Cambridge A Level Physics 9702 — 2024 May/June Paper 4 · Variant 2

1 (a) Define the radian. ................................................................................................................................................... ............................................................................................................................................. [1] (b) A circular metal disc spins horizontally about a vertical axis, as shown in Fig. 1.1. rotation metal disc axis modelling clay 9.3 cm 1.2 cm Fig. 1.1 (not to scale) A piece of modelling clay is attached to the disc. For the instant when the piece of modelling clay is in the position shown, draw on Fig. 1.1: (i) an arrow, labelled V, showing the direction of the velocity of the modelling clay [1] (ii) an arrow, labelled A, showing the direction of the acceleration of the modelling clay. [1] (c) The metal disc in Fig. 1.1 has a radius of 9.3 cm. The centre of gravity of the modelling clay is 1.2 cm from the rim of the disc and moves with a speed of 0.68 m s–1. (i) Calculate the angular speed ω of the disc. ω = .............................................. rad s–1 [2] (ii) Calculate the acceleration a of the centre of gravity of the modelling clay. a = ................................................ m s–2 [2] (d) A second piece of modelling clay is attached to the disc in the position shown in Fig. 1.2. second piece of modelling clay first piece of modelling clay Fig. 1.2 The second piece of modelling clay has a larger mass than the first piece. By placing one tick (3) in each row, complete Table 1.1 to show how the quantities indicated compare for the two pieces of modelling clay. Table 1.1 less for second piece greater for second piece quantity same for both pieces than first piece than first piece angular speed linear speed acceleration [3] [Total: 10]

Mark scheme: 1(a) angle (subtended at the centre of a circle) when arc (length) = radius B1 1(b)(i) arrow, labelled V, pointing in NE direction B1 1(b)(ii) arrow, labelled A, pointing in NW direction B1 1(c)(i) v = r C1  = 0.68 / (0.093 – 0.012) = 8.4 rad s–1 A1 1(c)(ii) a = v2 / r or a = r2 C1 a = 0.682 / (0.093 – 0.012) or (0.093 – 0.012)  8.42 = 5.7 m s–2 A1 1(d) angular speed: same for both pieces B1 linear speed: less for second piece than first piece B1 acceleration: less for second piece than first piece B1

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2 (a) With reference to thermal energy, state what is meant by two objects being in thermal equilibrium. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [1] (b) Two cylinders X and Y each contain a sample of an ideal gas. The samples are in thermal equilibrium with each other. X has a volume of 0.0260 m3 and contains 0.740 mol of gas at a pressure of 1.20 × 105 Pa. Y has a volume of 0.0430 m3 and contains gas at a pressure of 2.90 × 105 Pa. Data for the two cylinders are shown in Fig. 2.1. X Y 0.740 mol 1.20 × 105 Pa 2.90 × 105 Pa 0.0260 m3 0.0430 m3 Fig. 2.1 (i) Show that the temperature of the gas in X is 234 °C. [3] (ii) Determine the number N of molecules of the gas in Y. Explain your reasoning. N = ......................................................... [3] (iii) The gas in X consists of molecules that each have a mass that is four times the mass of a molecule of the gas in Y. Explain how the root-mean-square (r.m.s.) speed of the molecules in X compares with the r.m.s. speed of the molecules in Y. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] [Total: 10]

Mark scheme: 2(a) (if in thermal contact) no net transfer of (thermal) energy (between them) B1 2(b)(i) pV = nRT C1 T = (1.20  105  0.0260) / (0.740  8.31) ( = 507 K) M1 temperature = 507 – 273 = 234 °C A1 2(b)(ii) thermal equilibrium so temperatures (of X and Y) are equal B1 pV = NkT C1 N = (2.90  105  0.0430) / (1.38  10–23  507) = 1.78  1024 A1 2(b)(iii)  (molecular) kinetic energy is proportional to temperature or kinetic energy (of molecules) is same in both cylinders  kinetic energy proportional to mass  mean-square speed or temperature proportional to mass  mean-square speed or r.m.s. speed proportional to √(temperature / mass)  mean-square speed inversely proportional to mass or r.m.s. speed inversely proportional to √(mass) Any two bulleted points, 1 mark each B2 r.m.s. speed (of molecules) in X is half r.m.s. speed (of molecules) in Y B1

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3 (a) State what is meant by the internal energy of a system. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) With reference to molecular kinetic and potential energies, describe and explain how the internal energy of the system changes when: (i) a gas is heated at constant volume so that its temperature increases ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) a wire is stretched within its elastic limit at constant temperature. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] [Total: 8]

Mark scheme: 3(a) sum of potential energy and kinetic energy B1 (total) energy of random motion of particles B1 3(b)(i) no change in separation so no change in (molecular) potential energy B1 temperature increases so kinetic energy (of molecules) increases B1 kinetic energy increases and potential energy unchanged, so internal energy increases B1 3(b)(ii) temperature constant so no change in (molecular) kinetic energy B1 separation increases so potential energy (of molecules) increases B1 potential energy increases and kinetic energy unchanged, so internal energy increases B1

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4 A block of mass m oscillates vertically on a spring, as shown in Fig. 4.1. spring block oscillations equilibrium position Fig. 4.1 The acceleration a of the block varies with displacement x from its equilibrium position, as shown in Fig. 4.2. 2A a A 0 –3Y –2Y –Y 0 Y 2Y 3Y x –A –2A Fig. 4.2 The amplitude of the oscillations is 3Y and the maximum acceleration is 2A. (a) Explain how Fig. 4.2 shows that the oscillations of the block are simple harmonic. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Deduce expressions, in terms of some or all of m, A and Y, for: (i) the angular frequency ω of the oscillations ω = ......................................................... [1] (ii) the maximum speed v0 of the oscillations v0 = ......................................................... [2] (iii) the energy E of the oscillations. E = ......................................................... [2] (c) The period of the oscillations is 0.75 s and the value of 3Y is 1.8 cm. Determine an expression for x in terms of time t, where x is in cm and t is in seconds. x = ......................................................... [2] [Total: 9]

Mark scheme: 4(a) straight line through the origin shows that a is proportional to x B1 negative gradient shows that a and x are (always) in opposite directions B1 4(b)(i) a = –2x  = √(2A / 3Y) A1 4(b)(ii) v0 = x0 C1 = 3Y  √(2A / 3Y) = √(6AY) A1 4(b)(iii) E = ½ m2x02 C1 = ½ m  (2A / 3Y)  (3Y)2 = 3mAY A1 4(c)  = 2 / T ( = 2 / 0.75) C1 x = 1.8 sin (8.4 t) A1

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5 (a) Define electric potential at a point. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Two isolated charged metal spheres X and Y are near to each other in a vacuum. The centres of the spheres are 1.2 m apart, as shown in Fig. 5.1. X Y P x 1.2 m Fig. 5.1 (not to scale) Point P is on the line joining the centres of spheres X and Y and is at a variable distance x from the centre of X. Fig. 5.2 shows the variation with x of the total electric potential V due to the two spheres. V 0 0 0.2 0.4 0.6 0.8 1.0 1.2 x / m Fig. 5.2 State three conclusions that may be drawn about the spheres from Fig. 5.2. The conclusions may be qualitative or quantitative. 1 ................................................................................................................................................ ................................................................................................................................................... 2 ................................................................................................................................................ ................................................................................................................................................... 3 ................................................................................................................................................ ................................................................................................................................................... [3] (c) A proton is held at rest on the line joining the centres of the spheres in (b) at the position where x = 0.60 m. The proton is released. Describe and explain, without calculation, the subsequent motion of the proton. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 7]

Mark scheme: 5(a) work done per unit charge B1 work (done) moving positive charge from infinity (to the point) B1 5(b) Any three points from: Up to 2 points from:  radius of sphere X is 0.30 m  radius of sphere Y is 0.10 m  radius of X is treble the radius of Y Up to 2 points from:  charge on X is positive  charge on Y is positive  spheres X and Y carry charges of the same sign Up to 1 point from:  (magnitudes of) charges on the spheres are equal  charges on the spheres have the same magnitude B3 5(c) proton remains at rest (in the position of release) M1 potential energy of proton is (already) at its minimum or (electric) forces (from spheres) on proton are equal and opposite or no resultant (electric) force on proton or resultant electric field strength (at proton) is zero A1

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6 (a) Two capacitors X and Y are connected in series to a power supply of voltage V, as shown in Fig. 6.1. V X Y Fig. 6.1 The capacitance of X is CX and the capacitance of Y is CY. Derive an expression, in terms of CX and CY, for the combined capacitance CT of the capacitors in this circuit. Explain your reasoning. [3] (b) Two capacitors P and Q are connected in parallel to a power supply of voltage V. The capacitance of P is 200 μF. The capacitance CQ of Q can be varied between 0 and 400 μF. When CQ = 0, the total energy stored in the capacitors is 2.5 mJ. (i) Show that the supply voltage V is 5.0 V. [2] (ii) Calculate the total energy, in mJ, stored in the capacitors when CQ has its maximum value. total energy = ................................................... mJ [3] (iii) On Fig. 6.2, sketch the variation of the total energy E stored in the capacitors with CQ, as CQ varies from 0 to 400 μF. 10.0 E / mJ 7.5 5.0 2.5 0 0 100 200 300 400 CQ / μF Fig. 6.2 [2] [Total: 10]

Mark scheme: 6(a) equal charge on both capacitors B1 VX + VY = V M1 (Q / CX) + (Q / CY) = (Q / CT) leading to (1 / CX) + (1 / CY) = (1 / CT) or (VX/ Q) + (VY/ Q) = (V / Q) leading to (1 / CX) + (1 / CY) = (1 / CT) A1 6(b)(i) E = ½CV 2 C1 V = √[(2  2.5  10–3) / (200  10–6)] = 5.0 V A1 6(b)(ii) total capacitance = 600 F C1 E = ½  600  10–6  5.02 ( = 7.5  10–3 J) C1 = 7.5 mJ A1 6(b)(iii) line with positive gradient starting at (0, 2.5) B1 straight line passing through (400, 7.5) B1

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7 (a) State Faraday’s law of electromagnetic induction. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Fig. 7.1 shows a coil at rest in a uniform magnetic field that is parallel to the axis of the coil. coil magnetic field V Fig. 7.1 The coil is connected to a centre-zero voltmeter. The flux density B of the uniform magnetic field varies with time t as shown in Fig. 7.2. 8 B / mT 4 0 0 1 2 3 4 5 6 t / ms Fig. 7.2 The coil consists of 340 turns, each of cross-sectional area 3.2 × 10–4 m2. (i) Calculate the maximum magnetic flux through one turn of the coil. maximum magnetic flux = ................................................... Wb [2] (ii) Determine the maximum rate of change of magnetic flux linkage in the coil. maximum rate of change of flux linkage = ............................................. Wb s–1 [3] (iii) State the maximum electromotive force (e.m.f.) V0 induced across the coil. V0 = ...................................................... V [1] (iv) On Fig. 7.3, sketch the variation of the e.m.f. V induced across the coil with t from t = 0 to t = 6.0 ms. V0 V 0 0 1 2 3 4 5 6 t / ms –V0 Fig. 7.3 [3] (v) The variation of V with t can be described by V = A sin Bt where A and B are constants. Determine the values of A and B. Give units with your answers. A = .......................................unit ................ B = .......................................unit ................ [3] [Total: 14]

Mark scheme: 7(a) (induced) e.m.f. is (directly) proportional to rate M1 of change of (magnetic) flux (linkage) A1 7(b)(i)  = BA C1 = 7.2  10–3  3.2  10–4 = 2.3  10–6 Wb A1 7(b)(ii) tangent drawn at steepest point on Fig. 7.2 C1 evidence of multiplication by 340 C1 maximum rate of change of flux = 0.82 Wb s–1 A1 7(b)(iii) V0 = 0.82 V or V0 given as identical numerical answer to the answer in (b)(ii) A1 7(b)(iv) sinusoidal curve of period 2.0 ms from t = 0 to t = 6.0 ms B1 all peaks at +V0 and all troughs at –V0 B1 line showing V = 0 at (and only at) t = 0, 1.0, 2.0, 3.0, 4.0, 5.0 and 6.0 ms B1 7(b)(v) A = 0.82 V or A has same numerical value as answer in (b)(iii), with unit V A1 B = 2 / (2.0  10–3) C1 = 3100 rad s–1 A1

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8 Fig. 8.1 shows part of the emission spectrum of visible radiation emitted by hydrogen gas in a star in a distant galaxy. increasing frequency Fig. 8.1 The galaxy is moving away from the Earth at a speed of 6.2 × 106 m s–1. (a) (i) Explain how the positions of the lines in the emission spectrum seen by an observer on the Earth differ from the positions shown in Fig. 8.1. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) On Fig. 8.1, draw the three lines in possible positions in the spectrum seen by the observer. [2] (b) The lines in Fig. 8.1 correspond to electron transitions down to the energy level –3.40 eV. One of the lines represents emitted radiation of wavelength 488 nm. (i) Calculate the energy of a photon of this radiation. photon energy = ...................................................... J [2] (ii) Determine the energy, in eV, of the energy level from which the electron transition originates to cause the emission of this radiation. energy level = .................................................... eV [2] (iii) Determine the wavelength, in nm, of this radiation as detected by the observer on the Earth. wavelength = ................................................... nm [2] (c) A value for the Hubble constant is 2.3 × 10–18 s–1. Determine the distance of the galaxy from the Earth. distance = ..................................................... m [2] [Total: 12]

Mark scheme: 8(a)(i) movement of star causes change in (observed) frequency or movement of star causes redshift B1 observed frequency is lower (than emitted frequency) B1 8(a)(ii) all three lines shown to left of corresponding printed lines B1 distance between drawn line and corresponding printed line approximately the same for all three lines B1 8(b)(i) E = hf and  = c / f C1 E = (6.63  10–34  3.00  108) / (488  10–9) = 4.08  10–19 J A1 8(b)(ii) photon energy = (4.08  10–19) / (1.60  10–19) = 2.55 eV C1 energy level = –3.40 + 2.55 = –0.85 eV A1 8(b)(iii)  =   (v / c) = (488  6.2  106) / (3.00  108) ( = 10 nm) C1 observed wavelength = 488 +  = 488 + 10 = 498 nm A1 Question Answer Marks 8(c) v = H0d C1 d = (6.2  106) / (2.3  10–18) = 2.7  1024 m A1

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9 (a) State what is meant by the binding energy of a nucleus. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Table 9.1 shows the masses of two sub-atomic particles and a polonium-212 (21284Po) nucleus. Table 9.1 mass / u proton 1.007 276 neutron 1.008 665 polonium-212 nucleus 211.942 749 For the polonium-212 nucleus, determine: (i) the mass defect Δm, in kg Δm = .................................................... kg [3] (ii) the binding energy binding energy = ...................................................... J [2] (iii) the binding energy per nucleon. binding energy per nucleon = ...................................................... J [1] (c) (i) On Fig. 9.1, sketch the variation with nucleon number A of binding energy per nucleon for values of A from 1 to 250. binding energy per nucleon 0 1 250 A Fig. 9.1 [2] (ii) On your line in Fig. 9.1, draw an X to show the approximate position of polonium-212. [1] (iii) Polonium-212 is radioactive and undergoes alpha-decay. Suggest and explain, with reference to Fig. 9.1, why the alpha-decay of polonium-212 results in a release of energy. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 13]

Mark scheme: 9(a) energy required to separate (all) the nucleons (in the nucleus) M1 to infinity A1 9(b)(i) m = {[(84  1.007276) + (128  1.008665)] – 211.942749} (u) ( = 1.778 u) C1 = 1.778  1.66  10–27 (kg) C1 = 2.95  10–27 kg A1 9(b)(ii) E = ()mc2 C1 binding energy = 2.95  10–27  (3.00  108)2 = 2.66  10–10 J A1 9(b)(iii) binding energy per nucleon = (2.66  10–10) / 212 = 1.25  10–12 J A1 9(c)(i) line rising to a single peak that is to the left of the ‘9’ in the Fig. 9.1 label and then continually decreasing B1 steep positive gradient on the left of the peak and shallow negative gradient on the right B1 9(c)(ii) X shown on the line at a value of A that is to the right of the left-hand edge of the ‘A’ in the axis label, and to the left of ‘2’ in the 250 label B1 9(c)(iii) nucleus formed (as a result of the decay) has a lower nucleon number B1 (nucleus formed has a) greater binding energy per nucleon B1

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10 (a) Describe how reflected ultrasound pulses may be used to obtain diagnostic information about internal structures. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) (i) Define specific acoustic impedance of a medium. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Table 10.1 shows some data for water and for glass. Table 10.1 density / kg m–3 speed of sound / m s–1 water 1000 1420 glass 2500 4560 Determine the intensity reflection coefficient for ultrasound that is incident on a water–glass boundary. intensity reflection coefficient = ......................................................... [3] [Total: 7]

Mark scheme: 10(a) time gives information about depth (of boundary) B1 intensity gives information about nature of boundary B1 10(b)(i) product of density and speed M1 speed of ultrasound in medium (and density of medium) A1 10(b)(ii) Zwater = 1000  1420 (= 1.42  106 kg m–2 s–1) and Zglass = 2500  4560 (= 11.4  106 kg m–2 s–1) C1 intensity reflection coefficient = (11.4 – 1.42)2 / (11.4 + 1.42)2 C1 = 0.61 A1

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