Cambridge A Level Physics 9702 — 2014 Oct/Nov Paper 2 · Variant 1

1 (a) Mass, length and time are SІ base quantities. State two other base quantities. 1. .............................................................................................................................................. 2. .............................................................................................................................................. [2] (b) A mass m is placed on the end of a spring that is hanging vertically, as shown in Fig. 1.1. spring mass m Fig. 1.1 The mass is made to oscillate vertically. The time period of the oscillations of the mass is T. The period T is given by m T = C k where C is a constant and k is the spring constant. Show that C has no units. [3]

Mark scheme: 1 (a) temperature B1 current B1 [2] (allow amount of substance and luminous intensity) (b) base units of force constant: kg m s–2 m–1 or kg s–2 B1 base units of time and mass: s and kg C1 base units of C: s (kg s–2 / kg)1/2 cancelling to show no units B1 [3]

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2 (a) Define pressure. .............................................................................................................................................. [1] (b) A cylinder is placed on a horizontal surface, as shown in Fig. 2.1. diameter cylinder Fig. 2.1 The following measurements were made on the cylinder: mass = 5.09 ± 0.01 kg diameter = 9.4 ± 0.1 cm. (i) Calculate the pressure produced by the cylinder on the surface. pressure = .................................................... Pa [3] (ii) Calculate the actual uncertainty in the pressure. actual uncertainty = .................................................... Pa [3] (iii) State the pressure, with its actual uncertainty. pressure = ........................................... ± ........................................... Pa [1]

Mark scheme: 2 (a) pressure = force / area (normal to the force) [clear ratio essential] B1 [1] (b) (i) P = mg / A = (5.09 × 9.81) / A C1 A = (πd2 / 4) = π × (9.4 × 10–2)2 / 4 (= 0.00694 m2) C1 P = 49.93 / 0.00694 = 7200 (7195) Pa (minimum of 2 s.f. required) A1 [3] (ii) ∆P / P = ∆m / m + 2∆d / d C1 = 0.01 / 5.09 + (2 × 0.1) / 9.4 (= 0.0020 + 0.021 or 2.3%) C1 ∆P = 170 (165 to 167) Pa A1 [3] (iii) P = 7200 ± 200 Pa A1 [1]

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3 The resistance R of a uniform metal wire is measured for different lengths l of the wire. The variation with l of R is shown in Fig. 3.1. 4.0 3.0 R / 1 2.0 1.0 0 0 0.20 0.40 0.60 0.80 1.00 l / m Fig. 3.1 (a) The points shown in Fig. 3.1 do not lie on the best-fit line. Suggest a reason for this. ................................................................................................................................................... .............................................................................................................................................. [1] (b) Determine the gradient of the line shown in Fig. 3.1. gradient = .......................................................... [2] (c) The cross-sectional area of the wire is 0.12 mm2. Use your answer in (b) to determine the resistivity of the metal of the wire. resistivity = .................................................. Ω m [3] (d) The resistance R of different wires is measured. The wires are of the same metal and same length but have different cross-sectional areas A. On Fig. 3.2, sketch a graph to show the variation with A of R. R 0 0 A Fig. 3.2 [2]

Mark scheme: 3 (a) random error (in the measurements) of the length OR resistance B1 [1] (b) gradient = (3.6 – 1.9 ) / (0.8 – 0.4) C1 = 4.25 A1 [2] (c) R = ρl / A C1 ρ = gradient × area = 4.25 × 0.12 × 10–6 C1 = 5.1(0) × 10–7 Ω m A1 [3] (d) resistance decreasing with increasing area B1 correct shape with curve being asymptote to both axes B1 [2]

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4 A trolley moves down a slope, as shown in Fig. 4.1. trolley v 25° horizontal Fig. 4.1 The slope makes an angle of 25° with the horizontal. A constant resistive force FR acts up the slope on the trolley. At time t = 0, the trolley has velocity v = 0.50 m s−1 down the slope. At time t = 4.0 s, v = 12 m s−1 down the slope. (a) (i) Show that the acceleration of the trolley down the slope is approximately 3 m s−2. [2] (ii) Calculate the distance x moved by the trolley down the slope from time t = 0 to t = 4.0 s. x = ..................................................... m [2] (iii) On Fig. 4.2, sketch the variation with time t of distance x moved by the trolley. x 0 0 4.0 t / s Fig. 4.2 [2] (b) The mass of the trolley is 2.0 kg. (i) Show that the component of the weight of the trolley down the slope is 8.3 N. [1] (ii) Calculate the resistive force FR. FR = ...................................................... N [2]

Mark scheme: 4 (a) (i) acceleration = (v – u) / t or (12 – 0.5) / 4 C1 = (12 – 0.5) / 4 = 2.9 (2.875) (= approximately 3 m s–2) M1 [2] (ii) x = (u + v) t / 2 = [(12 + 0.5) × 4] / 2 C1 = 25 m A1 [2] (iii) line with increasing gradient M1 non-zero gradient at origin A1 [2] (b) (i) weight down slope = 2 × 9.81 × sin 25° = 8.29 / 8.3 M1 [1] (ii) (F = ma) 8.3 – FR = 2 × 2.9 C1 FR = 2.5 (2.3 if 3 used for a) N A1 [2] 2

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5 A motor is used to move bricks vertically upwards, as shown in Fig. 5.1. motor bricks container Fig. 5.1 The bricks start from rest and accelerate for 2.0 s. The bricks then travel at a constant speed of 0.64 m s−1 for 25 s. Finally the bricks are brought to rest in a further 3.0 s. The total mass of the bricks is 25 kg. (a) Determine the change in kinetic energy of the bricks (i) in the first 2.0 s, change in kinetic energy = ...................................................... J [2] (ii) in the next 25 s, change in kinetic energy = ...................................................... J [1] (iii) in the final 3.0 s. change in kinetic energy = ...................................................... J [1] (b) The bricks are in a container. The weight of the container and bricks is 350 N. Calculate, for the lifting of the bricks and container when travelling at constant speed, (i) the gain in potential energy, energy gain = ...................................................... J [3] (ii) the power required. power = ..................................................... W [2]

Mark scheme: 5 (a) (i) change in kinetic energy = ½ mv2 C1 = 0.5 × 25 × (0.64)2 = 5.1(2) J A1 [2] (ii) zero A1 [1] (iii) (–) 5.1(2) J A1 [1] (b) (i) PE = mgh C1 = 350 × 0.64 × 25 C1 = 5600 J A1 [3] (If full length used allow 1/3) (ii) P = Fv or gain in PE / t, EP / t or work done / t, W / t C1 = 350 × 0.64 or 5600 / 25 = 220 (224) W A1 [2]

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6 Distinguish between melting and evaporation. melting: ............................................................................................................................................. .......................................................................................................................................................... .......................................................................................................................................................... evaporation: ...................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... [4]

Mark scheme: 6 melting: solid to liquid B1 at a specific / one temperature / at the melting point B1 evaporation: liquid to vapour / gas OR molecules escape from surface of liquid B1 at all temperatures B1 [4]

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7 (a) A cell with internal resistance supplies a current. Explain why the terminal potential difference (p.d.) is less than the electromotive force (e.m.f.) of the cell. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [1] (b) A battery of e.m.f. 12 V and internal resistance 0.50 Ω is connected to a variable resistor X and a resistor Y of constant resistance, as shown in Fig. 7.1. 12 V 0.50 1 X Y Fig. 7.1 The resistance R of X is increased from 2.0 Ω to 16 Ω. The variation with R of the current І in the circuit is shown in Fig. 7.2. 3.0 I / A 2.0 1.0 0 0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 R / 1 Fig. 7.2 Calculate, for І = 1.2 A, (i) the p.d. across X, p.d. = ...................................................... V [2] (ii) the resistance of Y, resistance = ..................................................... Ω [3] (iii) the power dissipated in the battery. power = ..................................................... W [2] (c) Use Fig. 7.2 to explain the variation in the terminal p.d. of the battery as the resistance R of X is increased. ................................................................................................................................................... .............................................................................................................................................. [1]

Mark scheme: 7 (a) due to the lost volts in internal resistance / cell or energy losses in the internal resistance / cell B1 [1] (b) (i) V = ІR C1 = 1.2 × 6 = 7.2 V A1 [2] (ii) p.d. across Y and internal resistance r = 4.8 (V) [12 – 7.2] C1 resistance of Y + r = 4.8 / 1.2 = 4 (Ω) C1 resistance of Y = 4 – 0.5 = 3.5 Ω A1 [3] or Rtotal = 12 / 1.2 = 10 (Ω) (C1) X + r = 6.5 (Ω) (C1) resistance of Y = 3.5 Ω (A1) (iii) P = І2r C1 = (1.2)2 × 0.5 = 0.72 W A1 [2] (c) terminal p.d. increases as R is increased current decreases so there are less lost volts B1 [1]

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8 (a) Explain how stationary waves are formed. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) The arrangement of apparatus used to determine the wavelength of a sound wave is shown in Fig. 8.1. microphone loudspeaker metal plate signal generator c.r.o. Fig. 8.1 The loudspeaker emits sound of one frequency. The microphone is connected to a cathode-ray oscilloscope (c.r.o.). The waveform obtained on the c.r.o. for one position of the microphone is shown in Fig. 8.2. 1.0 cm 1.0 cm Fig. 8.2 The time-base setting of the c.r.o. is 0.20 ms cm−1. (i) Use Fig. 8.2 to show that the frequency of the sound is approximately 1300 Hz. [2] (ii) Explain how the apparatus is used to determine the wavelength of the sound. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (iii) The wavelength of the sound wave is 0.26 m. Calculate the speed of sound in this experiment. speed = ................................................ m s−1 [2]

Mark scheme: 8 (a) two waves (of the same kind) travelling in opposite directions overlap B1 waves have same frequency / wavelength and speed B1 [2] (b) (i) T = 0.8 (ms) C1 f = 1 / (0.8 × 10–3) = 1250 (Hz) A1 [2] (ii) microphone is moved from plate to loudspeaker or vice versa B1 wavelength is the twice the distance between adjacent maxima or minima (seen on c.r.o.) B1 [2] (iii) v = fλ C1 = 1250 × 0.26 = 330 (325) m s–1 A1 [2]

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