Cambridge A Level Physics 9702 — 2011 May/June Paper 2 · Variant 2

1 (a) Distinguish between scalar quantities and vector quantities. .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (b) In the following list, underline all the scalar quantities. acceleration force kinetic energy mass power weight [1] (c) A stone is thrown with a horizontal velocity of 20 m s–1 from the top of a cliff 15 m high. The path of the stone is shown in Fig. 1.1. 20 m s–1 cliff 15 m ground Fig. 1.1 Air resistance is negligible. For this stone, (i) calculate the time to fall 15 m, time = .............................................. s [2] (ii) calculate the magnitude of the resultant velocity after falling 15 m, resultant velocity = ........................................ m s–1 [3] (iii) describe the difference between the displacement of the stone and the distance For that it travels. Examiner’s Use .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [2]

Mark scheme: 1 (a) scalar has only magnitude B1 vector has magnitude and direction B1 [2] (b) kinetic energy, mass, power all three underlined B1 [1] (c) (i) s = ut + ½ at2 15 = 0.5 × 9.81 × t2 C1 T = 1.7 s A1 [2] if g = 10 is used then –1 but only once on paper (ii) vertical component vv: vv2 = u2 + 2as = 0 + 2 × 9.81 × 15 or vv = u + at = 9.81 × 1.7(5) vv =17.16 C1 resultant velocity: v2 = (17.16)2 + (20)2 C1 v = 26 m s–1 A1 [3] If u = 20 is used instead of u = 0 then 0/3 Allow the solution using: initial (potential energy + kinetic energy) = final kinetic energy (iii) distance is the actual path travelled B1 displacement is the straight line distance between start and finish points (in that direction) / minimum distance B1 [2]

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2 (a) A sphere of radius R is moving through a fluid with constant speed v. There is a frictional For force F acting on the sphere, which is given by the expression Examiner’s Use F = 6πDRv where D depends on the fluid. (i) Show that the SI base units of the quantity D are kg m–1 s–1. [3] (ii) A raindrop of radius 1.5 mm falls vertically in air at a velocity of 3.7 m s–1. The value of D for air is 6.6 × 10–4 kg m–1 s–1. The density of water is 1000 kg m–3. Calculate 1. the magnitude of the frictional force F, F = ............................................. N [1] 2. the acceleration of the raindrop. acceleration = ........................................ m s–2 [3] (b) The variation with time t of the speed v of the raindrop in (a) is shown in Fig. 2.1. For Examiner’s Use v 0 0 t Fig. 2.1 (i) State the variation with time of the acceleration of the raindrop. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [3] (ii) A second raindrop has a radius that is smaller than that given in (a). On Fig. 2.1, sketch the variation of speed with time for this second raindrop. [2]

Mark scheme: 2 (a) (i) base units of D: force: kg m s–2 B1 radius: m velocity: m s–1 B1 base units of D: [F / (R × v)] kg m s–2 / (m × m s–1) M1 = kg m–1 s–1 A0 [3] (ii) 1. F = 6π × D × R × v = [6π × 6.6 × 10–4 × 1.5 × 10–3 × 3.7] = 6.9 × 10–5 N A1 [1] 2. mg – F = ma hence a = g – [F / m] m = ρ × V = ρ × 4/3 π R3 = (1.4 × 10–5) C1 a = 9.81 – [6.9 × 10–5] / ρ × 4/3 π × (1.5 × 10–3)3 (9.81 – 4.88) M1 a = 4.9(3) m s–2 A1 [3] (b) (i) a = g at time t = 0 B1 a decreases (as time increases) B1 a goes to zero B1 [3] (ii) Correct shape below original line M1 sketch goes to terminal velocity earlier A1 [2] GCE AS/A LEVEL – May/June 2011 9702 22

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3 (a) (i) Explain what is meant by work done. For Examiner’s .................................................................................................................................. Use ............................................................................................................................. [1] (ii) Define power. .................................................................................................................................. ............................................................................................................................. [1] (b) Fig. 3.1 shows part of a fairground ride with a carriage on rails. 4.1 m 9.5 m s–1 30° Fig. 3.1 The carriage and passengers have a total mass of 600 kg. The carriage is travelling at a speed of 9.5 m s–1 towards a slope inclined at 30° to the horizontal. The carriage comes to rest after travelling up the slope to a vertical height of 4.1 m. (i) Calculate the kinetic energy, in kJ, of the carriage and passengers as they travel towards the slope. kinetic energy = ............................................ kJ [3] (ii) Show that the gain in potential energy of the carriage and passengers is 24 kJ. [2] (iii) Calculate the work done against the resistive force as the carriage moves up the For slope. Examiner’s Use work done = ............................................ kJ [1] (iv) Use your answer in (iii) to calculate the resistive force acting against the carriage as it moves up the slope. resistive force = ............................................. N [2]

Mark scheme: 3 (a) (i) work done equals force × distance moved / displacement in the direction of the force B1 [1] (ii) power is the rate of doing work / work done per unit time B1 [1] (b) (i) kinetic energy = ½ mv2 C1 = 0.5 × 600 (9.5)2 C1 = 27075 (J) = 27 kJ A1 [3] (ii) potential energy = mgh = 600 × 9.81 × 4.1 M1 = 24132 (J) A1 = 24 kJ A0 [2] (iii) work done = 27 – 24 = 3.0 kJ A1 [1] (iv) resistive force = 3000 / 8.2 (distance along slope = 4.1 / sin 30°) C1 = 366 N A1 [2]

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4 A student measures the Young modulus of a metal in the form of a wire. For Examiner’s (a) Describe, with the aid of a diagram, the apparatus that could be used. Use .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (b) Describe the method used to obtain the required measurements. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [4] (c) Describe how the measurements taken can be used to determine the Young modulus. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [4]

Mark scheme: 4 (a) clamped horizontal wire over pulley or vertical wire attached to ceiling with mass attached B1 details: reference mark on wire with fixed scale alongside B1 [2] (b) measure original length of wire to reference mark with metre ruler / tape (B1) measure diameter with micrometer / digital calipers (B1) measure initial and final reading (for extension) with metre ruler or other suitable scale (B1) measure / record mass or weight used for the extension (B1) good physics method: measure diameter in several places / remove load and check wire returns to original length / take several readings with different loads (B1) MAX of 4 points B4 [4] (c) determine extension from final and initial readings (B1) plot a graph of force against extension (B1) determine gradient of graph for F / e (B1) calculate area from πd2 / 4 (B1) calculate E from E = F l / e A or gradient × l / A (B1) MAX of 4 points B4 [4] GCE AS/A LEVEL – May/June 2011 9702 22

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5 (a) For a cell, explain the terms For Examiner’s (i) electromotive force (e.m.f.), Use .................................................................................................................................. ............................................................................................................................. [1] (ii) internal resistance. .................................................................................................................................. ............................................................................................................................. [1] (b) The circuit of Fig. 5.1 shows two batteries A and B and a resistor R connected in series. R 3.0 V 12 V A B 0.10 Ω 0.20 Ω Fig. 5.1 Battery A has an e.m.f. of 3.0 V and an internal resistance of 0.10 Ω. Battery B has an e.m.f. of 12 V and an internal resistance of 0.20 Ω. Resistor R has a resistance of 3.3 Ω. (i) Apply Kirchhoff’s second law to calculate the current in the circuit. current = .............................................. A [2] (ii) Calculate the power transformed by battery B. power = ............................................. W [2] (iii) Calculate the total energy lost per second in resistor R and the internal For resistances. Examiner’s Use energy lost per second = ......................................... J s–1 [2] (c) The circuit of Fig. 5.1 may be used to store energy in battery A. Suggest how your answers in (b) support this statement. .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [1]

Mark scheme: 5 (a) (i) energy converted from chemical to electrical when charge flows through cell or round complete circuit B1 (ii) (resistance of the cell) causing loss of voltage or energy loss in cell B1 [2] (b) (i) EB – EA = I (R + rB + rA) 12 – 3 = I (3.3 + 0.1 + 0.2) C1 I = 2.5 A A1 [2] (ii) Power = E × I = 12 × 2.5 C1 = 30 W A1 [2] (iii) P = I2 × R or P = V2 / R or P = VI = (2.5)2 × 3 = 92 / 3.6 = 9 × 2.5 C1 = 22.5 J s–1 A1 [2] (c) power supplied from cell B is greater than energy lost per second in circuit B1 [1]

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6 (a) Apparatus used to produce interference fringes is shown in Fig. 6.1. The apparatus is For not drawn to scale. Examiner’s Use two slits B bright fringe P dark fringe LASER C bright fringe screen Fig. 6.1 (not to scale) Laser light is incident on two slits. The laser provides light of a single wavelength. The light from the two slits produces a fringe pattern on the screen. A bright fringe is produced at C and the next bright fringe is at B. A dark fringe is produced at P. (i) Explain why one laser and two slits are used, instead of two lasers, to produce a visible fringe pattern on the screen. .................................................................................................................................. ............................................................................................................................. [1] (ii) State the phase difference between the waves that meet at 1. B ............................................. [1] 2. P ............................................. [1] (iii) 1. State the principle of superposition. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [2] 2. Use the principle of superposition to explain the dark fringe at P. .................................................................................................................................. ............................................................................................................................. [1] (b) In Fig. 6.1 the distance from the two slits to the screen is 1.8 m. The distance CP is For 2.3 mm and the distance between the slits is 0.25 mm. Examiner’s Calculate the wavelength of the light provided by the laser. Use wavelength = ........................................... nm [3]

Mark scheme: 6 (a) (i) to produce coherent sources or constant phase difference B1 [1] (ii) 1. 360° / 2π rad allow n × 360° or n × 2π (unit missing –1) B1 [1] 2. 180° / π rad allow (n × 360°) – 180° or (n × 2π) – π B1 [1] (iii) 1. waves overlap / meet B1 (resultant) displacement is sum of displacements of each wave B1 [2] 2. at P crest on trough (OWTTE) B1 [1] (b) λ = ax / D C1 = 2 × 2.3 × 10–3 × 0.25 ×10–3 / 1.8 C1 = 639 nm A1 [3]

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