Cambridge A Level Physics 9702 — 2015 Oct/Nov Paper 2 · Variant 1

1 (a) State two SΙ base quantities other than mass, length and time. 1. ............................................................................................................................................... 2. ............................................................................................................................................... [2] (b) A beam is clamped at one end and an object X is attached to the other end of the beam, as shown in Fig. 1.1. l oscillation of X clamp beam object X Fig. 1.1 The object X is made to oscillate vertically. The time period T of the oscillations is given by Ml 3 T = K E where M is the mass of X, l is the length between the clamp and X, E is the Young modulus of the material of the beam and K is a constant. (i) 1. Show that the SΙ base units of the Young modulus are kg m–1 s–2. [1] 2. Determine the SΙ base units of K. SΙ base units of K .......................................................... [2] (ii) Data in SΙ units for the oscillations of X are shown in Fig. 1.2. quantity value uncertainty T 0.45 ± 2.0% l 0.892 ± 0.2% M 0.2068 ± 0.1% K 1.48 × 105 ± 1.5% Fig. 1.2 Calculate E and its actual uncertainty. E = ..................................... ± ..................................... kg m–1 s–2 [4]

Mark scheme: 1 (a) temperature B1 current B1 [2] (allow amount of substance, luminous intensity) (b) (i) 1. E = (stress / strain =) [force / area] / [extension / original length] units of stress: kg m s–2 / m2 and no units for strain B1 units of E: kg m–1 s–2 A0 [1] 2. units for T: s, l: m and M: kg K 2 = T 2 E / M l 3 hence units: s2 kg m–1 s–2 / kg3 (= m–4) C1 units of K: m–2 A1 [2] (ii) % uncertainty in E = 4% (for T 2) + 0.6% (for l 3) + 0.1% (for M) + 3% (for K 2) = 7.7% B1 E = [(1.48 × 105)2 × 0.2068 × (0.892)3] / (0.45)2 = 1.588 × 1010 C1 7.7% of E = 1.22 × 109 C1 E = (1.6 ± 0.1) × 1010 kg m–1 s–2 A1 [4] 12 12

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2 The signal from a microwave detector is recorded on a cathode-ray oscilloscope (c.r.o.), as shown in Fig. 2.1. 1 cm 1 cm Fig. 2.1 The time-base setting on the c.r.o. is 50 ps cm–1. (a) Using Fig. 2.1, determine the wavelength of the microwaves. wavelength = ........................................................ m [4] (b) The signal from a radio wave detector is recorded on the same c.r.o. The wavelength of the radio waves is 1.5 × 103 m. Determine the time-base setting required to display the same number of oscillations on the c.r.o. as shown in Fig. 2.1. time-base setting = ....................................... unit........................ [2]

Mark scheme: 2 (a) ps = 10–12 (s) or T = 4 × 50 × 10–12 (s) B1 v = fλ or v = λ / T C1 λ = 3.0 × 108 × 4 × 50 × 10–12 C1 = 0.06(0) m A1 [4] (b) 1500 = 3.0 × 108 × 4 × time-base setting or T = 5 × 10–6 s C1 time-base setting = 1.3 (1.25) µs cm–1 A1 [2]

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3 (a) An object is moved from point P to point R either by a direct path or by the path P to Q to R, as shown in Fig. 3.1. R vertical object horizontal P Q Fig. 3.1 P and Q are on the same horizontal level. R is vertically above Q. Explain whether the work done moving the object against the gravitational field is the same or different along paths PR and PQR. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) A ball is thrown with an initial velocity V at an angle θ to the horizontal, as shown in Fig. 3.2. path of ball V e horizontal Fig. 3.2 (not to scale) The variation with time t of the height h of the ball is shown in Fig. 3.3. 12.0 10.0 h / m 8.0 6.0 4.0 2.0 0 0 1.00 2.00 3.00 t / s Fig. 3.3 Air resistance is negligible. (i) Use the time to reach maximum height to determine the vertical component Vv of the velocity of the ball for time t = 0. Vv = ........................................................ m s–1 [2] (ii) The horizontal displacement of the ball at t = 3.00 s is 25.5 m. On Fig. 3.4, draw the variation with t of the horizontal displacement x of the ball. 30 x / m 20 10 0 0 1.00 2.00 3.00 t / s Fig. 3.4 [1] (iii) For the ball at maximum height, calculate the ratio potential energy of the ball . kinetic energy of the ball ratio = .......................................................... [3] (iv) In practice, air resistance is not negligible. State and explain the effect of air resistance on the time taken for the ball to reach maximum height. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2]

Mark scheme: 3 (a) work done is force × distance moved in direction of force or no work done along PQ as no displacement/distance moved in direction of force B1 work done is same in vertical direction as same distance moved in direction of force B1 [2] (b) (i) at maximum height t = 1.5 (s) or s = ½(u + v) t, s = 11 m and t = 1.5 s C1 Vv = 0 + 9.81 × 1.5 Vv = (11 × 2) / 1.5 = 15 (14.7) m s–1 A1 [2] (ii) straight line from (0,0) to (3.00, 25.5) B1 [1] (iii) at maximum height Vh = 25.5 / 3 (= 8.5 m s–1) B1 ratio = mgh / ½ mv2 C1 = (2 × 9.81 × 11.0) / (8.5)2 = 3.0 (2.99) A1 [3] (iv) deceleration is greater/resultant force (weight and friction force) is greater M1 time is less A1 [2]

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4 Fig. 4.1 shows a metal cylinder of height 4.5 cm and base area 24 cm2. metal cylinder 4.5 cm base area 24 cm2 Fig. 4.1 The density of the metal is 7900 kg m–3. (a) Show that the mass of the cylinder is 0.85 kg. [2] (b) The cylinder is placed on a plank, as shown in Fig. 4.2. plank cylinder 40° horizontal Fig. 4.2 The plank is at an angle of 40° to the horizontal. Calculate the pressure on the plank due to the cylinder. pressure = .................................................... Pa [3] (c) The cylinder then slides down the plank with a constant acceleration of 3.8 m s–2. A constant frictional force f acts on the cylinder. Calculate the frictional force f. f = ...................................................... N [3]

Mark scheme: 4 (a) density = mass / volume C1 mass = 7900 × 4.5 × 24 × 10–6 = 0.85 (0.853) kg M1 [2] (b) pressure = force / area C1 force = W cos 40° C1 pressure = (0.85 × 9.81 cos 40°) / 24 × 10–4 = 2.7 (2.66) × 103 Pa A1 [3] (c) F = ma C1 W sin 40° – f = ma C1 0.85 × 9.81 × sin 40° – f = 0.85 × 3.8 f (= 5.36 – 3.23) = 2.1 N [5.38 – 3.242 if 0.8532 kg is used for the mass] A1 [3]

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5 (a) A progressive wave transfers energy. A stationary wave does not transfer energy. State two other differences between progressive waves and stationary waves. 1. ............................................................................................................................................... ................................................................................................................................................... 2. ............................................................................................................................................... ................................................................................................................................................... [2] (b) A stationary wave is formed on a stretched string between two fixed points A and B. The variation of the displacement y of particles of the string with distance x along the string for the wave at time t = 0 is shown on Fig. 5.1. 10 position of particles at t = 0 5 y / mm A B 0 0 1.0 2.0 x / m –5 –10 Fig. 5.1 The wave has a period of 20 ms and a wavelength of 1.2 m. The maximum amplitude of the particles of the string is 5.0 mm. (i) On Fig. 5.1, draw a line to represent the position of the string at t = 5.0 ms. [2] (ii) State the phase difference between the particles of the string at x = 0.40 m and at x = 0.80 m. phase difference = ......................... unit .................... [1] (iii) State and explain the change in the kinetic energy of a particle at an antinode between t = 0 and t = 5.0 ms. A numerical value is not required. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2]

Mark scheme: 5 (a) progressive: all particles have same amplitude stationary: no nodes or antinodes or maximum to minimum/zero amplitude B1 progressive: adjacent particles are not in phase stationary: waves particles are in phase (between adjacent nodes) B1 [2] (b) (i) wavelength 1.2 m (zero displacement at 0.0, 0.60 m, 1.2 m, 1.8 m, 2.4 m) either peaks at 0.30 m and 1.5 m and troughs at 0.90 m and 2.1 m or vice versa (but not both) B1 maximum amplitude 5.0 mm B1 [2] (ii) 180° or π rad A1 [1] (iii) at t = 0 particle has kinetic energy as particle is moving B1 at t = 5.0 ms no kinetic energy as particle is stationary so decrease in kinetic energy (between t = 0 and t = 5.0 ms) B1 [2]

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6 (a) Define electromotive force (e.m.f.) for a battery. ................................................................................................................................................... .............................................................................................................................................. [1] (b) A battery of e.m.f. 6.0 V and internal resistance 0.50 Ω is connected in series with two resistors X and Y, as shown in Fig. 6.1. 6.0 V 0.50 1 4.0 1 12 1 X Y Fig. 6.1 The resistance of X is 4.0 Ω and the resistance of Y is 12 Ω. Calculate (i) the current in the circuit, current = ....................................................... A [2] (ii) the terminal potential difference (p.d.) across the battery. p.d. = ....................................................... V [1] (c) A resistor Z is now connected in parallel with resistor Y in the circuit in (b). The new arrangement is shown in Fig. 6.2. 6.0 V 0.50 1 4.0 1 12 1 X Y Z Fig. 6.2 Resistor Y is made from a wire of length l and diameter d. Resistor Z is a wire made from the same material as Y. The length of the wire for Z is l / 2 and the diameter is d / 2. (i) Calculate the resistance R of the combination of resistors Y and Z. R = ....................................................... Ω [3] (ii) State and explain the effect on the terminal p.d. across the battery. A numerical value is not required. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (d) For the circuits given in (b) and (c), show that the ratio power developed in the external circuit in Fig. 6.1 power developed in the external circuit in Fig. 6.2 is approximately 0.8. [3]

Mark scheme: 6 (a) energy converted from chemical to electrical per unit charge B1 [1] (b) (i) current = E / (R + r) C1 = 6.0 / (16 + 0.5) = 0.36 (0.364) A A1 [2] (ii) terminal p.d. = (0.36 × 16) = 5.8 V or (6 – 0.36 × 0.5) = 5.8 V A1 [1] (c) (i) use of R = ρl / A or proportionality with length and inverse proportionality with area or d 2 C1 d / 2 and l / 2 gives resistance of Z = 2RY = 24 (Ω) C1 R = resistance of parallel combination = [1/24 + 1/12]–1 = 8(.0) (Ω) A1 [3] (ii) resistance of circuit less therefore current larger B1 lost volts greater therefore terminal p.d. less B1 [2] (d) power = I 2 R or VI or V 2 / R C1 current in second circuit (= 6.0 / 12.5) = 0.48 (A) B1 ratio = [(0.36)2 × 16] / [(0.48)2 × 12] = 0.75 [0.77 if full s.f. used] B1 [3]

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7 Two parallel, vertical metal plates in a vacuum are connected to a power supply and a switch, as shown in Fig. 7.1. path of _-particles metal metal radioactive source + – power supply Fig. 7.1 A radioactive source emitting α-particles is placed below the plates. The path of the α-particles is shown on Fig. 7.1. The switch is closed producing a potential difference (p.d.) across the plates. This gives rise to a uniform electric field between the plates. The separation of the plates is 12 mm. (a) (i) On Fig. 7.1, draw the path of the α-particles. [1] (ii) Explain why the metal plates are placed in a vacuum. ........................................................................................................................................... ...................................................................................................................................... [1] (iii) Calculate the p.d. required to produce an electric field of 140 MV m–1. p.d. = ................................................... MV [2] (b) The α-particle source is replaced by a β-particle source. By reference to the properties of α-radiation and β-radiation, suggest three possible differences in the deflection observed with β-particles. 1. ............................................................................................................................................... ................................................................................................................................................... 2. ............................................................................................................................................... ................................................................................................................................................... 3. ............................................................................................................................................... ................................................................................................................................................... [3] (c) Complete Fig. 7.2 to show the changes in the proton number Z and the nucleon number A of different radioactive nuclei when either an α-particle or a β-particle is emitted. emitted particle change in Z change in A α-particle β-particle Fig. 7.2 [1]

Mark scheme: 7 (a) (i) curved path towards negative (–) plate (right-hand side) B1 [1] (ii) range of α-particle is only few cm in air/loss of energy of the α-particles due to collision with air molecules/ionisation of the air molecules B1 [1] (iii) V = E × d C1 = 140 × 106 × 12 × 10–3 = 1.7 (1.68) MV A1 [2] (b) β have opposite charge to α therefore deflection in opposite direction B1 β has a range of velocities/energies hence number of different deflections B1 β have less mass or q / m is larger hence deflection is greater or β with (very) high speed (may) have less deflection B1 [3] (c) emitted particle change in Z change in A α-particle –2 –4 β-particle +1 0 A1 [1]

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