the a diverging apart focal after mirror also shows reflection — they three diverge, rays, but appearing INTRODUCTION TO IM A GING this to have point. It is best not to draw the mirrors Ray diagrams mirror. They are used use rays to determine whose the image behaviour is position formed by a as curved surfaces — they are effectively flat compared with known. the distances between the mirror, Figure C.1.1(c) shows three such rays for a converging mirror. The image and object. Figure C.1.2 three rays are predictable; in other words, they follow set rules. shows the symbols often used for the two types of mirror. On a ray diagram: Arrows are conventionally drawn on the rays to show the direction A similar convention is used for • ray X travels first parallel to the of travel. lenses (Figure C.1.3). principal axis and after reflection Rays from innity are imagined as goes through F originating from the same point on • ray Y arrives through F and then the object and travelling such a long reflects parallel to the principal axis distance that they become parallel to each other. • ray Z goes through C and then returns along its original path. Figure C.1.2(a) shows rays for a converging mirror. The image formed Real images can be formed on a is smaller than the object (diminished) and is real. screen. They form where rays meet Figure C.1.2(b) diminished shows virtual the image ray is diagram always for a diverging produced with lens. this A type and cross. of mirror. Vir tual images cannot be formed on a screen. They are formed by rays that appear to have come from a point. A focusing system is needed to view them. image C F F C image This image is smaller than the This image is smaller than the object (diminished), object (diminished), upside the right way up (erect) and virtual (because the down (inverted) and real rays appear tohave come from the image and (because the rays cross). do not cross). (a) Figure C.1.2. (b) Ray diagrams for (a) converging (b) diverging mirrors Magnication M is a proper ty of both angles subtended at the mirror by the mirror and lens systems. It is the ratio rays from the top of the image and of image size to object size: object, respectively. Magnification does not always mean that the image is larger height of image Magnication has no units. M = than the object — it is the ratio of height of object For simple systems image size : object size. So it can image distance where v and u are the distances from = θ v = i be less than one. In this case, the = the mirror to the image and object, object distance respectively, and θ and θ u θ term diminished can be used to o are the o Spherical spherical mirror the mirrors distort aberrations. where all describe the image. images. Aberration rays parallel These can to be the distortions eliminated principal are known using axis are a as parabolic reflected to focus. 183 C IM A GING Example C.1.1 Construct the where object ray diagram for a converging mirror (b) object the a) between b) at F is: and C c) C d) at F between the the mirror pole and C of F F. Solution image (a) (c) image C object F object C F (d) object Converging The physics of refraction at a mirrors. single plane interface between two the and Figure diverging C.1.3 image F C lenses shows the have a technical important language quantities similar associated to with lenses. media is covered in Topic 4.4. The lenses refract the light. As a plane wavefront moves into the lens, Effects here are extensions to the wavefront: earlier work as there are two interfaces to the lens and both are • slows • must down in the lens curved. be continuous at the interface. rays from innity optical centre focal point O F F principal axis O F (a) This means point principal axis (b) Figure C.1.3. curved interface F The terms for a (a) converging lens and (b) a diverging lens that which is the the wavefronts focus curve (Figure in C.1.4). the The lens effect and is converge reinforced to a when wavefronts the wavefront exits the second surface of the lens. ray ray The assumption thin. Figure C.1.4. Wavefronts modified by a single curved interface lens This used in emphasised (FigureC.1.3). equations lenses. 184 is must be the by When used DP the the to physics symbols lens allow has for course used is for significant this. that the the two lenses types thickness, Predictable rays are of different exist for the C .1 Ray ① parallel (diverging to lens: the principal moving axis away Ray ② through the optical Ray ③ through the focal is from centre refracted the of the focal lens through the focal INTRODUCTION TO IM A GING point point). is Make sure that you can draw all undeviated. these diagrams confidently and point travels parallel to the principal axis accurately. The same advice after refraction (for a diverging lens, a ray aimed at the focal point is applies to the more complicated refracted parallel to the principal axis). diagrams in Option C.2. There is Figure C.1.5 diverging shows the ray diagrams for both converging and advice there on how to remember lenses. the constructions but you might consider a similar approach to (a) object (b) image your learning here. image 2F O F F 2F object object O F F (c) image object object O F F image image Figure C.1.5. There are system Ray diagrams (a) and (b) for a converging lens, and (c) a diverging lens also using numerical the methods thin-lens to predict the behaviour of a lens equation The thin-lens equation is The thin-lens equation links to the ideas of wavefront modification by 1 the lens. The more strongly curved the surfaces, the greater the 1 = to the wavefront Therefore, a shape, small and radius of the smaller curvature the value means a of f for small f. the This 1 change f lens. leads to + u v the where u is the distance of the power of a lens to alter waves; a powerful lens has a small f object (from the optical centre of The thin-lens distinguish and equation requires mathematically between the focal the use between lengths of of real a sign and convention virtual converging and 75 cm an objects diverging the lens), v the image distance and to and images f the focal length. lenses. Example C.1.2 A lens of power +4D is placed from object of height 5.0 cm. The thin-lens equation involves reciprocals and a sign convention. Calculate: Take care with both, par ticularly a) the position of the image formed with the reciprocal at the end of b) the nature of the image. the calculation. 185 C IM A GING The sign convention used in the DP Solution 1 physics course is known as ‘real is a) The focal length f is and the lens is converging. f = +0.25 m. 4 positive’. The object distance is 0.75 m. • Real objects and images are taken to be positive. 1 1 1 = Rearranging f • Vir tual objects and images are 1 + 1 gives u 1 = v − v 1 1 +0.25 0.75 = f v taken to be negative. • = 1 2.6 6 = • Converging lenses have positive 0.375 focal lengths and positive powers. So the image distance is +0.375 m = 37.5 cm from the lens. • Diverging lenses have negative b) The answer to part a) was positive, so the image is real. focal lengths and negative powers. 0 Its magnification 375 1 is = 0 1 , 75 so the image height is 5 × = 2 2.5 cm 2 The power P of a lens is defined as 1 P ∝ radius of curvature of surfaces Figure C.1.5 virtual (b) image. shows This is how the a converging magnifying lens glass, produces one of the a magnified oldest optical 1 = instruments. f A magnifying glass can be used: The power is measured in dioptres 1 (D) and has the unit m . A lens with • with the object at f and the image at infinity; this gives an angular θ f = 0.25 m has a power +4D. i m magnification = where θ and θ I are the angles subtended at the o θ o D eye by the image and object respectively , leading to: M = infinity f • with the object closer than f and the image at the near point; this D gives an angular magnification: M = near + 1 point f Example C.1.3 The human eye has a near point. It is the closest distance D at which A simple we can see an object without strain. by The near point is taken to be 25 cm a magnifying person with magnification of a glass near the has point object a of focal length 0.30 m. when the of 0.10 m. Calculate image is the formed It is used angular at the for the normal eye. person’s near point. Solution D M = The angle than that Lenses, have 0 30 0 10 + 1 = f from for like two + 1 = top the × bottom of the image is four times greater object. mirrors, forms to 4 of are prone aberration: to aberration. spherical (due Unlike to the mirrors, physical lenses shape spherical aberration of Figure C.1.6. by a lens 186 Aberrations produced the with lens) and chromatic wavelength form. in the (due lens). to the Figure variation C.1.6 of shows refractive how the index aberrations C .1 Chromatic with blue aberrations wavelength light different colour in the shapes fringing The aberration two lenses — a They are required cancel are (colour) lens, and at so can be of the give the made from the red reduced the variation Red blue the using and a light refractive wave travels focal faster lengths. an achromatic indices are in and so opposite IM A GING than with This are leads to different. doublet lens — are TO speed emerge magnifications diverging aberrations in wavefronts different by lens the light. because different while by and slightly image converging power caused INTRODUCTION where combined. produce directions the and out. S AMPLE STUDENT ANS WER A lamp is located 6.0 m from a screen. Somewhere between the 6.0 m lamp and the screen, a lens is placed so that it produces a real inver ted image on the screen. The lamp image produced is 4.0 times larger than the lamp. a) Determine the distance between the lamp and the lens. This answer could have achieved 2/3 [3] marks: ▼ The answer begins with an v unfortunate −v M = v + u = slip. When 4 then = u 6 m u 4u = v, not the other way round. −v −4 = u ▲ However is 4v = u 5v → = correct. = total and 1.2 m the error U = 6 − 1.2 = distance screen which must b) Calculate the focal length of the lens. answer could have is the achieved 1/1 be 1 between that 5v the lamp (following examiner will the carry and hence u, in this case, marks: the use of the thin- 1 equation is not affected by the + error v in scores 48 solution [1] lens = the shows 4.8 m. ▲ Fortunately, 1 of 4 .8 m forward) This rest answer 6 m the V the The 1.2 f = the full previous question and marks. 0.96 m ▼ There is detail missing here. The c) The lens is moved to a second position where the image on the screen facts needed are: Is the image real is again focused. The lamp–screen distance does not change. Compare or the characteristics of this new image with the original image. Is This answer could have achieved 1/2 marks: virtual? Is it In it magnied this case, new image is still real and or the omitted — the T he upright or inverted? [2] diminished? last point answer is is that the inverted. 1 magnication is . 4 187 C IM A GING C . 2 I M A G I N G I N S T R U M E N TAT I O N You must know: ✔ how optical You should be able to: compound telescopes form images ✔ construct and compound ✔ how astronomical refracting telescopes ✔ construct or complete astronomical how astronomical reecting telescopes the in ray diagrams normal for a adjustment form images ✔ interpret microscope the refracting ray diagram telescope in for an normal form adjustment images ✔ ✔ what and are the meant by the Cassegrain Newtonian mounting for investigate the compound microscope performance of an optical mounting and an astronomical reecting refracting telescope telescopes ✔ ✔ the operation and resolution of a single solve problems magnication radio the and the angular resolution of an optical telescope compound ✔ involving dish denition of a microscope radio-interferometer ✔ solve problems involving the angular telescope magnication ✔ how of to describe Earth-based the and comparative performance satellite-based point is discussed in Option C.1. optical astronomical telescopes. The The meaning of the term near of telescopes. magnifying Small focal curvatures The optical that there the required means compound is more is C.2.1 normal that in Option for one lens. shows only at the the ray C.1 has aberrations this; near to point diagram objective know of for a the the and are compound converging need disadvantages. magnifications solves lens — a You formed large spherical microscope than eyepiece image Figure eyepiece described are required converging when glass lengths a the large problem. means lens normal and observer ’s compound a adjustment eye. microscope in adjustment. objective lens O f f o e • The object is to the left of the objective lens which forms a real I 1 magnified image I of the object at a point between the focal 1 O point • D The of the eyepiece eyepiece acts as and a the eyepiece magnifying itself. glass viewing I to produce a 1 virtual, Figure C.2.1. magnified image of the intermediate I Compound microscope in 1 normal adjustment • The the distance between observer ’s eye the should virtual be image placed as and close the to eyepiece the is D; eyepiece as possible. The angular multiplying v magnification the D of the magnifications compound of the microscope objective and is given eyepiece by lenses. DL o This is + 1 u o where L is ≈ f the e length f f o of e the microscope linear overall magnification v v o o 188 o ≈ u f o defined magnification = using angular magnification × of and tube, eyepiece of objective C.2 The microscope criterion) sin θ just = resolves , 1.22 two θ where is images the when angle (following subtended at the the IM A GING I N S T R U M E N TAT I O N Rayleigh eye by the d Learn the function of the images and d is the effective diameter of the aperture (this is usually individual elements in the optical the objective). light and as Resolution wide an is improved aperture as by using short wavelengths of instruments. For example, a compound microscope can be possible. regarded as a modification of the magnifying glass (the eyepiece) Example C.2.1 with the objective lens providing a two-stage magnification process A compound microscope and lenses is in normal adjustment with the objective that avoids the magnifying glass eyepiece separated by 23 cm. The object is 6.2 mm from limitations. the objective the eyepiece which is has a focal length of 6.0 mm. The focal length of To draw the instruments in an 50 mm. examination, begin with the final Determine the magnification of the microscope. image and work backwards. Add Solution An x the intermediate images at this intermediate from the 1 1 + 6.2 image is formed in the microscope tube at distance point, and then the initial rays. objective: Make sure that you practice drawing them. 1 = x 6 1 So, x = 1 1 − = 6 6.2 186 mm from the objective. 186 The linear magnification of the objective is = 30 × 6.2 And the angular magnification of the eyepiece The the the The magnification astronomical lenses. of overall The focal eyepiece length of refracting length and, the in tube of telescope the is + f f • astronomical The object is telescope at infinity , = . is consists longer adjustment, Figure 1 = 6 × 50 180 × also objective normal o an is 30 × 6 + So 250 is C.2.2 the of two than focal shows the converging focal points the ray objective eyepiece length coincide. diagram for f f o o f e o e in so normal the rays adjustment. incident on the objective image lens θ e at ∞ are • parallel. These rays are focused by the objective lens at its focal length f to o Figure C.2.2. form a real image of the object at the Astronomical focus. telescope in normal adjustment • This real image This lens forms is a real object for the eyepiece lens of focal length f . e • The two lenses a final image magnify the at infinity object from this separately . object. The overall angular f θ o o magnification and magnification of the telescope is M = = θ f e e 189 C IM A GING Example C.2.2 An astronomical focal a) lengths and in normal adjustment has two lenses of +12 cm. Calculate: i) the length ii) the The at the is the telescope magnification used to view of the telescope. the Moon which the image of subtends an angle of Earth. Determine the of angular telescope 0.52° b) telescope +96 cm the diameter of the Moon as formed by telescope. Solution a) i) The total length = f + f o = 108 cm e f 96 o ii) Angular magnification = = = f 8 × 12 e b) The f × diameter tan(angle of the image subtended is: by object) = 96 × tan(0.52) = 0.871 cm = 8.7 mm o Another than type of refracting telescope are mounting astronomical lenses. required (Figure Two for telescope forms the C.2.3a)) DP and of uses the astronomical physics the reflecting course: Cassegrain the mirrors rather reflecting Newtonian mounting (Figure C.2.3b)). (a) (b) Figure C.2.3. Astronomical reflecting telescope (a) Newtonian mounting and (b) Cassegrain mounting Rays to a from focus placed at a on the Newtonian side, the plane 190 distant the principal focus, mount: rays mirror. object an When can image been flat focused axis. image The having This are is be a primary or converging photographic mirror plate is seen. by does a screen viewed turned surface by 90° not through from the modify an eyepiece principal the from axis by magnification. the a C.2 Cassegrain a small lengthens image These mount: A curved hole is in the the main viewed prepared. tube using telescopes lens-based primary do a not imaging is However, hyperbolic mirror. and The increases converging suffer used). the from Only surfaces mirror the rays main I N S T R U M E N TAT I O N through effectively the final lens. chromatic one the mirror magnification. Again, eyepiece are sends secondary IM A GING aberration curved vulnerable (when surface and easily has no to further be damaged. Example C.2.3 A Cassegrain a) Compare with b) a telescope the effectiveness human eye with The effective focal and the length focal Calculate primary the a angular the pupil length of of mirror of the has a telescope diameter the diameter telescope eyepiece is magnification at of of 0.80 m. collecting energy 8.0 cm. main mirror is 2.8 m 7.0 cm. of the telescope. Solution 2 a) The the area of the telescope mirror collects is (10) 100 × = as 100 times much f the energy area every of the eye, so second. 280 o b) Angular magnification = = = f 7 40 × 0 e Radio for telescopes the radiation Single-dish radiation focus of principal the dish, also distant parabolic axis the are reflecting than reflectors from a are rather collect All focused greater the they use radio electromagnetic objects dish. telescopes; wavelengths light. at and rays the energy reflect it parallel focus. The collected to to Interferometer the Signals the larger per from telescopes The second. a telescopes number of emulates a a small (geographically system are recent single-dish separated) large development. single are dish combined. of baselineB λ The resolution is given by sin θ = 1.22 λ , so the larger The resolution The exact is given by sin θ approximately d d the better the resolved images. of large d of Disadvantages a very dish are the small retaining the parabolic shape in a large the problems Telescopes • Stars This • The are emit is now a steering routinely radiation not some problem atmosphere changes of and in a massive placed of which on is dishes. depends There are on the proposals arrangement radio telescopes that for combine very the large signals structure from and resolution problems baseline of ≈ arrays of dish telescopes in different continents. object. satellites. absorbed by the atmosphere. space. distorts telescope images due to refractive index turbulence. • Extra-terrestrial • International telescopes collaboration are is a immune feature to of light pollution satellite from cities. development. Example C.2.4 6 Two stars that are 2 × wavelength 630 nm, which primary The has a light-year is a 10 are light-years imaged mirror measure of of by apart, the 2.40 m emitting Hubble light Space of telescope diameter. distance. 191 C IM A GING a) Determine the b) the angle between two images just resolved by instrument. Two stars light years Deduce, can in still the Andromeda from in be galaxy are both about three million Earth. light-years, the separation of these stars so that they resolved. Solution λ a) sin θ Using = 1.22 leads to d 7 6.3 × 10 1 θ = 7 sin = 1.22 × 3.2 × 10 rad 2.4 7 b) The distance must be 3.2 × 10 6 × 3 × 10 = 1 light-year S AMPLE STUDENT ANS WER Both optical refracting telescopes and compound microscopes consist of two converging lenses. a) Compare the focal lengths needed for the objective lens in a refracting telescope and in a compound microscope. This ▲ The student receives a answer the showing focal understanding length ratios for have achieved 1/1 marks: mark Focal for could [1] lengths of the objective lens are larger than its eyepiece of lens both for telescopes whilst focal length of objective lens is smaller instruments. than eyepiece lens focal length for microscopes. b) A student has four converging lenses of focal length 5, 20, 150 and 500 mm. Determine the maximum magnication that can be obtained with a refracting telescope using two of the lenses. [1] The question is “determine”. It This answer could have achieved 1/1 marks: would have been good to see a statement such as “the maximum f 500mm o value for magnification is obtained M = = 100 = f f 5 mm e o is greatest”. when the ratio f e c) There are optical telescopes that have diameters about 10 m. There are radio telescopes with single dishes of diameters at least 10 times greater. i) Discuss why, for the same number of incident photos per unit area, radio telescopes need to be much larger than optical telescopes. This ▼ Long wavelengths identied as leading answer the The real to a small energy . is needed radiation to collect 0/1 marks: photons A larger radio the telescopes have radio waves of wavelengths that each that are long and would result in poor resolution. T hus to area compensate is achieved poor answer long-wavelength have have are Because resolution. could [1] for this, radio telescopes have to be larger in equivalent power. diameter . ii) Outline how is it possible for radio telescopes to achieve diameters of ▼ The question needs an answer the order of a thousand kilometres. that refers to single-dish together the use radio into arrangement. an of telescopes linked This answer The many together ’ not T hey phrase small mark. 192 have achieved 0/1 marks: could clear create the dish of large diameter in a huge earth dishes crater a could interferometer ‘combining is [1] multiple enough for and by combining many small dishes together . C.3 C . 3 F I B R E the structure of You should be able to: an optic bre ✔ explain total ✔ the denition graded-index of a step-index bre and the denition ✔ and ✔ the of material meaning waveguide (modal) solve problems total the denition attenuation An optical fibre of the decibel consists of a (dB) very in an the internal ✔ solve problems ✔ describe the optical and bre critical context reection involving advantages twisted-pair ✔ of reection in terms of angle of bre and optics critical angle dispersion dispersion of action a bre of the internal using ✔ O P T I CS O P T I C S You must know: ✔ FIBRE and coaxial attenuation of bre optics over cables. scale. thin core of transparent material. The basic physics behind optical It is surrounded core. The cladding electromagnetic red by radiation cladding is usually radiation are with is a covered shone commonly lower with along used), refractive total a the index protective core internal (both than sheath. light reflection the fibres was covered in Topic 4.4. When and infra- occurs at For total internal reflection, the 1 core–cladding interface so that none of the radiation is lost from the the equation n = , where c sin c core. The glass has very low attenuation at the wavelength used. is the critical angle, follows from Early optical fibres were step-index fibres. However, a graded-index fibre, sinθ n 1 2 = with a gradual reduction in refractive index from the centre to the n , which you met in sinθ 2 outside of the core, has advantages for the transmission process. 1 This Topic 4.4. reduces the effects of waveguide (modal) dispersion bre (a) Step-index bres have a constant cross-section refractive index n in the core with an abrupt change of n between the core and cladding. Figure C.3.1(a) shows the variation bre of n and the passage of two light n rays through a straight bre. One ray is along the axis the other ray distance travels a much longer distance than the rst ray. A single pulse of light (b) will be broadened in time when it reaches the end (Figure C.3.2). This is waveguide (modal) dispersion bre n Graded-index bres can correct waveguide dispersion. The refractive index is not constant in distance the core but varies with distance from the centre, as shown in Figure Figure C.3.1. (a) Step-index and (b) graded-index optical fibres C.3.1(b). The centre of the core has input output power power a larger refractive index compared with the outer par t of the core. Large-angle rays now travel faster when in the outer region of the core, whereas rays in the centre travel more slowly. Smaller core diameters and graded- index materials reduce waveguide time time dispersion signicantly. Figure C.3.2. Changes in the profile of a pulse as it passes along an optical fibre 193 C IM A GING Tra nsmitted Dispersion signals ca n broadens als o the be pulse dist ort e d and, when th roug h one material pulse dispersion overlaps the next, The cause of material dispersion electronic systems cannot separate the two pieces of information. links to chromatic aberration and the basic ideas of refraction. Use Material dispersion arises because blue light that entered the core earlier. these common ideas to help build the refractive index of the core your understanding of these topics. depends on wavelength. Suppose the Material dispersion leads to a spread in the pulse width. The problem digital signal is transmitted using a can be reduced by restricting the ray of white light. The refractive index wavelengths used in the core. for red light is less than that for blue, This restricts the bandwidth and, and so red light travels faster in the ultimately, the number of channels glass. This means that, the red light available in the bre. will exit rst and could ‘catch up’ with The signal overcome as a ratio needs the amplification effects using the of Bel at regular attenuation. The intervals change in along the intensity fibre is to expressed scale The Bel scale is dened as the logarithm to base 10 (log Attenuation in decibel (dB) ) of the 10 I = 10log ratio of the intensity (or power) of a 10 I 0 signal to a reference level of 10 dB is a ten-fold change in intensity. the signal. A ratio of two in intensity is about 3 dB. I Attenuation in bel = log 10 Because intensity is propor tional to I 2 0 amplitude , attenuation can also be An attenuation of 5 bel is a power A 5 ratio of 10 written as: , which is large, so the 20log 10 and are two principal A decibel is frequently used. There 0 causes of attenuation in optical fibres: absorption scattering. Example C.3.1 1 An optical needs to fibre be has an amplified attenuation when the loss power of in 2.6 dB km the signal . The has signal been 15 attenuated The input Deduce optical to 6.0 power the × 10 to W. the maximum optical fibre distance between ratio is 2 each = kilometre of fibre 116 = every 2 194 for this 6 44 7 ≡ 5 × 10 10 log 15 6 as amplifiers 3 power 10 So, 25 mW. fibre. Solution The is 45 km 0 × 10 loses 116 dB . 2.6 dB, an amplifier is needed C.3 Optical fibres physical coaxial are links. external and will, Coaxial carry They the have norm for communication significant advantages channels over O P T I CS involving twisted-pair and cables. A twisted-pair An now FIBRE electrical therefore, cables weak modern arrangement be give signals. optical signal gives will cancelled good only generate similar immunity emfs in from both noise. wires out. immunity However, moderate the to electrical cable is bulky noise. and They are expensive, used to unlike fibres. S AMPLE STUDENT ANS WER Optical fibres can be classified, based on the way the light travels through them, as single-mode or multimode fibres. Multimode fibres can be classified as step-index or graded-index fibres. a) State the main physical dierences between step-index and graded-index bres. This answer could [1] have achieved 1/1 marks: ▲ The Step-index bres have a refractive index for the core and answer description refractive varying index for refractive the cladding. indexes that Graded-index decrease bres outwards bre have from the of the cladding. describing b) Explain why graded-index bres help reduce waveguide dispersion. [2] poorly could have achieved 2/2 of good n. rst that core the The have n is greater than speeds travels since graded long pathways index bres are have made to refractive travel at indices cladding faster that ▲ A complete decrease outwards from the core and as refractive index proportional to speed of light, it causes light which with variation explain travel extra distances to travel at higher speeds and thus have light arrive with the same arrival times, to reduce the that refractive correctly how distance) the the answer is deals inversely are added n core which – marks: that Light – indices cladding could bre answer refractive and and sentence step-index expressed. for the the the means different answer a graded-index core is This the variation ▼ However, to has a the travel high-angle and axial more rays index goes rays on to (shorter slowly (longer than physical smearing distance). of pulses and reduce waveguide dispersion. 195 C IM A GING C . 4 M E D I C A L I M A G I N G You must know: ✔ how medical X-ray You should be able to: X-rays images are ( A H L ) are detected and how the ✔ explain recorded X-ray and half-value ✔ techniques for improving sharpness solve imaging problems involving thickness, in the context attenuation linear and of coefcient, mass absorption and coefcients contrast in the X-ray image ✔ ✔ how ultrasound is generated and explain, the detected in medical in choice how nuclear magnetic resonance is used inside the between how in ✔ to explain solve advantages, ultrasound methods a the use of a gradient eld problems, speed NMR the simple and and disadvantages nuclear be of medical the use ultrasound, of gel and the A and B scans in the context of medical body ultrasound, ✔ context frequency , to ✔ image of contexts difference ✔ the subsequently able assessment to of and magnetic discuss risks of ✔ scanning them involving ultrasound relative intensity explain the spin including of and nuclear levels origin how acoustic through of this magnetic of the leads impedance, tissue and air, and ultrasound relaxation to an resonance of proton emitted signal in (NMR). risk. In X-ray medical imaging, radiation is incident on the patient and Photoelectricity and pair selectively absorbed by bone and tissue. The contrast in transmitted production are described in intensity Topic 12.1. greater X-rays by a is the are recorded density of produced heavy metal on a the photographic material, when target transferred to internal in the of X-ray as they the more high-energy (typically , energy in plate using radiation electrons tungsten). the or target, are Most but a of a computer. is absorbed. rapidly the small The decelerated kinetic amount energy is is emitted The intensity I of a monochromatic form photons. These photons are attenuated or scattered X-ray beam after attenuation is µx I = I pass through the patient or other parts of the equipment. e 0 where I is the original intensity These are some of the mechanisms Compton scattering. At high energies, for attenuation and scattering. an X-ray photon removes an outer- 0 shell electron from an atom and a x is the thickness of the material Photoelectric eect. The photons lower energy photon is emitted. remove inner-shell electrons from µ is the linear absorption atoms. Light is emitted, as other Pair production. Electron–positron electrons lose energy to occupy the pairs can be produced with very shell. Photoelectric scattering provides energetic photons. coecient of the material. The unit of linear absorption 1 coecient is m contrast between tissue and bone. Beam divergence. This causes the The amount of attenuation is I Coherent scattering. This involves intensity of the beam to decrease low-energy X-ray photons. Steps are with distance from the X-ray source. 1 , where an initial = 10log I 10 often taken to remove these photons I 0 intensity I 0 as they degrade image contrast. has decreased to I 1 This is similar to attenuation with The probability of an individual photon being absorbed or scattered optic bres, as covered in Option C.3. is a related to material. its It chance is of interacting analogous to with radioactive an atom. decay , This and is leads constant to a for similar equation. The linear material for µ absorption state; because for coefficient example, their ice densities is and differ. not useful. steam The It have mass depends very on the different absorption values coefficient µ l depends 196 m on the absorber element and is not dependent on density . C.4 The equation that connects µ IM A GING (AHL) Again, by analogy with radioactive and µ l MEDIC AL m µ µx1 decay, this is given by = ln2. I is µ 2 = , where ρ is density. m ρ ln2 x1 This leads to ln2 = = for the 2 µ The unit of mass absorption 2 ρµ l m 1 kg coecient is m two coecients. Half-thickness is dened for a x1 Values of and µ can be obtained l 2 material; it is the thickness required from graphs that show the variation of to reduce the intensity of the X-ray ln I with x beam by one-half. V ery penetrating X-rays, X-rays with with large wavelengths When the material, of and the (µ I = I , small absorption of coefficients coefficients are soft X-rays, with two linear or more thicknesses absorption of coefficients, 1 x l2 2 ) . with different µ 2 +µ are hard 10 pm. 1 nm. penetrates µ and l1 interfaces x l1 absorption x 1 when have wavelengths about beam x X-rays typical are There plane is a and parallel, similar the expression final for intensity mass , and l2 will be absorption e 0 coefficients. Many techniques formed • • by a Fluorescent thin by photons have are elements risk time, Risk are is such hard a the patient large to distances the patient intercontinental radiation than to above by the X-ray absorb and image, low-energy below the photons. patient to remove off- image. to improve the barium contrast photographic are used to using X-ray photons plate. improve contrast in some image. and and whether computation: quick technique However, X-rays that are costs absorbers should be for average kept to in example, chest minimize a than represent short exposure exposure. though; much less and uses X-ray perspective, provides far ionizing radiographer. A radiographer flight, the plate with as to scans. and or used direct sophisticated to are improve plate the interacted that imaging grids blur screens not Heavy more to metal lead that tissues X-ray used photographic Collimation that • a Filtration axis • on an higher dose of X-ray . Example C.4.1 A parallel thickness The beam of X-rays is normally incident on tissue of x incident intensity is I . The intensity leaving the tissue is I 1 The a) half-value Calculate this b) thickness the linear of 2 the tissue attenuation is 2.5 cm. coefficient of the X-rays for tissue. The X-ray beam is incident on a different tissue type. I 2 is smaller for the same x with the second tissue. I 1 197 C IM A GING Compare that c) in the part linear coefficient for this tissue with a). Explain, with drinks liquid a attenuation reference to attenuation containing barium to coefficient, help image why the a patient stomach. Solution µx ⎛ ⎞ I 2 a) Taking logs of the expression: I = I e 2 ⇒ µx = ln 1 ⎟ ⎜ I ⎝ ⎠ 1 ln ( 2 ) 1 So µ = = 0.28 cm 2.5 b) The half-value than c) The the first; stomach therefore, well. and, and is liquid showing Ultrasound linear and small Barium in thickness its similar values a the it µ coats in as metal outline frequencies the tissues for heavy form, for second attenuation of the tissue will be smaller coefficient will be larger. have the large tissue with a high the stomach the organ range 2-20 values does for x the MHz and, attenuate attenuation wall, on not X-rays coefficient absorbing X-rays image. can be used to image the A-scan. The equipment plots a body non-invasively . graph of the variation with time of the reected signal strength. The Ultrasound is generated using the piezoelectric effect. A crystal distance d of an interface from the deforms transducer is related to time t the scan and the speed of sound c in the tissue by d = when a potential difference is applied across it. When the pd on c × 2t . The alternates and at high contracts at frequency , the same the crystal frequency to (or ceramic produce a material) expands longitudinal wave. factor 2 is because the wave travels Ultrasound obeys the normal rules for longitudinal waves and is to and from the reector. reflected, absorbed and attenuated by matter. B-scan. The operator rocks the To transducer from side to side to illuminate all the internal surfaces. a form the image piezoelectric (called a transducer scan) is of placed the in internal contact organs with the of a patient, skin. A gel A computer builds up an image of a between the skin and transducer prevents significant energy loss at slice through the patient from the the air interface. A single pulse of ultrasound is transmitted by the resulting series of A-scans. transducer; pulses T wo reflected varieties Body tissues acoustic The the of transmission by scan reflect impedance intensity between the tissue are and is Z values used: to at an inside and transducer receives patient. the to more complex different B-scan extents. The tissues. interface two the the ultrasounds compare the and the A-scan absorb for stops interfaces used reflected then depends media on the differences concerned. The acoustic impedance Z ρ depends on c and the density of The the tissue: Z = operator cannot simply go to the highest frequency available ρc because attenuation of ultrasound increases with frequency . There is a The unit of acoustic impedance is compromise 2 kg m between image resolution and reflected signal strength. 1 s Other The ratio incident intensity I : reected 0 and uses blood for ultrasound speed using in medicine Doppler shift include and the detection of enhancement blood of flow blood- 2 vessel I intensity I (Z − Z (Z + Z 2 r 1 using microbubbles of gas. = is 2 r I 0 where Z images ) 2 1 ) is the acoustic impedance Ad vantages of ultrasound techniques Disad vantages of ultrasound techniques • Excellent for imaging soft tissue • Limited resolution • Non-invasive • Cannot transmit through bone • Quick and inexpensive • Lungs and digestive system cannot be 1 of the tissue that the wave leaves and Z is the acoustic impedance of 2 the tissue that the wave enters. imaged as the gas in them strongly reflects • No known harmful side-effects 198 C.4 MEDIC AL IM A GING (AHL) Example C.4.2 Topic 9.4 discusses how resolution depends on wavelength. Data of about these the velocity materials are of sound in some materials 1 Gel the density 3 Sound velocity / m s Air and provided. Density / kg m 330 1.3 1400 980 I r The Muscle 1600 equation will be given to I 1100 0 you in an examination as it is not Demonstrate transmitter that and it is the necessary skin of a to use a gel between an ultrasound provided in the data booklet. patient. Solution Using Z ρc, = 2 in kg m the acoustic , 1.4 for air, gel and = 999 muscle, are 6 430, impedances 1 s × 6 10 and 1.8 × 10 respectively . 2 6 I (1 I r 4 × 10 − 430 + 430 ) r from air → muscle is = 0 2 I 6 I 0 0 So almost all of the 4 × 10 1 ( incident energy is ) reflected back to the transmitter. 2 6 I I r (1.8 × 10 (1.8 × 10 6 − 1.4 × 10 ) r from gel → muscle is = = 0 0156 2 I 6 I 0 0 So most Medical of energy magnetic resonance are the (NMR) hydrogen is transmitted resonance to imaging produce (water) 6 + 1.4 × 10 into the (MRI) detailed muscle. uses images ) of nuclear parts of magnetic the body that rich. NMR • Protons • The have spins magnetic and • the field. the a and protons are magnetic radio-frequency Larmor spin Imposing individual When as of charge frequency a behave arranged strong fields (rf) as line magnets. randomly magnetic field — is f and so field there aligns is no net proton spins up. of a applied, particular some frequency — known protons flip so their L magnetic Larmor field is reversed frequency is and directly they enter a proportional high-energy to the state. magnetic The field 7 strength (f /Hz = 4.26 × 10 B, where B is in tesla). L • • The protons now changing, an When rf the low-energy emf field state, precess will is be and, because induced removed, emitting the in a their magnetic conductor high-energy electromagnetic field is nearby . protons signals as revert they to do their so. MRI MRI provides A magnetic original positional gradient field plus data field the is about added gradient origin to field the are of the NMR original arranged signals. uniform so that field. there is The a 199 C IM A GING linear value the variation of the patient. of magnetic Larmor Specific field frequency , values of strength therefore, are f B across also associated the varies with patient. linearly specific The across positions L within The the fields switched patient. are off switched to allow electromagnetic signal strength the signal with on is to protons to acquired allows f initiate a proton relax. by precession As nearby computer to they and relax, coils. The recover then the variation information of the about L the number The of protons computer can emitting then construct Ad vantages of MRI scans • Images with good resolution (down to the a signal spatial at each image point of in the the patient. proton density . Disad vantages of MRI scans • millimetres) Strong magnetic fields can affect hear t pacemakers and metal implants can prevent a good image being produced • No exposure to radiation • The rf currents can give rise to local heating in tissues • No known risk from high-strength • The MRI scanners are very noisy as the magnetic fields switch on and off • The space for the patient is small and can be claustrophobic magnetic fields S AMPLE STUDENT ANS WER ▲ This is an incomplete answer a) Outline the formation of a B-scan in medical ultrasound imaging. that correctly between terms of of B the the A- and B-scans This in answer variant. description obtaining from of the There the is also method signals different of is for the take series them This A computer the of into use multiple A-scans) the of process marks: visual and is omitted 2-dimensional version are sent waves are used to the target tissue and has to create There is a A-scan. the repeated 1-dimensional Ultrasound reections from the from different images. angles, a When 2-D this image can T he ultrasound is created by vibrating crystals using (a AC current which is called piezoelectricity. is image. b) The attenuation values for fat and muscle at dierent X-ray energies an between in this A are shown. [3] answer sentence Muscle attenuation telling Energy of X-rays / keV −1 coefficient / cm the examiner generated in the how but this question. information ultrasound was Do that is not not not −1 coefficient / cm is required 1 2030.9767 3947.2808 5 18.4899 43.8253 10 2.3560 5.5720 20 0.4499 0.8490 give required. A monochromatic X-ray beam of energy 20 keV and intensity I penetrates 0 5.00 cm of fat and then 4.00 cm of muscle. fat 5.00 cm X-ray beam, muscle 4.00 cm l o Calculate the final beam intensity that emerges from the muscle. 200 be is Fat attenuation here. of translate computing difference and 2/3 required reections important B is and nal a waves formed. to achieved (‘repeated angles’). endpoint missing. have a process ▼ The could two-dimensionality B-scan the [3] distinguishes C.4 This answer could have achieved 2/3 ▲ The marks: stages MEDIC AL solution and the is IM A GING carried second out (AHL) in two evaluation −Yx I = I e is correct after an error carried o 3 20 × 10 forward −0.4499×4 × e = in the second line (the 3307 eV interim answer should be 2110). −0.8490×4 T hen I = 3307 × e = 110.81 eV ▼ The use of 20 000 for the intensity 110.81eV ≈ of −1 = shows a misunderstanding 110 eV 1.1 × 10 what is intensity keV as a required. should fraction of I The have . nal been There quoted should not 0 have been any units quoted either. Practice problems for Option C Problem 1 a) Explain the features of the graphs that show An object is placed 12.0 cm from a diverging mirror that attenuation of the signal. has a focal length of 8.0 cm. b) The width of the pulse increases with time. a) Construct a scaled ray diagram for this object and Outline reasons for this increased width. mirror. c) Suggest, with reference to the diagram, why there is b) Estimate, using your diagram, the linear a limit of the pulse frequency that can be transmitted magnification of the image. along a length of optical fibre. c) Comment on the advantages that a parabolic mirror Problem 5 has over a spherical mirror. a) Equal intensities of 15 keV and 30 keV X-rays are Problem 2 incident on a sheet of aluminium. Monochromatic light from a distant point object is 15 keV X-ray half thickness = 0.70 mm 30 keV X-ray half thickness = 3.5 mm incident on a lens and the image is formed on the principal axis. a) Outline the Rayleigh criterion for the resolution of two point sources by an astronomical telescope. Determine the ratio of the intensities of these X-ray beams after passing through an aluminium sheet of thickness 6.0 mm. b) Explain why telescopes with high resolution are usually reflecting rather than refracting b) Explain why low-energy X-radiation is filtered out of a beam for medical use. instruments. Problem 6 Problem 3 A radio telescope has a dish diameter of 40 m. It is used to observe a wavelength of 20 cm. Calculate the smallest distance between two point sources on the Sun that can be distinguished at this a) Draw a ray diagram to show how a converging lens is used as a magnifying glass. b) Explain why the image cannot be formed on a screen. wavelength by the telescope. c) Suggest why a magnifying glass is likely to be better 8 Sun–Ear th distance = 1.5 × 10 km in blue light than red light. d) The converging lens is now used with an illuminated Problem 4 The graph shows the input and output signal powers of object to produce an image on a screen that is four times larger than the object. Determine the focal an optical fibre. length of the lens. rewop 0 0 time 201 A ST R O P H YS I C S D D . 1 S T E L L A R Q U A N T I T I E S You must know: ✔ the objects comets, in the You should be able to: solar constellations, planetary system including nebulae, planets ✔ identify ✔ describe denitions of a single star, binary stars, ✔ (open denition of and denitions ✔ what of galaxies, of balance in a between gravitational force star the astronomical (ly) and unit (AU), the light year the parsec (pc) clusters of galaxies and meant the comment method on its of stellar parallax and limitations galaxies astronomical is use describe ✔ how the pressure nebulae superclusters ✔ Universe globular) ✔ ✔ the stellar ✔ clusters in systems and ✔ objects and by distances are luminosity solve dened and problems brightness and involving luminosity , apparent distance. apparent brightness. The Solar System is a collection of objects held together by gravity . Example D.1.1 Object Outline the nature of a • orbital to orbit the thousands Most and consist frozen approach Sun, frozen a and from include: Sun, the which, planets as a young roughly spinning 4.5 billion star, years had a gas disc that evolved ago • the terrestrial • the gas • a • six planets Mercury , V enus, Earth and Mars giant planets Jupiter, Saturn, Uranus and Neptune rock As Kuiper belt that consists of dwarf planets, including Pluto they planets that have moons the material the gaseous away system with • comet tail the the asteroids, between which Mars and are rocky objects that orbit the Sun in a belt Jupiter that • points solar years years. dust, matter. gases develops Sun from of of the previously releases the into periods the comet. Solution Comets in comets, which are irregular objects consisting of frozen materials, Sun. rock have and dust; highly thousands Stars The form energy begin. As of dust the energy and by with of released, in a energy the Sun’s periodic nebula of energy , enough amounts is gas internal high Large orbits potential and is trapped gravitational times varying field from and years for there is is nuclear are an condensed material forming the energy the a is protostar. fusion released in outwards by gravity . transferred of Eventually , hydrogen the to form radiation of to the helium photons. pressure The processes of fusion are opposing the gravitational force inwards. The star is now stable. covered in Topic 7.2. It 202 remains to years. when temperature are elliptical gravitational kinetic to most stable, on the main sequence, for up to billions of years. D .1 When the hydrogen is used up, other processes take over, and S T E LL A R QUANTITIE S the Stars form various groupings. temperature of the star (and, therefore, its colour) changes. The Binary stars — two stars that eventual endpoint of the star is determined by its initial mass. rotate about a common centre of Nebulae form. are regions Origins for of intergalactic nebulae dust and gas clouds in which stars mass — are thought to make up about half of the stars near to us. are: Stellar clusters are groups of • gas clouds formed 380 000 years after the Big Bang when positive stars held together by gravity. The nuclei attracted electrons to produce hydrogen number in the cluster varies from a • matter ejected from a supernova few dozen to millions. explosion. Open clusters are groups of a few Galaxies are collections of s ta rs , ga s an d dus t gr a vi t a t i ona l l y bou n d . hundred young stars with gas and There are billions of s ta rs an d pla nets wi t h i n each one . Most dust lying between them. galaxies occur thousands for about of in clusters gala x ies . 90% of a ll containing Superclusters g a la xies a nd anyth in g of these for m a between galactic n e t work doz en s c lu s ters of and a ccount f i l ame n t s Globular clusters are much older than open clusters — they were a nd probably formed about 11 billion sheets. Between th e n etwork , sp a ce is a pp a re n t l y e mp ty. years ago. They have many stars and are spherically shaped as their Parallax measurements are used to determine distances to the nearest name implies. stars. As month the Earth period, ‘background’ moves the of across positions fixed of distant a diameter the nearest stars. The of its stars orbit move distance over a six- relative across the to Constellations are groups of stars the baseline is that form a pattern as seen from Ear th. There is no connection two astronomical units (2 AU) so the parallax angle (half the six-month between the stars, gravitational or 1 variation) p is related to the distance d to the star by d , = where d is in otherwise. p parsec and p is in Stellar-parallax arc-seconds. measurements made from the surface of the Earth The distances in astronomy are allow distance estimates up to about 100 pc because turbulence in the very large and involve large powers atmosphere limits the smallest angle that can be measured. When an of ten. Non-SI units are frequently orbiting satellite outside the atmosphere is used, the distance measured used to avoid this. They include the by parallax goes up to 10 000 light years (ly). light year the astronomical unit and the parsec. The light year (ly) is the distance Example D.1.2 travelled by light in one year; A star has a parallax a) Outline what b) Calculate, angle this from parallax Earth angle of 15 0.419 arc-seconds. 1 ly = 9.46 × 10 m. The astronomical unit (AU) is the means. average distance between the Ear th in light years, the distance to the star. 11 and the Sun; 1 AU = 1.50 × 10 c) State stars why less the terrestrial than a few parallax hundred method parsecs can only be used for away . m. The parsec (pc) is dened using parallax angle; a star that is 1 pc from Ear th will subtend a parallax Solution angle of 1 arc-second. a) This is half six-month the angle period subtended when the by Earth is 1 b) For the parallax angle p, d = For the larger distances, distortions the two = by at the Earth extremes of over its a orbit. = 2.39 pc ≡ 7.78 ly. 0.419 parallax introduced at star 1 p c) the the angle becomes atmosphere small produce and large The data booklet form fractional errors in the result. of this equation is 1 d The output intensity at power a of a distance star d is from known the star as is its luminosity known as its L. The star ’s (parsec) = to p (arc-second) apparent remind you of the correct units for brightness b. the quantities d and p 203 D A S T R O P H YS I CS Luminosity L brightness b Example D.1.3 and apparent are connected by the 8 The apparent brightness of star X is 4.6 × 10 2 W m . X has a L equation b luminosity = that is 420 times that of the Sun. 2 4 πd Determine, Knowledge of b and in parsec, the distance of X from the Sun. d allow an estimate of luminosity. 26 Luminosity of the Sun = 3.8 × 10 W Solution 26 The luminosity of X = (3.8 × 10 × 420). 26 L The conversions parsec → light Rearranging b L = gives d 3.8 × 10 = × 420 = 2 8 4 πd year and light year → 4 πb 4π metre are × 4.6 × 10 both given in the data booklet . 17 = 5.3 × 10 m 17 5 3 × 10 15 Use 1 ly ≡ 9.46 × 10 m to convert from m to = ly: 56 ly 15 9 46 × 10 = 17 pc Topic 4.3 shows that the intensity I 56 of a wave at a distance d from a Use 1 pc ≡ 3.26 ly to convert from ly to pc: P source of power P is I 3 26 = 2 4 πd Spica Topic 8.2 shows that the power 4 P = . eσ AT Using luminosity rather than 4 power, the P 1.00 ytisnetni dezilamron output P from a black body of temperature T is (23,000 K) 0.75 Antares (5800 K) 0.25 equation ∝ T (3400 K) the Sun 0.50 4 gives L , since e = 1 for a = σ AT 0 (black body) star. For a spherical 0 star that can be treated as a black 2 body with radius R, L 4 πσ R = 500 1000 1500 2000 4 wavelength/nm T Figure D.1.1. Normalised intensity–wavelength curves for three stars with different temperatures Example D.1.4 Figure with The luminosity of Antares times that of the shows normalised plotted for graphs the Sun, of the Spica variation and of Antares; intensity these graphs is peak 98 000 D.1.1 wavelength at different maximum wavelengths. This links to the Wien Sun. displacement law and gives an indication of surface temperature. R Deduce , where R is The the assumption that a star is a black body is reasonable, and leads to R radius of conclusions Antares and R is in the radius of the Option about star size and surface temperature that are explored D.2. Sun. S AMPLE STUDENT ANS WER Solution Alpha Centauri A and B is a binary star system in the main sequence. 2 Rearrange L = 4 πσ R 4 T to Alpha Centauri A Alpha Centauri B L give R = 4 Luminosity 1.5 L 4 πσ T 0.5 L ⊙ ⊙ 4 R L So, T = L T a) State what is meant by a binary star system. 4 = 98 000 × 5800 3400 This binary radius of Antares is 900 204 have achieved star is a normal the Sun. which is formed fusion of H → from × the of could 0/1 [1] marks: ▼ A binary star is a system in which two fusion process stars orbit each other. The 910 star that answer A The 5300 × R = 5800 Surface temperature / K He. is true for all stars. D. 2 b S T E LL A R C H A R A C T E R I S T I CS AND S T E LL A R E VOLUTION apparent brightness of Alpha Centauri A A = b) i) Calculate [2] b apparent brightness of Alpha Centauri B B This answer could have achieved 0/2 marks: ▼ The student should have L 5 L L b × 5800 × 5300 used 58 the relationship b = 2 = = 4 πd 3 × 2 4 nd 5 L 53 for the two stars. The distance d is b A the 2 same for both, so b ∝ L . The =3.28 Wm temperature is not required at this b B point in the question. 25 ii) The luminosity of the Sun is 3.8 × 10 W. Calculate the radius of alpha Centauri A. This [2] answer could have achieved 0/2 marks: ▼ The student apparent is confusing the brightness–luminosity 25 3.8 × 10 relationship b = 3.28 × b A 3.28 × b B 2 L C H A R A C T E R I S T I C S S T E L L A R EV O L U T I O N stellar what a ✔ spectra Hertzspring–Russell (HR) explain how obtained ✔ explain the ✔ what ✔ the is HR diagram meant meaning by of a indicates Cepheid black hole, stellar variable red surface from how evidence how leads directly A N D temperature the spectrum of of a the star can star diagram shows ✔ which T You should be able to: be ✔ equation 4 R. S T E L L A R You must know: about 4 πσ R = to ✔ the 2 4π d D . 2 with = B giant, stellar for the spectra chemical can provide composition of stars evolution ✔ apply the ✔ sketch mass–luminosity relation star and interpret Hertzsprung–Russell (HR) neutron diagrams star and white dwarf ✔ ✔ the mass–luminosity relationship for identify and sequence the main regions of the HR diagram main describe the properties of stars in these stars regions ✔ the Chandrasekhar limit for the maximum ✔ mass of a white dwarf describe variable ✔ the Oppenheimer–V olkoff limit for the mass neutron star that is not to become a black how to describe the evolution of stars off sequence and the variation of Cepheid determine the role of mass in distance using data on Cepheid stars the ✔ main for hole variable ✔ reason stars of ✔ a the star sketch and interpret evolution pathways of stellar stars with reference to the HR diagram. evolution. Most This stars regions are emit radiation which often fail completely the continuous to of ground state cooler, wavelengths the show ionized spectrum through absorbs characteristic stars of a passes chemical and therefore, absorb before elements absorption and, from their low-density lines has photons internal in the regions. star ’s outer re-emission. Absorption in the because no hot gas cooler the electrons from the gas. The hydrogen to be star ’s lines hottest gas The formation of atomic emission and absorption spectra is covered in Topic 7.1. is promoted out interior. 205 D A S T R O P H YS I CS Spectral analysis of a star is difficult because: Wien’s displacement law links the surface temperature of the star and • the peak wavelength: there can be many superimposed on elements each present and the absorption lines are other 3 λ T = 2.9 × 10 , where T is max • Doppler broadening of the lines occurs because the atoms move the kelvin temperature of the star • surface and λ stars often rotate; one limb of the star approaches an observer while is measured in max the opposite limb moves away , which also causes Doppler shift. metres. Cepheid (the variable light slowly stars curve before show rises the a regular quickly sequence to change maximum repeats). in their emitted luminosity These stars and have intensity then moved falls off the Cepheid variables are standard main sequence D.2.1 includes into the instability strip of the HR diagram. Example candles because they allow an estimate of the distance of the of variable star from Ear th. The the an explanation of the reasons for the variation in output star. variation with pulsation period of the luminosity of Cepheid variables Example D.2.1 is known. When the pulsation Explain the periodic changes in the luminosity of a Cepheid variable. period and apparent brightness b are measured, the distance to the Solution star can be estimated. The luminosity contractions occurs in variation the caused layers of by the periodic star. The expansions pulsation and process because: ① a ② the layer are closer the temperature ③ outer is layer of gas in the becomes star is pulled compressed in and by gravity more opaque because the ions together of the layer increases because more radiation is retained ④ the ⑤ as internal the layer radiation ⑥ the layer between The ① ⑥ pressure expands and falls it and becomes the more layer is pushed transparent, outwards absorbs less cools inwards radiation cycle increases by gravity pressure and as the hydrostatic gravity is equilibrium disturbed. repeats. 6 10 10 R The 1000 R 100 R 5 Hertzsprung–Russell patterns blue giants of stellar (HR) behaviour diagram and represents evolution. The HR 10 . . plot of luminosity against temperature (Figure D.2.1) supergiants 4 with 10 stars features of the according to type, shows the main diagram. red giants Dwarfs 3 grouped 1 R 10 The main sequence consists of stars producing energy giants 2 L / ytisonimul by fusing hydrogen and other light nuclei. About 10 90% of all stars are on the main sequence. They move 1 10 0.1 R along it throughout temperature their change. At life the as their bottom luminosity right and of the HR top left are the Sun 0 10 diagram large are hot small blue cool stars. red The stars. present At the position of is –1 0.01 R 10 shown with an X. .. –2 red dwarfs 10 Red giant stars luminosities have than lower the temperatures Sun — their but surface higher areas and –3 10 0.001 R white dwarfs diameters are much larger than the Sun. –4 10 Supergiant 40000 20000 10000 5000 stars are rare, very brigh t a nd much l a rge r 2300 than red giants. A ty p ical sup erg ia n t e mi t s temperature / K 5 10 Figure D.2.1. red 206 times the power of th e The Her tzsprung–Russell diagram giants or superg ia n ts . Sun . A bou t 1% of st a r s a re D. 2 White They dwarf have down and stars low will are very dense luminosity take and billions of and constitute small years surface to do the area. so. S T E LL A R remains They About of are 9% of C H A R A C T E R I S T I CS old AND S T E LL A R E VOLUTION stars. cooling all stars are You need to be familiar with all the white dwarfs. details of the HR diagram. Note, The instability strip is a region where variable stars are for example, the unusual axes. found. The temperature scale is reversed Lines of constant radius are a set of diagonal lines for constant stellar from the usual direction – it runs radius. The constant lines go from the upper left to the lower right of from high to low. Both axes are the diagram and indicate stars of the same physical size. logarithmic. On Figure D.2.1, the Stars and spend high than a • the • to different times temperature small, more cool can star. massive be This the on main expected is star, the sequence. to burn Stars their with fuel large more mass quickly because: the relative intensity axis goes up in factors of 10 and the temperature axis halves every division. greater the gravitational compression Examiners will expect accuracy in your sketches of HR diagrams. achieve equilibrium, the radiation pressure must be greater to Make sure that the Sun’s position match this compression is correct (it has a relative • the • this temperature must, therefore, be greater intensity of 1 and a temperature of about 5700 K). You do not need makes the rate of fusion greater as the fusion probability increases to draw the stars in the main because the internal energy enables the nucleons to approach closer. sequence as a series of points; The mass–luminosity relationship reflects this argument. a band will do to indicate the 4 Option D.1 outlined the processes by which fusion in a star ) position from about (20 000, 10 begins. 4 As the fusion as protostar to long begin. as it has gains The mass, star its joins hydrogen to temperature the main convert increases sequence to and sufficiently remains down to (2500, 10 for there ). for helium. The mass–luminosity (M–L) When most of the core hydrogen is used, the star moves off the main the inward relationship can be written as: sequence. Outward radiation pressure is no longer equal to 3.5 gravitational forces, and the star shrinks. Another temperature increase L M 3.5 L occurs, allowing the remaining hydrogen in the outer layers to ∝ or M fuse and expand The core, so that the size of the star increases L = M again. where L is the luminosity of the however, continues to shrink and heat up so that heavier Sun and M is its mass. elements — such as continues most to (the the in most star the carbon stable) depends massive iron on and its and oxygen — can stars nickel. so that From form they this by fusion. produce point on, Fusion elements the up evolution of mass. Stars up to 4 solar masses Stars greater than 4 solar masses The core temperature is not high In the red-giant phase of these stars, the core is still large and at a high temperature so that nuclei fuse to enough for fusion beyond carbon. create elements heavier than carbon. As the helium becomes exhausted, The star ends its red-giant phase as a layered structure with elements of decreasing proton number from the the core shrinks while still radiating. centre to the outside. Outer layers of the star are blown Gravitational attraction is opposed by electron degeneracy pressure, but this cannot now stabilize. away as a planetary nebula. With a core larger than the Chandrasekhar limit, electrons and protons combine to produce neutrons and Eventually, the core will have neutrinos. reduced to about the size of Ear th and will contain carbon and oxygen The star collapses and the neutrons rush together to approach as closely as in a nucleus. ions and free electrons. The outer layers collapse inwards too, but when they meet the core they bounce outwards again forming a Electron degeneracy pressure The star is now a white dwarf with 9 a high density (≈10 supernova. The effects of this are to blow the outer layers away leaving what remains of the core as a neutron star. prevents fur ther shrinkage. Neutron degeneracy pressure opposes any gravitational collapse. 3 kg m ) and When the mass of the neutron star is greater than the Oppenheimer–Volkoff limit, then the star will collapse gradually cools. gravitationally forming a black hole. 207 D A S T R O P H YS I CS Black holes form when large neutron stars collapse. Nothing can escape The Chandrasekhar limit states that from a black hole, including photons (hence the name). Matter is the mass of a white dwarf cannot attracted by , and spirals into, a black hole so that the mass of the black be more than 1.4 times the mass of hole increases with time. Observations that may confirm the existence the Sun. of black holes include: The Oppenheimer–Volko limit states that there is a maximum • value for the mass of a neutron radiation emitted because, as the matter spirals it heats up, emitting X-rays star to resist gravitational collapse. • the emission of giant jets of matter by some galaxies; it is suggested The present limit is estimated to be that these are caused by rotating black holes between 1.5 and 3.0 solar masses. Neutron stars with a greater mass • the modification of the trajectories of a star near a black hole by the than the limit will form black holes. gravitational field of the black hole. Example D.2.2 Black holes are discussed in more detail in Option A .5. A main sequence star X has a mass of 2.2 M . The luminosity of the 26 Sun is 3.8 × 10 a) Determine b) i) W. the luminosity of the star. Electron degeneracy pressure Suggest why the time T a star spends on the main sequence is arises because of the Pauli M exclusion principle. Two electrons proportional where to M is the mass of the star. L cannot be in identical quantum ii) Compare the time that X is likely to spend on the main states, so the electrons provide sequence with the time that the Sun is likely to spend on the a repulsion that counters the main sequence. gravitational attraction that attempts to collapse the star. Solution 3.5 L = a) L L which, 3 So with a substitution, 2 5 2 = 16 L the i) star The of has a luminosity number hydrogen fusion The The of total possible atoms energy of transferred by So T total E atoms luminosity the that and is 16 fusions there is × that depends a fixed time energy L and, is the is available a is therefore, measure at available which energy 2 the star for is the of the release released 2M energy initial release number from the sequence likely than mass at initial M which energy E M ∝ L L is rate the = 16 L Sun initial to M = L The the by for on Sun. star. given M ii) the proportional = rate of reaction. number 208 becomes M = b) M to the spend star. 7 3L seven times longer on the main is D. 2 S T E LL A R C H A R A C T E R I S T I CS AND S T E LL A R E VOLUTION S AMPLE STUDENT ANS WER The first graph shows the variation of apparent brightness of a Cepheid star with time. ssenthgirb tnerappa 0 2 4 6 8 10 12 14 16 18 20 22 time / days The second graph shows the average luminosity with period for Cepheid stars. 100000 seitisonimul ralos / ytisonimul 20000 10000 2000 1000 200 100 1 2 5 10 20 50 100 period / days Determine the distance from Ear th to the Cepheid star in parsecs. 25 The luminosity of the Sun is 3.8 × 10 W. 9 The average apparent brightness of the Cepheid star is 1.1 × 10 2 W m . [3] ▼ You This answer could have achieved 0/3 way marks: Here, that often through you need a to plan problem need to professional in follow your Option the D. steps astronomers 26 L 3.8 × use 10 to determine the distance to a 9 b = 1.1 × 10 = Cepheid 2 variable. 2 4 πd 4 πd Use the rst graph to estimate the 26 3.8 × 10 periodic 9 d = 1.1 × 10 2 time of the star. 26 4πd = 3.8×10 8 1.38 × 10 Then 26 3.8 × the second luminosity the Sun) graph (relative to to nd that of 10 2 d use the = for this time period; and 9 1.1 × 10 × 4π hence the luminosity in W 17 d = 1.65 × 10 m = 0.07 pc You know now L and b so you can star and L calculate d using b = 2 4 πd The answer Sun quantities to use the here confuses and graphs does at not appear all. 209 D A S T R O P H YS I CS D . 3 C O S M O L O G Y You must know: You should be able to: ✔ ✔ Hubble’s law describe Big ✔ how to origin describe of the the Big Bang model of space what is meant background ✔ what that ✔ the is meant redshift by ✔ cosmic (CMB) Z denition by is of accelerating consequence cosmic as originating with the describe CMB radiation evidence for and a explain Hot Big how Bang it model microwave radiation the a time the Universe provides ✔ and Bang scale Universe of and ✔ solve ✔ estimate this factor the problems the involving age expansion of rate the of Hubble’s Universe the law, Z and assuming Universe is R that constant. R American astronomer Edwin Hubble compared galactic light spectra Hubble’s law states that the with spectra obtained in an Earth laboratory . The galactic spectra recessional speed v of a galaxy were shifted to red wavelengths as expected from the Doppler effect. away from the Ear th is directly However, when distances, he he used Cepheid variables to determine the galactic propor tional to its distance d from Ear th: distance. v = H found This led that to the amount Hubble’s of redshift depended on the law d 0 where H Hubble is the constant of showed that galaxies are moving apart. The present model 0 propor tionality, called the Hubble of the Universe 13.7 billion constant. was is years immensely that ago space in small a and Hot (less time Big than came Bang. the At size of into this an existence instant, atom) about the and at Universe a Galactic speeds are usually 32 temperature 1 measured in km s of 10 K. Within one second, the Universe had cooled with distances 10 in Mpc; this leads to a modern value 1 for H of about 70 km s to 10 K and was rapidly expanding. Since then, the Universe has 1 Mpc continued to cool to its present temperature of 2.8 K. 0 Hubble’s and • • law assumptions Hubble’s the light • the law to than) the an the estimate estimating true reach recessional less in is from Universe The speed here is a recessional allows for most all of the the age of Universe the age Universe. The steps are: times distant galaxy has taken the age of the us speed speed of of this galaxy from Earth is constant T is and (just light. speed, not an actual speed. The Hubble equation becomes = 0 ( cT ), where the age of the Always use the term recessional 1 in this context. Do not fall into the and T Universe = H 0 trap of imagining that the Ear th is stationary. Special relativity tells 1 With the value for H of 70 km s 1 Mpc , the estimate for the age of 0 us that there is no place in the Universe is about 14 billion years. Universe where absolute speed can be considered to be zero. Example D.3.1 The variation clusters a) is Identify galactic b) 210 with distance of the recessional speed of galactic plotted. one method for determining the distance to cluster. Determine, using the graph, the age of the Universe. a the D. 3 C O S M O LO G Y Solution a) A suitable the use a (The a standard Cepheid use of 15 be candle variable parallax smk star. of as might 1– such method is for distances 01× 3 unsuitable 10 to galaxies.) 5 v b) v = H d ⇒ H 0 , = which 0 d 0 is the gradient of the graph 0 2 4 provided. 6 24 distance / × 10 m 6 (16 × 10 − 0 ) 18 1 The = is gradient/s 2 7 × 10 24 (6 × 10 − 0 ) 1 17 The age of the Universe is approximately = 3.8 × 10 s. H 0 Further evidence background According for (CMB) to the the Hot Big radiation. Bang This model was is the predicted cosmic in the microwave 1940s. model: You should be clear about the • about 400 000 temperature years had after fallen the to Universe formed, the nature of the expansion. The Universe Universe is not expanding into 3000 K anything — the fabric of space is • charged ionic matter was able to form neutral atoms with changing. The galaxies move apar t free electrons for the first time — space became transparent to because the space between them electromagnetic radiation instead absorbed and photons could escape in all directions is stretching. Electromagnetic of being radiation takes time to arrive • since then, longer the wavelength wavelength intensity peak at a and of this the photons black-body wavelength of has shifted radiation 0.07 m in the to now a much has microwave an region. from a distant galaxy and, during that time, the light wavelengths stretch so that the redshift is cosmological in origin. For this Satellite measurements have since confirmed that the CMB radiation reason, take care to use the term exists, and that it matches the expected profile of black-body radiation. cosmological redshift The (it radiation comes is from However, homogeneous all small directions variations and and in is the almost the CMB completely same are from now all isotropic directions). known to exist. Example D.3.2 The variation microwave of intensity background with wavelength (CMB) is shown for the cosmic radiation. 1 ytisnetni evitaler 0.5 0 0 1 2 3 wavelength / mm a) Estimate the assumption temperature you make in of the this CMB radiation. State one estimate. 211 D A S T R O P H YS I CS b) Explain how Hot Bang Big this temperature supports the theory of the model. Solution a) Assume The that peak Using the spectrum wavelength Wien’s is represents about displacement black-body 1.1 mm. law: −3 3 −3 2.90 × 10 2.90 × 10 λ radiation. T and (metre) = 2.90 × 10 = = max 3 T λ (kelvin) 1.1 × 10 max This be b) gives the The model was at a as is scale of the the a at that of low As this time the and of cosmological an Big wavelengths black-body has the due to of time, 3 K will Universe Universe peak to its to increased wavelength wavelengths the space the the corresponding radiation decreased predictions effect estimate, Bang, with increased. A long temperature the the then, temperature. As to is answer. the the Universe a corresponds is that, 2.6 K. the temperature meant to of specify frequencies lengthened, Redshift high This equivalent which to suggests photon the temperature way very expanded. the a best have present value model. stretching. It is not related Example D.3.3 to the relative Doppler A galaxy has a relative 4 of 4.6 × to the in astrophysics. wavelength ∆λ the size of relative when observer electromagnetic (unlike Doppler the standard The ratio of the wavelength equation change to the can be original is given the symbol z: the to was light v its = ≈ λ present from c the As galaxy the and the z size so, source km s Determine Universe Even the Earth 1 10 effect). of recessional used velocity velocities the Universe expands, all distances are increased by the same emitted. cosmic scale factor R. In other words, light of λ wavelength emitted by 0 Solution For this a galaxy , galaxy has will be changed received from R to with a wavelength λ as the cosmic scale factor R 0 7 v z ≈ 4 6 × 10 3 0 × 10 ∆λ = = 0 153 So z = ∆R R = − 1 = 8 c λ R R 0 The R Rearranging z = existence 0 of CMB radiation was confirmed by Penzias and Wilson − 1 gives when R a microwave signal, picked up by their microwave antenna, 0 proved R R R = signal = = to be was CMB due to radiation a fault in from their space. They equipment. originally Serendipity thought is that the sometimes 0.87 R 0 ( z + 1) Therefore, Universe present emitted 1.153 the was value the very size of the 87% of its when light. the important There is now accelerating. galaxy estimate the for mass gravitational a has 212 evidence Type Ia up source been to tend energy named dark 1 Gpc; the attraction of the objects. of would that expansion supernovae these acceleration Normal be science. distances determined for in be of cosmological These the used as Universe is standard candles redshifts measurements can also provide to be evidence expansion. to so, that can if slow the we energy down any expansion have not yet expansion is through accelerating, discovered or there must observed. This D. 3 C O S M O LO G Y S AMPLE STUDENT ANS WER ▼ z Light reaching Ear th from quasar 3C273 has z is certainly that reects the cosmological a) Outline what is meant by z. answer could have factor existence redshift; of however, z [1] is This a = 0.16. achieved 0/1 marks: not the It is the the change in wavelength the change in dimensions universe Red-shift redshift. to the present ratio due of of to the value of wavelength. b) Calculate the ratio of the size of the Universe when light was emitted by the quasar to the present size of the Universe. This answer could have achieved 1/1 [1] marks: 15 ▲ The 1.3 × 10 R z = answer is correct. − 1 15 1.2 × 10 o −15 (0.16 + 1)(1.2 × 10 ▼ However, 13 ) = R R = 1.39 × the solution is 10 poorly to laid invent a out. There value for R is no as need only the c) Calculate the distance of 3C273 from Ear th using 1 H = 68 km s R 1 Mpc . [2] ratio o is required. R This answer could have achieved 2/2 marks: 0 V z = G 8 0.16 × 3.00 × 10 = ▲ The v easy V = H to solution is correct and follow. d o 8 0.16 × 3.00 × 10 68000 d) Explain how cosmic microwave background (CMB) radiation provides suppor t for the Hot Big Bang model. This answer CMB could have radiation achieved provides [2] 1/2 marks:: support for the Hot Big Bang model ▲ The because the microwaves are getting larger proving that CMB support Universe the is expanding Universe was very and very the hot CMB when are it proof started ▼ The idea expanding only fall gain into scheme ideas. what do a not sentence the one are the start and of the for of the is in expansion it of is a provides the consequence of the Universe. is a to mark about their The clear expansion Universe same easy a microwaves give the is can separate issues increases. not the but It Points also that start point Universe twice, longer ’— it does one once. trap. written: ‘get the awarded wavelength that that mark There is at as expanding. occurs this are that radiation the were last view and at time. 213 D A S T R O P H YS I CS D . 4 S T E L L A R P R O C E S S E S You should be able to: You must know: ✔ the ✔ about off Jeans the the criterion process main for of star formation nucleosynthesis occurring ✔ apply ✔ apply the various place in qualitative for nuclear stars off fusion the main description neutron of reactions sequence the s and that r about to type Jeans the the criterion to the mass–luminosity of stars on the formation equation main to sequence of a star compare relative Sun take including ✔ a describe the reactions processes different taking types place off of the nuclear main fusion sequence capture ✔ ✔ the lifetimes sequence to ✔ ( A H L ) Ia and distinguish type II between supernovae and describe in how stars, how including temperature them. Stars form from dust and elements gas heavier details of than the iron form required increases. clouds that can be stable for millions The Jeans criterion can be expressed of in terms of the Jeans mass M years until some event disturbs them, such as an interaction with below J another cloud. A large cloud then becomes unstable and collapses. As which stars cannot form, or in terms it does so, the temperature rises and, if it rises enough, nuclear fusion of the energies involved. will begin. The magnitude of the gravitational potential energy of the gas cloud For a small must be greater than the kinetic (travelling energy of the cloud for the gas cloud quickly to collapse. In practice, a cold dense gas (small kinetic, large gravitational long and for before cloud, at when speed restore the the the waves sound collapse of stability . to reach has begins, sound) time can When the to the region. restore pressure cross the cloud is Gravity waves in collapsing large, will it the gas region will take compress too the gas stability . potential energy) will be more likely This is incorporated in the Jeans criterion for a critical mass below which to collapse than a hot, low-density gas (high kinetic, small gravitational a star cannot form. potential energy). In For a cold dense gas, the mass limit stars the larger CNO than cycle the occurs. Sun, The when CNO the core cycle has temperature 6 exceeds 20 MK, steps: will be lower than for the hot diuse ① a proton fuses ② the unstable ③ the C-13 ④ the N-14 ⑤ the unstable ⑥ the N-15 with a carbon-12 nucleus to give nitrogen-13 (N-13) cloud and the lower the mass limit, the more likely star formation will be. N-13 fuses decays with fuses a with via proton a positron to proton to give give emission into carbon-13 (C-13) N-14 oxygen-15 (O-15) In Sun-like stars, the fusion O-15 decays via positron emission into N-15 process is the proton–proton chain. This has already been described in fuses with a proton to give C-12 emitting He-4. Topic 7.2. At each stage, energy is The overall The C-12 reappears change The C-12 in is again the fusion of four protons to give star used He-4. released. These fusion processes are connected to the plot of the the hydrogen and shrinking as is what temperature CNO its the causes at the cycle core is will arises the the to star process. because the transformed pressure expansion move of mostly radiation the end the off falls and, red-giant the main to has helium. therefore, phase. up The it The its core heats is up. increase This in sequence. magnitude of binding energy per Further nucleosynthes is now begins. nucleon against nucleon number. Iron and nickel are at the peak of • Two helium nuclei fuse to produce beryllium-8 • Another helium fuses with the beryllium • Another helium fuses with the carbon (Be-8). the plot and represent the most to form carbon-12. stable nucleons. Use this work in Option D as an oppor tunity to form oxygen-16. to review and consolidate your This process continues with heavier and heavier elements being understanding of Topic 7.2. produced 214 until the most stable nuclei (iron-56 and nickel-62) are reached. D. 4 The process elements capture. beyond increases a the nucleus decay Further nickel. are nucleus via to negative neutron be by number). captures to can via The decay . absorbed elements one It be are electrostatic captured de-excite. beta must heaviest subject number proton photon energy The not and nucleon (its gamma-ray to because Neutrons approach the stops the but new may This nucleus be a by change will new (AHL) neutron and nuclear unstable, forms produce created not P R O C E SS E S synthesise attraction strong does to S T E LL A R can force. the generally allowing This nature a of emit a neutron element. increasingly heavy elements. Slow neutron capture (s process) There are two processes: the s process and the r process. occurs in massive stars. It leads to the production of heavy nuclides up Example D.4.1 to those of bismuth-209. Massive A star with a mass of the Sun moves off the main sequence. stars have only a small neutron flux arising from the fusion of carbon, a) Outline the nucleosynthesis processes that occur in the core of silicon and oxygen. Unstable the star before and after it leaves the main sequence. nuclides produced by neutron b) Outline the subsequent evolution of: capture thus have time to decay by negative beta emission before i) this star fur ther neutron capture occurs. ii) a star with a much larger mass. Rapid neutron capture(r process) Solution occurs when the new nuclei do not a) Before leaving the main sequence, the core is fusing hydrogen have time to undergo beta decay to form of the helium. Leaving the main sequence occurs when much before neutron capture happens for hydrogen fuel is consumed. After leaving, the core is a second time. The heavier nucleons fusing helium to form the element carbon (after a chain of fusion rapidly build up, one at a time. Type reactions). II supernovae have a high neutron b) i) The star planetary will form nebula a red giant. followed by a This will white eventually become flux and they can produce nuclides a heavier than bismuth-209 in a few dwarf. minutes before beta decay has ii) A much more massive star will form a red supergiant. This time to act. The high neutrino flux will become a supernova and will then become either a neutron in a supernova can also lead to the star or a black hole. creation of new elements via the weak interaction and conversion of a The more massive the star, the shorter is its lifetime. This sounds neutron into a proton. counter-intuitive but the lifetime T of a star of mass M compared with 2.5 T that of the (lifetime T Sun with M mass ) is . = T of the the Sun’s reason and radiation The be is Ia attracts dwarf. (1.4 into and be expected a When solar the shorter can its to to have type occur gravity causes such a in or when white on a higher lifetime a the dwarf it high to core used night of 0.3% of that temperature no up The more fusion rate quickly . sequence. sky . They brightness suddenly before. II white usually is main was type the equilibrium. core the there I companion, generating the time where as the maintain in regularly accreting masses), nickel, a classified from star, hydrogen position supernovae the required the observed at are mass larger spends are bright Supernovae the pressure star Supernovae appear that higher larger T ype will 10 Sun. The must mass A star M times M a dwarf giant reaches collapse. radiation in star the a binary or star another white Chandrasekhar Carbon pressure and that oxygen the star limit fuse is 10 blown apart. It reaches luminosities of around 10 times that of the Sun. 215 D A S T R O P H YS I CS The reaction always occurs at this limiting mass and, therefore, we know Type I supernovae have no the luminosity . A comparison with its apparent brightness provides an hydrogen in their line spectrum. estimate of the distance to the galaxy that contains the supernovae. They are the product of old stars of low mass. The type I supernovae The are fur ther classified as Ia, Ib, Ic, years, remnants and so on, depending on other material of the eventually for new supernovae merging with expand the outwards interstellar for thousands material to of provide the stars. spectral features. Type Type II supernovae have hydrogen to II the supernovae type 1a explode through a completely different mechanism process. in their line spectrum because they ① After all the core hydrogen has been fused, hydrogen can only fuse are young stars with large mass. in a shell point surrounding where the the (now) temperature helium will core. sustain The helium core collapses fusion to to carbon a and 6 oxygen — it Example D.4.2 ② Describe how a type The core can be 10 years collapses to again, use all the allowing helium. carbon fusion to even heavier Ia elements supernova takes used with the usual temperature increase. This stage takes about to 4 10 determine galactic years. distances. ③ The repetition of collapse–fusion continues with shorter time periods Solution A Type Ia massive supernova star in a is each time. Eventually , now there is silicon fuses to iron-56 taking just a few days. By a almost no radiation pressure and gravity is the dominant distant force. galaxy and that has has luminosity on Earth. that The brightness be of supernovae mass explode have is large ④ observed apparent The when Type Ia to do star all a means Chandrasekhar cannot neutrons that wave. so, prevent resists As heavy surface, this and the is neutrinos wave passes are reached, collapse; collapse elements the limit further leading in the temperature rises As to star to form through formed. electron the a of the the neutron an star, and degeneracy outwards-moving taking shock 20 000 K degeneracy implodes, wave the a few hours reaches star the explodes as supernova. luminosity . The This the pressure shock the stars therefore, same can have the When pressure producing event always and, the a this measured. same exploded acquired that they two types of supernovae can be distinguished by observation. act 10 Type as standard candles; Ia have brightness typically 10 that of the Sun. Their the brightness apparent luminosities quickly reaches its maximum and then falls gradually over and 9 about this luminosity can be Sun. to calculate to the the six Their supernova in is Type brightness II have initially luminosities peaks and typically then falls to 10 a that of slightly the lower distance plateau galaxy months. used which for some days, and then falls rapidly . the located. S AMPLE STUDENT ANS WER a) Describe how some white dwarf stars become type la supernovae. ▼ The on the answer binary system and needs nature the way to of in focus the This the from orbiting the other stars which accretes until Chandrasekhar it answer could have achieved 0/3 marks: star one If of [3] the star has a mass of more than 3Mo, it has the chance of mass exceeds the becoming a type la supernovae. limit. b) Hence, explain why a type la supernova is used as a standard ▼ There standard is luminosity as a no sense candle and standard has that that a it amount candle. a predictable can of be This compared brightness of estimation is 216 an with a could have achieved 0/2 marks: energy . the object, possible. answer used It When [2] is used as a standard candle because of its great mass and apparent distance size and due to its brightness, other stars are compared to it. D. 5 D . 5 F U R T H E R C O S M O L O G Y the ✔ cosmological models of what meant is relates ✔ what to is the the principle and its role ✔ in describe in Universe by a mass meant by C O S M O LO G Y (AHL) ( A H L ) You should be able to: You must know: ✔ FURTHER rotation of a dark curve and how ✔ it describe dark galaxy ✔ matter the models cosmological of the why principle and its role Universe rotation curves are evidence for matter derive rotational velocity from Newtonian gravitation ✔ that there ✔ the ✔ what are uctuations cosmological is meant origin by in of critical the CMB ✔ derive ✔ describe what ✔ how is meant by dark sketch variation Einstein of and interpret cosmic extended his from Newtonian gravitation the and interpret the observed anisotropies CMB energy ✔ to density density in ✔ critical redshift scale graphs factor general of with theory of describe the energy how affects the the presence value of or otherwise the cosmic of dark scale factor. time. relativity to cosmology by The cosmological principle is that making two simplifying assumptions that have since been shown to the Universe is: apply on a large scale. This cosmological principle, together with the ① general theory of relativity , can be used to show that matter homogeneous (which means it distorts is the same everywhere, which is spacetime in one of only three possible ways. true ignoring the relatively small ① Positive to a curvature, return spacetime to the where original world as travel position deforming to through in a the Universe spacetime could (imagine a flat presence of galaxies) lead 2D ② sphere). isotropic (which means it appears the same in whichever direction we look). ② Negative lead to a curvature, return to the surface deforming ③ curvature to where travel original a through position in the Universe spacetime would (imagine a never flat saddle) The critical density is given by 2 Zero where travel through the Universe would never 3H 0 ρ return (a flat 2D surface remains = c flat). 8πG The appropriate model depends on the density of matter in The ratio of the actual Universe the density to the critical density is Universe and its value relative to a critical density value ρ that c ρ maintains zero curvature. given by Ω = 0 ρ c The theory of the Hot Big Bang model suggests that, after an initial • If Ω = 1, a at Universe results 0 inflationary period following the Big Bang, the expansion rate of the and the Universe continues to Universe has been decreasing. However, data from type 1a supernovae expand to a maximum limit at a suggest dark that there may be an acceleration energy . in the expansion caused by decreasing expansion rate. 4 • If Ω accelerated < 1, the Universe would be 0 the critical Figure with the of from various is important density D.5.1 time factor it shows R (the for an the the to variation D.3) It know Universe. cosmic Option scenarios. curve of scale for these includes )R( rotcaf elacs cimsoc Therefore, Universe 3 open and would expand forever. Ω < 1 0 • If Ω 2 Ω = 1 0 eventually stop, followed by a collapse and a Big Crunch. 1 Ω > 1 0 accelerated A simple derivation of the 0 –10 Universe in which the effect of > 1, the Universe would 0 be closed and expansion would now 10 20 the 30 critical density is given in example 9 (hypothetical) dark energy time / 10 exceeds years D.5.1. A rigorous derivation of ρ c that of the baryonic gravitational matter and effects dark of matter. Figure D.5.1. Variation of R for different density parameters requires general relativity theory and is beyond the scope of the DP physics course. 217 D A S T R O P H YS I CS As Example space law Show, by equating the wavelength of radiation also expands. gravitational states that λ T is max constant. There is no evidence that kinetic Wien’s the shape 1 of and expands, D.5.1 the black-body spectrum has changed with time so T . ∝ The potential λ 2 3H wavelength will scale by R together with all other dimensions and 0 energies, that ρ = for c 1 8 πG therefore T the ∝ Universe. R Evidence for the existence of dark matter comes from galactic rotation Solution curves. The Universe will expand providing arms. enough make the kinetic total energy E = the within the bulk of the galaxy or in one of the Newton’s law of gravitation can be used to less determine the energy , limiting rotation speed of the star for both options. to E Assuming T, positive; lie there predicted is can continue dense to Stars case is at a distance spherical r from the shape for centre, 0. will within its orbital radius galactic be hub, affected by the the star, when galactic orbiting mass 3 4 T the ρ, πr ie ρ where is the galactic density . 3 For the E Universe, = E T + E K 2 P E = T mv 2 = r . So v ∝ r 3 r r . − gives v and 2 2 1 and 4 πG ρ mv GMm Equating GMm 2 r For The radius of the a star outside the galactic mass in a spiral arm, on the other hand, Universe 1 is given by v = H r and the v 0 4 mass πr is ∝ r 3 ρm therefore 3 3 4 ρ πr 2 1 mH 2 r = two speed with results give a predicted graph for the variation of orbital and G 0 2 These m c 3 radius, as shown in Figure D.5.2. r 2 3H The observed result — the rotation curve for a real galaxy — is different. 0 ρ = c Stars well outside the galaxy are moving with the same speed as those 8 πG inside. at the This A proposed outer rim material of explanation the cannot is that dark matter forms a shell or halo galaxy . be detected as it does not radiate. 200 Other evidence for dark matter is: obser ved 150 • Some galaxies s mk / v 1– masses of far bright in these clusters galaxies orbit to be each other. estimated. Observations However, the allow the galaxies are 100 less than the mass measurements indicate. predicted • 50 Radiation lensed. 0 would 0 10 20 30 40 be through from distant expected from or near quasars the massive is much luminous objects more mass Observed versus predicted rotation curves for a galaxy galaxy . • It has more Images of hot bound gas than The to ratio be of mass elliptical expected gravitationally contained than within the to visible galaxies the for than the expected. made galaxies. gas mass to to The be dark in the X-ray galaxies trapped mass in in the region must this show have more halos of mass way . galaxies is thought 1 : 9. Candidates particles for dark (WIMPs) matter, and at present, massive are compact weakly halo interacting objects WIMPs massive (MACHOs). MACHOs Non-baryonic subatomic par ticles that have These include the neutron stars, black holes different proper ties from ordinary matter. and small brown dwarf stars. These are They must interact only weakly with normal high-density objects and can be detected by matter and there needs to be considerable gravitational lensing, but they are unlikely to quantities of WIMP material. The theory relies be present in large enough numbers to provide on hypothetical par ticles not yet observed. the amount of dark matter required throughout the Universe. 218 is distorted 60 radius / kpc Figure D.5.2. passing Light D. 5 Evidence Universe to dark that Some and, 5% The of the 68% of space the the incorporate Satellite suggest of as the images of CMB These dark the to be energy to this into is of radiation temperature a present the of appeared during of be due suggest dark be this model theories of to model. years cosmic could required Bang the spacetime This However, 20 in to that being energy . existing Big (AHL) energy . may past fluctuations 27% C O S M O LO G Y early data property accelerate. over the thought with dark hypotheses the is dark Modification new that provided matter amount Universe variations has thought definitive. of today; baryonic that does suggestions anisotropies — small radiation. so development these is than suggests mission Universe expands, regarded or slowly Universe expansion be gravity more supernovae ESA Planck the astronomers cannot distant expanded and as cause very energy . only matter from FURTHER have revealed In Option D.3 there was the background suggestion that the CMB was largely inflation — a 36 period of accelerated expansion that took place from 10 s to roughly isotropic. This section looks more 32 10 s after occurred have seen radiation mark Big during become patterns the the and Bang. this epoch magnified in the when that and into satellite the have Quantum these the frozen (at galactic images Universe been fluctuations are time) clusters the thought minute observed differences became into that are that transparent fabric of the at to closely at that assumption. have differences today . were the The present in 400 000-year Universe. S AMPLE STUDENT ANS WER The graph shows the observed orbital velocities of stars in a galaxy against their distance from the centre of the galaxy. The core of the galaxy has a radius of 4.0 kpc. 1– s mk / yticolev radius / kpc ▼ The a) Calculate the rotation velocity of stars 4.0 kpc from the centre of the 21 galaxy. The average density of the galaxy is 5.0 × 10 correct . [2] , answer could have achieved 0/2 = −11 v r = ν × 6.67 × 10 errors 5.0 × 10 there × is × 30 10 ν = 1.39 × 10 its 4 × the 4 kpc converted unit quoted to for metres, the to the factor 4 kpc does not the [2] could have achieved 0/2 give evidence incorporated. answer does not rotation curves and and the express predicted (assuming the for dark matter because they the that can addition of contribution. velocities between observed no dark rotation marks: curves cur ves been difference matter) answer have 10 matter. the use: not and ▼ The show number 3 × b) Explain why the rotation curves are evidence for the existence of dark Rotation a 3 15 This are 4 speed, 3.5 in is no appear = there −21 × = 3 v but 3 distance 4π quoted, r marks: of 4 πGρ is 4 πG ρ v This equation 3 kg m rotation signals emitted by the be a modelled dark The fact with matter that the dark predicted is the curves are not observed evidence. matter . 219 D A S T R O P H YS I CS Practice problems for Option D Problem 1 Problem 4 The cosmic microwave background (CMB) radiation Iron is one of the most stable elements in terms of its corresponds to a temperature of 2.8 K . nuclear instability. a) Estimate the peak wavelength of this radiation. Explain how neutron capture can lead to the production of elements with proton numbers greater than b) Identify t wo other features of the CMB radiation that that of iron. the Hot Big Bang model predicts. Problem 5 c) The cosmic scale factor has changed by about 1000 a) Describe two characteristics of a red supergiant since the emission of the CMB radiation. star. Estimate, for the time when the CMB was emitted, the b) Explain what is meant by a constellation. wavelength of a spectral line of present wavelength Problem 6 21 cm. The luminosity of star X is 98 000 times that of the Sun Problem 2 7 and it has an apparent brightness of 1.1 × 10 The present-day value of the Hubble constant H 2 W m is 0 1 72 km s 1 Mpc 26 Luminosity of Sun = 3.9 × 10 . W 1 a) Outline the significance of . a) Calculate, in pc, the distance of X from Ear th. H 0 b) State an appropriate method for measuring the b) A galaxy emits light of wavelength 500 nm. This light distance of X from Ear th. is observed on Ear th to have a wavelength of 430 nm. c) Another star Y has a luminosity that is the same Deduce the distance of the galaxy from Ear th. as X and is on the main sequence. Problem 3 Deduce the mass of star Y relative to the mass of the a) State the Jeans criterion for the formation of a Sun M star. . d) Suggest the subsequent evolution of star Y. b) Outline why a cold dense cloud of interstellar gas is more likely to form new stars than a hot diffuse cloud. 220 INTERNAL During your assessment course project you (IA). will do This an ASSESSMENT internal project takes Choosing a topic and personal engagement around You 10 hours and is an integral part of the need to engagement It accounts for 20% of your total mark in the your with your IA. have a high level of or You express may show scientific creativity concepts in a course. type of project is flexible. It might experimental data collection from laboratory work you investigation • spreadsheet and way . a project you if will you investigate that be are a some is based more on likely woodwind aspect of a to personal succeed. player, the you physics of analysis this processing then example, might database engaged choose interest, For • and be: If • you thinking, personal The that DP in physics show assessment. with data you instrument. The key is to think creatively . have Sometimes, the aspects can be developed in collected laboratory • the investigation of an experimental simulation. data on what Collaboration between students is allowed in stages of the IA when data are being you must carry out the data Internet others have (words that Where you earn as many of as possible, you should pay the to following the assessment but do not books. use suitable Look anyone at data) without giving full else’s credit to try before to carry the out IA time preliminary begins. This available work should appear with evaluative particular comments attention or possible, preliminary marks done, in be author. experiments ensure published may analysis independently . To or there collected. the However, the alternatively , the work early work; criteria, given in in your final report. the table. Criterion Weighting Personal engagement 8% Exploration 25% When writing your introduction, ensure that Analysis 25% Evaluation 25% Communication ✔ your personal engagement is clear ✔ a research question is given that is meaningful, relevant and achievable. 17% ✔ you use technical language ✔ you include any useful diagrams ✔ you do not assume any specialist prior knowledge and any symbols given are defined. You may be produced allowed in your to see school examples in previous of IA work years. If so, Try look carefully at the quality of work that to make pervades high marks and, in contrast, identify why that or gained you so of are low seeing work thinking marks may and be the scored final some of missing. poorly . report the after a two reflect the refinements in both good methodology that occurred as personal and that engagement this throughout the engagement is report. Exploration report need to decide on a research question. It must thinking be and your work weeks You should that your Remember developmental However, all students clear who sure gained the focused and identify all the issues linked to it. project The physics concepts and techniques should arise progressed. naturally from the topic and not be forced into it. 221 INTERN AL When you question have is A SS E SS M E N T chosen well your topic, make sure your Here Bad some ▲ question What your Good are drop happens through when spheres How water? does sphere will question be should investigating, dependent variable the vary terminal with speed of yourself you propose the to temperature? • Are clearly with identify the what independent Is the other and and Can and Is there variable(s) variable(s)? how variables will you measure questions about Is the Can what What will and you keep will you achieve this? continuous vary it? the How or discrete? independent will you variable measure the variable? experimental safety of is an it appropriate too easy or standard too What achieve leave your time plans for a in do be scope for method safe in all respects? improved? the of by predict a as sensible the outcome? Is this theory? hard)? time thorough refinement you informed (in What graphs prediction available negative analysis? confirms • independent do. words, still consider. expressed. following physics you and dependent • • the How you • • might a • Ask you question • Your your dependent constant What things defined. • ▼ are the IA as is will help you to know correct/incorrect? result your is just as significant prediction. An that Notice as IA does the that one not a that “fail” it because the research question has not been progresses? confirmed. • Are your firm • Will proposed experiments likely to lead to conclusions? you be as engaged with the project and its These can be summarised as follows. outcomes on the last day as on the first? ✔ If you can answer ‘yes’ to all these, then Make a prediction and know what graphs/display your techniques will demonstrate it research task is suitable for you. ✔ Planning an IA involves choosing the apparatus methodology . outset If you your and You modify choose research an much and should these the in the will than predictions light of topic, allow your ensure the clearly and effectively ✔ from appropriate data. Will it allow you Manipulate data accurately and use them to verify/ the deny your prediction, leading to a new refined work. prediction. that collection This of Make measurements safely and display them just experimental make experimental question more to access cycle is sometimes known as the scientific a method: continuously sensible variable independent dependent variable? variable Can your with a variables analyse be made to vary over a wide range so that draw there data is a significant highest difference between the lowest conclusion and state values? problem Scientic collect method data An independent variable is one that is changed and formulate manipulated by you. hypothesis A dependent variable is the one that you will measure design or observe. experiment Control variables are the ones that are held constant during the measurement. Here as are you some work guidelines through that your you should consider project. Continuous variables are ones that can take an infinite number of values (such as an ammeter reading); discrete variables can only take a fixed number (such as surfaces on which a block slides: glass, sandpaper, concrete, and so on). • Keep all references scientists for an back 222 accurate daily use as your you find of record work at your them. hard-backed electronic up notes on the work Many notebooks a working but computer end of including every or if you tablet, session. go • Spend some reviewing the time the analysis during work of the every work you have done. day’s data, plot session Carry any out graphs When analysing your data: from these data, and plan out the objectives ✔ and targets for the following session. It be careful to compare the outcome with the original may prediction be appropriate to do these analyses at home ✔ as laboratory time during a project is if appropriate, make use of a spreadsheet program precious. for analysing the data and offering a possible Remember: IB learners are reflective. equation to model them. Analysis Data from any experiment or investigation need to Evaluation be processed, analysed and interpreted. The analysis It criterion focuses on this aspect. The research is important you must lead to a detailed and valid answer. There are a enough data to justify your conclusion and form a view of the experimental data. In essence, the analysis uncertainties must be the answering the research with a curved trend variable varies with the this, graphs including Manipulate straight the your give exact the change basics in data, if at all of is the of the in relationship. possible, to produce their a If and errors the obvious these where or (experimental arose compare as your published recognise and (if you should and see if there are any ability beginning to undertake uncertainties can same apparatus of possible, values you internal your a full analysis of you results work problems in of other your suggest how they could work, be the were seek able to ability to repeat identify to do your any this is work). shortcomings a strength not a be in random your or data. measurement). and errors. the data Make to decide realistic (or carefully on carry a in Group short to end rather science; that is 4 project than unlikely your to is about through ability happen. to You from invent your teacher with evidence of your should progress Remember systematic Look assessment to the in in If work; provide that that weakness. trends. important errors it. these Scientists than new It work trend. beyond hidden the issues how The Look of any another. information nature whether scientist. question. show more and accepted overcome Straight-line experiment, directed discuss one an professional in scientists. Graphs quality analytical) with towards a you completed your or must for or to evaluate student Consider be to question the estimates the project. both at the magnitude of them possible. Here are some possible transformations to allow you to obtain a straight trend. Intercept on Plot as Equation Gradient Predicted Notes y-axis relationship y = kx + c y-axis x-axis y x y = mx + c k c This reduces the 2 y = kx y x y = x k k 0 impact of errors in x 2 2 + c y = kx y x y = y x c c 0 c 1 xy = c k x n y = x In y In x In y = nIn x n n/a HL only In y x In y = In A + kx k In A HL only kx y = Ae 223 INTERN AL • Write it • down A SS E SS M E N T everything day-by-day (you can edit • an outline • details of producing a rough project draft three-quarters of the way through of any the IA work. This will a complete allow you to identify implement when you still have time account • them. details of preliminary work full on and ideas show is that your comments are aligned you work, did; which use an may of your data collection (these a may need complete to be in analysis an of to the physics quoted the results earlier a conclusion and graph safety that presents the findings with and unambiguously together with plots. your Consider work this developed. sure data nature and linked clearly your the for how • Make the results that your of format to appendix) Give work simple depend improvements preliminary the appropriate • underlying at • about • physics later). Consider • the issues and whether they evaluation and reflections on the whole investigation. were adequate. • Draw attention possible, the to anomalies reasons for in the results and, if them. Diagrams can and should be used where appropriate. • Discuss the extent to which your results support Any diagrams should be fully annotated. your conclusion, linked Finally , be to your critical which research of your should be strongly question. own work. There is no Referencing such thing scientist is as perfect the science. ability to be A hallmark of a good self-critical. You are sources ▼ Bad ▲ evaluation Good statements that just Focuses on the extent to is that it the is trend hard to or state draw the a research question your and report, reference academic encouraged but any theft to you work take to use must that is make not someone secondary sure your else’s you own. work and which pass repeat in correctly evaluation It Simple allowed can it off as your own. be answered. Although the IB does not recommend a particular conclusion. Discussion Description of practical of the of the limitations experimental reference be that had overcome, than to and rather consideration realistic improving issues of method for citing material is the Harvard style. for method. In the main any preliminary (Homer, text 2019, you p. add 22). a This citation using reference the cites this format: give all the work experimental common of Identies fundamental suggestions the a method, scientific issues style, that could have been particular page in this book. method. carried out. At the end references Web of your you must the together. references author report and the are similar; date if give possible, the and name also of the the date Your evaluation must focus strongly on the research on which you accessed the reference, for example, question and the extent to which you feel you have (Garza, 2014). been able to answer it. Communication Scientific Aim you for writing the can achieve. be about be placed intrude 6–12 in into be concise standard The pages. an main Large of part main text and as effective. presentation of the quantities appendix—do the scientific the References papers are as and web pages the formats in follows: REFERENCES should highest For not they text of allow will that should data can them disrupt to Homer, D R 2019, IB Prepared Physics, Oxford University Press, Oxford. Garza, Celina 2014, Academic honesty – principles to practice, International Baccalaureate Organisation, viewed 15 October 2018 the <https://www.ibo.org/ flow that you want to achieve. contentassets/71f2f66b529f48a8a61223070887373a/ academic-honesty.-principles-into-practice---celina-garza.pdf> A report • a • the 224 should statement research contain: to put the question work in context Checklist Here that is a you Evaluation checklist are that you addressing all may of wish the to use marking to check criteria. Does my conclusion relate to Have I described and justied my data? the conclusion in enough detail? Do my data suppor t my research question? Personal engagement Is there evidence that I have Is this evidence clear and linked the IA to my personal does it show signicant engagement? thought, initiative and Does my conclusion relate Have I justied my to my scientic context as investigation in the context of summarised earlier in the the accepted science using a repor t? relevant approach? Have I considered the Have I discussed issues such strengths and weaknesses of as data limitations and error my investigation? sources and shown that I creativity? Have I justied my research Has my work shown personal question and the topic in signicance and interest? understand their relevance to general? my conclusion Have I shown a personal input Is this evident in the design and initiative? phase, the implementation and the presentation of my Have I suggested fur ther work Have I said how I would arising directly from my IA? ex tend the work is I had more time? What realistic work? and relevant improvements Exploration would I make if I were doing the IA again? Have I identied my topic and Is my description relevant research question? and fully focused? Have I provided background Is this entirely appropriate, information? concise and relevant and does it enhance a reader ’s Communication Have I given thought in my repor t to: • presentation Does my presentation help to make my work clear to a understanding of the context reader? Have I eliminated as Have I made my experimental Is the method highly method clear? appropriate to the investigation? many errors as possible? • structure Is my repor t clear and well structured? Is the Have I repor ted the factors Have I discussed the that inuenced my data relevance and reliability of collection? my data and whether there information on my focus, what I did, and how I analysed my data coherent and well are enough data? linked? Does my personal engagement pervade the Have I commented on other Have I discussed the safety relevant issues that have an of my work for myself and impact on the investigation? others? If necessary, have I repor t and is it a genuine engagement? considered the ethical and • relevance environmental impacts of my Is my repor t concise and relevant so that a reader work? can easily understand my focus, my analysis and my Analysis conclusions? Have I included my raw data Are my data sucient to either in the main text or in an reach a detailed and valid • technical language? Have I used all my physics terminology correctly and appendix? conclusion? Is my data processing Have I checked my data? repor ted appropriately? Are they appropriate and appropriately? suciently accurate for me to address my research question in a consistent way? Have I considered Is my analysis of error uncer tainties? consistent, full and appropriate? Have I provided an Are my conclusions correct, interpretation of my valid and in enough detail for processed data? the reader? 225 P R A CT I C E At this point, Physics approach. papers, course. you syllabus. 1, It 2 is will now and 3, Answers have re-familiarized Additionally , time with to to put the these you will these same papers structure are yourself have skills E XA M with picked up the test; as the external at content some to available in the this PA P E R S key from the techniques section you assessment will you topics and find will and skills options to refine practice complete of the your IB exam examination at the end of the DP www.oxfordsecondary.com/ib-prepared-support Paper 1 SL: 45 HL: 1 minutes hour Instructions • Answer • For all each sheet to candidates the questions. question, (provided copy of at choose the answer you consider to be the best and indicate your choice on the answer www.oxfordsecondary.com/ib-prepared-support). • A clean the physics data • The maximum mark for the SL • The maximum mark for the HL booklet is required examination paper examination for is [30 paper this paper. marks] is [40 marks] SL candidates: answer questions 1–30 only HL candidates: answer all questions. 1. The length width is of a rectangle is 2.34 cm and 4. the 5.6 cm. Mass P slides collides with immediately What is the area appropriate of the number rectangle of to significant 1 × 10 a horizontal identical before the mass surface Q. collision The is and speed of +u an figures? What after 1 A. along an are the the velocities of P and Q immediately collision? 2 cm 2 B. 13 cm C. 13.1 cm D. 13.10 cm V elocity of P V elocity of Q 2 2 A. u u + 2. Which list contains only fundamental 2 2 u +u (base) B. units? C. A. kilogram, mole, B. kilogram, coulomb, C. ampere, D. coulomb, 0 D. mole, volt mole, A boy fires time acceleration of a pellet of mass m a vertical Celsius distance The 0 u ampere 5. 3. +u kelvin an object varies linearly d from into the the air instant using the a boy rubber band. releases the The pellet from until it leaves the rubber band is Δt. 2 zero to 30 m s in 20 s. What What is the change in speed of the object is the power developed by the after band as it fires the pellet? 20 s? m gΔ ∆t A. 1 A. 1.5 m s 3.0 m s d 1 D. mg m gd 300 m s B. D. ∆t Δ 226 C. 150 m s 1 B. mgdΔt 1 C. dΔ ∆t rubber P 6. A constant force the variation the work W of acts on kinetic done on a mass. energy the What of the describes mass 11. with A fixed mass density ρ, E B. E C. E D. E is constant A. V ∝ W B. V , ∝ W C. V , T ∝ W D. ρ, T = an ideal temperature gas T has and pressure volume p, V mass? What A. of always proportional to p? ke ρ ke 2 ke 2 ke 7. A boy throws across ground a now same a throws are the horizontal pebble t an the it with after with pebble d was identical height time The distance after distance hits horizontally ground. time starting What pebble horizontal projection He a horizontal from initial release travelled the v 12. the point What ideal is from and the when the after release Distance 13. travelled A. frequency B. mean C. total D. mean 2d C. t 2d D. 2t d 14. Newton’s change of collision momentum velocityof executes second law states momentum ∝ external is on force acting change of displacement amplitude C. kinetic D. time force In a of period standing force = momentum force acting object on a molecules molecules molecules harmonic motion. to the restoring force that from the of of equilibrium oscillation the the system oscillation wave: the kinetic energy to of the the wave amplitude is of oscillation acceleration per unit time the medium = each part of the wave oscillates with a acting frequency and phase from every ∝mass C. An other molecules acting other 9. the the simple energy different D. with of that: B. external an system? B. of C. of the proportional the proportional B. energy A. A. A. of kinetic A system acts d 2t body , of 2v second B. a temperature the speed t For the ground? A. 8. to of What Time proportional gas? released. pebble an speed strikes planet drops from rest through the part amplitude along the wave varies with a time distance of 4.0 m in a time of 1.0 s. Air resistance D. is no energy is transferred along the wave negligible. What is the acceleration of free fall? 15. A transverse wave travels along a string. Two 2 A. 2.0 m s points on a string are separated by half a 2 B. 4.0 m s C. 8.0 m s D. 12.0 m s wavelength. 2 Which 2 these 10. An of electric mass liquid m. heater In a changes transfers time by t the a power P temperature to of a liquid the statement points is about displacements of true? A. they are constant B. they are the ΔT. the same as each other π C. they are rad out of phase with each other 2 What is the specific heat capacity of the liquid? D. they are π out of phase with each other Pt A. 16. A pipe is open at one end and closed at the ΔT m∆ other. The length of the pipe is D Pm B. tΔ ∆T What is the standing wavelength wave for this of the third harmonic pipe? mΔ ∆T C. Pt 3D A. P C. 3D 2 D. ΔT m∆ 2D D B. D. 3 3 227 PRACTICE 17. A beam single a of slit What PA P E R S monochromatic of distance E X AM width D from changes in b. A pattern the b light is incident forms on a on a screen slit. and constant object to an increase in D, the has makes carried out width of the of Change the M circular with complete a horizontal path revolution radius in path. of time R The and T What is the direction of the resultant force that on the object? pattern? b in mass a central acts maximum a one on separately , 22. lead speed Change A. outwards B. upwards C. towards D. along from the centre of the circular path D in A. increase increase B. increase decrease C. decrease increase D. decrease decrease 23. What is along the the the the centre velocity net force string of the vector acting circular of on the the path mass mass? 2 4π 18. What is the definition of electromotive force mr 2 of A. 4π Tmr C. 2 a T device? 2 4π mr 2 B. A. power supplied by the device per D. 4π 2 T mr unit T current B. force that charge C. the device carriers energy in supplied provides the by to drive circuit the device per unit 24. Which pair are both classes of hadrons? current D. electric the 19. Three to a eld acting on a charge moving identical of emf lamps 6.0 V X, Y and and Z are negligible mesons, baryons C. mesons, leptons D. bosons, components are in parallel. The labelled Lamp What Y ‘6.0 V , breaks happens so What is true A. the current B. the current C. the current D. the lamps in does lamps not X conduct Electrons in a the proton (Z), neutron (N) and (A) numbers? and Z = A N B. A = Z N C. A = N D. Z = A Z? them increases in them stays in them decreases immediately A. current. the burn horizontal beam Z + N same 26. What are evidence is Geiger–Marsden out A. 20. for 1.2 W’. and to mesons lamps nucleon are bosons connected resistance 25. all leptons, B. device cell that A. in moving alpha provided by the Rutherford– experiment? particles have discrete amounts of due energy north. What is the direction of the magnetic field B. due to the electron beam vertically above the positive charge concentrated A. to the east B. to the west C. upwards D. downwards Two wires, C. nuclei D. gold A and B, have a circular cross-section an initial a an small protons have energy A radioacitve has in contain atoms binding 27. 21. of atom is it? large per nuclide activity volume and neutrons magnitudes for the nucleon with A at a half-life time t = of 1 hour 0. 0 and identical lengths and resistances. Wire A has When: twice the diameter of wire B. • resistivity What of t = 1 hour, the activity is A wir 1 is ? resistivity of wire B • t = 2 hours, • t = 3 hours the activity is A 2 A. 0.25 B. 0.5 C. 2 D. 4 the activity is A 3 A 0 What is ? A 2 The following information is needed for questions 22 A A and 23. An object is suspended by a string from a fixed 3 point 228 on a ceiling. The object is made to move at a A 3 A 0 0 D. C. B. A A 0 1 A. 2 3 28. What solar are the overall heating panels energy and transformations photovoltaic in 33. cells? An object with performs time period T simple and harmonic amplitude x motion . The 0 displacement Solar heating panels Photovoltaic of the object is at a maximum at cells time t = 0. T A. solar to thermal solar to thermal What is the displacement when t = ? 4 B. solar to electrical solar to thermal x 0 A. x C. 0 C. solar to thermal solar to electrical D. solar to electrical solar to electrical 2 B. D. 0 x 0 34. 29. A black body emits radiation that has a Two isolated spheres gravitational wavelength at λ at a maximum intensity I p temperature of have the same peak . potential at their surfaces. The p the black body is then What increased. is identical for the two spheres? 3 What are the changes to the peak wavelength radius radius A. and the maximum C. intensity? mass mass 2 Change to peak Change radius to wavelength A. greater intensity greater λ than than mass I p p 35. B. less than greater λ than I Two parallel by distance a charged greater less λ than than than in a X and vacuum. X Y is are separated and Y is negatively positively charged; both plates p have less s plates I p D. metal p p C. radius D. B. less λ than I with p p the same charge gaining q magnitude is kinetic of accelerated energy E charge. from as it A particle rest at reaches X to Y Y . ke 30. A power station burns fuel with an overall What efficiency of E. The energy density of the fuel is the strength D and the its mass specific of fuel energy is is S; consumed the is rate at magnitude of the electric between X and Y? which qs E M ke A. C. E qs What is the power output of the field is power ke station? q sE ke D. B. M × S sE q ke M × S × E A. C. E M × M × B. D × E 36. D A transformer has an efficiency of 100%. The D. primary E coil has N turns and a power P is input p to it. The number of turns on the secondary coil output by the transformer? The following questions are for HL candidates only is N s 31. Two astronomical objects have an angular What separation images are at the just Earth of resolved 0.50 mrad. in the is the power Their 2.5 cm N s wavelength with a radio P A. telescope. C. P N p What is the diameter of the circular telescope N p aperture? 1 P B. D. N P s A. 6 km B. 600 m C. 60 m D. 6 m 37. 32. The wavelength of a spectral line measured on A 0.10 mF Charge is capacitor made to is initially flow onto uncharged. the capacitor at 1 Earth an is 500 nm. astronomical The same source line is moving observed away from from c Earth a constant the rate capacitor difference of 10 µC s breaks across . The down the dielectric when capacitor the plates in potential is 10 kV . at 10 What What is the wavelength from the source is breaks measured on the charging time before the capacitor when down? Earth? 8 A. A. 50 nm C. 550 nm B. 450 nm D. 5000 nm 10 2 5 s B. 10 s C. 10 5 s D. 10 s 229 PRACTICE 38. A coil is is E X AM connected placed along the to PA P E R S an axis ammeter. of the A bar magnet 39. coil. An a electron potential magnet and coil are moved as The magnet and coil are moved of this V. from The electron rest de is through Broglie λ follows. What ① accelerated difference wavelength The is with the is the de Broglie wavelength when the same electron is accelerated through 6 V? velocity . ② The magnet is moved towards the stationary coil. λ λ B. A. 6 ③ The coil is moved towards the C. 3λ 6λ D. 3 stationary magnet. 40. In which on the situations will a current be indicated An electron neutrino is emitted during an interaction. ammeter? What A. ① and ② C. ② B. ① and ③ D. ①, and ② ③ and what is is the the charge nature on of an the electron neutrino and interaction? ③ Charge on electron Interaction neutrino + A. 0 During β emission B. 0 During β emission + 230 C. +e During β emission D. +e During β emission Paper 2 Instructions • Answers to candidates must provided at be written within the answer boxes (answer sheets are www.oxfordsecondary.com/ib-prepared-support). • A calculator is • A clean • The maximum mark for the SL • The maximum mark for the HL copy required of the for this physics paper. data booklet is required examination paper examination for this is [50 paper paper. marks] is [90 marks] SL candidates: answer questions 1–5 only HL candidates: answer all questions. 1. An object The is graph time t. Air released, shows from the resistance rest, variation is not in of air above the speed the v Earth’s of the surface. object with negligible. 25 20 s m / v 1– 15 10 5 0 2 0 4 6 8 10 12 14 t / s a) Explain on b) the Outline, reaches c) The the a graph increases with Estimate reaches d) how object reference terminal the its of to the i) capacity Show that about 4.8 kJ. 15 N, ii) Estimate iii) Explain 330 J kg the resistance acting [3] forces acting, why the object its starting point at which the object [3] falls total through 350 m. It has a specic 1 K energy of the object has decreased by [2] the one air speed. weight of the [3] from 1 heat that time. speed. distance terminal object, indicates with increase in temperature assumption that is of made the in object. part d) [2] ii). [1] Question 1(e) is for HL candidates only e) The object forms a Calculate brought 2. a) Outline A string to i) the to Draw the Earth’s surface and, during the impact, deep. average force that acts on the object as it is [3] ways in which a standing wave differs from [2] length in its the 1.8 m is xed third-harmonic standing wave at both ends. The string is Explain at both made mode. produced. Label all the nodes present. ii) a wave. of vibrate to 12 cm rest. two travelling b) falls crater [2] how ends. a standing wave forms on a string fixed [3] 231 PRACTICE E X AM PA P E R S iii) When the harmonic frequency standing Determine 3. Two conductors, characteristics the speed A and as is B, increased wave of forms the have by on wave 23 Hz, the on the fourth- string. the string. [3] potential difference–current variation of (V–I) shown. 4.0 3.0 V / V 2.0 B A 1.0 0 0 0.2 0.4 0.6 0.8 1.0 I / A a) Sketch a graph potential axes b) of A and and in your B i) are B is the it. No the numbers resistance are of required A with on the [2] in a resistance. the Calculate in the across connected Determine ii) show graph. negligible A and to difference series The circuit energy with a battery transferred each of emf ε second same. ε. [4] the total power transferred circuit. [2] Question 3(c) is for HL candidates only c) 4. a) Outline whether Outline what is A or B meant is by an ohmic the conductor. binding energy of [2] a nucleus. [2] 239 b) One possible nuclear fission 239 1 Pu + Calculate ii) Determine of more with Pu 94 ) undergoes 1 Ba + x 56 n 0 [1] the binding than power energy nucleus = reference neutrons A fossil-fuel ( x. plutonium Explain, plutonium 146 Sr + 38 i) Mass 5. → 0 when is 91 n 94 c) reaction to nucleon for plutonium. 239.052157 u nuclear protons station per in burns forces, the coal. [3] why plutonium has nucleus. Its [3] energy is transferred to consumers. a) Outline b) Describe and 232 what the dynamo is meant energy of the by the transfers power specic that station. energy take of place the in fossil the fuel. [1] turbine [3] c) A fossil-fuel of station has a maximum power output 3.5 GW. A nuclear power power output power station. Determine in d) power the the nuclear A point 2.8 m. on The Calculate is minimum power the rotor rotor the station to throughout year annual the as same that mass of loss maximum the of fossil-fuel pure uranium-235 station. in the rotates linear provide the 60 [3] dynamo times speed of a the moves in a circle of radius second. point on the rotor. [2] The following questions are for HL candidates only 6. Yellow light grating of from spacing wavelengths a) the Determine be The a Explain 7. a) State b) A point few what +42 nC one is The light has separation normally two on close a diffraction spectral lines of of the two spectral lines in the [3] of for the grating is centimetres there is meant metal slits rst-order charge and incident 590 nm. number the why charged At the in is spectrum. diffraction placed and illuminated resolved c) 1.25 µm. angular second-order must spectrum 589 nm Calculate b) a is a the instant, images the of the two grating lines to that be just spectrum. from electric The has point diffraction [3] illuminated travelling sphere the the maximum by sphere. of by diffraction width potential directly charge charge is to at a of from a single slit grating. this slit. [3] point. towards magnitude a light [2] the the centre point of a charge is +3.5 µC. 2.5 m from the centre of the 1 sphere i) and Show of The that 2.5 m electric point ii) has a speed the from electric its is of it of potential centre potential charge towards is the about sphere 1.8 m s of the sphere at a distance 13 kV . is [2] 70 kV . The mass of the 0.23 g. Determine whether the point charge will collide with the sphere. 8. A metal at a aircraft constant vertically b) Faraday’s how an law rate Outline, of flying horizontally At the pole, the over the magnetic magnetic field north direction pole is upwards. Explain the is velocity . a) to [4] of for emf is induced suggests change the of that between the induced the wing emf is tips. [3] related ux. aircraft, what is meant by rate of change flux. [1] 1 c) The its speed the wingspan eld is strength Determine d) of Outline 35 m. at the how aircraft the The emf law to vertical magnetic induced Lenz’s relative ground component pole across applies the to is is of 270 m s the and magnetic 0.60 mT. the wingspan. this situation. [2] [3] 233 PRACTICE E X AM PA P E R S 9. No photoelectron surface when minimum a) emission incident is observed monochromatic from light a particular on it is metal below a frequency . Outline why the wave theory for light cannot explain this observation. [2] 14 b) Radiation surface. i) ii) of The The the Determine Outline 7.14 function threshold the × of 10 Hz the is metal frequency maximum kinetic incident is of on the metal 2.0 eV . the energy , metal. in eV , [2] of the electrons explanation described 234 work Calculate emitted c) frequency as a what by [4] Einstein paradigm is meant of shift by a the in photoelectric effect is often physics. paradigm shift. [1] Paper 3 SL: 1 HL: hour 1 hour 15 Instructions • Answers sheets minutes to candidates must are be written provided at within the answer boxes provided (answer www.oxfordsecondary.com/ib-prepared- support). • A calculator is required for this paper. • The maximum mark for the SL • The maximum mark for the HL examination paper examination is [35 paper marks] is [45 marks] Section A SL and HL candidates: answer all questions. 1. A student obtains volume for the V form of a a data gas. showing The graph. data Error the are bars variation given are as a shown of pressure p table on the of results with and in graph. 1.7 3 V / cm 1.6 p / MPa 10 1.30 15 1.08 25 0.76 30 0.64 35 0.54 40 0.46 1.5 1.4 1.3 1.2 aPM / p 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 3 V / cm a) Estimate, using the graph, the percentage uncertainty in the 3 value of p when V = 1.0 cm . [2] k b) The student proposes that the data obey the relationship p , = V where k is a relationship c) Estimate p constant. is Suggest, using the table, whether correct. this [3] when: 3 i) V = 2.0 cm ii) V = 0.50 cm [1] 3 d) Suggest more which reliable. [1] of your two estimates in part c) is [2] 235 PRACTICE E X AM PA P E R S 2. A student the uses horizontal took for x are Reading x / cm shown ruler of in a to measure small this table. the The dimensions x readings 3 4 5 49.5 50.7 49.8 50.2 49.9 Calculate The the mean mean value of value y the and y of with x its and its absolute absolute uncertainty . uncertainty the area of the table surface and its absolute uncertainty . State one using [2] is 0.4) cm. Determine c) of student table. 2 b) ± metre 1 a) (75.1 a surface a [3] possible metre systematic error that can occur when ruler. [1] Section B Answer all of the questions from one of the options. Option A: Relativity SL candidates: answer questions 3–6 only HL candidates: answer questions 3–8. 3. A free electron electrons the of free the in electron. a) Outline b) The what being A rocket and of length 0.85c. Calculate rest in 0.20c A pion is meant The of the the the free a by a a is a metal drift wire. speed stationary frame the in Conduction equal the to the frame speed of force space length space the of reference. acting on [1] the electron of from in with of the force in the frame of [4] station length is at of a speed the relative rocket is to 650 m the and the 6.5 km. the rocket according to an observer station. to a the an velocity decays the nature station moves moving Calculate of the time of the at shuttle time velocity decay to observer proper a [2] rocket of of rest in the relative with a speed station. to the to an rocket. 6. Explain, Jack to a) b) and reach Jill a relative pion as observer. measured by the 236 terms are Jill ii) Jack. Sketch a of twins. position i) show [2] in Calculate [2] 36 ns. 0.98c the station observer. b) of reference electron. proper according Calculate It the the A shuttle of is describes approaches proper a) observer explain of station at 5. to with magnetic. reference b) parallel move An observer State a) wire wire. as 4. moves the the 3.0 length Jack light journey contraction, leaves years time Earth, the effect according in to part Jill, at a). [3] 0.6c away . according to: [3] spacetime the diagram worldlines for in both Jill’s Jack reference and Jill. frame to [3] The following questions are for HL candidates only 7. An electron line is with and the 2.0 MeV . In annihilated 8. the and a) Calculate b) Determine, a) a positron same speed. collision, two the for energy ii) momentum. is photons one initial the minimum i) A ball approach The each electron with and of the along energy the identical speed photon, other kinetic of the the positron energy are same electron are produced. electron. [3] its: [2] thrown, initially horizontally , near to the surface of a planet. Explain, the b) with reference to spacetime, the trajectory of ball. [3] Calculate would the need radius in that order to an object become a with black the mass of the Earth hole. [2] 24 Mass Option B: of Earth = 6.0 Engineering × 10 kg. physics SL candidates: answer questions 9–11 only. HL candidates: answer questions 9–13. 9. A revolving revolution makes a door in further uniformly to 10. × 10 Deduce Estimate 11. the radius They A fixed complete The time the cylinder external Predict a total vertical motor is axis revolutions moment completing switched of off, while inertia of the one system decelerating the door is 2 b) ramp. the kg m a) A solid about When three rest. 3 8.0 rotates 18 s. which mass and are roll taken average a released down an the hollow the cylinder of for from rest gas is of the taken come acting the to on same rest. the [3] system. mass simultaneously without reach to torque cylinder ramp will ideal system frictional at [3] and the top of a sliding. bottom through of the the ramp first. following [3] cycle of processes. ① Isothermal ② Constant ③ Isothermal ④ Constant compression volume increase expansion volume at at low in pressure. high cooling to temperature. temperature. the original pressure, volume and temperature. a) Sketch b) Air the complete cycle on a pV diagram. 5 in a cylinder of volume 7.0 × 10 [3] 3 m has a pressure of 6 4.0 × 10 Pa A piston and at one a temperature end of the 4 a final volume of 1.3 × 10 of 340 K. cylinder is now allowed to expand to 3 m . At this point, the pressure in the 6 cylinder i) is 1.4 Deduce Show Pa. or not the expansion of the gas is an process. that 0.1 mol. 10 whether adiabatic ii) × the quantity [3] of gas in the cylinder is about [2] 237 PRACTICE E X AM PA P E R S iii) Calculate iv) Outline law of the the final temperature changes in the gas of the with gas. [2] reference to the first thermodynamics. [1] The following questions are for HL candidates only. 12. A solid cube of wood with sides of 0.25 m has a density 3 160 kg m with a) one . It is face Estimate under b) The placed the the 13. part A simple string seawater of density 1030 kg m and floats fraction of the height of the cube that is initially surface. wood Predict in horizontal. [2] absorbs dimensions to of 3 the of the effect water cube of with do this time. not Assume that the change. absorption on your answer a). [2] pendulum from a consists support. The of Q a solid factor sphere for the suspended system is 150 by a and the 1 angular a) frequency Discuss the is 8.0 rad s motion of the pendulum after it is set oscillating. b) The frequency Compare relative and Option C: [3] pendulum f = now oscillates horizontally with f the to support amplitude the of movement the of pendulum the support and its when f phase = 1.2 Hz 3.0 Hz. [3] Imaging SL candidates: answer questions 14–16 only HL candidates: answer questions 14–18. 14. Rays of light principal reached a) the The with a in diverging length +18 cm. the diverging lens parallel corresponding to these to the rays has lens. diagram, the passage of the wavefront through A compound a) State b) The lens the the focal the to how the lens subsequently [2] has a of lengths are focal the is in is a axis the normal image objective and 240 mm. –6.0 cm. this The lens second wavefront of lens, focal the rays again. lenses. [3] adjustment. formed and 90 mm of converging through between nal the 30 mm tube to principal the of length lens passing distance nature by the eyepiece respectively . An observer instrument. lenses The has a of [1] the length of near 25 cm. Determine the wavefront, diverging After microscope of the microscope microscope point of image. parallel Determine 238 a virtual from at of terms passes the a [2] Explain, emerge 15. on wavefront lens. forms c) incident The surface Describe, the b) are axis. the position observer ’s near of the point. object so that the image is [4] 16. An optical cladding a) fibre with Calculate has n = the a core with refractive index n = 1.56 and 1.38. critical angle for this core–cladding combination. b) A ray θ c) to is the central Calculate the reflection at A signal bre, [2] incident axis the of end the maximum the with which on a of the value has a of θ of of for an which interface 15.0 mW length with angle of core. core–cladding power bre is 5.70 km. total internal occurs. input The to [3] the optical attenuation of this 1 bre is 1.24 dB km Calculate the output power of the signal. [3] The following questions are for HL candidates only 17. When X-rays temperature metres travel T, in through kelvin, air the of pressure half-value p, in pascal, thickness is and given in by 5 1.8 × 10 x × T = 1 p 2 X-rays reach surface The the from average temperature the atmosphere 25 km above the Earth’s space. is Estimate b) Determine top of pressure a) the top the of of the atmosphere is 20 kPa and the average 240 K. half-value the the fraction thickness of the atmosphere for X-ray that is the atmosphere. intensity transmitted incident to the [1] on Earth’s surface. c) [2] Comment us on on Earth the from extent the to which incident the atmosphere X-rays at the top protects of the atmosphere. 18. a) Outline how [1] protons are used in nuclear magnetic resonance. b) Explain [3] the role of the gradient eld in magnetic resonance imaging. Option D: [3] Astrophysics SL candidates: answer questions 19–22 only HL candidates: answer questions 19–24. 19. The observed over d. a The a) b) Earth–star Draw d and The star a displacement period distance diagram to θ is and of the a star orbital viewed from diameter of Earth the Earth D is D show the relationship between θ, D. When State 20. angular six-month [1] is one Zeta measured different Puppis in parsec, consistent has a θ set surface is of measured units for temperature in D of arc-seconds. and θ. 42 400 K [1] and a 9 radius of 7.70 × 10 m. The parallax angle that Zeta Puppis subtends 3 from Earth is a) Calculate b) Deduce c) Calculate Zeta 3.40 the the × 10 arc-seconds. distance of luminosity the Puppis. peak Zeta of Puppis Zeta wavelength from Earth. Puppis. in the spectrum [1] [2] of [1] 239 PRACTICE E X AM PA P E R S 21. A red a) giant star Determine X has the a luminosity mass of 370 times that of the Sun. X. 30 Mass b) 22. The Y is of a Sun red = 2.0 × 10 supergiant i) Compare ii) Suggest the X could and Y the characteristics kg. [2] star. likely evolution circumstances of be the three Luminosity in of Y to which that the of X. [2] evolution of same. stars [2] are given in this table. Surface temperature / K 26 3.8 × 10 Sun W ≡ 1L 5700 80L Capella 5000 40L Vega 9600 a) Compare the surface b) Calculate the radius c) The area of of Capella and the Sun. [3] 8 apparent Calculate, in brightness parsec, [3] Capella. the of V ega is distance 2.2 from × 10 Earth 2 W m to V ega. [2] The following questions are for HL candidates only 23. a) b) Outline Star i) what A has until ii) 24. a) it Draw a a ii) an the graph z closed 100 a the times Oppenheimer–Volkoff that answer neutron eventual showing of to [1] Sun. a), the evolution of star. outcome the the part limit. A [2] of variation A. [1] with time t of the cosmic for: Universe how by your accelerating Explain suggestion 240 mass becomes factor i) meant using Deduce scale b) a Predict, is without Universe observations that the of Universe dark with energy dark energy . supernovae possesses have dark [3] led to energy . the [3] I n d e x Key terms are in bold radioactivity structure A-scan 198 atoms aberrations lenses absolute fibre 183 142, 144 absolute uncertainty absolute zero ac (alternative average 5, 6 stars 205 current) 10, 11, acceleration–time 34, B-scan generators 35, action–reaction air changes alpha 89, 193 11, 35 93 199 14–15 decay ammeters 168, 169, 73, barium, X-ray baryons 79, batteries 60 Bel 135 (ac) generators 118–22 scale oscillation 34, measure angular displacement angular momentum angular speed angular velocity 178 (rad) 164, 66 66 166 66 66, 92, 164 48, 51 stars Archimedes’ principle under a assessment graph iv–v , astronomical astrophysics Bang energy per 217, 189–92 203 black holes 161, Bohr, Niels 130 89, constant 29 wavelength 130, capacitance carbon 217–19 240–1 205–9 214, 123–6 216 Carbon-Nitrogen-Oxygen Carnot cycle stellar processes 214–16 centripetal acceleration stellar quantities centripetal force 202–5 81 Cassegrain wall Cepheid 77 mount, 72–83 fundamental charge levels forces 75 72–5 telescopes insulation cycle 214 versus 206, limit time charged bodies charged particles 191 88 67 67 variables Chandrasekhar energy (CNO) 170–3 cavity physics 135 122–6 121, 205–9 discrete 134, 141 characteristics atomic 212 30 evolution mass 211, 208 stellar atomic 77 195 capacitors questions freedom 219 nucleon stellar asymptotic 113 204 radiation law 138 204 binding 89, electron) 210–13, 203, 77 calculators 210–13, exam moving model stars Broglie 78 135–6, 73 energy de 73, 82, 86 on body 178 7 decay binding cables 221 unit 205 74 175 202–20 cosmology practice 203–4, 141 telescopes astronomical Big 175, graphs particles Albert Boyle’s brightness, radioactivity) particles Boltzmann 80 196 of equation black-body 42 antiquarks (unit lines, (force black 197 194 Bev binary 165 165, Betz, images 74 80 divergence beta-plus 59 acceleration apparent radiation 62 magnets Becquerel 170 200 beta-minus angular antinodes 29 bar 79 131, 58, radian antennae 198, best-fit current amplitude, constant beta-minus 57, 119 background beam 74 15 alpha-decay area graphs 10, 90 alternative 194 power Bernoulli particles angles, 118–22 42 resistance albedo 198, pairs radioactive adiabatic aerials graphs impedance activity, 78–82 200–1 Avogadro acceleration–displacement acoustic matter 26 lines, acceleration optics X-rays error absorption of 130 attenuation 186–7 mirrors 72, 72–5 209, 208, 211 215 graphs, capacitors 125 53 112–13 241 INDE X charge–pd Charles’s chromatic circuit climate 57, clusters, 187 171 cosmic 203 background colour concave terms and inelastic 52, 182, 183 54, 56, see 116–17 80 conservation of energy conservation of momentum constant of particle continuous length control rods control variables convection mirrors 154, 155 53, cosmic microwave 211, 212, 213, 182, 186, 183, 187 184 183 background principle redshift rate, law angle critical density curve, cycles, 107, 43, gradient at 112 decay point magnetic graphs nodes 72–5 35, 92, 93 graphs 37 50 184, mirrors 10, 35, 37, 92, effect Doppler equations 187 183 102–5 103, 212 interference decimal equation 45 places 15 52, 53 frequency , friction climate 74 on 178 oscillations 178, 179, 15 141 89 gravitational current Einstein, field losses 107 120 19 Albert cosmological general 93 11 185, 182, Doppler force 39, 50 graphs graphs diverging speed 37, 11 lenses drift 63 222 34, diverging drag force 125 levels antinodes 10, 155 dilation 125 10, efficiency 34 the equations displacement eddy oscillators a radiation 154, time Earth 217 oscillation 150, displacement dynamic 44 damped 101–2 121 variables see 47 211 217–19 radioactive critical critically 53, 217 104, 104, discrete driver cosmological Coulomb’s (CMB) 219 210–13, of double-slit 219 cosmological cosmology time 98, displacement–time 62 185, 45, 96–7 energy dp inflation 44–5, grating displacement 151, 66 cosmic interference 76 distance–time 184, 182, 2–3 displacement–distance 222 150, current mirrors 242 47 88 converging count 45, 222 6 discrete distance lenses cosine 53 222 converging convex 80 85 conventional variables bridges discharge interactions 174–5 variables of 21 135 3 gravitational discharge 203 interference equation contraction 168 proportionality constructive continuity 18, 134, 124 of direction rules, constellations diode 130, 174 units also 140–5 130 (dp) quantity dilation 141 questions vii 52, diffraction 116–17 6, projects 136 places diffraction 196 confinement conservation 192 graphs assessment wavelength definitions dielectrics 188–9, 219 internal constant deuterium 225 217, 219 Clinton destructive 224, 180 practical Broglie derived 80 skills mirrors 88, 22 v–vi scattering conductors points, dependent microscopes conduction data derived communication Compton 196 197 elastic compound 214 212, analysis, decay 179, 218, data density confinement command cycle 195 scattering collimation collisions, matter decimal cables coherent energy dark de microwave (Carbon-Nitrogen-Oxygen) coaxial dark Davisson, radiation CNO 178, data-based 90 galaxies see damping 66–8 statement 89, 123 58 motion Clausius capacitors 30 aberrations laws circular CMB graphs, law principle relativity 158–9, 217 160, 217 180 223, 225 INDE X Newtonian postulates photoelectric special elastic effect relativity collisions electric cells charge electric current electric electric fluid forced 123, 52, of rigid 55–9 of 52–4, 52, 107, 110 110, potential 107, electric potential difference also potential electrical electricity 109 107 difference resistance bodies enhanced 112–13 55–9, 119, 123 172, entropy formulation of state 29–30 of state for fields magnetic effects electromagnetic capacitance power of electric fields electromagnetic electromagnetic error 80, force current 147 116–27 electromagnetic waves force degeneracy electron–positron 39 38–9, (emf) 60, 116–17, 131, 208 structure capture quantum force spectra emissivity empirical 142 exchange external 15, 17–20, energy changes oscillator 67 density energy levels, energy losses energy pathways a 35 thermal 72 transformers 17, photon of state 120 18 72 222–4, 225 125 188, 189, 190 117 136 82 gravitational strength 106, fields; magnetic fields 110 106–15 work 109–14 of first minimum 48, 49, 50 position, single-slit diffraction 96, 97 76 points, Fleming’s Kelvin left-hand equations fluid dynamics fluid friction forces 106–8 197 harmonic forced 88–91 81, 193–5 scale rule 26 63, 117 86 174–8 15 174–8 fluorescent 84–7 transfer of capacitors 61 also fluids 84–91 energy energy 79 13 first flow ground in 162 diagrams optics fixed 88–91 85 production sources law lines fission 25–8 energy 158–9 142 evaluation Enrico filtration transfer 7, relativity microscopes/telescopes field at 30 energy general discharge, force fibre fields 72 cycles 108 226–39 particles eyepiece, field force 168 temperature of 161, vi, descriptions energy energy Excel see 89 laws thermal 134 132 electromotive emission papers Feynman 130 131, 29 4 exam Fermi, scattering function electrostatic energy 72 81 physics Rutherford see 119 gas 36 112 horizon Faraday’s electron emf speed exponential 118, 6, 35, 196 electrons wave escape experiments, 104–5 pressure pairs 118–22 surface principle, graphs ideal 34, 5–7 estimations transmission spectrum atomic 61–5 122–6 and bars, errors event electromagnetic electron 55–9 75 induction generation electromotive current an position electric 89 171 equation equivalence electric effect 173 equation equipotential of 164–7 168–73 greenhouse 52–65 effects 238–9 164–7 171, cells heating questions entropy electric 52–4 174–8 178–80 dynamics equilibrium 59–61 164–81 178–80 thermodynamics electric see exam 124 dynamics vibrations rotational 61–5 18, fluid resonance 54 strength fields 126 17, physics and practice 52, effects field stores engineering 157 22 effects magnetic energy 59–61 electric heating 149 128–9 screens vibrations 197 178–80 13–17 force–speed graphs 19 243 INDE X fossil-fuel stations fossil 84, fuels fractional free-body free 86, diagrams space, frictional forces resonance 16 78, of 53 rotation 34, 39 scales 15 galaxies Galileo V–I Galilei gamma constant gas laws gases 27, gravitational experiment relativity 158–62, orbit geosynchronous geothermal orbit gradient at gradient of a 111, 112 86 fibres 193 on straight a line 141 amplitude area against under binding time graphs data analysis data points discharge for for 6, distance–time 6, force–speed laws guidance graphs capacitor 125 123 force 68–71 69, 70, 107–8, 110, 14, 69–70, 67, frequency shifts 110 161, 162 107 160 108, 109, gravitational potential energy 111 gravitational redshift gravitational time 18, 114 104 dilation 160, 161 75 greenhouse effect greenhouse gases state 89 89 energy level 72 79, 80 74, 75, 136, rectifiers harmonic motion, harmonic series heat capacity heat engine, 35, 37 37, 93 Hot 11 HR Big 121 48–9, see 19 34, 36, 92–6 50 efficiency uncertainty fusion reaction 170–2, principle 76, (HR) 78, 131, diagram 206 217 model 210–13, Hertzsprung–Russell law hydroelectric 217, 219 diagram 210 systems 142 fusion reaction spectrum angle for electron 142 scattering 134 stellar 87 214, 215, 216 losses 207 120 223 IA see lines 78, 102 evolution hysteresis projects 76, 7, 142 internal assessment ideal ammeters ideal fluid 175 58 173 214 81 Bang Hubble’s 197 27 homogeneous 92, beams simple thermal boson 137 X-ray hydrogen assessment 141 fields Heisenberg’s 142 maximum–minimum 244 discharge 141 plotting nucleon 140–5 7, of 107 Higgs graphs 10, law strength difference half-wave 125 graphs 141, intercepts labels 178 against capacitor graphs 93 56 Hertzsprung–Russell intensity–diffraction internal oscillators 30 gradients graphs field half-thickness, 141 7, 27 potential helium, displacement–time gas 93 223 graphs bars 11, nucleon displacement–distance error 10, 141 per energy gravitational hadrons 77 versus charge–pd time, graph energy number charge a 11 with 107, half-lives graphs 35 141 graphs acceleration–time 218 potential ground curve of gravitational gravity 203 point a 217 130 clusters graded-index 79 111 stations Lester globular gravitational 170–1 10, variation Newton’s gravitational 28–33, galaxy motion characteristics gravitational 73 degrees 141 gravitation, 29–30 geostationary Germer, 149 29 Geiger–Marsden general 147, 146 particles gas 146, a graphs velocity–time transformations for harmonic transfers 203 Galilean curves temperature 104 137 for 179 speed–time 121 207 redshift decay curves 141 simple units galactic 169 damping 15 rectifiers 76, 6 oscillation/waves drag graphs radioactive 5, permittivity frictional full-wave pV uncertainty frequency, fusion 86 90 projects 132 INDE X ideal gases ideal voltmeters imaging fibre 29, 32 kinetic 58 optics 193–5 instrumentation imaging practice exam independent induced 43, force (emf) 117 infrared cosmic instability 165, 167 146, 147, 149–56 40, 41, beams intercepts, stars 206, 207 196 angle graphs, electron 142 telescopes 45, evaluation of exploration skills 223, 224, 222–4, 225 225 221, 225 absorption motion speed drag Lorentz of momentum 150, 151, 154, 21 155 coefficient 196 164 66 27 (current-carrying) waves 37, 38, transformations luminosity , stars 203–4, 39, wire 62 41 148–52 205, 216 26 60 transformations radiation see 73 149 magnetic fields magnetic flux 116, magnetic flux density 116, 117 magnetic flux linkage 116, 118 magnetic force 61, 63, and 169 168 sequence, energy) 18 scale 25, Kelvin statement third 26 171 law 112 186, of stars Ernest mass absorption mass defect current 61–5 188 4 206, 207 79 coefficient 196–7 77 compact halo objects (MACHOs) relationship system dispersion 35, 218 207 94–5 194 matter of 26 structure of 78–82 interaction maximum–minimum Maxwell, Kelvin 119 41 matter–radiation of 116, 183 glasses states changes objects 67 orders massive 112–13, electric magnitude, law 62, halo 63 118 magnifying material 168, 217 isovolumetric strength mass–luminosity 168–9 changes compact field mass–spring changes massive magnetic Marsden, 52–3 Kepler ’s 200 15 straight main 221 resistance (unit 199, 81 linear Malus’s energy joule 178 203 magnetism, 224 internal isothermal 58 117 79, magnification experiments internal isotropic 225 225 221–2, choice isobaric 221–5 221 engagement referencing ions 191 projects 225 analysis ionizing (IA) criteria communication inverse law MACHOs phenomena assessment assessment topic 57, 28 conservation Lorentz 7, wave interferometer personal 27, longitudinal 97–100 data 57 law 141 contraction liquids 47 graphs checklist 35 134 interference, internal 32, energy law circuit 176–7, frequency year liquid intensity–diffraction scattering 31, kinetic circuit linear long waves 25, 184–7 linear 89 intensity X-ray lenses light 219 strip, flow leptons frames radiation of Lenz’s 107 inflation, graphs heat length 22 of reference infinity law 222 76 collisions moment inertial second laminar 44 variables fission inelastic Kirchhoff ’s latent electromotive induced rotational first labels, 239–40 45 angle 22, Kirchhoff ’s Larmor waves incidence inertia, 196–201 questions 23 phase, also 18, 188–92 medical impulse in see 182–201 energy James measurements and 128–33 graphs 7, 142 147 2–9 uncertainties vectors Clerk lines, and errors scalars 5–7 7–9 245 INDE X mechanics energy forces 10–24 non-viscous 13–17 momentum motion power work non-renewable 17–20 21–4 10–13 19 imaging medical magnetic metal 79, 80, 196–201 of Albert modal number mass moment of spacetime of 27 21–4, patterns a 45, 98 magnetic experiment resonance 178, 179 curvature 90 217 force) 13, 15 neutron degeneracy neutron stars 215 speed of a metre, 188, diffraction 189 56 vi 203 of out of limit 216 193–5 111 98 magnitude 34–6, phase, 4 48–9, waves 178, 179, 45 oscillators production 131, 178 196 203 physics 72–83 energy levels fundamental forces radioactivity 72–5 of matter particles definition 208 gas 85 72–5 75 78–82 Newtonian postulates Newton’s first Newton’s law law Newton’s second Newton’s third of telescopes of 146, 147, motion gravitation law law of of 190 total 149 14, 165 68–71, motion motion 14, 14, Pascal’s 218 21, 165 215 Pauli 165 peak magnetic resonance of function power pendulums period, 157 157 131, principle exclusion percentage nuclear 30–1 difference energy wave 112–13 26 modelling potential mount, 208 43 spectrum charged Newtonian 51 133 particles capture 48, fibres orbital structure 136 neutron see optical parsec 53 (resultant 77 56 discrete negative nodes per densities particle charge 73, optical pair 203 neutrons law over-damped 186 negative 246 imaging 150–1 stations power per 188 lens oscillations gas force system resources orders 21 natural neutrinos energy 135 slits clusters order oscillations NMR 28–9 167 frequency , nickel open substance) natural nebulae 98 157 14, medical of 165, 148 Edward point 85 Oppenheimer–V olkoff inertia decay 134, lenses Ohm’s quantity 10–13, stations conductors online Morley , see object, ohmic 193 light net diagrams objective monochromatic muon see 29 momentum 78–82 76–8 129 interference of of 72–5 75 76–8 matter nucleus–alpha 132 174 principle 208 119 34, 94 uncertainty circular permittivity of 199 72–5 binding 72, (NMR) 133–9 levels forces reactions nucleus 85 modulation, 72–83, reactions power 84 135 resonance energy of sources 175 134, radioactivity nucleon, 188–9 method (unit physics nuclear 148 dispersion motion 62 58 diagrams moderators near 199–200 182–4 mixtures, MRI (MRI) 27 experiment Minkowski molar wire mixtures microscopes mirrors nuclear structure characteristics Millikan magnetic nuclear Michelson, mole imaging 82 current-carrying V–I nuclear nuclear resonance wire method density fundamental medical mesons fluid nuclear discrete 17–20 energy motion free 5 66 space 53 180 gratings 98 INDE X phase changes phase difference photocells photoelectric 72, Pitot effect 131, cells 128–9, 196 129 radius 50 129 53 positron–electron potential 106, also 217 pairs at a point potential difference electric potential-divider Pound, Robert 26, 19, generation power of power rating power stations 32, lens 185, 157 difference 85, exam images 29, 110 transmission giants 86, papers 90 28, law 29, 226–39 cells 60 primary energy functions, motion proper length proper time 84 43, physics 130, 12 150 constant of 53 157 hydroelectric systems 169 87 43, angle fluctuations physics matter–radiation probability quarks functions 130, 131 131 radian measure mechanics 104 (rad) radiation–matter special 66 128–33 158–62 questions mechanics diagrams relativity energy 55–9, resistivity 55 mass 156–8 152–6 156–8 sources 119, variable resolution, 148–52 237–8 84, 86 224 resistance wave 123 60, 61 phenomena 101–2 102 178–80 150, 157 force Reynolds (net number bodies rotational interaction 156–8 transformations rotational astronomy 53 exam rigid 192 43 force) 13, 15 177 164–7 root-mean-square 80–1 189–90, 99 relativity rotational radar 99 practice resultant 128–33 224 190–1 Lorentz rest interaction tunnelling 79, 219 128–39 projects 146–63 resonance quantum 149–56 99 permittivity resolvance quantum 212 44 48, refraction resistors, storage 147, 211, 99 report-writing interval 210, assessment telescopes renewable 150 proportionality, 131 43, waves spacetime quantum 146, telescopes relativistic sources 161, internal waves relativity 174 183 215 160, angle relativistic vii 30 primary quantum 206, refraction relative 215 32 frames refracting 186 process) 183 104, reflecting 118–22 (r 73 121 general pressure graphs gases real reflected 109, 3 pressure pumped real reflected 58 94, infinity referencing, 56 problems probability from refracted practice SI 101 rays redshift capture criterion 40 134 decay 185 rays 136–7 5 183, reflection and practice protons 109, 55 a projectile 59, 160 power neutron reference 53, potential power prefixes, 110 circuits energy potential 74, nucleus diagrams rectifiers gravitational a radioactive red potential; potential potential 196 110 electric also 131, 73, 72–5 random Rayleigh 41 charge of 192 44 errors ray curvature 42, random rapid experiment R 191, decay radioactivity 196 waves positive pV waves 86 positive see radio 76 polarization see telescopes 175 constant plasma radio equation standing tubes Planck 99 35 radioactive 81, photovoltaic pipes, 34, 129 photoelectric photons 26–7, (rms) dynamics kinetic motion 119 164–7 energy 165, 166, 167 164 Rutherford atomic Rutherford scattering model 79 134 247 INDE X sample exam sample student Sankey papers 226–39 answers diagrams stars vii characteristics 85 satellites angular orbits scalars scales, momentum 69–70, 111, graphs Erwin scientific notation scientific writing energy sign stream 84 strong 3 figures harmonic sine (sf) 3 36, suvat 94 single-slit reflectors diffraction neutron law 43, 96–7, photovoltaic solar power Solar System of 176, 75, 178 77 (galaxies) (stars) 141 176 203 206 215–16 40, 47, 11, 48 12 (s process) 215 tangent panels 86 cells 86 of waves 45–6, diagrams spacetime interval relativity specific energy specific heat latent analysis 156–8 heat order 27 27, 28 10, 11 power software Model, waves 186 fission time 76 decay 73 total 142 total particle total physics 79, 80 of 63 30 stations 88–91 85 88 systems of 168 168–73 170, 171, equation 104, 172 172 185, 150, 187 154, gravitational 165, strength) 25–33 168, law period torque 206 47–51 also field transfer radiation dilation see 183, radioactive standing models thermal thin-lens spontaneous Standard magnetic physics law 176 69 energy first 98 aberrations candles of thermal second nuclear standard (unit 25–8 15, thermal time graphs changes speed thermodynamics 206 of 89 thermodynamic spontaneous spreadsheet Earth objects thermal 11 speed–time spherical 150 capacity spectral 10, 152–6 85 specific 189–92 theoretical 103 spacetime spectrum, the terminal test 38, 5 66 energy 202 tesla sound errors temperature 62 27 11 3 systematic 86 solids 248 175, force equations telescopes solar speed 175, 98 44 heating solenoids 191 capture solar special 174, nuclear motion SI single-dish 51, 174, symbols pendulums Snell’s 129 gradient superposition 34, 66 slow tubes supernovae motion 89 193 potential line, supergiants 186 law 176 superclusters 92–6 simple 15 stars streamlines figures convention simple straight sources 203 26 fibres law stopping 2–3 significant see Stokes’s 60 significant matter step-index 3 secondary 202–5 clusters friction stellar... 161 224 cells of 205–9 Stefan–Boltzmann 222 secondary units 196 130 radius method SI static photons scientific prefixes 214–16 stellar 141 X-ray Schwarzschild SI processes states Schrödinger, see of quantities 7–9 scattering, sf 167 112 205–9 evolution 155 time dilation 34 166 energy of internal particle reflection potential train–tunnel a energy 43, 223 44 95 simultaneity transformations 157 problem 154, 155 INDE X transformers 119–20, translational transverse travelling triple waves of 85, turbulent twin waves point turbines Type II 176, water, 41 water 26 forces uncertainties 215–16 simple 216 198–9, 200 13 speed wave theory waveguide oscillators mass velocity unit 178 76 2–3 web definition 168 expansion travel of 210, through upthrust 212, 217 219 Wien’s wind 77 of vectors resistors 10, 66, velocity–time virtual 60, 61 93 graphs images viscosity 47–51 waves 57, 36–42 force interacting 75 massive (WIMPs) 218 vi 206, 207, displacement see particles weakly turbines 86, 215 law 89, 206, interacting 218 massive particles 87 93 V–I wire characteristics work 17–20 work done worldline 62 58 18 152, 153 183 176 voltmeters 193 wire 7–9 velocity dispersion 40–2 current-carrying variable 184 43–7 waves dwarfs WIMPs 175 uranium-238 218, 103, 37 resources white 98 39 (modal) nuclear weakly Universe 92–6 96–7, 129 40, behaviour weak motion 36–51 standing 141 measurement 97–100 37, travelling graphs 102–5 101–2 characteristics 13 130 92–105 diffraction wavelength waves units of wave wavefronts Heisenberg’s 132 harmonic single-slit 26 duality interference principle, atomic uniform 131, effect resolution of 86 phenomena 195 132 unified function Doppler 5–7 under-damped point wave-particle 154–5 cables triple 88 turbines wave 178 imaging unbalanced insulation wave supernovae uncertainty 38, 36–42 supernovae ultrasound 131, 37, water flow paradox I wall 13 86–7 twisted-pair Type 121 equilibrium X-ray medical imaging 196–8 58 zero curvature 217 249 O X F O R D I B P R E P A R E D PH YSI C S Offering an unparalleled level of assessment suppor t at SL and HL , IB Prepared: Physics has been developed directly with the IB to provide Author David Homer the most up-to-date and authoritative guidance on DP assessment. You can trust IB Prepared resources to: ➜ Consolidate essential knowledge and facilitate more effective exam preparation via concise summaries of course content FOR FIRST ASSESSMENT ➜ Ensure that learners fully understand assessment requirements IN 2016 with clear explanations of each component, past paper material and model answers ➜ Maximize assessment potential with strategic tips, highlighted common errors and sample answers annotated with exper t advice What's on the cover? ➜ Build students’ skills and confidence using exam-style questions, A visual representation of practice papers and worked solutions the Higgs boson par ticle 10 FIELDS (AHL) Figure 10.1.3 shows the gravitational field due to a spherical planet Key syllabus material is explained Points of the the alongside key definitions on the green sphere green surface , and surface on surface so have no which a are the overall charge at the same work or same distance potential. is mass done. can When This move from a gives the mass an without centre moves on equipotential work being transferred. Because work is done when a charge or mass moves along a field line, –80 V equipotentials must always meet field lines at 90 –90 V Example 10.1.1 Assessment tips offer guidance and –100 V A precipitation consists of system two large collects parallel dust particles vertical plates, in a chimney . separated by It 4.0 m, warn against common errors maintained Figure 10.1.3. a) Explain b) A small at potentials what dust is of meant particle 25 kV by an moves and 25 kV . equipotential vertically surface up the centre of the Field lines and chimney , midway between the plates.The charge on the dust equipotentials around a planet particle is 5.5 nC. Assessment questions and sample student i) Show that there is an electrostatic force on the particle of about 0.07 mN. responses provide practice oppor tunities ii) The up mass the of the centre of dust the particle chimney is 1.2 at a 10 kg constant and it vertical moves speed of 0.80 m s and useful feedback Calculate strikes the one of minimum them. Air length of the resistance is plates so that the particle negligible. Solution a) An equipotential means Also available, from Oxford that no surface work is is a done surface in of moving constant charge potential. around on This the surface. Vq b) 978 0 19 8392132 i) The the force on plates. particle The where qE potential difference 4 5.0 So × 10 is is the distance between 50 kV . 9 × 5.5 × 10 force = 6.875 × 10 N 4.0 Example 10.1.1 b) i) is a ‘show ii) The horizontal The particle is 6.875 mass 1.2 10 acceleration 0.573 m s that’ question. You must convince the examiner that you have force in the centre of the plates, 10 so has ut + at to move 2.0 m completed all the steps to carry horizontally to reach a plate. Using s = and knowing out the calculation. The way to do that this is to quote the final answer to the particle has no initial horizontal at least one more significant figure component 2 speed gives 2.0 = 0 × t + 0.573t so t of 2.0 2.63 m = (sf) than the question quoted. and, 0.573 Here it is quoted to 4 sf – and in therefore, the length must be 2.63 0.8 2.1 m. this situation this is fine. S AMPLE STUDENT ANS WER Explain what is meant by the gravitational potential at the surface of a planet. ▲ There question are two and two marks for points to This this It this answer done per taking be the ‘small’ from has unit mass in a innity them mass, (it both: and does potential to the [2] answer could have achieved 2/2 marks: make– is the work done per unit mass to bring a small test mass work the not idea have of from a point of innity (zero PE) to the surface of that planet to denition) (in the gravitational eld). surface. 108 I B D I P L O M A P R O G R A M M E Suppor t material available at www.oxfordsecondary.co.uk/ib-prepared-suppor t IS B N web 9 7 8 -0 -1 9 -8 4 2 3 7 1 -3 www.oxfordsecondary.com/ib 780198 423713
