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Physics 9702/42 May/June 2022: question paper and mark scheme

Cambridge AS & A Level Physics (9702), Paper 4 A Level Structured Questions, May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9702_s22_qp_429702_s22_ms_429702/42/M/J/22

Paper
Paper 4 A Level Structured Questions
Total marks
100
Time
2 hours
Questions
10
Code on the paper
9702/42/M/J/22
9702_s22_ms_42.pdf

9702/42 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 110 marks

Question

(a) (i) Define gravitational potential at a point. …… [2] (ii) Starting from the equation for the gravitational potential due to a point mass, show that the gravitational potential energy EP of a point mass m at a distance r from another point mass M is given by GMm EP = – r where G is the gravitational constant. [1] (b) Fig. 1.1 shows the path of a comet of mass 2.20 × 1014 kg as it passes around a star of mass 1.99 × 1030 kg. X 34.1 km s–1 star mass 1.99 × 1030 kg comet mass 2.20 × 1014 kg Y path of comet Fig. 1.1 (not to scale) At point X, the comet is 8.44 × 1011 m from the centre of the star and is moving at a speed of 34.1 km s–1. At point Y, the comet passes its point of closest approach to the star. At this point, the comet is a distance of 6.38 × 1010 m from the centre of the star. Both the comet and the star can be considered as point masses at their centres. (i) Calculate the magnitude of the change in the gravitational potential energy ΔEP of the comet as it moves from position X to position Y. ΔEP = …… J [2] (ii) State, with a reason, whether the change in gravitational potential energy in (b)(i) is an increase or a decrease. …… [1] (iii) Use your answer in (b)(i) to determine the speed, in km s–1, of the comet at point Y. speed = …… km s–1 [3] (c) A second comet passes point X with the same speed as the comet in (b) and travelling in the same direction. This comet is gradually losing mass. The mass of this comet when it passes point X is the same as the mass of the comet in (b). Suggest, with a reason, how the path of the second comet compares with the path shown in Fig. 1.1. …… [1] [Total: 10]

Mark scheme

1(a)(i) work (done) per unit mass B1 work (done on mass) in moving mass from infinity (to the point) B1 1(a)(ii) EP = ϕm B1 EP = (– GM / r) × m = – GMm / r or ϕ = – GM / r and EP = ϕm = – GMm / r 1(b)(i) ΔEP = 6.67 × 10–11 × 1.99 × 1030 × 2.20 × 1014 × [1 / (6.38 × 1010) – 1 / (8.44 × 1011)] C1 = 4.23 × 1023 J A1 1(b)(ii) (gravitational) force is attractive so decrease B1 or (gravitational) force does work so decrease 1(b)(iii) ΔEP = ½m(v22 – v12) C1 4.23 × 1023 = ½ × 2.20 × 1014 × (v2 – 34 1002) C1 v (= 70 800 m s–1) = 70.8 km s–1 A1 1(c) both PE and KE equations include m, so path is unchanged B1

Question 213 marks

Question

(a) State Coulomb’s law. …… [2] (b) Positronium is a system in which an electron and a positron orbit, with the same period, around their common centre of mass, as shown in Fig. 2.1. centre of mass r electron positron Fig. 2.1 (not to scale) The radius r of the orbit of both particles is 1.59 × 10–10 m. (i) Explain how the electric force between the electron and the positron causes the path of the moving particles to be circular. …… [2] (ii) Show that the magnitude of the electric force between the electron and the positron is 2.28 × 10–9 N. [2] (iii) Use the information in (b)(ii) to determine the period of the circular orbit of the two particles. period = …… s [3] (c) Positronium is highly unstable, and after a very short period of time it becomes gamma radiation. (i) Describe how gamma radiation is formed from the two particles in positronium. …… [3] (ii) State one medical application of the process described in (c)(i). …… [1] [Total: 13]

Mark scheme

2(a) (electric) force is (directly) proportional to product of charges B1 force (between point charges) is inversely proportional to the square of their separation B1 2(b)(i) (electric) force is perpendicular to velocity (of particles) B1 force (perpendicular to velocity) causes centripetal acceleration B1 or force does not change the speed of the particles or force has constant magnitude 2(b)(ii) F = e2 / 4πε0x2 C1 = (1.60 × 10–19)2 / [4π × 8.85 × 10–12 × (2 × 1.59 × 10–10)2] A1 = 2.28 × 10–9 N 2(b)(iii) F = mrω2 and ω = 2π / T C1 or F = mv2 / r and v = 2πr / T F = 4π2mr / T2 C1 T = √ [4π2 × 9.11 × 10–31 × 1.59 × 10–10 / (2.28 × 10–9)] = 1.58 × 10–15 s A1 2(c)(i) electron and positron interact B2 positron is anti-particle of electron (pair) annihilation occurs Any two points, 1 mark each mass of the electron and positron converted into photon energy B1 2(c)(ii) PET scanning B1

Question 312 marks

Question

(a) Define specific latent heat of vaporisation. …… [2] (b) The specific latent heat of vaporisation of water at atmospheric pressure of 1.0 × 105 Pa is 2.3 × 106 J kg–1. A mass of 0.37 kg of liquid water at 100 °C is provided with the thermal energy needed to vaporise all of the water at atmospheric pressure. (i) Calculate the thermal energy q supplied to the water. q = …… J [1] (ii) The mass of 1.0 mol of water is 18 g. Assume that water vapour can be considered to behave as an ideal gas. Show that the volume of water vapour produced is 0.64 m3. [3] (iii) Assume that the initial volume of the liquid water is negligible compared with the volume of water vapour produced. Determine the magnitude of the work done by the water in expanding against the atmosphere when it vaporises. work done = …… J [2] (iv) Use your answers in (b)(i) and (b)(iii) to determine the increase in internal energy of the water when it vaporises at 100 °C. Explain your reasoning. increase in internal energy = …… J [2] (c) Use the first law of thermodynamics to suggest, with a reason, how the specific latent heat of vaporisation of water at a pressure greater than atmospheric pressure compares with its value at atmospheric pressure. …… [2] [Total: 12]

Mark scheme

3(a) (thermal) energy per unit mass B1 energy to change state between liquid and gas at constant temperature B1 3(b)(i) q = mL = 0.37 × 2.3 × 106 A1 = 8.5 × 105 J 3(b)(ii) pV = nRT and T = 373 K C1 n = 370 / 18 C1 V = [(370 / 18) × 8.31 × 373] / (1.0 × 105) = 0.64 m3 A1 3(b)(iii) w = pΔV C1 = 1.0 × 105 × 0.64 A1 = 6.4 × 104 J 3(b)(iv) (water does work against atmosphere so) work done on water is negative B1 increase in internal energy = (8.5 – 0.64) × 105 = 7.9 × 105 J A1 3(c) valid reasoning of how work done by water is affected M1 correct use of first law to draw conclusion about effect on specific latent heat that is consistent with work done A1

Question 48 marks

Question

(a) State what is meant by resonance. …… [2] (b) Fig. 4.1 shows a heavy pendulum and a light pendulum, both suspended from the same piece of string. This string is secured at each end to fixed points. fixed points string heavy pendulum light pendulum Fig. 4.1 Both pendulums have the same natural frequency. The heavy pendulum is set oscillating perpendicular to the plane of the diagram. As it oscillates, it causes the light pendulum to oscillate. Fig. 4.2 shows the variation with time t of the displacements of the two pendulums for three oscillations. heavy displacement / cm 0 light 0 t/s Fig. 4.2 The variation with t of the displacement x of the light pendulum is given by x = 0.25 sin 5.0rt where x is in centimetres and t is in seconds. (i) Calculate the period T of the oscillations. T = …… s [2] (ii) On Fig. 4.2, label both of the axes with the correct scales. Use the space below for any additional working that you need. [2] (iii) Determine the magnitude of the phase difference φ between the oscillations of the light and heavy pendulums. Give a unit with your answer. φ = …… unit …… [2] [Total: 8]

Mark scheme

4(a) oscillations (of object) at maximum amplitude B1 when driving frequency equals natural frequency (of object) B1 4(b)(i) T = 2π / ω C1 = 2π / 5.0π A1 = 0.40 s 4(b)(ii) displacement scale labelled –1.0, –0.5, (0), 0.5, 1.0 on the 2 cm tick marks B1 t scale labelled 0.2, 0.4, 0.6, 0.8, 1.0, 1.2 on the 2 cm tick marks B1 4(b)(iii) ϕ = 2πΔt / T C1 = 2π × 0.10 / 0.40 or 2π × 0.30 / 0.40 = 1.6 rad or 4.7 rad A1

Question 511 marks

Question

(a) Define the capacitance of a parallel plate capacitor. …… [2] (b) Two capacitors, of capacitances C1 and C2, are connected in parallel to a power supply of electromotive force (e.m.f.) E, as shown in Fig. 5.1. E C1 C2 Fig. 5.1 Show that the combined capacitance CT of the two capacitors is given by CT = C1 + C2. Explain your reasoning. You may draw on Fig. 5.1 if you wish. [3] (c) Two capacitors of capacitances 22 μF and 47 μF, and a resistor of resistance 2.7 MΩ, are connected into the circuit of Fig. 5.2. 12 V X S 2.7 MΩ Y 22 μF 47 μF Fig. 5.2 The battery has an e.m.f. of 12 V. (i) Show that the combined capacitance of the two capacitors is 15 μF. [1] (ii) The two-way switch S is initially at position X, so that the capacitors are fully charged. Use the information in (c)(i) to calculate the total energy stored in the two capacitors. total energy = …… J [2] (iii) The two-way switch is now moved to position Y. Determine the time taken for the potential difference (p.d.) across the 22 μF capacitor to become 6.0 V. time = …… s [3] [Total: 11]

Mark scheme

5(a) charge / potential (difference) M1 charge is charge on one plate, and potential is p.d. across the plates A1 5(b) p.d. across both capacitors = E B1 QT = Q1 + Q2 B1 CTE = C1E + C2E hence CT = C1 + C2 B1 5(c)(i) [(1 / 22) + (1 / 47)]–1 = 15 μF A1 5(c)(ii) energy = ½CV 2 C1 = ½ × 15 × 10–6 × 122 A1 = 1.1 × 10–3 J 5(c)(iii) initial p.d. (across 22 μF) = 12 × (15 / 22) C1 = 8.2 V or final p.d. across both capacitors = 6.0 × (22 / 15) = 8.8 V V = V0 exp [– t / (2.7 × 106 × 15 × 10–6)] C1 6.0 = 8.2 exp [– t / (2.7 × 106 × 15 × 10–6)] A1 or 8.8 = 12 exp [– t / (2.7 × 106 × 15 × 10–6)] t = 13 s

Question 610 marks

Question

(a) State the two conditions that must be satisfied for a copper wire, placed in a magnetic field, to experience a magnetic force. 1 …… 2 …… [2] (b) A long air-cored solenoid is connected to a power supply, so that the solenoid creates a magnetic field. Fig. 6.1 shows a cross-section through the middle of the solenoid. Z section through solenoid wires Y W X Fig. 6.1 The direction of the magnetic field at point W is indicated by the arrow. Three other points are labelled X, Y and Z. (i) On Fig. 6.1, draw arrows to indicate the direction of the magnetic field at each of the points X, Y and Z. [3] (ii) Compare the magnitude of the flux density of the magnetic field: ● at X and at W …… ● at Y and at Z. …… [2] (c) Two long parallel current-carrying wires are placed near to each other in a vacuum. Explain why these wires exert a magnetic force on each other. You may draw a labelled diagram if you wish. …… [3] [Total: 10]

Mark scheme

6(a) there must be a current (in the wire) B1 (wire) must be at a non-zero angle to the magnetic field B1 6(b)(i) arrow from X pointing horizontally to the left B1 arrow from Y pointing diagonally upwards and to the left at about 45° B1 arrow from Z pointing horizontally to the right B1 6(b)(ii) (flux densities at W and X are approximately) equal B1 (flux density at) Y greater than (flux density at) Z B1 6(c) current in wire creates magnetic field around wire B1 (each) wire sits in the magnetic field created by the other B1 (for each wire,) current / wire is perpendicular to magnetic field (due to other wire), (so) experiences a (magnetic) force B1

Question 710 marks

Question

(a) State Faraday’s law of electromagnetic induction. …… [2] (b) Two coils are wound on an iron bar, as shown in Fig. 7.1. coil 2 V iron bar coil 1 V1 Fig. 7.1 Coil 1 is connected to a potential difference (p.d.) V1 that gives rise to a magnetic field in the iron bar. Fig. 7.2 shows the variation with time t of the magnetic flux density B in the iron bar. B 0 0 0.1 0.2 0.3 t / s 0.4 Fig. 7.2 A voltmeter measures the electromotive force (e.m.f.) V2 that is induced across coil 2. On Fig. 7.3, sketch the variation with t of V2 between t = 0 and t = 0.40 s. V2 0 0 0.1 0.2 0.3 t / s 0.4 Fig. 7.3 [4] (c) Coil 2 in (b) is now replaced with a copper ring that rests loosely on top of coil 1. The supply to coil 1 is replaced with a cell and a switch that is initially open, as shown in Fig. 7.4. iron bar copper ring coil 1 Fig. 7.4 (i) The switch is now closed. As it is closed, the copper ring is observed to jump upwards. Explain why this happens. …… [3] (ii) Suggest, with a reason, what would be the effect of repeating the procedure in (c)(i) with the terminals of the cell reversed. …… [1] [Total: 10]

Mark scheme

7(a) induced e.m.f. is (directly) proportional to rate M1 of change of (magnetic) flux (linkage) A1 7(b) V2 stepped, all at non-zero values, between t = 0 and t = 0.40 s B1 V2 shown with same non-zero magnitude up to t = 0.15 s and after t = 0.25 s but with a different magnitude between these B1 times V2 shown with a magnitude between t = 0.15 s and t = 0.25 s that is three times the magnitude before t = 0.15 s and after B1 t = 0.25 s V2 shown with same sign up to t = 0.15 s and after t = 0.25 s, and opposite sign in between B1 7(c)(i) changing current in coil causes changing (magnetic) field B1 or changing (magnetic) flux causes induced e.m.f. in ring induced e.m.f. in ring causes current in ring B1 (magnetic) field due to (induced) current in ring interacts with (coil’s) field to cause upwards force (on ring) B1 or (induced) current in ring perpendicular to (coil’s magnetic) field causes upwards force (on ring) 7(c)(ii) both magnetic fields reverse direction so ring still jumps up B1 or current (in ring) and (coil’s) field both reverse so ring still jumps up

Question 88 marks

Question

(a) State one piece of experimental evidence for: (i) the particulate nature of electromagnetic radiation …… [1] (ii) the wave nature of matter. …… [1] (b) (i) Calculate the de Broglie wavelength λ of an alpha-particle moving at a speed of 6.2 × 107 m s–1. λ = …… m [3] (ii) The speed v of the alpha-particle in (b)(i) is gradually reduced to zero. On Fig. 8.1, sketch the variation with v of λ. λ 0 0 6.2 v / 107 m s–1 Fig. 8.1 [2] (c) Suggest an explanation for why people are not observed to diffract when they walk through a doorway. …… [1] [Total: 8]

Mark scheme

8(a)(i) photoelectric effect B1 8(a)(ii) electron diffraction B1 8(b)(i) λ = h/p C1 p = 4 × 1.66 × 10–27 × 6.2 × 107 C1 ( = 4.1 × 10–19 N s) λ = 6.63 × 10–34 / 4.1 × 10–19 A1 = 1.6 × 10–15 m 8(b)(ii) line with negative gradient throughout B1 curve asymptotic to both axes with non-zero λ at v = 6.2 × 107 m s–1 B1 8(c) (de Broglie) wavelength negligible compared with width of doorway B1

Question 97 marks

Question

(a) (i) State Hubble’s law. …… [2] (ii) Explain how cosmologists use observations of emission spectra from stars in distant galaxies to determine that the Universe is expanding. …… [2] (b) Explain how Hubble’s law and the idea of the expanding Universe lead to the Big Bang theory of the origin of the Universe. …… [3] [Total: 7]

Mark scheme

9(a)(i) speed is (directly) proportional to distance M1 where speed is speed of recession of galaxy (from observer) and distance is distance of galaxy away from observer A1 9(a)(ii) wavelengths (of spectral lines) are greater (than their known values) B1 redshift shows stars (in distant galaxies) moving away from Earth B1 9(b) (all) parts of Universe moving away from each other B1 more distant objects are moving away faster B1 matter must have been close together / very dense in the past B1

Question 1011 marks

Question

(a) State what is meant by radioactive decay. …… [2] (b) A radioactive sample consists of an isotope X of half-life T that decays to form a stable product. Only X and the stable product are present in the sample. At time t = 0, the sample has an activity of A0 and contains N0 nuclei of X. (i) On Fig. 10.1, sketch the variation with t of the number N of nuclei of X present in the sample. Your line should extend from time t = 0 to time t = 3T. 1.00N0 N 0.75N0 0.50N0 0.25N0 0 0 T 2T 3T t Fig. 10.1 [3] (ii) On Fig. 10.2, sketch the variation with N of the activity A of the sample for values of N between N = 0 and N = N0. 1.0A0 A 0.5A0 0 0 0.5N0 1.0N0 N Fig. 10.2 [2] (c) State the name of the quantity represented by the gradient of your line in: (i) Fig. 10.1 …… [1] (ii) Fig. 10.2. …… [1] N (d) For the sample in (b), calculate the fraction N at time t = 1.70T. 0 N N0 = …… [2] [Total: 11]

Mark scheme

10(a) spontaneous emission of (ionising) radiation B1 emission from unstable nucleus B1 10(b)(i) curve with decreasing negative gradient passing through (0, N0) B1 curve passing through (T, 0.5N0) B1 curve passing through (2T, 0.25N0) and (3T, 0.125N0) B1 10(b)(ii) line through origin with positive gradient B1 straight line passing through (N0, A0) B1 10(c)(i) activity B1 10(c)(ii) decay constant B1 10(d) N = N0 exp (– ln 2 × 1.70T / T) C1 N / N0 = 0.31 A1

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Where can I find the 9702/42 May/June 2022 mark scheme?

On this page, free: the mark scheme for AS & A Level Physics 9702/42, May/June 2022 (9702_s22_ms_42), and the question paper (9702_s22_qp_42), to read or download, with each question's marking points written out underneath.

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Physics 9702/42 for May/June 2022 is out of 100 marks, with 2 hours to answer it, across 10 questions.

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