Cambridge A Level Physics 9702 — 2025 May/June Paper 2 · Variant 1

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level * 0 7 5 4 4 7 2 6 0 7 * PHYSICS 9702/21 Paper 2 AS Level Structured Questions May/June 2025 1 hour 15 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● 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 may use a calculator. ● You should show all your working and use appropriate units. INFORMATION ● The total mark for this paper is 60. ● The number of marks for each question or part question is shown in brackets [ ]. DC (DE/SG) 342399/3 © UCLES 2025 ,










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2 9702/21/M/J/25 © UCLES 2025 Data acceleration of free fall g = 9.81 m s–2 speed of light in free space c = 3.00 × 108 m s–1 elementary charge e = 1.60 × 10–19 C unified atomic mass unit 1 u = 1.66 × 10–27 kg rest mass of proton mp = 1.67 × 10–27 kg rest mass of electron me = 9.11 × 10–31 kg Avogadro constant NA = 6.02 × 1023 mol–1 molar gas constant R = 8.31 J K–1 mol–1 Boltzmann constant k = 1.38 × 10–23 J K–1 gravitational constant G = 6.67 × 10–11 N m2 kg–2 permittivity of free space ε0 = 8.85 × 10–12 F m–1 (4 1 0 rf = 8.99 × 109 m F–1) Planck constant h = 6.63 × 10–34 J s Stefan–Boltzmann constant σ = 5.67 × 10–8 W m–2 K–4 Formulae uniformly accelerated motion s = ut + 2 1 at 2 v 2 = u 2 + 2as hydrostatic pressure ∆p = ρg∆h upthrust F = ρgV Doppler effect for sound waves fo = v v f v s s ! electric current I = Anvq resistors in series R = R1 + R2 + ... resistors in parallel R 1 = R 1 1 + R 1 2 + ... * 0000800000002 * ,










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4 9702/21/M/J/25 © UCLES 2025 1 (a) Define acceleration. … … [1] (b) In an experiment, two objects A and B are released from the side of a building, as shown in Fig. 1.1. 3.6 m 14 m u A B building ground Fig. 1.1 (not to scale) Object A is released from rest at a height of 14 m above the horizontal ground. Object B is released with an initial upwards vertical velocity u at a height of 3.6 m above the ground. Both objects take the same time to reach the ground and they do not collide with each other. Air resistance is negligible. (i) Calculate the time taken for object A to reach the ground. time = … s [2] (ii) Use your answer in (b)(i) to calculate u. u = … m s−1 [2] * 0000800000004 * ,










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5 9702/21/M/J/25 © UCLES 2025 [Turn over (c) In a second experiment, object B is released from the same height and given the same initial speed as in (b) but at a release angle θ to the vertical, as shown in Fig. 1.2. u B building θ Fig. 1.2 (i) State and explain whether the time taken for object B to reach the ground is less than, the same as or greater than the time in (b)(i). … … … [2] (ii) By considering energy, state and explain the effect of the change in release angle on the speed at which B reaches the ground. … … … [2] [Total: 9] * 0000800000005 * ,










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6 9702/21/M/J/25 © UCLES 2025 2 (a) State the principle of moments. … … … [2] (b) Three objects A, B and C are placed on a horizontal beam. The beam is in equilibrium, as shown in Fig. 2.1. C beam pivot 6.0 m B x A Fig. 2.1 (not to scale) The beam is uniform and has length 6.0 m. The pivot is at the midpoint of the beam. Object A has mass 60 kg and is at one end of the beam. Object B has mass 45 kg and is at a distance x from the pivot. Object C has mass 80 kg and is at the other end of the beam. Calculate x. x = …m [3] * 0000800000006 * ,










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7 9702/21/M/J/25 © UCLES 2025 [Turn over (c) The beam is 0.80 m above horizontal ground. Object A is removed and replaced by a spring connected to the ground and the beam, as shown in Fig. 2.2. C beam spring pivot 0.80 m B ground Fig. 2.2 After the change, the beam is again horizontal and in equilibrium. The positions of B and C are unchanged. The spring has an unstretched length of 0.59 m and obeys Hooke’s law. (i) Calculate the spring constant of the spring. spring constant = … N m–1 [3] (ii) Calculate the elastic potential energy of the spring. elastic potential energy = … J [2] [Total: 10] * 0000800000007 * ,










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8 9702/21/M/J/25 © UCLES 2025 3 (a) Define power. … … [1] (b) An electric car is powered by a motor. The car is travelling at a constant speed of 35 m s−1 along a straight horizontal road, as shown in Fig. 3.1. 35 m s–1 1750 N road car Fig. 3.1 There is a total resistive force of 1750 N acting on the car. (i) Calculate the power transmitted to the wheels of the car by the motor. power = …W [2] (ii) Calculate the useful work done by the motor when the car travels a distance of 17 km. work done = … J [2] * 0000800000008 * ,










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9 9702/21/M/J/25 © UCLES 2025 [Turn over (iii) The potential difference (p.d.) across the motor has a constant value of 600 V and the motor has an efficiency of 85%. Calculate the current in the motor. current = …A [3] (c) The car in (b) now reaches a slope, as shown in Fig. 3.2. 35 m s–1 road car Fig. 3.2 The car continues down the slope at the same speed as in (b). State and explain the effect, if any, of the slope on: (i) the air resistance acting on the car … … [1] (ii) the current in the motor. … … [1] [Total: 10] * 0000800000009 * ,










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10 9702/21/M/J/25 © UCLES 2025 4 (a) State the principle of superposition. … … … [2] (b) Light of wavelength 7.2 × 10−7 m is incident normally on a double slit, as shown in Fig. 4.1. light, wavelength 7.2 × 10−7 m double slit screen 0.16 mm D Fig. 4.1 (not to scale) A screen is at a distance D from the double slit. The double slit and the screen are parallel. The separation of the slits in the double-slit arrangement is 0.16 mm. The resulting interference pattern on the screen contains nine dark fringes, as shown in Fig. 4.2. 3.2 cm dark fringes bright fringes Fig. 4.2 (not to scale) * 0000800000010 * ,










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11 9702/21/M/J/25 © UCLES 2025 [Turn over The distance between the centres of the first and ninth dark fringes is 3.2 cm. (i) Calculate D. D = …m [3] (ii) The slit separation is now gradually decreased from 0.16 mm to 0.04 mm. The distance between the centres of adjacent dark fringes is x. On Fig. 4.3, sketch the variation of x with slit separation. 0 0 0.4 0.8 1.2 1.6 0.04 0.08 0.12 slit separation / mm 0.16 x / cm Fig. 4.3 [3] [Total: 8] * 0000800000011 * ,









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12 9702/21/M/J/25 © UCLES 2025 5 (a) Use the definitions of speed v, frequency f and wavelength λ to derive the wave equation v = f λ . [2] (b) A source of sound waves of frequency 236 Hz is travelling at a constant velocity of 20 m s−1. A stationary observer has a microphone connected to a cathode-ray oscilloscope (CRO). The microphone detects the sound waves as the source moves directly towards the observer. The resulting trace on the CRO is shown in Fig. 5.1. 1 div. Fig. 5.1 The time-base on the CRO is set to 1.0 ms div−1. * 0000800000012 * ,









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13 9702/21/M/J/25 © UCLES 2025 [Turn over (i) Calculate the frequency of the sound waves detected by the microphone. frequency = … Hz [2] (ii) Determine the speed of the sound in air. speed of sound = … m s−1 [2] [Total: 6] * 0000800000013 * ,









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14 9702/21/M/J/25 © UCLES 2025 6 A nichrome wire X of length 45 cm and cross-sectional area 4.7 × 10−7 m2 is connected into the circuit shown in Fig. 6.1. A X Fig. 6.1 The resistance of X is 1.1 Ω. The cell has electromotive force (e.m.f.) 1.3 V and negligible internal resistance. (a) (i) Calculate the current in X. current = …A [1] (ii) The number density of charge carriers (electrons) in nichrome is 8.5 × 1028 m−3. Calculate the average drift speed of the charge carriers in X. average drift speed = … m s−1 [2] (iii) Calculate the resistivity of the nichrome. resistivity = … Ω m [3] * 0000800000014 * ,









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15 9702/21/M/J/25 © UCLES 2025 [Turn over (b) Wire Y is identical to wire X. Wire Y is added to the circuit in parallel with wire X, as shown in Fig. 6.2. A X Y Fig. 6.2 State and explain the effect, if any, of this change on: (i) the reading on the ammeter … … … [2] (ii) the average drift speed of the charge carriers in X. … … [1] [Total: 9] * 0000800000015 * ,









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16 9702/21/M/J/25 © UCLES 2025 Permission 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 www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. 7 (a) Nitrogen-12 (12 7N) is an unstable isotope of nitrogen that decays by the emission of radiation to carbon-12 (12 6C). Complete the full nuclear equation for the decay, including all the particles involved. 12 7N [3] (b) Hydrogen-1 (1 1H) is an isotope of hydrogen. An atom of hydrogen-1 comprises a proton with an orbiting electron. An antiparticle equivalent of hydrogen-1 comprises an antiproton with an orbiting positron. The antiquarks in the antiproton are the antiparticles of the quarks in a proton. (i) State the charge on the positron in terms of the elementary charge e. charge = …e [1] (ii) State the group (class) of fundamental particle to which the positron belongs. … [1] (iii) In Table 7.1, state the flavour and charge of the three antiquarks that comprise the antiproton. Table 7.1 flavour charge / e [3] [Total: 8] * 0000800000016 * ,









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This document consists of 14 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level PHYSICS 9702/21 Paper 2 AS Level Structured Questions May/June 2025 MARK SCHEME Maximum Mark: 60 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the May/June 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 14 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 3 of 14 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind. Science-Specific Marking Principles 1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks should not be awarded if the keywords are used incorrectly. 2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for any correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the question should be ignored. 3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation from other syllabus terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection). 4 The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used in a scientifically correct way, the candidate should be awarded these subsequent marking points. Further guidance will be included in the mark scheme where necessary and any exceptions to this general principle will be noted. 5 ‘List rule’ guidance For questions that require n responses (e.g. State two reasons …): • The response should be read as continuous prose, even when numbered answer spaces are provided. • Any response marked ignore in the mark scheme should not count towards n. • Incorrect responses should not be awarded credit but will still count towards n. • Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this should be treated as a single incorrect response. • Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 14 6 Calculation specific guidance Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show your working’. For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values. For answers given in standard form (e.g. a  10n) in which the convention of restricting the value of the coefficient (a) to a value between 1 and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme. Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded. Exceptions to this general principle will be noted in the mark scheme. 7 Guidance for chemical equations Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme. State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 14 Annotations guidance for centres Examiners use a system of annotations as a shorthand for communicating their marking decisions to one another. Examiners are trained during the standardisation process on how and when to use annotations. The purpose of annotations is to inform the standardisation and monitoring processes and guide the supervising examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and is understood by all examiners who mark the component. We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the response. The annotations listed below were available to examiners marking this component in this series. Annotations Annotation Meaning Information missing or insufficient for credit Arithmetic error Benefit of the doubt given Contradiction in response, mark not awarded Incorrect point or mark not awarded Error carried forward applied Ignore the response Mandatory mark not awarded Power of ten error Blank page seen Error in number of significant figures

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 14 Annotation Meaning Transcription error Correct point or mark awarded Incorrect physics

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 14 Abbreviations / Alternative and acceptable answers for the same marking point. ( ) Bracketed content indicates words which do not need to be explicitly seen to gain credit but which indicate the context for an answer. The context does not need to be seen but if a context is given that is incorrect then the mark should not be awarded. ___ Underlined content must be present in answer to award the mark. This means either the exact word or another word that has the same technical meaning. Mark categories B marks These are independent marks, which do not depend on other marks. For a B mark to be awarded, 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 later depend. For an M mark to be awarded, the point to which it refers must be seen specifically in the candidate’s answer. If a candidate is not awarded an M mark, then the later A mark cannot be awarded either. C marks These are compensatory marks which can be awarded 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 them. 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 the candidate knew the equation, then the C mark is awarded. If a correct answer is given to a numerical question, all of the preceding C marks are awarded automatically. It is only necessary to consider each of the C marks in turn when the numerical answer is not correct. A marks These are answer marks. They may depend on an M mark or allow a C mark to be awarded by implication.

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 14 Question Answer Marks 1(a) rate of change of velocity B1 1(b)(i) s = ut + ½at2 and u = 0 or s = ½at2 t = (2  14 / 9.81) C1 t = 1.7 s A1 OR v = (02 + 2  9.81  14) v = 17 (m s−1) 17 = (0 + 9.81)  t or 17 = 9.81  t or 14 = ½  (0 + 17)  t or 14 = ½  17  t (C1) t = 1.7 s (A1) 1(b)(ii) s = ut + ½at2 u = (3.6 – ½  9.81  1.72) / 1.7 C1 u = (–) 6.2 m s–1 A1 OR v = (3.6 + ½  9.81  1.72) / 1.7 v = 10 (m s−1) 10 = u + 9.81  1.7 or 102 = u2 + (2  9.81  3.6) or 3.6 = ½  (u + 10)  1.7 (C1) u = (−) 6.2 m s−1 (A1)

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 14 Question Answer Marks 1(c)(i) Less time (as it reaches a lower height) B1 (because) the initial vertical (component of the) velocity is smaller (than in part (b)) B1 1(c)(ii) The (total) initial energy is the same (as in part (b)) B1 change in gravitational potential energy is same, so speed is the same B1 Question Answer Marks 2(a) in (rotational) equilibrium B1 sum / total of clockwise moments about a point = sum / total of anticlockwise moments about the (same) point. B1 2(b) 80  9.81  3 or 60  9.81  3 or 45  9.81  x C1 80  9.81  3 = (60  9.81  3) + (45  9.81  x) C1 x = 1.3 m A1 2(c)(i) k = F / x C1 x = 0.80 – 0.59 = 0.21 m C1 k = (60  9.81) / 0.21 = 2800 N m−1 A1

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 14 Question Answer Marks 2(c)(ii) E = ½ kx2 or E = ½ Fx or E = ½ F 2/k = ½  2800  0.212 or = ½  60  9.81  0.21 or = ½  (60  9.81)2 / 2800 C1 = 62 J A1 Question Answer Marks 3(a) work done per unit time B1 3(b)(i) P = F  v = 1750  35 C1 = 6.1  104 W A1 3(b)(ii) W = Fs = 1750  17 000 C1 = 3.0  107 J A1 or W = Pt = 6.1  104  (17 000 / 35) (C1) = 3.0  107 J (A1)

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 14 Question Answer Marks 3(b)(iii) P = V  I Power in = 600  I C1 useful power output Efficiency = (total) power input C1 0.85 = 6.1  104 / (600  I) I = 120 A A1 3(c)(i) Air resistance is the same, as the speed is the same B1 3(c)(ii) The motor is producing less power (because of gravitational force / conversion of gravitational potential energy to kinetic energy) so the current will be smaller. B1 Question Answer Marks 4(a) (when two or more) waves meet / overlap (at a point) B1 (resultant) displacement is sum of the individual displacements B1 4(b)(i) Fringe width, x = 3.2  10−2 / 8 = 4.0  10−3 (m) C1 D = ax /  = (4.0  10−3  0.16  10−3) / 7.2  10−7 C1 = 0.89 m A1

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 14 Question Answer Marks 4(b)(ii) Curved line with a negative gradient of decreasing magnitude throughout, from slit separation 0.04 mm to 0.16 mm B1 Line of negative gradient ending at (0.16, 0.4), from slit separation 0.04 mm B1 Line of negative gradient passing through (0.08, 0.8) and (0.04, 1.6) B1 Question Answer Marks 5(a) wavelength: • wavelength = distance between successive / adjacent in phase points / wavefronts / crests / troughs •  = d / (number of) oscillations frequency: • frequency = (number of) oscillations / cycles crests / troughs / wavefronts (passing a point) per unit time • f = (number of) oscillations / t One correct point from either list B1 One correct point from both lists and speed = distance / time and one of: • wavelength  frequency (= distance per unit time) = speed • [(number of) oscillations / t ]  [d / (number of) oscillations] = f • v (= d / t) =  / (1/f) = f or v (= d / t) =  / T = f B1 5(b)(i) T = 4  10−3 f = 1 / T = 1 / 0.004 C1 f = 250 Hz A1

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 13 of 14 Question Answer Marks 5(b)(ii) fo = fs v / (v – vs) 250 = 236  v / (v – 20) v = (250  20) / (250 – 236) C1 = 360 m s−1 A1 Question Answer Marks 6(a)(i) I = 1.3 / 1.1 = 1.2 A A1 6(a)(ii) v = I / nqA = 1.2 / (8.5  1028  1.60  10−19  4.7  10−7) C1 = 1.9  10−4 m s−1 A1 6(a)(iii)  = RA / L C1 = (1.1  4.7  10−7) / 0.45 C1 = 1.1  10−6  m A1 6(b)(i) (Total) resistance decreases (and the potential difference stays the same) M1 (so the reading on the ammeter) increases A1 6(b)(ii) (The average drift speed will be) the same because the current is the same (in X). B1

9702/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 14 of 14 Question Answer Marks 7(a) ( ) ( ) 0 12 12 0 7 6 1 0 N C n + → + + beta-plus shown B1 neutrino shown B1 symbols, nucleon numbers and proton numbers all correct B1 7(b)(i) +1 B1 7(b)(ii) Lepton(s) B1 7(b)(iii) flavour charge / e up / u 2 – 3 up / u 2 – 3 down / d ( ) + 1 3 3 correct quark flavours B1 Charge on anti-up quark –⅔(e) B1 Charge on anti-down quark (+)⅓(e) B1