Cambridge A Level Physics 9702 — 2015 May/June Paper 2 · Variant 3

1 (a) The distance between the Sun and the Earth is 1.5 × 1011 m. State this distance in Gm. distance = ................................................... Gm [1] (b) The distance from the centre of the Earth to a satellite above the equator is 42.3 Mm. The radius of the Earth is 6380 km. A microwave signal is sent from a point on the Earth directly below the satellite. Calculate the time taken for the microwave signal to travel to the satellite and back. time = ....................................................... s [2] (c) The speed v of a sound wave through a gas of density ρ and pressure P is given by CP v = ρ where C is a constant. Show that C has no unit. [3] (d) Underline all the scalar quantities in the list below. acceleration energy momentum power weight [1] (e) A boat travels across a river in which the water is moving at a speed of 1.8 m s–1. The velocity vectors for the boat and the river water are shown to scale in Fig. 1.1. water velocity 1.8 m s–1 river boat velocity 3.0 m s–1 60° river bank Fig. 1.1 (shown to scale) In still water the speed of the boat is 3.0 m s–1. The boat is directed at an angle of 60° to the river bank. (i) On Fig. 1.1, draw a vector triangle or a scale diagram to show the resultant velocity of the boat. [2] (ii) Determine the magnitude of the resultant velocity of the boat. resultant velocity = ................................................ m s–1 [2]

Mark scheme: 1 (a) 150 or 1.5 × 102 Gm A1 [1] (b) distance = 2 × (42.3 – 6.38) × 106 (= 7.184 × 107 m) C1 (time =) 7.184 × 107 / (3.0 × 108) = 0.24 (0.239) s A1 [2] (c) units of pressure P: kg m s–2 / m2 = kg m–1 s–2 M1 units of density ρ: kg m–3 and speed v: m s–1 M1 simplification for units of C: C = v2 ρ / P units: (m2 s–2 kg m–3) / kg m–1 s–2 and cancelling to give no units for C A1 [3] (d) energy and power (both underlined and no others) A1 [1] (e) (i) vector triangle of correct orientation M1 three arrows for the velocities in the correct directions A1 [2] (ii) length measured from scale diagram 5.2 ± 0.2 cm or components of boat speed determined parallel and perpendicular to river flow C1 velocity = 2.6 m s–1 (allow ± 0.1 m s–1) A1 [2]

More questions on Scalars and vectors

2 The variation with time t of the velocity v of a ball is shown in Fig. 2.1. 5 v / m s–1 0 0 2 4 6 8 10 12 14 16 t / s ï Fig. 2.1 The ball moves in a straight line from a point P at t = 0. The mass of the ball is 400 g. (a) Use Fig. 2.1 to describe, without calculation, the velocity of the ball from t = 0 to t = 16 s. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) Use Fig. 2.1 to calculate, for the ball, (i) the displacement from P at t = 10 s, displacement = ...................................................... m [2] (ii) the acceleration at t = 10 s, acceleration = ................................................ m s–2 [2] (iii) the maximum kinetic energy. kinetic energy = ....................................................... J [2] (c) Use your answers in (b)(i) and (b)(ii) to determine the time from t = 0 for the ball to return to P. time = ....................................................... s [2]

Mark scheme: 2 (a) constant rate of increase in velocity/acceleration from t = 0 to t = 8 s B1 constant deceleration from t = 8 s to t = 16 s or constant rate of increase in velocity in the opposite direction from t = 10 s to t = 16 s B1 [2] (b) (i) area under lines to 10 s C1 (displacement =) (5.0 × 8.0) / 2 + (5.0 × 2.0) / 2 = 25 m or ½ (10.0 × 5.0) = 25 m A1 [2] (ii) a = (v – u) / t or gradient of line C1 = (–15.0 –5.0) / 8.0 = (–) 2.5 m s–2 A1 [2] (iii) KE = ½ m v2 C1 = 0.5 × 0.4 × (15.0)2 = 45 J A1 [2] (c) (distance =) 25 (m) (= ut + ½ at 2) = 0 + ½ × 2.5 × t 2 C1 (t = 4.5 (4.47) s therefore) time to return = 14.5 s A1 [2]

More questions on Equations of motion

3 (a) Define power. ................................................................................................................................................... ...............................................................................................................................................[1] (b) Fig. 3.1 shows a car travelling at a speed of 22 m s–1 on a horizontal road. speed 22 m s–1 1200 N resistive force horizontal road Fig. 3.1 The car has a mass of 1500 kg. A resistive force of 1200 N acts on the car. Calculate (i) the force F required from the car to produce an acceleration of 0.82 m s–2, F = ...................................................... N [3] (ii) the power required to produce this acceleration. power = ..................................................... W [2] (c) The resistive force on the car is proportional to v 2, where v is the speed of the car. Suggest why the car has a maximum speed. ................................................................................................................................................... ...............................................................................................................................................[1]

Mark scheme: 3 (a) (power =) work done / time (taken) or rate of work done A1 [1] (b) (i) F – R = ma C1 F = 1500 × 0.82 + 1200 C1 = 2400 (2430) N A1 [3] (ii) P = Fv C1 = (2430 × 22) = 53 000 (53 500) W A1 [2] (c) (there is maximum power from car and) resistive force = force produced by car hence no acceleration or suggestion in terms of power produced by car and power wasted to overcome resistive force B1 [1]

More questions on Momentum and Newton’s laws of motion

4 Fig. 4.1 shows the values obtained in an experiment to determine the Young modulus E of a metal in the form of a wire. quantity value instrument diameter d 0.48 mm length l 1.768 m 5.0 N to 30.0 N load F in 5.0 N steps extension e 0.25 mm to 1.50 mm Fig. 4.1 (a) (i) Complete Fig. 4.1 with the name of an instrument that could be used to measure each of the quantities. [3] (ii) Explain why a series of values of F, each with corresponding extension e, are measured. ........................................................................................................................................... .......................................................................................................................................[1] (b) Explain how a series of readings of the quantities given in Fig. 4.1 is used to determine the Young modulus of the metal. A numerical answer for E is not required. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2]

Mark scheme: 4 (a) (i) diameter and extension: micrometer (screw gauge) or digital calipers B1 length: tape measure or metre rule B1 load: spring balance or Newton meter B1 [3] (ii) to reduce the effect of random errors or to plot a graph to check for zero error in measurement of extension or to see if limit of proportionality is exceeded B1 [1] (b) plot a graph of F against e and determine the gradient B1 E = (gradient × l) / [πd 2 / 4] B1 [2]

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5 A uniform resistance wire AB has length 50 cm and diameter 0.36 mm. The resistivity of the metal of the wire is 5.1 × 10–7 Ω m. (a) Show that the resistance of the wire AB is 2.5 Ω. [2] (b) The wire AB is connected in series with a power supply E and a resistor R as shown in Fig. 5.1. E R M B A 2.5 1 C N D Fig. 5.1 The electromotive force (e.m.f.) of E is 6.0 V and its internal resistance is negligible. The resistance of R is 2.5 Ω. A second uniform wire CD is connected across the terminals of E. The wire CD has length 100 cm, diameter 0.18 mm and is made of the same metal as wire AB. Calculate (i) the current supplied by E, current = ...................................................... A [4] (ii) the power transformed in wire AB, power = ..................................................... W [2] (iii) the potential difference (p.d.) between the midpoint M of wire AB and the midpoint N of wire CD. p.d. = ...................................................... V [2]

Mark scheme: 5 (a) R = ρl / A C1 = (5.1 × 10−7 × 0.50) / π(0.18 × 10−3)2 = 2.5 (2.51) Ω M1 [2] (b) (i) resistance of CD = 8 × resistance of AB = 20 (Ω) C1 circuit resistance = [1 / 5.0 + 1 / 20]−1 = 4.0 (Ω) C1 current = V / R = 6.0 / 4.0 C1 = 1.5 A A1 [4] (ii) power in AB = I 2R or power = V 2 / R C1 = (1.2)2 × 2.5 = 3.6 W = (3.0)2 / 2.5 = 3.6 W A1 [2] (iii) potential drop A to M = 1.25 × 1.2 = 1.5 V M1 potential drop C to N = 3.0 V p.d. MN = 1.5 V A1 [2]

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6 (a) Two overlapping waves of the same type travel in the same direction. The variation with distance x of the displacement y of each wave is shown in Fig. 6.1. 3.0 y / cm 2.0 1.0 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 x / m Fig. 6.1 The speed of the waves is 240 m s–1. The waves are coherent and produce an interference pattern. (i) Explain the meaning of coherence and interference. coherence: ......................................................................................................................... ........................................................................................................................................... interference: ....................................................................................................................... ........................................................................................................................................... [2] (ii) Use Fig. 6.1 to determine the frequency of the waves. frequency = .................................................... Hz [2] (iii) State the phase difference between the waves. phase difference = ........................................................ ° [1] (iv) Use the principle of superposition to sketch, on Fig. 6.1, the resultant wave. [2] (b) An interference pattern is produced with the arrangement shown in Fig. 6.2. B S1 A laser 0.13 mm S2 85 cm screen Fig. 6.2 (not to scale) Laser light of wavelength λ of 546 nm is incident on the slits S1 and S2. The slits are a distance 0.13 mm apart. The distance between the slits and the screen is 85 cm. Two points on the screen are labelled A and B. The path difference between S1A and S2A is zero. The path difference between S1B and S2B is 2.5 λ. Maxima and minima of intensity of light are produced on the screen. (i) Calculate the distance AB. distance = ...................................................... m [3] (ii) The laser is replaced by a laser emitting blue light. State and explain the change in the distance between the maxima observed on the screen. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[1]

Mark scheme: 6 (a) (i) coherent: constant phase difference B1 interference is the (overlapping of waves and the) sum of/addition of displacement of two waves B1 [2] (ii) wavelength = 3.2 m (allow ± 0.05 m) M1 f (= v / λ = 240 / 3.2) = 75 Hz A1 [2] (iii) 90° (allow ± 2°) or π/2 rad A1 [1] (iv) sketch has amplitude 3.0 ± 0.1 cm M1 correct displacement values at previous peaks to produce correct shape A1 [2] (b) (i) λ = ax / D C1 x = (546 × 10–9 × 0.85) / 0.13 × 10–3 (= 3.57 × 10–3 m) C1 AB = 8.9 (8.93) × 10–3 m A1 [3] (ii) shorter wavelength for blue light so separation is less B1 [1]

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7 The equation represents the spontaneous radioactive decay of a nucleus of bismuth-212. 212 208 83 Bi X + 81 Tl + 6.2 MeV (a) (i) Explain the meaning of spontaneous radioactive decay. ........................................................................................................................................... .......................................................................................................................................[1] (ii) State the constituent particles of X. .......................................................................................................................................[1] (b) (i) Use the conservation of mass-energy to explain the release of 6.2 MeV of energy in this reaction. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) Calculate the energy, in joules, released in this reaction. energy = ....................................................... J [1]

Mark scheme: 7 (a) (i) (rate of decay) not affected by any external factors or changes in temperature and pressure etc. B1 [1] (ii) two protons and two neutrons B1 [1] (b) (i) (total) mass before decay/on left-hand side is greater than (total) mass M1 on right-hand side/after the decay the difference in mass is released as kinetic energy of the products A1 [2] (may also be some γ radiation) (to conserve mass-energy) (ii) (6.2 × 106 × 1.6 × 10−19 =) 9.9(2) × 10−13 J A1 [1]

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