Cambridge A Level Physics 9702 — 2021 May/June Paper 2 · Variant 3

1 (a) A property of a vector quantity, that is not a property of a scalar quantity, is direction. For example, velocity has direction but speed does not. (i) State two other scalar quantities and two other vector quantities. scalar quantities: .................................................... and .................................................... vector quantities: .................................................... and .................................................... [2] (ii) State two properties that are possessed by both scalar and vector physical quantities. 1. ....................................................................................................................................... 2. ....................................................................................................................................... [2] (b) A ship at sea is travelling with a velocity of 13 m s–1 in a direction 35° east of north in still water, as shown in Fig. 1.1. N N velocity 13 m s–1 35° W E S Fig. 1.1 (i) Determine the magnitudes of the components of the velocity of the ship in the north and the east directions. north component of velocity = ...................................................... m s–1 east component of velocity = ...................................................... m s–1 [2] (ii) The ship now experiences a tidal current. The water in the sea moves with a velocity of 2.7 m s–1 to the west. Calculate the resultant velocity component of the ship in the east direction. resultant east component of velocity = ................................................ m s–1 [1] (iii) Use your answers in (b)(i) and (b)(ii) to determine the magnitude of the resultant velocity of the ship. magnitude of resultant velocity = ................................................ m s–1 [2] (iv) Use your answers in (b)(i) and (b)(ii) to determine the angle between north and the resultant velocity of the ship. angle = ........................................................° [2] [Total: 11]

Mark scheme: 1(a)(i) two correct scalar quantities e.g. time, mass, distance, temperature B1 two correct vector quantities e.g. force, acceleration, velocity, displacement B1 1(a)(ii) magnitude B1 unit B1 1(b)(i) north component of velocity = 11 m s–1 A1 east component of velocity = 7.5 m s–1 A1 1(b)(ii) velocity = 7.5 – 2.7 = 4.8 m s–1 A1 1(b)(iii) velocity = √(112 + 4.82) C1 = 12 m s–1 A1 1(b)(iv) angle = tan–1 (4.8 / 11) C1 = 24° A1 Question Answer Marks

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2 (a) Define acceleration. ................................................................................................................................................... ............................................................................................................................................. [1] (b) A stone falls vertically from the top of a cliff. Fig. 2.1 shows the variation with time t of the velocity v of the stone. 40 v / m s–1 30 20 10 0 0 5 10 15 20 25 30 t / s Fig. 2.1 (i) Explain, with reference to forces acting on the stone, the shape of the curve in Fig. 2.1. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) Use Fig. 2.1 to determine the speed of the stone when the resultant force on it is zero. speed = ................................................ m s–1 [1] (iii) Use Fig. 2.1 to calculate the approximate height through which the stone falls between t = 0 and t = 30 s. height = ..................................................... m [3] (iv) On Fig. 2.2, sketch the variation with t of the acceleration a of the stone between t = 0 and t = 30 s. 20 a / m s–2 15 10 5 0 0 5 10 15 20 25 30 t / s Fig. 2.2 [3] [Total: 11]

Mark scheme: 2(a) change in velocity / time (taken) B1 2(b)(i) air resistance increases (with speed/with time) B1 resultant force decreases (as speed increases/with time) so acceleration decreases (as speed increases/with time) B1 when air resistance equals the weight the speed/velocity/v becomes constant B1 2(b)(ii) speed = 36 m s–1 A1 2(b)(iii) height given by area under the curve C1 height = 950 m Round to two significant figures and award 2 marks for a value in the range 920–980 m and 1 mark for a value in the range 900–910 m or 990–1000 m. A2 2(b)(iv) line starting at (0, 9.8) B1 curve with negative gradient between t = 0 and t = 20 s B1 line showing zero acceleration between t = 20 s and t = 30 s B1 Question Answer Marks

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3 (a) Define the moment of a force about a point. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Fig. 3.1 shows a type of balance that is used for measuring mass. fixed point P mm scale 200 spring 52.6 cm pan 1.8 cm pointer rod pivot 0 6.2 cm Fig. 3.1 (not to scale) A rigid rod is pivoted about a point 6.2 cm from the centre of a pan which is attached to one end. The object being measured is placed on the centre of this pan. A spring, attached to the rod 1.8 cm from the pivot, is attached at its other end to a fixed point P. The spring obeys Hooke’s law over the full range of operation of the balance. A pointer, on the other side of the pivot, is set against a millimetre scale which is a distance 52.6 cm from the pivot. When the system is in equilibrium with no mass on the pan, the rod is horizontal and the pointer indicates a reading on the scale of 86 mm. An object of mass 0.472 kg is now placed on the pan. As a result, the pointer moves to indicate a reading of 123 mm on the scale when the system is again in equilibrium. (i) Show that the increase in the length of the spring is approximately 1.3 mm. [2] (ii) Calculate the magnitude of the moment about the pivot of the weight of the object. moment = .................................................. N m [2] (iii) Use your answer in (b)(ii) to determine the increase in the tension in the spring due to the 0.472 kg mass. increase in tension = ..................................................... N [2] (iv) Use the information in (b)(i) and your answer in (b)(iii) to determine the spring constant k of the spring. Give a unit with your answer. k = ...................................... unit ............ [2] [Total: 10]

Mark scheme: 3(a) force × distance M1 perpendicular distance of (line of action of) force from the point A1 3(b)(i) distance moved by pointer = 123 – 86 (= 37 mm) C1 (extension =) 37 × (1.8 / 52.6) = 1.3 (mm) or sin or tan θ = 37 / 526 (so θ = 4.0° so extension =) sin or tan θ × 18 = 1.3 (mm) A1 3(b)(ii) moment = 0.472 × 9.81 × 6.2 × 10–2 C1 = 0.29 N m A1 3(b)(iii) (Δ)F × 1.8 × 10–2 = 0.29 C1 ΔF = 16 N A1 3(b)(iv) k = F / x C1 = 16 / (1.3 × 10–3) = 1.2 × 104 N m–1 A1 Question Answer Marks

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4 (a) State the principle of superposition. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Two waves, with intensities I and 4I, superpose. The waves have the same frequency. Determine, in terms of I, the maximum possible intensity of the resulting wave. maximum intensity = ....................................................... I [2] (c) Coherent light of wavelength 550 nm is incident normally on a double slit of slit separation 0.35 mm. A series of bright and dark fringes forms on a screen placed a distance of 1.2 m from the double slit, as shown in Fig. 4.1. The screen is parallel to the double slit. screen 1.2 m light 0.35 mm wavelength 550 nm double slit Fig. 4.1 (not to scale) (i) Determine the distance between the centres of adjacent bright fringes on the screen. distance = ..................................................... m [3] (ii) The light of wavelength 550 nm is replaced with red light of a single frequency. State and explain the change, if any, in the distance between the centres of adjacent bright fringes. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] [Total: 8]

Mark scheme: 4(a) (when two or more) waves meet/overlap (at a point) B1 (resultant) displacement is sum of the individual displacements B1 4(b) intensity ∝ amplitude2 C1 maximum intensity = 9I A1 4(c)(i) x = λD / a C1 = (550 × 10–9 × 1.2) / (0.35 × 10–3) C1 = 1.9 × 10–3 m A1 4(c)(ii) red light has longer wavelength (than 550 nm) so distance (between fringes) increases B1 Question Answer Marks

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5 (a) Define the electromotive force (e.m.f.) of a source. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) The circuit shown in Fig. 5.1 contains a battery of e.m.f. E that has internal resistance r, a variable resistor, a voltmeter and an ammeter. E r X Y A V I Fig. 5.1 Readings from the two meters are taken for different settings of the variable resistor. The variation with current I of the potential difference (p.d.) V across the terminals XY of the battery is shown in Fig. 5.2. 8 V / V 6 4 2 0 0 0.2 0.4 0.6 0.8 1.0 1.2 I / A Fig. 5.2 Explain why V is not constant. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (c) For the battery in (b), use Fig. 5.2 to determine: (i) the e.m.f. E E = ...................................................... V [1] (ii) the maximum current that the battery can supply maximum current = ...................................................... A [1] (iii) the internal resistance r. r = ..................................................... Ω [2] (d) On Fig. 5.2, sketch a line to show a possible variation with I of V for a battery with a lower e.m.f. and a lower internal resistance than the battery in (b). Your line should extend over at least the same range of currents as the original line. [2] [Total: 11]

Mark scheme: 5(a) energy per unit charge B1 energy transferred by source driving charge around the complete circuit or energy transferred from other forms to electrical energy B1 5(b) there is a p.d. across the internal resistance/r B1 change in current/I results in a change in p.d. across the internal resistance B1 V = E – p.d. across internal resistance or change in p.d. across r causes a change in V (as e.m.f. is constant) B1 5(c)(i) E = 7.4 V A1 5(c)(ii) maximum current = 0.92 A A1 5(c)(iii) r = E / IMAX or (–)gradient C1 e.g. r = 7.4 / 0.92 = 8.0 Ω A1 5(d) straight line with negative gradient that is smaller in magnitude than the original line B1 line which would have intercept on V-axis below the original line B1 Question Answer Marks

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6 (a) State the quark composition of: (i) a proton ..................................................................................................................................... [1] (ii) a neutron ..................................................................................................................................... [1] (iii) an alpha-particle. ........................................................................................................................................... ..................................................................................................................................... [2] (b) In the alpha-particle scattering experiment, alpha-particles were directed at a thin gold foil. State what may be inferred from: (i) the observation that most alpha-particles pass through the foil ..................................................................................................................................... [1] (ii) the observation that some alpha-particles are scattered through angles greater than 90°. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (c) A proton and an alpha-particle are moving in the same uniform electric field. Determine the ratio acceleration of proton due to the electric field . acceleration of alpha-particle due to the electric field ratio = ......................................................... [2] [Total: 9]

Mark scheme: 6(a)(i) up up down B1 6(a)(ii) up down down B1 6(a)(iii) (alpha-particle is) 2 protons and 2 neutrons C1 6 up, 6 down A1 6(b)(i) most of an atom is empty space or the nucleus (volume) is (very) small compared with the atom B1 6(b)(ii) the nucleus is charged B1 the majority of the mass of atom is in the nucleus B1 6(c) F = Eq and a = F / m C1 a = Eq / m ratio = (e / m) / (2e / 4m) = 2 A1

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