Cambridge A Level Physics 9702 — 2019 May/June Paper 2 · Variant 2

1 (a) The diameter d of a cylinder is measured as 0.0125 m ± 1.6%. Calculate the absolute uncertainty in this measurement. absolute uncertainty = ...................................................... m [1] (b) The cylinder in (a) stands on a horizontal surface. The pressure p exerted on the surface by the cylinder is given by 4 W p = 2 . π d The measured weight W of the cylinder is 0.38 N ± 2.8%. (i) Calculate the pressure p. p = ................................................ N m−2 [1] (ii) Determine the absolute uncertainty in the value of p. absolute uncertainty = ................................................ N m−2 [2] [Total: 4]

Mark scheme: 1(a) = 2 × 10–4 m A1 1(b)(i) p = (4 × 0.38) / (π × 0.01252) = 3100 N m–2 A1 1(b)(ii) percentage uncertainty = 2.8 + (2 × 1.6) (= 6%) or fractional uncertainty = 0.028 + (2 × 0.016) (= 0.06) C1 absolute uncertainty = 0.06 × 3100 = 190 N m–2 (allow to 1 significant figure) A1

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2 (a) State Newton’s second law of motion. ................................................................................................................................................... ...............................................................................................................................................[1] (b) A car of mass 850 kg tows a trailer in a straight line along a horizontal road, as shown in Fig. 2.1. car trailer tow-bar mass 850 kg horizontal road Fig. 2.1 The car and the trailer are connected by a horizontal tow-bar. The variation with time t of the velocity v of the car for a part of its journey is shown in Fig. 2.2. 15 14 v / m s–1 13 12 11 10 9 8 0 5 10 15 20 25 t / s Fig. 2.2 (i) Calculate the distance travelled by the car from time t = 0 to t = 10 s. distance = ...................................................... m [2] (ii) At time t = 10 s, the resistive force acting on the car due to air resistance and friction is 510 N. The tension in the tow-bar is 440 N. For the car at time t = 10 s: 1. use Fig. 2.2 to calculate the acceleration acceleration = ................................................ m s−2 [2] 2. use your answer to calculate the resultant force acting on the car resultant force = ...................................................... N [1] 3. show that a horizontal force of 1300 N is exerted on the car by its engine [1] 4. determine the useful output power of the engine. output power = ..................................................... W [2] (c) A short time later, the car in (b) is travelling at a constant speed and the tension in the tow-bar is 480 N. The tow-bar is a solid metal rod that obeys Hooke’s law. Some data for the tow-bar are listed below. Young modulus of metal = 2.2 × 1011 Pa original length of tow-bar = 0.48 m cross-sectional area of tow-bar = 3.0 × 10−4 m2 Determine the extension of the tow-bar. extension = ...................................................... m [3] (d) The driver of the car in (b) sees a pedestrian standing directly ahead in the distance. The driver operates the horn of the car from time t = 15 s to t = 17 s. The frequency of the sound heard by the pedestrian is 480 Hz. The speed of the sound in the air is 340 m s−1. Use Fig. 2.2 to calculate the frequency of the sound emitted by the horn. frequency = .................................................... Hz [2] [Total: 14]

Mark scheme: 2(a) (resultant) force proportional/equal to/is rate of change of momentum B1 2(b)(i) distance = area under graph or s = ½ (u + v) t = ½ × (9 + 13) × 10 or s = ut + ½at 2 = (9 × 10) + (½ × 0.40 × 102) or s = vt – ½at 2 = (13 × 10) – (½ × 0.40 × 102) or v 2 = u 2 + 2as 132 = 92 + (2 × 0.40 × s) C1 distance = 110 m A1 Question Answer Marks 2(b)(ii) 1. a = gradient or a = (v – u) / t or a = ∆v / (∆)t e.g. a = (14 – 9) / 12.5 or (13 – 9) / 10 C1 a = 0.40 m s–2 A1 2. resultant force = 850 × 0.40 = 340 N A1 3. (F =) 510 + 440 + 340 = 1300 (N) A1 4. P = Fv C1 = 1300 × 13 = 1.7 × 104 W A1 2(c) E = σ / ε C1 E = (F / A) / (∆L / L) or E = FL / A∆L C1 ∆L = (480 × 0.48) / (3.0 × 10–4 × 2.2 × 1011) = 3.5 × 10–6 m A1 2(d) fo = fs v / (v – vs) 480 = fs × 340 / (340 – 14) C1 fs = 460 Hz A1

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3 (a) State what is meant by the centre of gravity of a body. ................................................................................................................................................... ...............................................................................................................................................[1] (b) A uniform square sign with sides of length 0.68 m is fixed at its corner points A and B to a wall. The sign is also supported by a wire CD, as shown in Fig. 3.1. D wire 54 N 35° B C sign E wall 0.68 m W A 0.68 m Fig. 3.1 (not to scale) The sign has weight W and centre of gravity at point E. The sign is held in a vertical plane with side BC horizontal. The wire is at an angle of 35° to side BC. The tension in the wire is 54 N. The force exerted on the sign at B is only in the vertical direction. (i) Calculate the vertical component of the tension in the wire. vertical component of tension = ...................................................... N [1] (ii) Explain why the force on the sign at B does not have a moment about point A. ........................................................................................................................................... .......................................................................................................................................[1] (iii) By taking moments about point A, show that the weight W of the sign is 150 N. [2] (iv) Calculate the total vertical force exerted by the wall on the sign at points A and B. total vertical force = ...................................................... N [1] (c) The sign in (b) is held together by nuts and bolts. One of the nuts falls vertically from rest through a distance of 4.8 m to the pavement below. The nut lands on the pavement with a speed of 9.2 m s−1. Determine, for the nut falling from the sign to the pavement, the ratio change in gravitational potential energy . final kinetic energy ratio = .......................................................... [4] [Total: 10]

Mark scheme: 3(a) the point where (all) the weight (of the body) is taken to act B1 3(b)(i) vertical component = 54 sin 35° = 31 N A1 3(b)(ii) the (line of action of the) force (at B) passes through (point) A or the (line of action of the) force (at B) has zero (perpendicular) distance from (point) A B1 3(b)(iii) 54 sin 35° × 0.68 or 54 cos 35° × 0.68 or W × 0.34 C1 54 sin 35° × 0.68 + 54 cos 35° × 0.68 = W × 0.34 so W = 150 (N) A1 3(b)(iv) total vertical force = 150 – 31 = 120 N A1 3(c) (∆)E = mg(∆)h C1 E = ½mv 2 C1 ratio = (m × 9.81 × 4.8) / (½ × m × 9.22) or (9.81 × 4.8) / (½ × 9.22) C1 = 1.1 A1

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4 (a) For a progressive water wave, state what is meant by: (i) displacement ........................................................................................................................................... .......................................................................................................................................[1] (ii) amplitude. ........................................................................................................................................... .......................................................................................................................................[1] (b) Two coherent waves X and Y meet at a point and superpose. The phase difference between the waves at the point is 180°. Wave X has an amplitude of 1.2 cm and intensity I. Wave Y has an amplitude of 3.6 cm. Calculate, in terms of I, the resultant intensity at the meeting point. intensity = .......................................................... [2] (c) (i) Monochromatic light is incident on a diffraction grating. Describe the diffraction of the light waves as they pass through the grating. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) A parallel beam of light consists of two wavelengths 540 nm and 630 nm. The light is incident normally on a diffraction grating. Third-order diffraction maxima are produced for each of the two wavelengths. No higher orders are produced for either wavelength. Determine the smallest possible line spacing d of the diffraction grating. d = ...................................................... m [3] (iii) The beam of light in (c)(ii) is replaced by a beam of blue light incident on the same diffraction grating. State and explain whether a third-order diffraction maximum is produced for this blue light. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] [Total: 11]

Mark scheme: 4(a)(i) distance (in a specified direction of particle/point on wave) from the equilibrium position B1 4(a)(ii) the maximum distance (of particle/point on wave) from the equilibrium position or the maximum displacement (of particle/point on wave) B1 4(b) I ∝ A2 C1 IR / I = (3.6 – 1.2)2 / (1.2)2 resultant intensity = 4.0I A1 4(c)(i) as wave(s) pass through the slit(s) B1 wave(s) spread (into geometric shadow) B1 4(c)(ii) nλ = d sin θ C1 3λ = d sin 90° or 3λ = d C1 d = 3 × 630 × 10–9 = 1.9 × 10–6 m A1 4(c)(iii) wavelength of blue light is shorter (than 540 nm/630 nm/wavelengths of original light) M1 (so) third order diffraction maximum is produced A1

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5 (a) State Kirchhoff’s second law. ................................................................................................................................................... ...............................................................................................................................................[2] (b) A battery of electromotive force (e.m.f.) 5.6 V and internal resistance r is connected to two external resistors, as shown in Fig. 5.1. 5.6 V r V 90 18 Fig. 5.1 The reading on the voltmeter is 4.8 V. (i) Calculate: 1. the combined resistance of the two resistors connected in parallel combined resistance = ...................................................... Ω [2] 2. the current in the battery. current = ....................................................... A [2] (ii) Show that the internal resistance r is 2.5 Ω. [2] (iii) Determine the ratio power dissipated by internal resistance r . total power produced by battery ratio = .......................................................... [3] (c) The battery in (b) is now connected to a battery of e.m.f. 7.2 V and internal resistance 3.5 Ω. The new circuit is shown in Fig. 5.2. 5.6 V 2.5 3.5 7.2 V Fig. 5.2 Determine the current in the circuit. current = ....................................................... A [2] [Total: 13]

Mark scheme: 5(a) sum of e.m.f.(s) = sum of p.d.(s) M1 around a loop/around a closed circuit A1 5(b)(i) 1. 1 / R = 1 / R1 + 1 / R2 1 / R = 1 / 90 + 1 / 18 C1 R = 15 Ω A1 2. I = V / R C1 I = 4.8 / 15 or I = 4.8 / 90 + 4.8 / 18 I = 0.32 A A1 5(b)(ii) E = V + Ir or E = I(R + r) C1 5.6 = 4.8 + 0.32 r so r = 2.5 (Ω) or 5.6 = 0.32 × (15 + r) so r = 2.5 (Ω) A1 5(b)(iii) P = EI or P = VI or P = I2R or P = V2 / R C1 ratio = (0.322 × 2.5) / (5.6 × 0.32) or 0.256 / 1.792 C1 = 0.14 A1 Question Answer Marks 5(c) 7.2 – 5.6 – 2.5I – 3.5I = 0 C1 I = 0.27 A A1

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6 (a) State what is meant by a field line (line of force) in an electric field. ................................................................................................................................................... ...............................................................................................................................................[1] (b) An electric field has two different regions X and Y. The field strength in X is less than that in Y. Describe a difference between the pattern of field lines (lines of force) in X and in Y. ................................................................................................................................................... ...............................................................................................................................................[1] (c) A particle P has a mass of 0.15 u and a charge of −1e, where e is the elementary charge. (i) Particle P and an α-particle are in the same uniform electric field. Calculate the ratio magnitude of acceleration of particle P . magnitude of acceleration of α-particle ratio = .......................................................... [3] (ii) Particle P is a hadron composed of only two quarks. One of them is a down (d) quark. By considering charge, determine a possible type (flavour) of the other quark. Explain your working. ........................................................................................................................................... .......................................................................................................................................[3] [Total: 8]

Mark scheme: 6(a) path/direction in which a (free) positive charge will move B1 6(b) (lines) closer together in Y/further apart in X B1 6(c)(i) a = Eq / m or F = Eq and F = ma C1 ratio = (1e / 0.15 u) × (4 u / 2e) or 1 / 0.15 × 4 / 2 C1 ratio = 13 A1 6(c)(ii) down quark charge is –(1 / 3)e C1 – (1 / 3)e + q = –1e so q = –(2 / 3)e A1 (–(2 / 3)e is) anti-up / u (quark) (allow charm or top antiquark) B1

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