Cambridge A Level Physics 9702 — 2024 Oct/Nov Paper 4 · Variant 1

1 (a) State Newton’s law of gravitation. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A planet may be considered as a uniform sphere. A satellite is in circular orbit of period T around the planet at a height h above the surface. The height of the orbit can be adjusted by use of the satellite’s rocket engines. 2 Fig. 1.1 shows the variation with h of T 3 . 1600 1200 2 2 T 3 /s 3 800 400 0 0 2 4 6 8 10 12 h / 106 m Fig. 1.1 (i) By reference to forces, explain why the orbit of the satellite is circular. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Use Newton’s law of gravitation to show that h and T are related by GA 2 (h + B)3 = T 4π2 where G is the gravitational constant and A and B are constants that depend on the properties of the planet. [3] (iii) Use the gradient and intercept of the line in Fig. 1.1 to determine values for A and B. Give units with your answers. A = ....................................... unit .................. B = ....................................... unit .................. [5] [Total: 12]

Mark scheme: Question Answer Marks 1(a) (gravitational) force is (directly) proportional to product of masses B1 force (between point masses) is inversely proportional to the square of their separation B1 1(b)(i) (gravitational) force acts perpendicular to direction of motion B1 gravitational force provides centripetal acceleration B1 1(b)(ii) (F =) GMm / x2 = mx2 and = 2 / T M1 or GMm / x2 = 42mx / T2 completion of algebra leading to x3 = GMT2 / 42 A1 clear indication that B = radius of planet and that A = mass (of planet) B1 1(b)(iii) gradient = 3√(42 / GA) C1 e.g. (1280 – 360) / (12  106) = 3√(42 / [6.67  10–11  A]) C1 A = 1.3  1024 kg A1 intercept = gradient  B C1 e.g. 360 = ((1280 – 360)  B) / (12  106) A1 B = 4.7  106 m

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2 (a) Define specific heat capacity. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Two solid blocks X and Y are made from different metals. The blocks have different initial temperatures. Block Y is initially at room temperature. The blocks are placed in direct thermal contact with each other at time t = 0. Fig. 2.1 shows the variation with t of the temperatures of the two blocks. 100 75 X temperature / °C 50 25 Y 0 0 0.5 1.0 1.5 2.0 2.5 3.0 t / min Fig. 2.1 (i) State three conclusions that may be drawn from Fig. 2.1. The conclusions may be qualitative or quantitative. 1 ........................................................................................................................................ ........................................................................................................................................... 2 ........................................................................................................................................ ........................................................................................................................................... 3 ........................................................................................................................................ ........................................................................................................................................... [3] mass of block Y(ii) The ratio is equal to 1.3. mass of block X The metal in block Y has a specific heat capacity of 901 J kg–1 K–1. Determine the specific heat capacity of the metal in block X. specific heat capacity = .......................................... J kg–1 K–1 [3] [Total: 8]

Mark scheme: 2(a) (thermal) energy per unit mass (to change temperature) B1 (thermal) energy per unit change in temperature B1 2(b)(i) Any three bulleted points from: B3 • the blocks end up in thermal equilibrium • heat capacity of Y is larger than heat capacity of X • no heat loss to the surroundings Up to 2 points from these six: • initial temperature of X = 85 °C • initial temperature of Y = 25 °C • the temperature change of X = 45 °C • the temperature change of Y = 15 °C • the temperature change in X is three times that in Y • final temperature of both = 40 °C 2(b)(ii) = 45 °C for X and 15 °C for Y C1 mc  45 = 1.3  m  901  15 C1 c = 390 J kg K–1 A1

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3 (a) (i) State what is meant by the Avogadro constant. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] (ii) State the relationship between the Avogadro constant NA, the molar gas constant R and the Boltzmann constant k. [1] (b) Two samples X and Y of ideal gases are both at thermodynamic temperature T. Sample X has volume V and consists of N molecules, each of mass m. Sample Y has volume 2V and consists of 2N molecules, each of mass 2m. (i) Complete Table 3.1 by giving expressions, in terms of some or all of N, m, T, V and the constants in (a)(ii), for the quantities indicated. Table 3.1 sample X sample Y pressure amount of substance mean-square speed of molecules internal energy [4] (ii) The temperature of sample X is now varied. On Fig. 3.1, sketch the variation with thermodynamic temperature of the root-mean- square (r.m.s.) speed of the molecules of the gas. r.m.s. speed 0 0 thermodynamic temperature Fig. 3.1 [2] [Total: 8]

Mark scheme: 3(a)(i) number of particles per unit amount of substance B1 3(a)(ii) NA = R / k B1 3(b)(i) X pressure and Y pressure both = NkT / V B1 X amount = N / NA and Y amount = 2N / NA B1 X mean-square speed = 3kT / m and Y mean-square speed = 3kT / 2m B1 X internal energy = 3NkT / 2 and Y internal energy = 3NkT B1 3(b)(ii) line passing through the origin and not returning to either axis B1 curve with positive decreasing gradient B1

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4 (a) State what is meant by simple harmonic motion. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A block is suspended from a spring, as shown in Fig. 4.1. spring block h floor Fig. 4.1 The block is pulled down and released at time t = 0. It then oscillates vertically with simple harmonic motion. Fig. 4.2 shows the variation of the velocity v of the block with height h of the base of the block above the floor. 10 v / cm s–1 5 0 0 2 4 6 8 10 12 h / cm –5 –10 Fig. 4.2 (i) Determine the amplitude, in cm, of the oscillations. amplitude = .................................................... cm [1] (ii) Show that the angular frequency of the oscillations is 3.2 rad s–1. [2] (iii) Calculate the period T of the oscillations. T = ....................................................... s [2] (iv) On Fig. 4.3, sketch the variation of h with time t from t = 0 to t = 6.0 s. 10.0 h / cm 7.5 5.0 2.5 0 0 1 2 3 4 5 6 t / s Fig. 4.3 [4] [Total: 11]

Mark scheme: 4(a) (motion in which) acceleration is (directly) proportional to displacement B1 (motion in which): B1 acceleration is (always) in the opposite direction to displacement or acceleration is (always) directed towards a fixed point 4(b)(i) amplitude = (9.5 – 3.5) / 2 A1 = 3.0 cm 4(b)(ii)  = v0 / x0 C1 = 9.5 / 3.0 = 3.2 rad s–1 A1 4(b)(iii) T = 2 /  C1 = 2 / 3.2 A1 = 2.0 s 4(b)(iv) attempted sinusoidal curve starting with a minimum at t = 0 B1 sinusoidal curve of period 2.0 s from t = 0 to t = 6.0 s B1 all peaks shown at h = 9.5 cm B1 all troughs shown at h = 3.5 cm B1

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5 (a) State the relationship between electric field and electric potential. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Two charged isolated insulating spheres X and Y are near to each other, as shown in Fig. 5.1. X Y P Fig. 5.1 P is a point on the line joining the centres of the spheres. Explain why it is not possible for the total electric potential and the resultant electric field to simultaneously be zero at point P. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (c) The magnitudes of the charges on spheres X and Y in Fig. 5.1 are Q and 2Q respectively. The spheres may be considered as point charges at their centres. Point P is a distance x from the centre of sphere X. The electric potential at point P is zero. (i) Show that the distance y of point P from the centre of sphere Y is equal to 2x. [2] (ii) State an expression, in terms of Q, x and the permittivity of free space ε0, for the electric field strength EX at P due to sphere X. EX = ......................................................... [1] (iii) Determine an expression, in terms of Q, x and ε0, for the resultant electric field strength E at point P due to the two spheres. E = ......................................................... [2] [Total: 10]

Mark scheme: 5(a) (electric) field equals (electric) potential gradient M1 reference to minus sign A1 5(b) • for potential to be zero, one potential must be positive and the other potential must be negative B3 • for potential to be zero, the charges must have opposite sign • for field to be zero, the fields (due to X and Y) must be in opposite directions • for field to be zero, the charges must have the same sign • the signs of the charges cannot (simultaneously) be both the same and opposite (so not possible) Any three points, 1 mark each 5(c)(i) VX = (–) Q / 4ε0x and VY = (–) 2Q / 4ε0y C1 (VX + VY = 0 so) Q / 4ε0x = 2Q / 4ε0y leading to y = 2x A1 5(c)(ii) EX = Q / 4ε0x2 A1 5(c)(iii) EY = 2Q / 4ε0(2x)2 C1 ( = Q / 8ε0x2) (opposite charges so fields in same direction so magnitudes add): A1 E = (Q / 4ε0x2) + (Q / 8ε0x2) = 3Q / 8ε0x2

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6 (a) (i) State what is meant by rectification of an alternating voltage. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) State the difference between half-wave rectification and full-wave rectification. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (b) (i) Complete Fig. 6.1 to show a circuit that produces half-wave rectification of an alternating input voltage VIN to produce output voltage VOUT across the resistor R. VIN C R VOUT Fig. 6.1 [2] (ii) State the purpose of the capacitor C in the circuit of Fig. 6.1. ........................................................................................................................................... ..................................................................................................................................... [1] (c) The input voltage VIN in Fig. 6.1 is a square wave. Fig. 6.2 shows the variation of VIN with time t. +12 VIN / V 0 0 0.01 0.02 0.03 0.04 t / s –12 Fig. 6.2 Fig. 6.3 shows the variation of VOUT with t. 12 VOUT / V 8 4 0 0 0.01 0.02 0.03 0.04 t / s Fig. 6.3 The maximum energy stored in the capacitor is 0.041 J. (i) Show that the capacitance of C is 570 μF. [2] (ii) Determine the resistance of R. resistance = ..................................................... Ω [3] [Total: 11]

Mark scheme: 6(a)(i) conversion (from a.c.) to d.c. B1 6(a)(ii) half-wave: voltage in one direction is removed B1 full-wave: voltage in one direction is reversed B1 6(b)(i) one gap connected by a single diode and other gap connected directly B1 diode drawn (in a circuit) with correct circuit symbol B1 6(b)(ii) smoothing B1 6(c)(i) E = ½CV2 C1 C = 2  0.041 / 122 = 5.7  10–4 F = 570 F A1 6(c)(ii) 8.0 = 12.0 exp (– 0.010 / RC) C1 ln (8.0 / 12.0) = – 0.010 / (R  5.7  10–4) C1 R = 43  A1

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7 (a) Define magnetic flux density. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A long, straight wire carries a current into the page, as shown in Fig. 7.1. Fig. 7.1 On Fig. 7.1, draw four field lines to represent the magnetic field around the wire due to the current in it. [3] (c) Two identical wires X and Y are placed parallel to each other. The wires both carry current into the page, as shown in Fig. 7.2. X Y Fig. 7.2 (i) Explain why the two wires exert a magnetic force on each other. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) On Fig. 7.2, draw an arrow to show the direction of the magnetic force exerted on wire X. Label your arrow F. [1] (iii) The current in X is double the current in Y. State how the magnetic force exerted on wire Y compares with the magnetic force exerted on wire X. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (iv) The direction of the current in both wires is now reversed. State, with a reason, the effect of this change on the direction of the force on wire X. ........................................................................................................................................... ..................................................................................................................................... [1] [Total: 11]

Mark scheme: 7(a) • force per unit length B2 • force per unit current • length / current perpendicular to field 1 mark for any two points, 2 marks for all three points 7(b) concentric circles around the wire (at least two circles needed) B1 spacing between circles increases with distance from wire (at least four circles needed) B1 arrows showing direction of field is clockwise B1 7(c)(i) (each) wire sits in the (magnetic) field created by the other B1 current (in one wire) is perpendicular to (magnetic) field (due to other wire) so (magnetic) force acts (on wire) B1 7(c)(ii) arrow drawn, starting from X and pointing towards Y, labelled F B1 7(c)(iii) (forces have) equal magnitudes B1 (forces are in) opposite directions B1 7(c)(iv) no change (in the direction of the force) since both the current in X and the field due to Y have reversed B1

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8 A polished sheet of magnesium in a vacuum emits electrons when it is illuminated by ultraviolet radiation. (a) State the name of this phenomenon. ............................................................................................................................................. [1] (b) For emission of electrons to occur, the frequency of the ultraviolet radiation must be at least 8.8 × 1014 Hz. (i) Calculate the work function energy of magnesium. work function energy = ...................................................... J [2] (ii) For ultraviolet radiation with a frequency of 11 × 1014 Hz, calculate the maximum speed of the emitted electrons. maximum speed = ................................................ m s–1 [3] (c) The frequency f of the ultraviolet radiation incident on the magnesium sheet is varied between 8.0 × 1014 Hz and 11 × 1014 Hz. On Fig. 8.1, sketch the variation with f of the maximum kinetic energy EMAX of the emitted electrons. Use the space below for any working that you need. 2.0 1.5 EMAX / 10–19 J 1.0 0.5 0 8.0 8.5 9.0 9.5 10.0 10.5 11.0 f / 1014 Hz Fig. 8.1 [3] [Total: 9]

Mark scheme: 8(a) photoelectric effect B1 8(b)(i) E = hf C1 work function = 6.63  10–34  8.8  1014 A1 = 5.8  10–19 J 8(b)(ii) hf =  + ½ mvMAX2 C1 6.63  10–34  11  1014 = (5.8  10–19) + (½  9.11  10–31  vMAX2) C1 vMAX = 5.7  105 m s–1 A1 8(c) EMAX shown as zero from f = 8.0 to 8.8 and non-zero from f = 8.8 to 11 B1 all non-zero EMAX shown as a single straight line with a positive gradient B1 line passing through (11, 1.45) B1

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9 Fluorine-18 ( 9F) decays by beta-plus (β+) emission with a half-life of 110 minutes. (a) (i) State the name of the beta-plus particle. ..................................................................................................................................... [1] (ii) Show that the decay constant of fluorine-18 is 1.05 × 10–4 s–1. [1] (iii) Determine the activity of 2.1 × 10–12 kg of fluorine-18. activity = .................................................... Bq [3] (b) A small sample of fluorine-18 injected into the body acts as a tracer for use in medical imaging. (i) Describe how the interaction of a β+ particle with an electron in the body enables the formation of an image. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) Suggest why 110 minutes is a suitable half-life for a nuclide used as a tracer in medical diagnosis. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 10]

Mark scheme: 9(a)(i) positron B1 9(a)(ii) = ln 2 / (110  60) = 1.05  10–4 s–1 A1 9(a)(iii) N = M / (18 u) or (M in grams  NA / 18) C1 N = (2.1  10–12) / (18  1.66  10–27) or (2.1  10–9  6.02  1023) / 18 ( = 7.0  1013) A = N C1 = 1.05  10–4  7.0  1013 A1 = 7.4  109 Bq 9(b)(i) • (pair) annihilation occurs B3 • the mass of the two particles is converted into energy • two gamma photons are formed and travel in opposite directions or two gamma photons are formed and leave the body • difference in arrival times of photons (at detector) is processed Any three points, 1 mark each 9(b)(ii) with a shorter half-life: sample would (almost) fully decay before the test is complete B1 a longer half-life: exposes patient to harmful/ionising radiation unnecessarily B1 or with a longer half-life: a larger dose (of tracer) needed to produce detectable activity

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10 (a) Explain how redshift leads to the idea that the Universe is expanding. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (b) Stars in a distant galaxy emit radiation. The total luminosity of the stars in the galaxy is 1.90 × 1036 W. The emission spectrum of the radiation contains a line X at a wavelength of 658 nm. Radiation from the galaxy is observed on the Earth. The observed radiation has a radiant flux intensity of 8.42 × 10–16 W m–2. In the observed emission spectrum, line X is at a wavelength of 726 nm. Determine: (i) the distance d of the galaxy from the Earth d = ..................................................... m [2] (ii) the speed v of the galaxy relative to the Earth. v = ................................................ m s–1 [2] (c) Observations of many galaxies, such as the one in (b), lead to many pairs of values of d and v. Plotting these values reveals a trend. (i) On Fig. 10.1, sketch the variation of v with d. v 0 0 d Fig. 10.1 [2] (ii) State the name of the quantity represented by the gradient of the line in Fig. 10.1. ..................................................................................................................................... [1] [Total: 10]

Mark scheme: 10(a) • redshift is the increase in observed wavelength / decrease in observed frequency (caused by Doppler effect) B3 • radiation from distant galaxies is observed to be redshifted • redshift provides evidence that galaxies are moving apart • galaxies moving apart means Universe must be expanding Any three points, 1 mark each 10(b)(i) F = L / 4d2 C1 d = √(1.90  1036 / [4  8.42  10–16]) A1 = 1.34  1025 m 10(b)(ii)  / = v / c C1 (726 – 658) / 658 = v / (3.00  108) v = 3.1  107 m s–1 A1 10(c)(i) line with positive gradient passing through the origin B1 straight line with positive gradient B1 10(c)(ii) Hubble constant B1

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