GCE A Level Physics 1976-2003 Topic 18 D.C. circuits

Cambridge International AS & A LevelDC (JC/CT) 182587/3© UCLES 2020[Turn overThis document has 12 pages. Blank pages are indicated.*5745011940*PHYSICS 9702/33Paper 3 Advanced Practical Skills 1 February/March 20202 hoursYou must answer on the question paper.You will need: The materials and apparatus listed in the confidential instructionsINSTRUCTIONS●Answer all questions.●Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs.●Write your name, centre number and candidate number in the boxes at the top of the page. ●Write your answer to each question in the space provided. ●Do not use an erasable pen or correction fluid. ●Do not write on any bar codes.●You will be allowed to work with the apparatus for a maximum of 1 hour for each question.●You should record all your observations in the spaces provided in the question paper as soon as theseobservations are made. ●You may use a calculator.●You should show all your working and use appropriate units.INFORMATION●The total mark for this paper is 40.●The number of marks for each question or part question is shown in brackets [ ].For Examiner’s Use 12TotalBLANK PAGE © UCLES 20209702/33/F/M/209702/33/F/M/20© UCLES 2020[Turn overYou may not need to use all of the materials provided.1In this experiment you will investigate the oscillations of a rod.(a)• Assemble the apparatus as shown in Fig. 1.1. •A djust the apparatus until the two springs are approximately 15 cm apart. Each springshould be vertical and the same distance from the middle of the rod. The rod should beparallel to the bench.Fig. 1.1•T he distance between the two springs where they support the rod is x , as shown in Fig. 1.1. Measure and record x .x = ....................................................cm [1](b)•L ift one end of the rod a short distance and push the other end of the rod down a shortdistance. Release the rod so that it oscillates with a rocking motion, as shown in Fig. 1.2.FRONT VIEWFig. 1.2•T ake measurements to determine the period T of the oscillation.T = .......................................................s [2](c) •C hange x by moving the stands. Adjust the apparatus until the springs are vertical andthe rod is parallel to the bench. Measure x and T.•Repeat until you have six sets of values of x and T.•Record your results in a table. Include values of 1x in your table.[9](d) (i) Plot a graph of T on the y-axis against 1x on the x-axis. [3](ii) Draw the straight line of best fit. [1] (iii) Determine the gradient and y-intercept of this line.gradient = ...............................................................y-intercept = ...............................................................[2]9702/33/F/M/20© UCLES 2020© UCLES 2020[Turn over9702/33/F/M/20(e) It is suggested that the quantities T and x are related by the equationT = ax + bwhere a and b are constants.Use your answers in (d)(iii) to determine the values of a and b.Give appropriate units.a = ...............................................................b = ...............................................................[2][Total: 20]9702/33/F/M/20© UCLES 2020You may not need to use all of the materials provided.2 In this experiment you will investigate the magnetic field produced by an electrical current.(a) You are provided with a length of wire wrapped around a plastic channel to form a coil, asshown in Fig. 2.1.other groove)Fig. 2.1Count and record the number N of turns of wire in the coil.N = (1)© UCLES 2020[Turn over9702/33/F/M/20•S lide the compass into the plastic channel so that it is in the middle of the coil.(b)•C onnect the circuit as shown in Fig. 2.2.•R otate the channel on the bench until the arrow of the compass is perpendicular to the channel, as shown.clipFig. 2.2•T he distance between the first and last turns of wire is L, as shown in Fig. 2.2. Measure and record L.L = (1)(c) •C lose the switch. The compass arrow will rotate through an angle θ.•M easure and record θ.θ = .............................................................°•R ecord the ammeter reading I.I = ...............................................................•O pen the switch.[2]© UCLES 20209702/33/F/M/20(d)Estimate the percentage uncertainty in your value of θ.percentage uncertainty = (1)(e) Calculate the value of B usingB = μ0(N−1) ILwhere μ0 = 1.26 × 10–6 N A–2.B = .........................................N A–1 m–1 [1]9702/33/F/M/20© UCLES 2020[Turn over(f) •D isconnect the crocodile clips and remove the compass.•R emove the tape and the wire. Re-wind the wire in adjacent grooves (instead of every other groove), as shown in Fig. 2.3.•Re-fix the tape.Fig. 2.3•R epeat (a), (b), (c) and (e).N = ...............................................................L = ...............................................................θ = .............................................................°I = ...............................................................B = ...............................................N A–1 m–1[3]© UCLES 20209702/33/F/M/209702/33/F/M/20© UCLES 2020[Turn over(g) It is suggested that the relationship between θ and B istan θ = B kwhere k is a constant.(i) Using your data, calculate two values of k .first value of k = ............................................................... second value of k = ............................................................... [1] (ii)Justify the number of significant figures you have given for your values of k . ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] (iii)Explain whether your results in (g)(i) support the suggested relationship. ........................................................................................................................................... ........................................................................................................................................... . (1)9702/33/F/M/20© UCLES 2020Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity.To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at after the live examination series.Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge.(h) (i) Describe four sources of uncertainty or limitations of the procedure for this experiment.1 ...................................................................................................................................................................................................................................................................................2 ...................................................................................................................................................................................................................................................................................3 ...................................................................................................................................................................................................................................................................................4 ........................................................................................................................................ (4)(ii) Describe four improvements that could be made to this experiment. You may suggest the use of other apparatus or different procedures.1 ...................................................................................................................................................................................................................................................................................2 ...................................................................................................................................................................................................................................................................................3 ...................................................................................................................................................................................................................................................................................4 ........................................................................................................................................ (4)[Total: 20]。

Section3 particle physics1．The nuclear atom(1)Mass number and atomic number(2)Alpha particle scattering experiment(3)Electrons are accelerated by E and B(4)Write equations using standard nuclear notating(5)Explain why high energies are required to break particles into their constituents and to investigate fine structure(6) de Broglie wavelength λ=h/p(201306R-7)The de Broglie wavelength associated with electrons moving at 2.5 × 106 m s–1 isA .2.9 × 10–4mB .2.4 × 10–8 mC .2.9 × 10–10mD .2.4 × 10–39m (201306-10) The de Broglie wavelength for neutrons used to study crystal structure is 1.2 nm. mass of a neutron = 1.67 × 10–27kgThe speed of these neutrons would beA .3.0 × 106 m s–1 B.3.3 × 102m s–1 C.3.0 × 10–3m s–1 D.3.3 × 10–7m s–1 (201306R-8 )Which of the following is not a valid conclusion from Rutherford’s alpha particle scattering experiment?A .The atom is mainly empty space.B .The mass of the atom is mostly concentrated in the nucleus.C. The nucleus must be positively charged.D .The nucleus must be very small compared to the atom.(201306R-9) Select the row in the table that correctly identifies the composition of a 235U+ ion.92Number of protons Number of neutrons Number of electrons A9214391B9214392C9223591D9323592(201306-1)The nucleus of one of the isotopes of nickel is represented by 6028Ni. Which line correctly identifies a neutral atom of this isotope?Number of protons Number of neutrons Number of electrons A283228B283232C286028D602828（201201-10）The de Broglie wavelength of a moving tennis ball is calculated as 1× 10-33 m. This means that the moving tennis ballA .diffracts through a narrow slit. B.does not behave as a particle.C.does not display wave properties.D.is travelling at the speed of light.(201306R-10) .A transmissionelectron microscope passes abeam of electrons through atiny specimen to form an imageon a viewing screenDue to the wave nature ofelectrons, diffraction occurswhich can blur the image. Toreduce this effect whenviewing a smaller object thebeam must containA .more electrons per second.B .fewer electrons per second.C. faster moving electrons. D .slower moving electrons(201306-13).In an experiment to investigate the structureof the atom, α-particles are fired at a thin metal foil,which causes theα-particles to scatter.(a)(i) State the direction in which thenumber of α-particles detected will be amaximum(ii) State what this suggests about the structure of the atoms in the metal foil.(b) Some α-particles are scattered through180°.State what this suggests about the structure of the atoms in the metal foil.(c)The diagram shows the path of anα-particle passing near to a single nucleus inthe metal foil.(i)Name the force that causesthe deflection of the α-particle.(ii) On the diagram, draw anarrow to show the direction of the force acting on the α-particle at the point where the force is a maximum. Label the force F.(iii) The foil is replaced by a metal of greater proton number.Draw the path of an α-particle that has the same initial starting point and velocity as the one drawn in the diagram.(201301-11) Early in the twentieth century physicists observed the scattering of alpha particles after they had passed through a thin gold foil. This scattering experiment provided evidence for the structure of the atom.(a) State why it is necessary to remove the air from the apparatus that is used for this experiment.(b) From the results of such an experiment give two conclusions that can be deduced about the nucleus of an atom.Conclusion 1Conclusion 2(c) The diagram shows three α-particles ,allwith the same kinetic energy .The path followedby one of the particles is shown . Add to thediagram to show the paths followed by the othertwo particles.(201302-12 )The electron in a hydrogen atom can be described by a stationary wave which is confined within the atom.This means that the de Broglie wavelengthassociated with it must be similar to the size of the atom which is of the order of 10–10 m.(a)(i)Calculate the speed of an electron whose de Broglie wavelength is1.00 × 10–10m. (ii) Calculate the kinetic energy of this electron in electronvolts(b)When βradiation was first discovered ,it was suggested that there were electrons in the atomic nucleus, but it was soon realised that this was impossible because the energy of such an electron would be too great.Suggest why an electron confined within a nucleus would have a much greater energy than the energy calculated in (a)(ii).2．high-energy collisions(1)The role of E in linear particle accelerators(2)r=p/BQ in B(3)⊿E=c2⊿m(4)Units:MeV,GeV,MeV/c2 ,GeV/c2,u(201301-10)The tubes of a linear accelerator (linac) get progressively longer down its length becauseA .the accelerating particles become relativistic.B .the frequency of the applied potential difference changes.C .the accelerating particles must spend the same time in each tube.D .the accelerating particles gain mass（201206-10）As a particle accelerates in a linac, it passes through drift tubes of increasing lengths. This is so thatA.the particle can be given more energy within each tube.B.the frequency of the accelerating voltage can be constant.C.the accelerating voltage can be as high as possible.D.the time spent in the tube by the particle is longer(201306R-6)Charged particles are travelling at a speed v, at right angles to a magnetic field of flux density B. Each particle has a mass m and a charge Q.Which of the following changes would cause a decrease in the radius of the circular path of the particles?A .an increase inB B .an increase in mC .an increase in vD .a decrease in Q (201306-6)A muon has a mass of 106 MeV/c2. The mass of a muon, to two significant figures, isA .1.7 × 10–11 kgB .5.7 × 10–20 kgC .1.9 × 10–28 kg D.1.9 × 10–34 kg (201201-7)The rest mass of a proton is 1.67 × 10–27 kg. This mass, in MeV/c2 is approximatelyA .2.4 × 10–20 B.3.1 × 10–6 C.1.0 D .940(201106-10)The Large Hadron Collider is designed to accelerate protons to very high energies for particle physics experiments. Very high energies are required toA .annihilate hadrons. B.collide hadrons.C .create particles with large mass. D.produce individual quarks(201206-13) An electron and a positron annihilate with the emission of two photons of equal energy. Calculate the wavelength of the photons.(201201-17) (d) The technology suggested in the science fiction series, Star Trek, for powering the Starship Enterprise relied on antimatter. When an anti-hydrogen atom meets a hydrogen atom, they annihilate and produce energy.(i)How much energy, in joules, would be produced by the annihilation ofjust 1 milligram of anti-hydrogen atoms?(ii) Anti-protons are required to produce anti-hydrogen atoms. The total production of anti-protons on Earth over the past 25 years adds up to only a few nanograms.Suggest why so little anti-matter has been created3. Particle accelerators(1) E and B in cyclotron particle accelerators(2) Relativistic effects at speeds near c(201306-9)A cyclotron is a type of particle accelerator. It consists of two metal Dees which are connected to a high frequency voltage supply and are in a strong magnetic field. The particles change their speed becauseA.of the magnetic field they are in.B.the voltage supply is alternating.C.there is a potential difference between the two Dees.D.the magnetic field is at right angles to the Dees.(201306R-11)*11 The diagram shows the basic structure of a cyclotron.With reference to the magnetic field and the alternating potential difference explain how the cyclotron produces a beam of high speed particles.4. Particle theory(1) E and B in particle detectors(2) Charge energy and momentum are conserved in interactions between particles andhence interpret records of particle tracks(3) In the standard quark-lepton model each particle has a corresponding anti-particle meson:quark-anti-quark pairs baryon :quark triplets(4) Write and interpret equations using Standard symbols(201206-9)A pion can decay to produce two leptons. Which one of the following is possible?A.π+→e+ + v e B .π0→e– + v e C.π+→e++ e– D π0→π++ e–(201201-8)A positive kaon (K+) is a meson which includes a strange quark. Its structure could beA . u B. us C. D .usds s d d(201201-9) The K+ is likely to decay tov vA.π+ + μ– + μB. π++π0Cπ++π– D π0 + μ– +μ(201306-7 )The diagram shows the tracks from an event at a point P in a bubble chamber. A magnetic field is directed into the page. The tracks cannot show the production of a proton-antiproton pair with equal kinetic energies becauseA .the curvature is perpendicular to the magnetic field.B .the tracks curve in different directions.C. the tracks have different curvatures.D .there is no track before point P(201306-11)Scientists studying anti-matter recently observed the creation of a nucleus of anti-helium 4, which consists of two anti-protons and two anti-neutrons. The diagram represents the path of a proton through a magnetic field starting at pointX.Add to the diagram the path of an anti-helium 4 nucleus also startingat point X and initially travelling at the same velocity as theproton.Explain any differences between the paths.(201306-14)The photograph shows tracks in aparticle detector.(a) Explain the role of a magnetic field in a particledetector(b) Explain how you can tell that track XY was produced by a particle moving fromX to Y rather than from Y to X.(c)The particle that produced track XY was a π+.Deduce the direction of the magnetic field in the photograph(d) At Y, the π+ decayed into a positively charged muon (μ+) and a muon neutrino. The μ+ has a very short range before decaying into various particles, including a positron which produced the final spiral.(i) Give two reasons why you can deduce that the muon neutrino is neutral.(ii) Explain the evidence from the photograph for the production of the muon neutrino at Y.(201306R-14) .Hadrons are a group of particles composed of quarks. Hadrons can be either baryons or mesons.(a) (i) State the quark structure of a baryon.(ii) State the quark structure of a meson.(b) State one similarity and one difference between a particle and its antiparticle. Similarity:Difference :(c) (i) The table gives some of the properties of up, down and strange quarks.Type of quark Charge/e Strangenessu+2/30d-1/30s-1/3-1One or more of these quarks combine to form aΚ+, a meson with a strangeness of +1.Write down the quark combination of theΚ+..(ii)TheΚ+can decay in the following way Κ+→ µ++ vµK–is the antiparticle of the K+ . Complete the equation below by changing each particle to its corresponding antiparticle in order to show an allowed decay for the K–meson.Κ－→(iii)The rest mass of theΚ+ is 494 MeV/c2Calculate, in joules, how much energy is released if a Κ+ meets and annihilates a K–(201306-12)The table gives some of the properties of the up, down and strange quarks.Type of quark Charge/e Strangenessu+2/30d-1/30s-1/3-1There are nine possible ways of combining u, d and s quarks and their antiquarks to make nine different mesons. These are listed belowu d s d u s s u du u u d d d s s s (a) From the list select the four strange mesons and state the charge and strangeness of each of them.Meson Charge/e Strangeness(b) Some of the mesons in the list have zero charge and zero strangeness.Suggest what might distinguish these mesons from each other.*14 A bubble chamber is a particle detector which makes use of electric and magnetic fields. Explain the role of electric and magnetic fields in a particle detector. (201301-17)In 2011 physicists at the Relativistic Heavy Ion Collider (RHIC) announced the creation of nuclei of anti-helium-4 which consists of anti-protons and anti-neutrons instead of protons and neutrons(a) ‘Ordinary’ helium-4 is written as 42He.What do the numbers 4 and 2 represent?(b) In the RHIC experiment, nuclei of gold 19779Au travelling at speeds greater than2.99 × 108m s–1, in opposite directions, collided, releasing energies of up to 200 GeV. After billions of collisions, 18 anti-helium nuclei had been detected.(i) What is meant by ‘relativistic’ in the collider’s name?(ii) State why it is necessary to use very high energies in experiments such as these. (iii) Show that the mass of a stationary anti-helium nucleus is about 4 GeV/c2(iv) State why the small number of anti-helium nuclei produced only survive for a fraction of a second.(v) A slow moving anti-helium nucleus meets a slow moving helium nucleus. If they were to combine to produce 2 high energy gamma rays, calculate the frequency of each gamma ray.(c) There are two families of hadrons, called baryons and mesons. Baryons such asprotons are made of three quarks.(i) Describe the structure of a meson(ii) Up quarks have a charge of +2/3e and down quarks a charge of –1/3e.Describe the quark composition of anti-protons and anti-neutrons and use this to deduce the charge on each of these particles.（201206-18）18 (a) Physicists were able to confidently predict the existence of a sixth quark. State why(b) The mass of the top quark was determined by an experiment. Collisions between protons and anti-protons occasionally produce two top quarks.(i) How do the properties of a proton and an anti-proton compare?(ii) After the collision the two top quarks move in opposite directions with the same speed. Explain why.(c) The two top quarks decay rapidly into two muons and four jets of particles. These can be detected and their momenta measuredThe diagram shows an end-on view of the directions of the four jets (J1 to J4) of particles. The two muons are shown as μ1and μ2. A muon neutrino is also produced but cannot be detected, so is not shown. Each momentum is measured inGeV/c.(i)Explain why GeV/c is a valid unit for momentum.(ii) The vector diagram shown below is not complete. Add to the diagram arrows to represent the momenta of J3 and J4.(iii) Complete the diagram to determine the magnitude of the momentum of the muon neutrino.(iv) Show that the total energy of all the products of this event is about 300 GeV. (v) Deduce the mass of a top quark in GeV/c2.(vi) Suggest why it took a long time to find experimental evidence for the top quark.。

Turn over*P49594A0124*P49594A©2016 Pearson Education Ltd.1/1/1/1/1/1/1/1Instructions• Use black ink or ball-point pen.•Fill in the boxes at the top of this page with your name,centre number and candidate number.• A nswer all questions in Section A.•In Section B, answer all of questions 6(a) to 6(e) and one question from 6(f) or 6(g).•A nswer the questions in the spaces provided– there may be more space than you need.Information•The total mark for this paper is 80.•T he marks for each question are shown in brackets– use this as a guide as to how much time to spend on each question.•C alculators may be used.Advice•Read each question carefully before you start to answer it.•Check your answers if you have time at the end.*P49594A0224*2*P49594A0324*Turn over3*P49594A0424*4*P49594A0524*Turn over5*P49594A0624*6*P49594A0724*Turn over7BLANK PAGEQUESTION 6 BEGINS ON THE NEXT PAGE.*P49594A0824*8*P49594A0924*Turn over9Extract AProblems facing UK first time buyersOver a third of aspiring first time buyers in the UK have given up hope of ever being able to save for a deposit to buy a property, new research has found. A deposit is the minimum amount that must be paid upfront when buying a property, which is a proportion of the price of the property.The UK Government has a Help to Buy scheme designed to help people buy a home. The government subsidises the mortgage (loan) needed to buy a property, so buyers require a smaller amount saved as a deposit. Under this scheme, first time buyers need as little as a 5% deposit to qualify for a mortgage. This is due to end in the first few months of 2017. The research from mortgage insurer Genworth says this will mean a return to 20% deposits, which would see the average time needed to save for a deposit rise from three years to over 10 years.“Trying to buy your first home in the current climate is like chasing a runaway train. Even with good salaries that could comfortably support a mortgage, thousands of aspiring first time buyers can only save modest sums, especially those who are already paying rent. This deposit trap is why many feel they are left with the all or nothing choice of borrowing from family or waving goodbye to ever owning a home,” said Simon Crone, a vice president at Genworth.“Help to Buy has significantly improved access to mortgages with deposits that areactually realistic to save. The numbers using the scheme may be modest, but it has made significant inroads in the short-term to boost access at the lower end of the property market,” he pointed out.(Source: adapted from /news/europe/uk-first-time-buyers-201409309641.html September 2014)Extract BHousing and flood defencesA radical new approach to housing and a £2.3 billion of investment in flood defences were announced as part of the UK Government’s new National Infrastructure Plan 2014.One key proposal is for the government to plan, build and sell homes. An initial programme on a government-owned former airfield near Cambridge will see the development of 10 000 homes. This approach will fast track the development by providing certainty and making new homes available more quickly.The government will make the initial investment but expects that later costs will be met through the sale of land and homes. It will also evaluate the feasibility and economiceffects of rolling out this model on a wider scale, to support and accelerate housing supply.The plan also commits to £2.3 billion of capital investment to over 1400 flood defence projects in a 6-year programme of investment. As a result, over 300 000 homes will be better protected and over £30 billion of economic damages prevented. Majorprojects that will benefit include £42 million for the Oxford Flood Alleviation Scheme, £80 million for the Humber Estuary and over £17 million for Tonbridge, Yalding and the surrounding communities.(Source: adapted from https:///government/news/ambitious-plans-for-housing-flood-defence-and-roads-set-out-in-national-infrastructure-plan-2014)510152051510*P49594A01024*106 (a) With reference to Figure 1, explain one possible reason for the differences in priceelasticity of supply of new housing between countries.(5)(b) With reference to Extract A, assess the likely impact of the Help to Buy scheme onthe market for rented properties in the UK.(10)(c) Explain the likely impact on producer surplus of an increase in the demand forhousing. Use a diagram to illustrate your answer.(6)(d) With reference to Figure 2, calculate the percentage change in house pricesbetween the first quarter of 2009 and the first quarter of 2015. You are advised to show your working.(4)(e) With reference to Figures 1 and 2 and your own knowledge, discuss the functionsof the price mechanism in allocating housing.(15)Answer EITHER(f) Evaluate the likely microeconomic effects of government intervention in the UKhousing market.(20)OR(g) Evaluate the case for government provision of goods and services such as flooddefence schemes or housing.(20)(5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(b) With reference to Extract A, assess the likely impact of the Help to Buy scheme onthe market for rented properties in the UK.(10) .................................................................................................................................................................................................................................................................................... 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.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(c) Explain the likely impact on producer surplus of an increase in the demand forhousing. Use a diagram to illustrate your answer.(6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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.................................................................................................................................................................................................................................................................................... ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(d) With reference to Figure 2, calculate the percentage change in house pricesbetween the first quarter of 2009 and the first quarter of 2015. You are advised toshow your working.(4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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.................................................................................................................................................................................................................................................................................... .........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(e) With reference to Figures 1 and 2 and your own knowledge, discuss the functionsof the price mechanism in allocating housing.(15) 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AP® Physics C1976 Free Response QuestionsThe materials included in these files are intended for use by AP teachers for course and exam preparation in the classroom; permission for any other use must be sought from the Advanced Placement Program®. Teachers may reproduce them, in whole or in part, in limited quantities, for face-to-face teaching purposes but may not mass distribute the materials, electronically or otherwise. These materials and any copies made of them may not be resold, and the copyright notices must be retained as they appear here. This permission does not apply to any third-party copyrights contained herein.These materials were produced by Educational Testing Service® (ETS®), which develops and administers the examinations of the Advanced Placement Program for the College Board. The College Board and Educational Testing Service (ETS) are dedicated to the principle of equal opportunity, and their programs, services, and employment policies are guided by that principle.The College Board is a national nonprofit membership association dedicated to preparing, inspiring, and connecting students to college and opportunity. Founded in 1900, the association is composed of more than 4,200 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 22,000 high schools, and 3,500 colleges, through major programs and services in college admission, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, thePSAT/NMSQT®, and the Advanced Placement Program® (AP®). The College Board is committed to the principles of equity andexcellence, and that commitment is embodied in all of its programs, services, activities, and concerns.APIEL is a trademark owned by the College Entrance Examination Board. PSAT/NMSQT is a registered trademark jointly owned by the College Entrance Examination Board and the National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service.1976M1. A small block of mass m slides on a horizontal frictionless surface as it travels around the inside of a hoop of radius R. The coefficient of friction between the block and the wall is μ; therefore, the speed v of the block decreases. In terms of m, R. μ, and v, find expressions for each of the following.a. The frictional force on the blockb. The block's tangential acceleration dv/dtc. The time required to reduce the speed of the block from an initial value v0 to v o/31976M2. A cloth tape is wound around the outside of a uniform solid cylinder (mass M, radius R) and fastened to the ceiling as shown in the diagram above. The cylinder is held with the tape vertical and then released from rest. As the cylinder descends, it unwinds from the tape without slipping. The moment of inertia of a uniform solid cylinder about its center is ½MR2.a. On the circle below draw vectors showing all the forces acting on the cylinder after it is released. Labeleach force clearly.b. In terms of g, find the downward acceleration of the center of the cylinder as it unrolls from the tape.c. While descending, does the center of the cylinder move toward the left, toward the right, or straightdown? Explain.1976M3. A bullet of mass m and velocity v o is fired toward a block of thickness L o and mass M. The block is initially at rest on a frictionless surface. The bullet emerges from the block with velocity v o/3.a. Determine the final speed of block M.b. If, instead, the block is held fixed and not allowed to slide, the bullet emerges from the block with aspeed v o/2. Determine the loss of kinetic energy of the bulletc. Assume that the retarding force that the block material exerts on the bullet is constant. In terms ofL o, what minimum thickness L should a fixed block of similar material have in order to stop the bullet?d. When the block is held fixed, the bullet emerges from the block with a greater speed than when theblock is free to move. Explain.1976E1. A solid metal sphere of radius R has charge +2Q. A hollow spherical shell of radius 3R placed concentric with the first sphere has net charge -Q.a. On the diagram below, make a sketch of the electric field lines inside and outside the spheres.b. Use Gauss's law to find an expression for the magnitude of the electric field between the spheres at adistance r from the center of the inner sphere (R < r < 3R).c. Calculate the potential difference between the two spheres.d. What would be the final distribution of the charge if the spheres were joined by a conducting wire?1976E2. A conducting bar of mass M slides without friction down two vertical conducting rails which are separated by a distance L and are joined at the top through an unknown resistance R. The bar maintains electrical contact with the rails at all times. There is a uniform magnetic field B, directed into the page as shown above. The bar is observed to fall with a constant terminal speed v0.a. On the diagram below, draw and label all the forces acting on the bar.b. Determine the magnitude of the induced current I in the bar as it falls with constant speed v0 interms of B, L, g, v0, and M.c. Determine the voltage induced in the bar in terms of B, L, g, v0, and M.d. Determine the resistance R in terms of B, L, g, v0, and M.1976E3. An ion of mass m and charge of known magnitude q is observed to move in a straight line through a region of space in which a uniform magnetic field B points out of the paper and a uniform electric field E points toward the top edge of the paper, as shown in region I above. The particle travels into region II in which the same magnetic field is present, but the electric field is zero. In region II the ion moves in a circular path of radius R as shown.a. Indicate on the diagram below the direction of the force on the ion at point P2, in region II.b. Is the ion positively or negatively charged? Explain clearly the reasoning on which you base yourconclusion.c. Indicate and label on the diagram below the forces which act on the ion at point P1 in region I.P1d. Find an expression for the ion’s speed v at point P1 in terms of E and B.e. Starting with Newton’s law, derive an expression for the mass m of the ion in terms of B, E, q, and R.。

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONSGCE Advanced Subsidiary Level and GCE Advanced LevelMARK SCHEME for the June 2004 question papers9702 PHYSICS9702/01 Paper 1 (Multiple Choice (AS)), maximum mark 409702/02 Paper 2 (Structured Questions (AS)), maximum mark 609702/03 Paper 3 (Practical (AS)), maximum mark 259702/04 Paper 4 (Structured Questions (A2 Core)), maximum mark 609702/05 Paper 5 (Practical (A2)), maximum mark 309702/06 Paper 6 (Options (A2)), maximum mark 40These mark schemes are published as an aid to teachers and students, to indicate the requirements of the examination. They show the basis on which Examiners were initially instructed to award marks. They do not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination.All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated.Mark schemes must be read in conjunction with the question papers and the Report on the Examination. •CIE will not enter into discussion or correspondence in connection with these mark schemes.CIE is publishing the mark schemes for the June 2004 question papers for most IG CSE and G CE Advanced Level syllabuses.Grade thresholds taken for Syllabus 9702 (Physics) in the June 2004 examination.minimum mark required for grade:maximummarkA B Eavailable40 34 32 22Component160 45 41 272ComponentComponent 3 25 19 17 1160 40 33 17Component430 24 22 145Component40 21 18 106ComponentThe thresholds (minimum marks) for Grades C and D are normally set by dividing the mark range between the B and the E thresholds into three. For example, if the difference between the B and the E threshold is 24 marks, the C threshold is set 8 marks below the B threshold and the D threshold is set another 8 marks down. If dividing the interval by three results in a fraction of a mark, then the threshold is normally rounded down.June 2004GCE ADVANCED SUBSIDIARY LEVEL AND ADVANCED LEVELMARK SCHEMEMAXIMUM MARK: 40SYLLABUS/COMPONENT: 9702/01PHYSICSPaper 1 (Multiple Choice (AS))Page 1 Mark Scheme Syllabus Paper A/AS LEVEL EXAMINATIONS - JUNE 2004 9702 01Question Number KeyQuestionNumberKey1 B 21 C2 A 22 A3 A 23 C4 C 24 B5 C 25 A6 C 26 B7 B 27 C8 D 28 D9 D 29 D10 B 30 A11 A 31 D12 C 32 B13 A 33 C14 B 34 A15 D 35 D16 B 36 B17 A 37 D18 C 38 C19 A 39 C20 D 40 DJune 2004GCE ADVANCED SUBSIDIARY LEVEL AND ADVANCED LEVELMARK SCHEMEMAXIMUM MARK: 60SYLLABUS/COMPONENT: 9702/02PHYSICSPaper 2 (Structured Questions (AS))Categorisation of marksThe marking scheme categorises marks on the MACB scheme.B marks: These are awarded as independent marks, which do not depend on other marks. For a B-mark to be scored, the point to which it refers must be seen specifically in the candidate’s answer.M marks: These are method marks upon which A-marks (accuracy marks) later depend. For anM-mark to be scored, the point to which it refers must be seen in the candidate’s answer. If a candidate fails to score a particular M-mark, then none of the dependent A-marks can be scored.C marks: These are compensatory method marks which can be scored even if the points to which they refer are not written down by the candidate, providing subsequent working gives evidence that they must have known it. For example, if an equation carries a C-mark and the candidate does not write down the actual equation but does correct working which shows he/she knew the equation, then theC-mark is awarded.A marks: These are accuracy or answer marks which either depend on an M-mark, or allow a C-mark to be scored.Conventions within the marking schemeBRACKETSWhere brackets are shown in the marking scheme, the candidate is not required to give the bracketed information in order to earn the available marks.UNDERLININGIn the marking scheme, underlining indicates information that is essential for marks to be awarded.1 (a) scalar: magnitude onlyB1 vector: magnitude and direction (allow scalar with direction) B1 [2] (allow 1 mark for scalar has no direction, vector has direction)(b) diagram has correct shapeM1 with arrows in correct directionsA1 resultant = 13.2 ± 0.2 N (allow 2 sig. fig) A2 [4](for 12.8 → 13.0 and 13.4 → 13.6, allow 1 mark)(calculated answer with a correct sketch, allow max 4 marks) (calculated answer with no sketch – no marks)Total [6] 2 (a) (i) λ = 0.6 m B1(ii) frequency (= v /λ) = 330/0.60 = 550 HzC1A1 [3] (use of c = 3 x 108 ms -1scores no marks) (b) amplitude shown as greater than a but less than 2a and constant B1 correct phase B1 [2] (wave to be at least three half-periods, otherwise -1 overall) Total [5] 3 (a) (i) scatter of points (about the line) B1(ii) intercept (on t 2axis) B1 [2] (note that answers must relate to the graph) (b) (i) gradient = ∆y /∆x = (100 – 0)/(10.0 – 0.6) C1gradient = 10.6 (cm s -2) (allow ±0.2) A1 [2] (Read points to within ±21square. Allow 1 mark for 11 cm s -2i.e. 2 sig fig, -1. Answer of 10 scores 0/2 marks)(ii) s = ut + 21at 2 B1so acceleration = 2 x gradient B1acceleration = 0.212 m s -2B1 [3] Total [7]4 (a) (i) (p =) mv B1 (ii) E k = 21mv 2 B1algebra leading to M1E k = p 2/2m A0 [3] (b) (i) ∆p = 0.035 (4.5 + 3.5) OR a = (4.5 + 3.5)/0.14 C1= 0.28 N s = 57.1 m s -2force = ∆p/∆t (= 0.28/0.14) OR F = ma (= 0.035 x 575.1) (allow e.c.f.) C1 = 2.0 N A1 Note: candidate may add mg = 0.34 N to this answer, deduct 1 markupwards B1 [4](ii) loss = 21x 0.035 (4.52 – 3.52) C1= 0.14 J A1 [2](No credit for 0.282/(2 x 0.035) = 1.12 J) (c) e.g. plate (and Earth) gain momentum i.e. discusses a 'system' B1 equal and opposite to the change for the ball i.e. discusses force/momentum M1 so momentum is conserved i.e. discusses consequence A1 [3] Total[12]5 (a) (i) distance = 2πnr B1(ii) work done = F x 2 πnr(accept e.c.f.)B1 [2](b) total work done = 2 x F x 2πnr B1but torque T = 2Fr B1hence work done = T x 2πn A0 [2](c) power = work done/time (= 470 x 2π x 2400)/60)= 1.2 x 105 W A1 [2]Total[6]6 (a) When two (or more) waves meet (not 'superpose' or 'interfere') B1resultant displacement M1is the sum of individual (displacements) A1 [3](b) (i) any correct line through points of intersection of crests B1(ii) any correct line through intersections of a crest and a trough B1 [2](c) (i) λ = ax/D OR λ = a sin θ and θ = x/D C1650 x 10-9 = (a x 0.70 x 10-3)/1.2 C1a = 1.1 x 10-3 m A1 [3]1 no change B1(ii)2 brighter B1[3]3 no change (accept stay/remain dark) B1Total [11]7 (a) (i) P = VI C1current = 60/240 = 0.25 A A1(ii) R ( = V/I) = 240/0.25 M1960ΩA0 [3]=(b) R = ρL/A(wrong formula, 0/3)C1960 = (7.9 x 10-7 x L)/(π x {6.0 x 10-6}2) C1L = 0.137 m A1 [3](use of A = 2πr, then allow 1/3 marks only for resistivity formula)(c) e.g. the filament must be coiled/it is long for a lamp B1 [1](allow any sensible comment based on candidate's answer for L)Total[7]8 (a) V/E = R/R tot or 0.5l x 3900 C1=1.0/1.5 = R/(R + 3900) or 1.0 = 0.5R/3900 M1R = 7800Ω.or R = 7800ΩA0 [2](b) V = 1.5 x (7800/{7800 + 1250}) or I = 1.5/(7800 + 1250) C1= 1.29 V..or V = IR = 1.29 V A1 [2](c) Combined resistance of R and voltmeter is 3900 ΩC1reading at 0 °C is 0.75 V A1 [2]Total [6]June 2004GCE ADVANCED SUBSIDIARY LEVEL AND ADVANCED LEVELMARK SCHEMEMAXIMUM MARK: 25SYLLABUS/COMPONENT: 9702/03PHYSICSPaper 3 (Practical (AS))Page 1 Mark Scheme Syllabus Paper A/AS LEVEL EXAMINATIONS - JUNE 2004 9702 03 (a) Pointer B reading to the nearest half millimetre or millimetre 1Extension correct and to nearest millimetreCondone negative values (i.e. do not penalise 'upside down' rule)(b) Calculation of spring constant to 2 or 3 sf 1k = 0.98/x answer must be given in N m-1.Ignore any negative signs. Do not allow fractions(c) (i) Diameter of one mass to at least 3 sf 1Accept value ± 0.2 mm of Supervisor’s value(ii) Percentage uncertainty in diameter 2 One mark for d∆ (either 0.1 mm or 0.2 mm).One mark for correct ratio and multiplication by 100.(iii) Cross-sectional area 2 One mark for A = πr2.One mark for correct substitution into A = πr2. ECF from (c)(i).Do not allow the second mark if diameter substituted into A = πr2.Wrong formula scores zero in this section.(d) (iv) Measurements 2Expect to see six sets of results in the table (one mark).l must be correct; check a value (one mark).If correct, then tick. If incorrect, then do not award the second mark, and write inthe correct value. If pointer reading not shown then this mark cannot be scored.Minor help given by Supervisor, -1. Major help, then -2.Column headings for d and l (one mark for each correct heading). 2Expect to see a quantity and a correct unit.There must be a distinguishing feature between the quantity and the unit.Consistency of d and l readings. 2Values should be given to the nearest mm.One mark each.(e) (iii) Gradient is negative. 1No ecf from misread rule if gradient is positive.Gradient calculation. 1∆ used must be greater than half the length of the drawn line.Check the read-offs (must be correct to half a small square).Ratio must be correct (i.e. ∆y/∆x and not ∆x/∆y).Graph Axes 1 Scales must be such that the plotted points occupy at least half the graph grid inboth the x and y directions (i.e. at least 6 large squares on the longer side of thegrid and at least 4 squares on the shorter side of the grid).Scales must be labelled. Do not allow awkward scales (e.g. 3:10, 6:10 etc.).Allow reversed axes (penalise in section (f))Plotting of points 1Count the number of plots and write as a ringed total on the graph grid.All the observations must be plotted or this mark cannot be scored.Check a suspect plot. Circle and tick if correct.If incorrect, show correct position with arrow, and -1.Work to half a small square.Line of best fit 1There must be at least 5 trend plots for this mark to be scored.There must be a reasonable balance of points about the line of best fit.Curved trend cannot score this mark. Quality of results1Judge by scatter of points about the line of best fit.There must be at least 5 trend plots for this mark to be scored. Incorrect trend (i.e. positive gradient) will not score this mark.(f)Gradient equated withkAgw ρ−. Condone misuse of negative sign. 1Value in range 800 – 1200 kg m -3(or 0.80 to 1.20g cm -3)1 This mark cannot be scored if the gradient has not been used.This mark will not be scored if there is a Power Of Ten error in the working or reversed axes.Unit correct (kg m -3)1If another unit has been given then it must be consistent with the value.Significant figures in w ρ1Accept 2 or 3 sf only. Ignore trailing zeros (except w ρ = 1000)(g) Difficulty1e.g. hard to see the water surface/surface tension problems/refraction effects/parallax errors. Do not allow vague 'human error'.Improvement 1 e.g. use calibrated beakers or masses/paper behind/mirror behind/travelling microscopeDo not allow 'use dye'/repeat readings.25 marks in totalJune 2004GCE ADVANCED SUBSIDIARY LEVEL AND ADVANCED LEVELMARK SCHEMEMAXIMUM MARK: 60SYLLABUS/COMPONENT: 9702/04PHYSICSPaper 4 (Structured Questions (A2 Core))Categorisation of marksThe marking scheme categorises marks on the MACB scheme.B marks: These are awarded as independent marks, which do not depend on other marks. For a B-mark to be scored, the point to which it refers must be seen specifically in the candidate’s answer.M marks: These are method marks upon which A-marks (accuracy marks) later depend. For anM-mark to be scored, the point to which it refers must be seen in the candidate’s answer. If a candidate fails to score a particular M-mark, then none of the dependent A-marks can be scored.C marks: These are compensatory method marks which can be scored even if the points to which they refer are not written down by the candidate, providing subsequent working gives evidence that they must have known it. For example, if an equation carries a C-mark and the candidate does not write down the actual equation but does correct working which shows he/she knew the equation, then the C-mark is awarded.A marks: These are accuracy or answer marks which either depend on an M-mark, or allow aC-mark to be scored.Conventions within the marking schemeBRACKETSWhere brackets are shown in the marking scheme, the candidate is not required to give the bracketed information in order to earn the available marks.UNDERLININGIn the marking scheme, underlining indicates information that is essential for marks to be awarded.1 (a) charge is quantised/enabled electron charge to be measured B1 [1](b) all are (approximately) n x (1.6 x 10-19C) so e = 1.6 x 10-19C (allow 2 sig. fig. only M1A1 [2]summing charges and dividing ten, without explanation scores 1/2 Total [3]2 (a) mean (value of the) square M1 of the speeds (velocities) of the atoms/particles/molecules A1 [2] (b) (i) p = 31 ><2c ρC1<c 2> = 3 x 2 x 105/2.4 = 2.5 x 105C1r.m.s speed = 500 ms -1A1 [3](ii) new <c 2> = 1.0 x 106 or <c 2> increases by factor of 4 C1<c 2> ∝ T or 3/2 kT = 1/2 m<c 2>C1 T = {(1.0 x 106) / (2.5 x 105)} x 300 = 1200 K A1 [3] Total [8]3 (a) (i) (force) = GM 1M 2/(R 1 + R 2)2B1(ii) (force) = M 1R 1ω2 or M 2R 2ω2B1 [2](b) ω = 2π/(1.26 x 108) or 2π/T C1= 4.99 x 10-8 rad s -1A1[2] allow 2 s.f.: 1.59π x 10-8scores 1/2 (c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω2 = M 2R 2ω2B1hence M 1/M 2 = R 2/R 1 A0[2] (ii) R 2 = 3/4 x 3.2 x 1011 m = 2.4 x 1011m A1R 1 = (3.2 x 1011) – R 2 = 8.0 x 1010m (allow vice versa) A1 [2] if values are both wrong but have ratio of four to three, then allow 1/2(d) (i) M 2 = {(R 1 + R 2)2 x R 1 x ω2} I G (any subject for equation) C1= (3.2 x 1011)2 x 8.0 x 1010 x (4.99 x 10-8)2/(6.67 x 10-11) C1= 3.06 x 1029kg A1 (ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4](9.17 x 1029kg for more massive star) Total [12] 4 (a) e.g. amplitude is not constant or wave is damped B1 do not allow 'displacement constant' should be (-)cos, (not sin) B1 [2] (b) T = 0.60 s C1ω = 2π/T = 10.5 rad s -1(allow 10.4 → 10.6) A1 [2] (c) same period B1 displacement always less M1 amplitude reducing appropriately A1 [3]for 2nd and 3rdmarks, ignore the first quarter period Total [7]5 (a) the (value of the) direct current M1 that dissipates (heat) energy at the same rate (in a resistor) A1 [2] allow 'same power' and 'same heating effect' (b) √2I rms = I 0 B1 [1] (c) (i) power ∝ I 2 or P = I 2R or P = V I C1 ratio = 2.0 (allow 1 s.f.) A1 [2] (ii) advantage: e.g. easy to change the voltage B1disadvantage: e.g. cables require greater insulation....... rectification – with some justificationB1 [2] (d) (i) 3.0 A (allow 1 s.f.) A1 (ii) 3.0 A (allow 1 s.f.) A1[2] Total[9] 6 0 - + (-1 for each error) B2 + + 0 (-1 for each error) B2 + + 0 (-1 for each error) B2[6]Total[6] 7 (a) λ = h /p or λ = h /mvM1with λ, h and (or mv) p identified A1 [2](b) E = 21mv 2C1= p 2/2m or v = √(2E /m ), hence M1 λ = h /√(2mE ) A0 [2] (c) E = qV C1(0.4 x 10-9)2 x 2 x 9.11 x 10-31 x 1.6 x 10-19 x V = (6.63 x 10-34)2C1 V = 9.4 V (2 s.f. scores 2/3) A1 [3] Total [7] 8 (a) S shown at the peak B1 [1] (b) (i) Kr and U on right of peak in correct relative positions B1 [1](ii)1 binding energy of U-235 = 2.8649 x 10-10Jbinding energy of Ba-144 = 1.9211 x 10-10Jbinding energy of Kr-90 = 1.2478 x 10-10J C2energy release = 3.04 x 10-11J (-1 if 1 or 2 s.f.) A1 [3]2E = mc 2C1m = (3.04 x 10-11)/3.0 x 108)2 = 3.38 x 10-28kg (ignore s.f.) A1 [2] (iii) e.g. neutrons are single particles, neutrons have no binding energy per nucleon B1 [1] Total [8]June 2004GCE ADVANCED SUBSIDIARY LEVEL AND ADVANCED LEVELMARK SCHEMEMAXIMUM MARK: 30SYLLABUS/COMPONENT: 9702/05PHYSICSPaper 5 (Practical (A2))Question 1(a) (v) Sensible use of fiducial marker placed at centreposition/ 1oscillation/meanofpositionequilibrium(a) (vi) Measurements 36 sets scores one mark. Allow more than 6 sets without penalty.Write the number of readings as a ringed total by the table.Choose a row in the table. Check values for T 2d & d2. Tick if correct.One mark each. If incorrect, write in correct values. Ignore small rounding errors.Impossible values of d or t, -1. Misread stopwatch –1.Minor help from the Supervisor, -1. Major help, then -2.Repeats 1 Expect to see at least two sets of readings of raw times.At least half the raw times > 20 s 1Column heading for T2d1 The column heading must contain a quantity and a unit (e.g. s2 m or s2 cm).There must be some distinguishing mark between the quantity and the unit.Consistency 1 tod (all values of d must be given to the nearest millimetre).ApplySF in d 21 Check by row in the table; compare with raw values of d.The number of significant figures in d2 must be the same as, or one better than,the number of significant figures in d.(a) (vii) Justification of sf in d 2 1Answer must relate the number of sf in d.Do not allow answers in terms of decimal places.(b) (i) Axes 1 The axes must be labelled with the quantities plotted. Ignore units on the axes. The plotted points must occupy at least half the graph grid in both the x and ydirections (i.e. 4 large squares in the x -direction and 6 large squares in the y -direction). Do not allow more than 3 large squares between the labels on an axis. Do not allow awkward scales (e.g. 3:10, 6:10, 8:10 etc.). If axes reversed (i.e. d 2 against T 2d ) then zero and ecf.Plotting of points 1 All the observations must be plotted. Do not allow plots in the margin area.Check one suspect plot. Circle this plot. Tick if correct. If incorrect, mark the correct position with a small cross and use an arrow to indicate where the plot should have been, and score zero. Allow errors up to and including half a small square.Line of best fit 1 Only a drawn straight line through a linear trend is allowable for this mark. This mark can only be awarded for 5 or more plots on the grid.There must be a reasonable balance of points about the drawn line. Do not allow a line of thickness greater than half a small square. Quality of results 1Judge by scatter of points about the line of best fit.5 trend plots can score this mark. Curved trend scores zero.This mark can only be scored if a graph of d 2 against T 2d or T 2d against d 2 has been plotted.(b) (iii) Gradient 1 Ignore any units given with the value. Hypotenuse of ∆ must be > half the length of line drawn. Check the read-offs. Work to half a small square. ∆x /∆y gets zero.Values taken from the table that lie on the line to within half a small square are acceptable. y -intercept 1 The value must be read to the nearest half square. Allow calculation from y = mx + c (c) k = gradient of line of best fit 1 A numerical value is expected. Substitution method scores zero. A = candidate’s value for the y -intercept 1 A numerical value is expected. Substitution method scores zero. Unit of A correct and consistent with value (e.g. s 2 m or s 2 cm) 1 If incorrect allow ecf from column heading in table. (d) Value of T w hen d = 1.0 cm 1 Must be in range 3 – 8 s. A power of ten error anywhere in the working will result in this mark not being scored. Working must be checked. Bald answer scores zero.20 marks in totalQuestion 2basicOK 1ideaandequipmentchoiceA1 Sensibleoffield/detectorSource/magneticInappropriate choice of apparatus cannot score this mark.Ignore lead or aluminium plates at this stage.A2 Method of measuring angle of deflection 1(e.g. detector at edge of large protractor/lengths & trig ratio used)Do not allow vague ‘use a protractor’.This mark can be awarded even if the detector has not been specified.A3 Use Hall probe/search coil/current balance to measure field strength 1 Allow Helmholtz coils expression if Helmholtz coils used.Allow a current or voltage measurement as indication of field strength (as I α B)B1 Method of removing α radiation or statement that α radiation almost undeflected 1 Use paper or distance to detector > few cm/air to absorb alphaCould be shown on the diagram. Do not allow lead/aluminium plate.Allow α to be shown deflecting in the opposite direction to β on the diagram.B2 γ-radiation undeflected/deflect beta particles using electric field 1 Can be shown on diagram. Do not allow ‘absorb gamma with lead plate’.B3 Workablefields 1uniformforprocedureMeasure deflection and field strength; change current in coils and repeat.C1/2 Any two safety precautions 2e.g. use source handling toolstore source in lead lined box when not in usedo not point source at people/do not look directly at sourceplace lead sheet at ‘end of experiment’ to absorb unwanted raysD1/2Any good/further detail. Examples of creditworthy points might be: 2 Type of detector (GM tube/film/screen/scintillation counter). N/a cloud chamber/CRORepeat readings to allow for randomness of activityCorrect deflection of beta on diagram/left hand rule ideas (diagram or written)Separation of coils = radius of coils for uniform fieldDiscussion of count rate (and not just count)Plane of semiconductor slice is perpendicular to field linesCalibrate Hall probeDetail of calibrationCollimation ideasAllow other valid points. Any two, one mark each.B1 = B2 = B3 = 0 if lead or aluminium plate is placed in front of the source. Allow thin(less than 1 mm) sheet or foil10 marks in total.June 2004GCE ADVANCED SUBSIDIARY LEVEL AND ADVANCED LEVELMARK SCHEMEMAXIMUM MARK: 40SYLLABUS/COMPONENT: 9702/06PHYSICSPaper 6 (Options (A2))Categorisation of marksThe marking scheme categorises marks on the MACB scheme.B marks: These are awarded as independent marks, which do not depend on other marks. For a B-mark to be scored, the point to which it refers must be seen specifically in the candidate’s answer.M marks: These are method marks upon which A-marks (accuracy marks) later depend. For anM-mark to be scored, the point to which it refers must be seen in the candidate’s answer. If a candidate fails to score a particular M-mark, then none of the dependent A-marks can be scored.C marks: These are compensatory method marks which can be scored even if the points to which they refer are not written down by the candidate, providing subsequent working gives evidence that they must have known it. For example, if an equation carries a C-mark and the candidate does not write down the actual equation but does correct working which shows he/she knew the equation, then the C-mark is awarded.A marks: These are accuracy or answer marks which either depend on an M-mark, or allow a C-mark to be scored.Conventions within the marking schemeBRACKETSWhere brackets are shown in the marking scheme, the candidate is not required to give the bracketed information in order to earn the available marks.UNDERLININGIn the marking scheme, underlining indicates information that is essential for marks to be awarded.© University of Cambridge International Examinations 2004Option A – Astrophysics and Cosmology1 (a) In an infinite and static Universe M1every line of sight should end on a star M1(or spherical shells argument)so sky at night should be bright A1 [3](b) For expanding Universefinite age limits size (1)light from distant galaxies is red-shifted out of visible (1)light from distant young stars not yet reached Earth (1)Any two points, maximum 2 B2 [2]Total [5]2 (a) 1 pc = 3.26 ly (allow 3.3 ly) C1distance = 16/3.26 = 4.9 pc A1 [2](b) base line is 2 AU C1angle = 2 x 1/4.9= 0.41 arc sec B1 [2]Total[4]3 (a) Universe is same everywhere/homogeneous/isotropic M1when considered on a sufficiently large scale A1 [2](b) characteristic of (black body) 3 K radiation B1CMB is highly isotropic/same from all directions M1This indicates that the Universe is highly uniform A1 [3]Total[5]4 (a) e.g. planet observed by reflected light B1this is too faint (against the starlight) B1small B1tooe.g.physicallyto be resolved (at such great distances) B1 [4](any sensible suggestion (B1) with some further comment (B1) – max 4)(b) e.g. change in intensity of starlight M1as the star is eclipsed A2 [2]e.g. wobble in position of star (M1)as planet orbits star (A1)(any sensible suggestion plus some further comment – max 2)Total [6]Option F – The Physics of Fluids5 (a) force = upthrust – weight of polystyrene in air C1=V x (1000 – 15) x 9.8 C125V = 2.6 x 10-3 m3 A1[3](b) boat will tend to right itself/float higher in the water M1if at positions B A1 [2]Total [5]6 (a) if air is streamline B1air above car moves faster than air below M1so (by Bernoulli) pressure above is lower than below M1and car experiences an upward force A1 [4](b) the spoiler causes turbulence M1turbulence prevents the lift force from developing A1 [2]Total [6]7 (a) symmetrical pattern on above/below sphere M1© University of Cambridge International Examinations 2004© University of Cambridge International Examinations 2004lines closer near top and bottom of sphere A1 [2](b) (i) force on particle = 4/3 πr 3(w ρρ−)gC1 = 4/3 x π x (4.5 x 10-7)3 x (2.9 x 103) x 9.8= 1.08(5) x 10-14NC1 1.085 x 10-14 = 6 x π x (4.5 x 10-7) x 9.5 x 10-4x vC1 v = 1.35 x 10-6 m s -1A1[4] (ii) in 1.0 hours, particles move 1.35 x 10-6 x 3600 (= 4.85 x 10-3m) B1 fraction = (8.0 – 4.85)/8.0 C1 = 0.39 A1 [3] (allow 2/3 for answer of 0.61) Total [9] Option M – Medical Physics8 (a) piezo-electric/quartz crystal B1 across which is applied an alternating voltage B1 crystal vibrates B1 at its resonant frequency B1 [4] (b) (i) trace length = 4.0 mm C1distance = speed x time = 1450 x 0.4 x 10 x 10-6= 5.8 x 10-3m C1 thickness = 0.29 cm A1 [3] (ii) trace length = 5.2 cm C1 thickness = 4.1 cm A1 [2] Total [9] 9 (a) ability of eye to form focused images M1 of objects at different distances from the eye A1 [2] (b) (i) 25 cm (allow ± 5 cm) to infinity B1 [1] (ii) (for close-up vision), power = 1/0.25 – 1/1.2 C1 = 3.17 D A1 (for distance vision), power = -0.25D A1 [3] (iii) use bifocal lenses B1 further detail e.g. region of lens identified B1 [2] Total [8] 10 loss of hearing at higher frequencies B1 loss of sensitivity (at about 3 kHz) B1 further comment on either e .g. upper limit should be about 15 kHz, at 3 kHz, I.L. should be about 10 dB (or less) B1 [3] Total [3] Option P – Environmental Physics11 (a) (i) Sun's energy incident per unit time per unit area M1 on the cross-sectional area of the Earth A1 [2](ii) solar constant = (3.9 x 1026)/(4π x {1.5 x 1011}2)C1 = 1380 W m -2A1[2] (b) at C, greater thickness of atmosphere so more absorption B1 also larger area (for beam of a particular width) B1 explanation of 'larger area' (e.g. diagram or 1/cos θ, with θ clear) B1 [3] Total [7] 12 (a) e.g. daily variations as industry opens up/closes down daily variations with TV programmes, cooking meals, lighting seasonal variations with heating/AC, length of day (any reasonable response, 1 for daily, 1 for seasonal plus 1 more)1 each, max 3 B3 [3](b) power demand may change suddenly B1 pumped water scheme can be brought onto full load in a short time B1 can use surplus energy at times of low demand to pump water 'back up' TotalB1 [3][6]。

9702 Physics June 2005 CONTENTS FOREWORD (1)PHYSICS (2)GCE Advanced Level and GCE Advanced Subsidiary Level (2)Paper 9702/01 Multiple Choice (2)Paper 9702/02 Structured Questions (3)Paper 9702/03 Practical 1 (6)Paper 9702/04 Core (7)Paper 9702/05 Practical 2 (10)Paper 9702/06 Options (12)FOREWORDThis booklet contains reports written by Examiners on the work of candidates in certain papers. Its contents are primarily for the information of the subject teachers concerned.PHYSICSGCE Advanced Level and GCE Advanced Subsidiary LevelPaper 9702/01Multiple ChoiceQuestion Number KeyQuestionNumberKey1 C 21 D2 C 22 D3 B 23 A4 C 24 B5 A 25 B6 D 26 B7 B 27 C8 D 28 B9 A 29 B10 A 30 D11 A 31 C12 C 32 C13 A 33 D14 D 34 D15 B 35 B16 C 36 D17 B 37 B18 A 38 C19 C 39 B20 A 40 CGeneral commentsThe paper provided good syllabus coverage and proved to have good discrimination and a relatively high overall facility. The mean mark of 24.7 with a standard deviation of 6.6 showed that while some candidates found many of the questions difficult there were enough straightforward questions for the majority to find plenty to do. This year there were some very able candidates, with around 5% able to score 90% or more. This is an outstanding achievement, especially bearing in mind the pressure of time in answering forty questions in only one hour. At the bottom end of the ability range only 8.4% scored less than 40%. This again shows that the vast majority of candidates were well prepared for the examination and had reasonable knowledge of the topics which were being tested. There is a tendency with a multiple choice examination for candidates to work on numerical questions just with their calculator. They do need to see that this is a dangerous habit as many of the distractors are obtained simply by manipulating the data given in a plausible way. Candidates do need to work carefully and use units as a check if they are to avoid pitfalls. It is difficult to ascertain how the timing of the paper seemed to the candidates, but there was no direct evidence of candidates being unable to complete the paper in the allotted time.Comments on specific questionsThe questions which generally did not cause candidates any problems were Questions1, 2, 6*, 12, 17, 19*, 23, 30, 31, 35*, 38* and 40* where the marks were correspondingly high. The starred questions have correct answers from more than 90% of candidates. This is particularly good for questions such as 38 and 40 as it indicates that candidates are familiar with standard terminology. It does however, mean that the discrimination is low for these questions.The following questions were answered poorly.Question 4, where candidates showed that they were often unable to work with percentage uncertainties. A 2% uncertainty in the time and a ¼% uncertainty in the distance gives a total uncertainty of 2¼% in the speed which is therefore 16.0 ± 0. 4.Question 7, where candidates do not seem to appreciate that a body under conditions of free fall has a constant acceleration of g, and even when stationary at the top of its path its acceleration is still g because its velocity is still changing at the same rate.Question 13, where particular care is required to get the moment of the 20 N force and to subtract the moments provides by the 5 N and 10 N forces. Many candidates ignored the 10 N force altogether.Question 33, where C was the most popular answer. This implies that many candidates ignored the fact that if the length is doubled and the volume remains the same then the area of cross-section must be reduced to half its former value, giving a new resistance of 4R.Question 39 was one of a very few questions where one of the distractors (A) had as many selecting it as the correct answer, but this may be because by this stage candidates were in a hurry and resorted to guessing.Paper 9702/02Structured QuestionsGeneral commentsThere were sections of questions that were accessible to all candidates. On the other hand, some parts provided a challenge to the more-able candidates. There were some excellent scripts and several Centres where large numbers of candidates achieved a uniformly high standard.Question 7 highlighted once again two areas of general weakness. First, the drawing of unjustified inferences from quoted formulae by considering two variables in isolation. Second, a lack of appreciation of significant figures both as regards premature ‘rounding’ at intermediate stages in calculations and also the number of significant figures that can reasonably be quoted in a final answer. Both aspects have been mentioned previously in reports.There was no evidence that candidates were short of time. In scripts where the last two questions had not been completed, then almost invariably there were gaps in answers to earlier questions.Comments on specific questionsQuestion 1The speed of sound produced the most reasonable estimates and the density of air the least. For all four quantities there was a very wide range of inadequate estimates, many of which showed that the candidates had little or no appreciation for the physical magnitudes of the various quantities.(a) Estimates ranged from 1.5 m s–1 to 3 × 1022 m s–1.(b) Estimates ranged from 1.7 × 10–31 kg m–3 to 8 × 1023 kg m–3.(c) Estimates ranged from 0.5 mg to 800 kg.(d) Estimates ranged from 10–6 cm3 to 1012 cm3.(a) It seemed that few candidates had actually observed the Brownian motion of smoke particles.Although most candidates did make a reference to ‘random’ or ‘haphazard’ motion, they frequently then went on to state that this motion was due to collisions between smoke particles or between smoke particles and the walls of the container. Only rarely did a candidate mention that specks of light are observed, not the smoke particles themselves.(b) Most answers did include a statement that the motion would still be random, but slower. Often, noexplanation was given. Where explanation was provided, this was usually confined to a statement that greater mass would lead to smaller velocity changes. Very few candidates made any reference to greater surface area and that, because of the random distribution of velocities of air molecules, the effects of collisions of the smoke particle with air molecules would tend to average out. This greater rate of collision would lead to a smaller, rather than a larger, randomness of collision and hence motion of the smoke particle.Question 3(a)(i) Most candidates were able to calculate the change in energy. E rrors were mostly due toinappropriate units for mass and/or vertical displacement.(ii) In most answers, the change in gravitational potential energy was associated correctly with thekinetic energy of the block and bullet. A few candidates obtained the correct answer by unjustified use of equations representing uniform motion in a straight line.(b) Candidates were instructed to use the principle of conservation of momentum and consequently,most answers were correct.(c)(i) Very few candidates failed to calculate a value for the kinetic energy, based on their answers to (b).(ii) The majority of answers included a correct deduction that there would be a loss of kinetic energyand thus the collision is inelastic. However, some asserted that the evidence for a loss of kinetic energy was that some had been converted into potential energy of the block and bullet.Answers : (a)(i) 0.51 J; (b) 390 ms –1; (c)(i) 152 J.Question 4 (a) With few exceptions, the glass was said to be brittle.(b)(i) This calculation presented very few problems for candidates. The most common error wasarithmetical.(ii) E xpressions quoted for the Young modulus were usually correct. Again, most errors werearithmetical, particularly amongst those candidates who did not make a separate calculation of the strain.(iii) The majority did calculate an appropriate area of the graph provided or did use an equivalentcorrect formula. However, a significant minority either used the expression21 × stress × strain ,without realising that this is energy per unit volume, or merely calculated the maximum strain. (c) In general, this was poorly answered. Despite the explicit information given in the question, many answers were based on an assumption that either the mass, the density or the kinetic energy would be different. Frequently, wording was imprecise. The hard ball was said to ‘have more force’ or the soft ball ‘to use up its force in squashing’. Nevertheless, there were some good answers, based on either the times of collision affecting the rate of change of momentum and thus the force on the glass or the conversion of kinetic energy to strain energy in the balls affecting the strain energy within the glass.Answers : (b)(i) 7.6 x 107 Pa, (ii) 6.1 x 1010 Pa, (iii) 9.0 x 10–3 J.(a) Candidates should be encouraged to be as precise as possible with definitions and explanations.Most answers included a statement, with varying degrees of clarity, that diffraction is the bending or spreading of waves through a narrow gap or at an edge. However, wording was frequently ambiguous so that the explanation could apply to refraction. Statements such as ‘change of direction when meeting an obstacle’ are not acceptable.(b)(i) This simple calculation caused many problems. Not only were there many errors involvingpowers-of-ten in otherwise correct calculations but also, many candidates thought that they must use the grating formula given in the question.(ii)In most scripts where d had been calculated correctly, then the maximum value of n was also correct. However, candidates who had made errors in d by as much as factors of up to 1012 usually also calculated equally ridiculous answers for n, without comment.(iii) Very few candidates recognised that the correct use of the formula relies on the fact that light incident on the grating has no path difference. Most candidates repeated the question by stating that the light is not normal to the grating.(c) Many candidates gave at least one relevant difference, based on either angles of diffraction orintensity. A surprisingly large number of answers ignored the fact that the question specified the two wavelengths involved. Consequently, answers referred to different wavelengths, frequencies or colour. In others, there were vague mentions of angles, without a clear indication as to which angle reference was being made.Answers: (b)(i) 1.33 x 10–6 m, (ii) 2.Question 6(a)(i) Candidates should be encouraged to use a ruler when drawing straight lines. In this case, it wasexpected that, by eye, the lines would be straight and equally spaced. A significant number of diagrams were unacceptable free-hand sketches.(ii) Almost all candidates were able to derive the given result.(b)(i) A surprisingly large proportion of candidates drew arrows that were either normal to the electricfield or normal to the axis of the particle.(ii)Although the majority of calculations were correct, there were significant numbers of answers with incorrect physics. Calculations of the Coulombic forces between two charged particles were not uncommon. Others introduced either a sine or a cosine term into the calculation.(iii) Most candidates did multiply the force calculated in (ii) by a distance. However, very frequently this was not the perpendicular distance between the forces. In most of these scripts, this was not a matter of confusion between use of sine or cosine terms.(iv) In the vast majority of scripts, it was realised that the forces would cause rotation. However, most answers gave the impression that the rotation would be continuous, with relatively few stating that the particle would align itself along the direction of the field.Answers: (b)(ii) 2.4 x 10–12 N, (iii) 3.4(4) x 10–15 Nm.Question 7(a) Many candidates gave unsatisfactory answers in terms of ‘the hindrance’ or ‘the opposition to thecurrent’. Of those who did attempt a definition in terms of a ratio, many were imprecise, either defining the unit rather than the property or, frequently, using the unqualified term ‘voltage’ when what was intended is the potential difference across the resistor.(b)(i) The vast majority of answers were correct for the data point.(ii)With few exceptions, the calculation was completed successfully.(iii) This part of the calculation was more challenging and there were many well-expressed correct solutions. However, there were significant numbers of answers where work was laid out poorly. In such cases, many candidates failed to appreciate whether they were dealing with the e.m.f. of the battery, the p.d. across the resistor or the p.d. across the internal resistor. A small but significant number of candidates treated the external and internal resistors as if they were in parallel.(c)This part was poorly answered with very few showing clarity of thought. Many candidates made nomention of internal resistance, simply quoting a formula for power dissipation, and concluding that the lower value of the p.d. across R would give a smaller power, ignoring the fact that both the value of the current and the resistance would change. Others stated that the larger p.d. across R would give greater power dissipation in R thus the power dissipation in the inernal resistor must be smaller. A minority of candidates argued, quite correctly, that a larger p.d. across R would give a smaller p.d. across the internal resistance and since internal resistance is constant and P = V2/r, then the power dissipation in the internal resistor would be lower.Answers: (b)(i) 1.13 W, 1.50 V, (ii) 1.99 Ω, (iii) 1.99 Ω.Question 8(a) Few candidates failed to plot correctly the position for the isotope of protactinium.(b) The values of A and Z were given and consequently, almost all candidates who attempted this taskdid plot the position correctly. It was realised that the daughter product of plutonium would have the same value for A. However, opinion was divided as to whether Z should be 93 or 95.Paper 9702/03Practical 1General commentsThe overall standard of the work produced by the candidates was generally good, although as in previous years the performance variation was mainly by Centre (i.e. some Centres continue to prepare their candidates very well for this examination). It was pleasing to see fewer low scores (< 12) than in previous years and there were quite a lot of strong candidates scoring 21+. Most Centres had no difficulty with the apparatus requirements, although there were some cases where Centres had used spring balances calibrated in grams instead of Newtons. There were very few reports of Supervisors giving assistance to candidates. There was no evidence that candidates were short of time and no problems with the rubric. Comments on specific questionsQuestion 1In this question candidates were required to investigate how the force required to maintain the equilibrium of a suspended mass depends on the angle between the line of action of the force and the horizontal.In (a)(ii) many candidates stated the uncertainty in θ to be 0.5°,which was considered to be unrealistic as it was difficult to place the protractor in the correct position when measuring the angle.In (b) the difficulties mentioned by candidates included thick string/difficulty of alignment of the Newton meter/holding the protractor steady whilst measuring θ/keeping AB horizontal, all of which were credited. Some answers were vague (e.g. ‘it was difficult to read the scale on the newton meter’ or ‘parallax error’ with no clarification). Oscillation of the string or the mass was not accepted.Most candidates were able to set up the apparatus correctly and use it to obtain six sets of readings for F and θ. A number of candidates misread the scale on the protractor and obtained values of θ which were acute instead of obtuse. A few candidates calculated 1/sin θ using θ in radians instead of degrees.Virtually all candidates presented the results in tabular form. Raw values of θ were sometimes given to an unreasonable degree of precision (i.e. to one or two decimal places). A number of candidates had impossibly large values for F , presumably because they had used spring balances instead of newton meters. These candidates were not penalised.Candidates were required to plot a graph of F against 1/sin θ. Common errors made by the weaker candidates included poor choice of scales (i.e. where the plotted points occupied less than half the graph grid in both the x and y directions) or where the scales were awkward (e.g. one large square on the grid corresponding to three units). Points were usually plotted correctly, although it was sometimes difficult to see where the points had been plotted. It is expected that small crosses will be used. When plotting errors occurred it was usually because awkward scales had been chosen. It is expected that candidates will plot six points since six observations have been made. Candidates who did not plot all their observations were penalised. Most of the better candidates were able to determine a value for the gradient of the line correctly. When the mark was not awarded it was usually because the triangle that had been used was too small or an error had been made in the read-offs (particularly when awkward scales had been used). The y -intercept was often read incorrectly from the graph because a ‘false origin’ had been used (i.e. the value was read from a line were x ≠ 0). The more able candidates substituted values from a point on the line, together with the gradient value, into y = mx + c .Two marks were available for the ‘quality of results’. This was judged on the scatter of points about the line of best fit. Candidates who had done the experiment carefully were able to score here if the scatter of points about a line of best fit was small. Candidates lost marks if they used a narrow range of angles (a spread of < 10° was common).The analysis section continues to differentiate well between candidates. The weaker candidates often did not attempt this section. The better candidates were able to equate mg with the gradient of the graph and kwith the y -intercept. A significant number of candidates equated θsin mgto the gradient. Candidates wereinstructed to use their answers from (e) to determine values for m and k . Many of the weaker candidates did not do this and attempted to substitute two sets of values into the given equation and solve the resulting equations simultaneously. Work of this kind was not credited.It was expected that m and k would be given to a sensible number of significant figures (i.e. two or three significant figures) and that units would also be given. A large number of candidates failed to recognise that the unit of k is the Newton, and gave the value of k to one significant figure only.It was pleasing to see many of the more able candidates scoring full marks in the analysis section.Paper 9702/04CoreGeneral commentsAll questions were accessible to better candidates. Weaker candidates tended to score the majority of their marks on the first four questions.It was pleasing to note that the number of scripts where work is laid out well and adequate explanation is given is increasing. There were some outstanding scripts from a number of Centres.Candidates appeared to have sufficient time to complete their answers. However, there were some instances where it appeared as if the candidate did not realise that a question was printed on the back page of the script, possibly because the penultimate question ended half-way down page 15 or because the subject material of the questions was not in syllabus order. Candidates should be encouraged to read carefully the question paper. On the cover page it stated that there are 16 printed pages in the question paper and on page 15, the instruction ‘[Turn over ’ was given.Comments on specific questionsQuestion 1(a)(i) Most candidates successfully calculated the magnitude of the angular velocity. A minority didconfuse ‘speed’ with ‘angular velocity’.(ii) A correct formula was usually given. There were very few arithmetical errors. Some did substitute w for v in the expression F = mv2/r.(b)(i)It was expected that candidates would mention gravitational force. Some did merely state ‘the Sun’but others gave more than required by stating that the gravitational force between the Sun and the Earth provides the centripetal force.(ii)Most candidates did quote a correct expression for the gravitational force and, generally, the arithmetic was correct. A minority of candidates used the expression GM = r3ω2. This alternative approach could be awarded full credit.Answers: (a)(i) 1.96 x 10–7 rads–1, (ii) 3.46 x 1022 N; (b)(ii) 1.95 x 1030 kg.Question 2(a) Most candidates mentioned either the universal gas equation or individual gas laws. However, itwas disappointing to note that very few stated that, for an ideal gas, the law(s) must be obeyed at all values of pressure, volume and temperature.(b)(i) A significant number stated that <c2> represented the root-mean-square speed. Very few referredto the square of the mean speed.(ii)There were some very pleasing derivations, with clear explanation at all stages. On the other hand, the work of some candidates lacked all meaning. It was common to find that density was defined as M/V, with M being used in the expression for the mean kinetic energy of an atom. The term N was then either ignored, cancelled or stated to be unity.(c)(i)Generally, this calculation was completed successfully. The most common error was a failure tosquare the value for the speed.(ii)It was pleasing to note that more-able candidates made reference to a distribution of speeds. A significant number of candidates thought that atoms with a lower speed would escape. There wasa number of ingenious explanations, including the possibility of isotopes.Answer: (c)(i) 1.9 x 104 K.Question 3(a) Definitions tended to lack precision. It was common to find that any reference to unit mass wasomitted. Furthermore, weaker candidates tended to fail to state what is meant by fusion or thought that energy is required to convert liquid to solid.(b)(i)Although the majority of answers were correct, a significant minority considered only the energyrequired to warm the ice or to melt it.(ii)Most candidates did attempt the calculation, using their answer in (i). However, very few took into account the energy required to heat the melted ice from 0°C to the final temperature.Answers: (b)(i) 8700 J, (ii) 16°C.Question 4(a) In general, definitions were adequate. The most common failing was either to omit to give the relative directions of the displacement and the acceleration or to express them poorly.(b)There were some very clearly expressed derivations. Weaker candidates tended to give some relevant expressions but were unable to link them. Candidates should always be encouraged to give as much relevant information as they can, since credit is often given for such expressions.(c)(i)The majority of answers did include an acceptable value for either the period or the frequency of the oscillations. However, there were many answers where it appeared that the expression for the area was unknown. Besides confusing diameter and radius, it was not uncommon to find the expression A = 4πr 2. Candidates should be advised to use data given in the question paper. The use of g = 10 ms –2 is not acceptable where data for a calculation is given to two or more significant figures.(ii)It was pleasing to note that very few candidates referred to displacement, rather than amplitude or peak height.Answer : (c)(i) 0.0384 kg. Question 5(a)Answers were disappointing. Very few made reference to potential gradient or x V ∆∆/. Most who attempted this part of the question stated E = V/x . This is, of course, incorrect since V is defined as potential and x as the distance from the centre of the sphere.(b)Surprisingly few candidates knew that the electric field within the conductor must be zero. Most candidates drew a sketch similar to that in Fig. 5.2 or merely a curve starting at x = O.Question 6(a)(i)Generally satisfactory but it was evident that a minority of candidates had no real appreciation of concepts involving electromagnetic induction.(ii)Most answers were based, quite correctly, on the equation P = VI . However, many failed to state that the output power would need to be constant.(b)(i) Generally correct.(ii)Most sketches did show an appropriate sinusoidal wave with the correct frequency. However, veryfew indicated the correct phase. Most indicated no phase difference with a few showing a π rad change between Fig. 6.3 and Fig. 6.4.(iii) It was common to find that this section was not attempted. Of those who did give an answer, themajority failed to state a unit for the angle.Answer : (b)(iii) π21or 90°.Question 7(a) Surprisingly, many graphs were drawn without a scale on the y -axis. Consequently, only the general shape of the line could be given.(b)(i)The initial number was calculated correctly in the majority of scripts. However, a significant number of candidates appeared to have little idea as to how to proceed with this basic calculation.(ii)A significant number of candidates thought that they had to use an equation involving exponential decay. Of those who did use the correct equation, it was pleasing to note that most did determine the decay constant in s –1.(c)Many candidates did not appear to know how to proceed with this problem. Of those who did, the majority took the ratio N/N o to be 1/9, rather than 1/10.Answers : (b)(i) 1.5 x 1016, (ii) 1.11 x 1012 Bq; (c) 8.63 hours.Question 8 (a) Definitions were disappointingly poor. Many failed to make it clear that a ratio is involved.Consequently, in such statements it appeared that capacitance is an electric charge.(b)(i) Most answers included, in some way, that there would be charge separation. However, very rarelywas any explanation given as to how this charge separation leads to the storage of electrical energy.(ii) Correct answers were in a minority. Many quoted 221CV as the energy stored in a chargedcapacitor but then assumed ()2122V V − is equal to ()212V V −. Others used the expression QV 21but assumed that Q would remain constant when V changed.Answer : (b)(ii) 1.4 J.Paper 9702/05 Practical 2General commentsThe general standard of the work done by the candidates was similar to last year, with quite a wide spread of marks. Question 1 was relatively straightforward, although some of the weaker candidates found the analysis section challenging and gave very brief answers to Question 2. As in previous years, there was a significant range in performance. It would be helpful to candidates generally if attention could be drawn to the published mark schemes.There were no reported difficulties from Centres in obtaining the necessary equipment for Question 1. Very little help was given to candidates from Supervisors in setting up the apparatus in Question 1. Supervisors are reminded that under no circumstances should help be given with the recording of results, graphical work or analysis.A small number of weaker candidates appeared to be short of time. Answers to Question 2 from these candidates was often very brief, or finished in mid-sentence.There were no common misinterpretations of the rubric.Comments on specific questionsQuestion 1In this question candidates were required to investigate how the period of oscillation of a loaded steel blade varies with the length of the blade. (a) Virtually all candidates were able to set up the equipment without help from the Supervisor. (b) In this part many candidates did not repeat the readings of raw times. (c) Virtually all candidates were able to record six measurements of d and t . A few candidates did notdivide the raw times by the number of oscillations and calculated log t instead of log T . Some of the raw times were too small (i.e. less than 10 seconds) because candidates did not allow the blade to perform a sufficient number of oscillations. Values of d were usually given to the nearest millimetre, although weaker candidates recorded their values of d to the nearest centimetre. It is expected that d will be recorded to the nearest millimetre as a millimetre scale had been used to make the measurement.。

Turn over P43321A©2013 Pearson Education Ltd.1/1/1/1/C2*P43321A0132*Instructions•Use black ink or ball-point pen.•Fill in the boxes at the top of this page with your name,centre number and candidate number.•Answer all questions.•Answer the questions in the spaces provided– there may be more space than you need.Information•The total mark for this paper is 80.•The marks for each question are shown in brackets– use this as a guide as to how much time to spend on each question.•Questions labelled with an asterisk (*) are ones where the quality of yourwritten communication will be assessed– you should take particular care with your spelling, punctuation and grammar, aswell as the clarity of expression, on these questions.•The list of data, formulae and relationships is printed at the end of this booklet.•Candidates may use a scientific calculator.Advice•Read each question carefully before you start to answer it.•Keep an eye on the time.•Try to answer every question.•Check your answers if you have time at the end.2*P43321A0232*BLANK PAGE3*P43321A0332*Turn over4*P43321A0432*5*P43321A0532*Turn over6*P43321A0632*7*P43321A0732*Turn over8*P43321A0832*9*P43321A0932*Turn over10*P43321A01032*Explain the meaning of the terms hard, stiff and high tensile strength.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................*(b) It is important that a piano wire has a high elastic limit.Explain why this is important.(3) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................(total for Question 12 = 6 marks)The strip of paper is shown below. The start and the end of the journey are indicated.Using measurements from the tape show that the final velocity of the trolley is.................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................(ii) Hence calculate the average acceleration of the trolley.(2) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................Average acceleration = ..................................................................(b) Using a ticker tape timer is one method of measuring the speed of a moving object ina laboratory. Another method is to use a light gate with a data logger and computer.Suggest an advantage of using the light gate method rather than using a ticker tapetimer.(1) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................(total for Question 13 = 5 marks)14 The picture shows a track for racing toy electric cars. A guide pin fits in a groove in thetrack to keep the car on the track. A small electric motor in the car is controlled, with ahand-controller, via contacts in the track.A child places a car of mass 95 g on the track. She adjusts the controller to a power of4.2 W so the car accelerates from rest for 0.40 s.(a) (i) Show that the energy transferred by the motor in 0.40 s is about 2 J.(2) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................(ii) Calculate the speed of the car at 0.40 s.(2) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................Speed = ........................................................(iii) Suggest why the actual speed of the car is less than the calculated speed.(1) .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................(b) At high speed the guide pin may become disengaged from the groove.Use Newton’s first law to explain why the car would then leave the track at a corner.(2) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................(total for Question 14 = 7 marks)(b) (i) Use Stoke’s law to show that the SI unit of viscosity is Pa s.(2) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................(ii) A small sphere is dropped into a large volume of ethanol at 24 °C.Show that, if the drag were due to viscous forces alone, the terminal velocitywould be about 4 ms–1.Assume that upthrust is negligible.radius of sphere = 5.0 × 10−4 mroom temperature = 24 °Cmass of sphere = 4.0 × 10−6 kg(3) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................*(c) Diesel is used as the fuel in some vehicles. Diesel is not renewable, so alternatives are being researched. Biodiesel is a fuel made from vegetable oil; biodiesel on its own is not suitable for use in vehicles.The table gives some information about diesel, biodiesel and ethanol.Viscosity / mPa sat 0 °c Viscosity / mPa sat 40 °cEnergy /MJ kg–1Freezingpoint / °cDiesel 4.9 2.643 –30Biodiesel17.3 4.639 –12Ethanol 1.80.927–114Blends of biodiesel with ethanol are being researched as a renewable alternative todiesel fuels for use in vehicles all year round.Using the information in the table, suggest why these blends are being researched.(3) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................(total for Question 15 = 10 marks)BLANK PAGE16 The photograph shows an athlete performing a long jump.At take-off his horizontal speed is 8.0 m s–1 and his vertical speed is 2.8 m s–1.(a) Show that the total time the athlete spends in the air is about 0.6 s.Assume that his centre of gravity is at the same height at take-off and landing.(3) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................(b) Calculate the horizontal distance jumped by the athlete.(2) .................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................. ..................................................................................................................................................................................................................................................Horizontal distance = ..................................................................21*P43321A02132*Turn over(c) In reality, when the athlete lands his centre of gravity is 50 cm lower than its position at take-off. Calculate the extra horizontal distance this enables the athlete to jump.(4)....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Extra horizontal distance = ..................................................................(total for Question 16 = 9 marks)22*P43321A02232*17 Pile drivers have been used for centuries to push piles into the ground for use as foundations of buildings and other structures. A large mass (the driver) is raised and then dropped onto an object (the pile) which is pushed into the ground.The picture shows the pile driver that was used to build a London bridge in the 17th century.(a) (i) The driver on the pile driver above had a mass of 810 kg and could be dropped a maximum distance of 6.0 m onto the pile.Show that the energy transferred from the driver is about 50 kJ.(2)............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ (ii) In one instance, 40% of this energy is used usefully to drive in the pile. The pile moves 0.20 m into the ground.Determine the average resistive force acting on the pile as it moves through the ground.(3)....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Average resistive force = .................................................................driverpile23*P43321A02332*Turn over24*P43321A02432*25*P43321A02532*Turn over *(iii) T he graph shows how the compression of the wooden cushion varies with force, as the force is applied and removed during an impact.Use the graph to explain the following: 1. the wooden cushion has to be replaced after a few hundred impacts,(2)........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................2. with each impact the temperature of the wooden cushion rises slightly.(1).................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... (total for Question 17 = 13 marks)ForceCompressionForce appliedForce removed。

Give your students the best global opportunitiesA school’s guide to EdexcelA level qualificationsThe A level landscapeis changingThe A level landscape is changing. This booklet will walk you through the latest developments, our Edexcel A level qualifications and the continuing support available to you.In 2013 the Office of Qualifications and Examinations Regulation (Ofqual) in England announced changes to GCE A level, including the immediate removal of January examinations and the proposed move from modular assessment to linear from September 2015, with the introduction of a new suite ofGCE A levels.We know how important January examinations are to many of our international customers, so we have developed an additional suite of A level qualifications - International Advanced Levels - that contain both January and June examinations.Find the right solution for your schoolInternational centres are now in the unique position of being able to offer Edexcel GCE A levels and Edexcel International Advanced Levels totheir students.Written to the same world-class standards as all Pearson qualifications, our two Edexcel A level qualifications equip students with the knowledge and skills to realise their ambitions and open doors to the university of their choice. Whether you choose to deliver Edexcel GCE A level, Edexcel International Advanced Level or both qualifications at your centre, you will receive the same comprehensive support services to help you deliver an outstanding learning experience, with confidence.Better technology.Better support.We understand that delivering first-class qualifications takes time and careful planning, which is why we strive to provide you with an unparalleled level of support alongside our high-quality Edexcel qualifications.Support with exam preparation and assessment›Past papers›Mark schemes›Examiner reports›ResultsPlus Mock Analysis Teaching and delivery support›Ask the Expert›Subject Advisors›Subject community forums›Training solutionsResources›Range of free teaching resources including: pastexam papers, schemes ofwork and more.›Range of paid-for digital and print resources Results/postresults support›ResultsPlus›Certification›Exam feedback sessions ›Grade boundariesResultsPlusProvides centres with instant and detailed analysis of your students’ GCE A level and International Advanced Level exam and mock performance. Widely used by teachers across the world, it helps you identify the areas where students could benefit from extra support and guidance, driving attainment.Find out more at: /resultsplusResultsPlus helps me by looking at individual students and analysing their performance, finding the exact topic they either did well and succeeded in, or the ones that need improvement... we then look at setting up revision or booster classes, some sort of intervention, so we can improve their result for the re-sit. Panayiotis Hadjisergis, student, The Nicosia Grammar School, Cyprus.Edexcel GCE A levelThe international gold standard in academic qualifications. British standardsRegulated by Ofqual, Edexcel GCE A Levels are the exact same as those offered in the UK. Demanding, rigorous and empowering, they are among the most widely known and respected academic qualifications in the world.Latest news on GCE A level reform - timelineOfqual and the UK Government are in the process of reforming A levels in England. The proposed changes, which will affect Edexcel GCE A levels, include:To keep up to date with the proposed changes visit/edexcel-alevel-reformThe Edexcel A level programme is great for me as this system demands consistency and will assist me towards my dream of studying Journalism.Natalie, student, Kuala Lumpur* GCE A levels to move from a modular assessment structure, to linear. All examinations to be taken at the end of the two-year course.From 2016 New GCE A levels for Mathematics, Further Mathematics, Geography and Modern Foreign Languages for first teaching from September 2016.From 2015New GCE A levels willbe introduced for firstteaching in September2015 in the followingsubjects: Art & Design,Biology, BusinessStudies, Chemistry,Computing Science,Economics, English(Language, Literatureand Language &Literature), History,Physics, Psychologyand Sociology.Subject areas available: ›Business and Economics ›Design and Technology ›Performing Arts ›English ›Humanities ›Languages ›Mathematics ›Music ›Physical Education ›Sciences Key Features:›level 3 qualifications›available in over 30 subjects›usually studied full-time by 16 – 19 year olds at school or college›modular structure until September 2015*›AS level a stand-alone qualification from September 2015*›annual examinations in June›typically, a Edexcel GCE AS level is taken over one year and a Edexcel GCE A level over two years›regulated by Ofqual.Key documents - available online:›Specifications›Sample Assessment Materials›Teacher Support Materials›Mapping documents›Examiner Reports›Question papers and mark schemes.Education that opens doorsAvailable in over 30 subjects, Edexcel GCE A levels equip learners with the knowledge and skills they need to access the very best global higher education and employment opportunities.Edexcel A levels offer excellent preparation for university, not only in terms of the learning material but also the level of intensity. The approaches we used to tackle analytical problems or written assignments for our A levels still come in useful at the university level.Lien Tran, student,The Nicosia Grammar School, Cyprus.Edexcel International Advanced Level (IAL)Flexible approach to international learningDesigned for the global learnerEdexcel International Advanced Levels have been developed to the same high standards as Edexcel GCE A Levels for the unique needs of the global learner. They provide the flexibility to teach a modular A level qualification that is 100% externally assessed with exams in both January and June.Initially available in eight popular subjects, IALs can be taken alongside Edexcel GCE A levels, or BTEC and LCCI qualifications to provide a comprehensive suite of subjects for students to choose from.Advance to the topDeveloped by education specialists specifically for international students, Edexcel International Advanced Levels prepares them to access to the world’s top universities and employment opportunities.Key documents - available online:›Specifications›Sample Assessment Materials›Teacher Support Materials›Mapping documents›Examiner reports›Question Papers and Mark Schemes.Edexcel International Advanced Levels are available in the following popular subjects: ›Accounting ›Biology ›Business Studies›Chemistry›Economics›English Language ›English Literature ›History ›Law ›Mathematics ›Physics ›PsychologyKey Features:›level 3 qualifications›usually studied full-time by 16 – 19 year olds at school or college›flexible, modular structure›January and June examinations›100% external assessment with no coursework›regulated by Pearson›specifications can be delivered in an international context›AS can be used to complete an Advanced Level or taken as a stand-alone qualification›written to the same rigorous standards as all Edexcel qualifications, for global recognition.Whateveryour A levelrequirements,we have asolution for you NEW subjects for first teaching fromSeptember 2016:›Applied ICT›Arabic›French›German ›Greek ›Geography ›SpanishTeach Edexcel A levels›Contact your regional office›We will work with you through your application›Together we choose the programme that is right foryou and your students›Your centre is approved to offer Edexcel qualificationsLearn more at/edexcelLearn more at/teachedexcelAbout PearsonAt the core of everything we do is the desireto make a measurable impact on improvingpeople’s lives through learning. Pearsonaspires to be the world’s leading learningcompany. From primary to secondaryschool, through to professional certification;our qualifications, curriculum materials,multimedia learning tools and testingprogrammes help to educate millions ofpeople worldwide.12/15。