Cambridge A Level Physics 9702 — 2017 May/June Paper 2 · Variant 3

1 (a) Two forces, with magnitudes 5.0 N and 12 N, act from the same point on an object. Calculate the magnitude of the resultant force R for the forces acting (i) in opposite directions, R = ....................................................... N [1] (ii) at right angles to each other. R = ....................................................... N [1] (b) An object X rests on a smooth horizontal surface. Two horizontal forces act on X as shown in Fig. 1.1. 18 N 115° X 55 N Fig. 1.1 (not to scale) A force of 55 N is applied to the right. A force of 18 N is applied at an angle of 115° to the direction of the 55 N force. (i) Use the resolution of forces or a scale diagram to show that the magnitude of the resultant force acting on X is 65 N. [2] (ii) Determine the angle between the resultant force and the 55 N force. angle = ........................................................ ° [2] (c) A third force of 80 N is now applied to X in the opposite direction to the resultant force in (b). The mass of X is 2.7 kg. Calculate the magnitude of the acceleration of X. acceleration = ..................................................m s–2 [3] [Total: 9]

Mark scheme: 1(a)(i) R = 7(.0) N B1 1(a)(ii) R = 13 N B1 1(b)(i) forces resolved: 18 sin 65° (vertical) and 55 + 18 cos 65° (horizontal) or scale drawing: correct triangle drawn for forces B1 F = [(18 sin 65°)2 + (55 + 18 cos 65°)2]1/2 = 65 (64.7) N or scale drawing: scale given, length of resultant given correctly, ± 1 N A1 1(b)(ii) angle = tan–1 [18 sin 65° / (55 + 18 cos 65°)] = tan–1 (16.3 / 62.6) or scale drawing: correct angle measured/direction correct on diagram below the 55 N force C1 angle = 15 (14.6)° (below the 55 N force) or scale drawing: angle = 15° ± 1° A1 1(c) (resultant) force = mass × acceleration C1 80 − 65 = 2.7a C1 a = 5.6 m s–2 [5.7 if 64.7 N used from (i)] A1

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2 (a) State Newton’s second law of motion. ................................................................................................................................................... ...............................................................................................................................................[1] (b) A constant resultant force F acts on an object A. The variation with time t of the velocity v for the motion of A is shown in Fig. 2.1. 9.0 v / m s–1 8.0 7.0 6.0 5.0 4.0 0 1.0 2.0 3.0 4.0 t / s Fig. 2.1 The mass of A is 840 g. Calculate, for the time t = 0 to t = 4.0 s, (i) the change in momentum of A, change in momentum = ............................................. kg m s–1 [2] (ii) the force F. F = ....................................................... N [1] (c) The force F is removed at t = 4.0 s. Object A continues at constant velocity before colliding with an object B, as illustrated in Fig. 2.2. A B 840 g 730 g at rest Fig. 2.2 Object B is initially at rest. The mass of B is 730 g. The objects A and B join together and have a velocity of 4.7 m s–1. (i) By calculation, show that the changes in momentum of A and of B during the collision are equal and opposite. [2] (ii) Explain how the answers obtained in (i) support Newton’s third law. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (iii) By reference to the speeds of A and B, explain whether the collision is elastic. ........................................................................................................................................... .......................................................................................................................................[1] [Total: 9]

Mark scheme: 2(a) (resultant) force is proportional/equal to the rate of change of momentum B1 2(b)(i) change in momentum = m(v2 − v1) = 0.84 × (8.8 − 4.2) C1 = 3.9 (3.86) kg m s–1 A1 2(b)(ii) F = (3.9 / 4.0) = 0.97 (0.965) N A1 2(c)(i) change in momentum for A: 0.84 × (4.7 − 8.8) = −3.4 (3.44) change in momentum for B: 0.73 × (4.7 − 0) = 3.4 (3.43) M1 change in momentum for B is equal and opposite to A A1 2(c)(ii) change in momentum equal (for A and B) M1 force is change in momentum / time and time (of collision) is the same hence force on A and B equal and opposite as for Newton’s third law A1 2(c)(iii) inelastic as relative speed of approach not equal to relative speed of separation B1

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3 (a) Define electric field strength. ................................................................................................................................................... ...............................................................................................................................................[1] (b) An electron is accelerated from point A to point B by a uniform electric field, as illustrated in Fig. 3.1. electric field A electron B Fig. 3.1 The distance between A and B is 12 mm. The velocity of the electron at A is 2.5 km s–1 and at B is 18 Mm s–1. Calculate (i) the acceleration of the electron, acceleration = ..................................................m s–2 [2] (ii) the change in kinetic energy of the electron, change in kinetic energy = ........................................................J [3] (iii) the electric field strength. electric field strength = .................................................V m–1 [3] (c) An α-particle moves from A to B in the electric field in (b). Describe and explain how the change in the kinetic energy of the α-particle compares with that of the electron. Numerical values are not required. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] [Total: 12]

Mark scheme: 3(a) force per unit (positive) charge B1 3(b)(i) a = (v 2 − u 2) / 2s = [(18 × 106)2 − (2.5 × 103)2] / (2 × 12 × 10–3) B1 = 1.3 (1.35) × 1016 m s–2 A1 3(b)(ii) KE = ½ mv 2 or ½ m(v2 – u2) C1 change in KE = 0.5 × 9.11 × 10–31 × [(18 × 106)2 − (2.5 × 103)2] B1 = 1.5 (1.48) × 10–16J A1 3(b)(iii) E = F / e = ma / e or eV = ∆KE so E = ∆KE / (e × d) C1 E = (9.11 × 10–31 × 1.35 × 1016) / 1.60 × 10–19 or E = (1.48 × 10–16) / (12 × 10–3 × 1.60 × 10–19) C1 = 7.7 (7.69) × 104 V m–1 A1 3(c) charge on α opposite to electron/charge on α is positive B1 ∆KE is negative/KE reduced B1 charge of α greater/twice that of electron causes larger/twice ∆KE (in magnitude) B1

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4 A spring is supported so that it hangs vertically, as shown in Fig. 4.1. spring mass M Fig. 4.1 Different masses are attached to the lower end of the spring. The extension x of the spring is measured for each mass M. The variation with x of M is shown in Fig. 4.2. 150 M / g 100 50 0 0 40 80 120 160 200 x / mm Fig. 4.2 (a) State and explain whether the spring obeys Hooke’s law. ................................................................................................................................................... ...............................................................................................................................................[1] (b) State the form of energy stored in the spring due to the addition of the masses. ...............................................................................................................................................[1] (c) Describe how to determine whether the extension of the spring is elastic. ................................................................................................................................................... ...............................................................................................................................................[1] (d) Calculate the work done on the spring as it is extended from x = 40.0 mm to x = 160 mm. work done = ........................................................J [3] [Total: 6]

Mark scheme: 4(a) the straight line does not go through the origin/the force is not proportional to extension (so does not obey Hooke’s law) A1 4(b) elastic potential energy B1 4(c) remove the force/masses and the spring returns to its original length if elastic B1 4(d) work done is represented by/linked to area under the line (× g) C1 work = ½ (145 + 70) × 10–3 × 9.81 × 120 × 10–3 C1 = 0.13 (0.127) J A1

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5 (a) A diffraction grating is used to determine the wavelength of light. (i) Describe the diffraction of light at a diffraction grating. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) By reference to interference, explain 1. the zero order maximum, .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... 2. the first order maximum. .................................................................................................................................... .................................................................................................................................... [3] (b) A diffraction grating is used with different wavelengths of light. The angle θ of the second order maximum is measured for each wavelength. The variation with wavelength λ of sin θ is shown in Fig. 5.1. 0.60 sinθ 0.50 0.40 0.30 0.20 0.10 300 350 400 450 500 550 λ/ nm Fig. 5.1 (i) Determine the gradient of the line shown in Fig. 5.1. gradient = ...........................................................[2] (ii) Use the gradient determined in (i) to calculate the slit separation d of the diffraction grating. d = .......................................................m [2] (iii) On Fig. 5.1, sketch a line to show the results that would be obtained for the first order maxima. [1] [Total: 10]

Mark scheme: 5(a)(i) waves at the elements/slits B1 waves spread (into the geometric shadow) B1 5(a)(ii) 1. waves (from each element/slit) overlap/meet/superpose B1 with a phase difference/path difference of zero B1 2. phase difference is 360°/path difference of λ B1 5(b)(i) e.g. gradient = (0.40 − 0.32) / [(500 − 400) × 10–9] C1 = 8(.0) × 105 A1 5(b)(ii) d sinθ = nλ d = n / gradient C1 = 2 / 8.0 × 105 = 2.5 × 10–6m A1 5(b)(iii) straight line drawn with lower gradient (about ½) and all points lower B1

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6 (a) Describe the I–V characteristic of (i) a metallic conductor at constant temperature, ........................................................................................................................................... .......................................................................................................................................[1] (ii) a semiconductor diode. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (b) Two identical filament lamps are connected in series and then in parallel to a battery of electromotive force (e.m.f.) 12 V and negligible internal resistance, as shown in Fig. 6.1a and Fig. 6.1b. 12 V 12 V Fig. 6.1a Fig. 6.1b The I–V characteristic of each lamp is shown in Fig. 6.2. 6.0 I / A 4.0 2.0 0 0 2.0 4.0 6.0 8.0 10.0 12.0 V / V Fig. 6.2 (i) Use the information shown in Fig. 6.2 to determine the current through the battery in 1. the circuit of Fig. 6.1a, current = .............................................................A 2. the circuit of Fig. 6.1b. current = .............................................................A [3] (ii) Calculate the total resistance in 1. the circuit of Fig. 6.1a, resistance = ............................................................Ω 2. the circuit of Fig. 6.1b. resistance = ............................................................Ω [3] (iii) Calculate the ratio power dissipated in a lamp in the circuit of Fig. 6.1a . power dissipated in a lamp in the circuit of Fig. 6.1b ratio = ...........................................................[2] [Total: 11]

Mark scheme: 6(a)(i) straight line through the origin B1 6(a)(ii) zero current for one direction (–ve V) up to zero or a few tenths of volt (+ve V) B1 straight line positive gradient/increasing gradient (+ve V) B1 6(b)(i) 1. current = 2.8 A A1 2. 4(.0) A for each lamp C1 current in circuit = 8(.0) A A1 6(b)(ii) use of R = V / I with correct values of V from graph for each arrangement C1 1. series resistance (= 2.1 + 2.1) = 4.2 or 4.3 Ω or (12 / 2.8) = 4.3 Ω A1 2. parallel resistance 1.5 Ω (each lamp 3.0 Ω) or (12 / 8.0) = 1.5 Ω A1 6(b)(iii) power = IV or V 2 / R or I2R C1 ratio = (2.8 × 6.0) / (4.0 × 12) = 0.35 A1

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7 (a) The following particles are used to describe the structure of an atom. electron neutron proton quark Underline the fundamental particles in the above list. [1] (b) The following equation represents the decay of a nucleus of 6207Co to form nucleus Q by β– emission. 6 2 07Co → ABQ + β– + x (i) Complete Fig. 7.1. value A B Fig. 7.1 [1] (ii) State the name of the particle x. .......................................................................................................................................[1] [Total: 3]

Mark scheme: 7(a) electron and quark both underlined/clearly indicated and no others B1 7(b)(i) value A 60 B 28 both correct B1 7(b)(ii) (electron) antineutrino B1

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