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

1 (a) The Young modulus of the metal of a wire is 1.8 × 1011 Pa. The wire is extended and the strain produced is 8.2 × 10–4. Calculate the stress in GPa. stress = ...................................................GPa [2] (b) An electromagnetic wave has frequency 12 THz. (i) Calculate the wavelength in μm. wavelength = .....................................................μm [2] (ii) State the name of the region of the electromagnetic spectrum for this frequency. .......................................................................................................................................[1] (c) An object B is on a horizontal surface. Two forces act on B in this horizontal plane. A vector diagram for these forces is shown to scale in Fig. 1.1. N 2.5 N B 30° W E S 7.5 N Fig. 1.1 A force of 7.5 N towards north and a force of 2.5 N from 30° north of east act on B. The mass of B is 750 g. (i) On Fig. 1.1, draw an arrow to show the approximate direction of the resultant of these two forces. [1] (ii) 1. Show that the magnitude of the resultant force on B is 6.6 N. [1] 2. Calculate the magnitude of the acceleration of B produced by this resultant force. magnitude = ................................................ m s–2 [2] (iii) Determine the angle between the direction of the acceleration and the direction of the 7.5 N force. angle = ........................................................ ° [1]

Mark scheme: 1 (a) stress = Young modulus × strain = 1.8 × 1011 × 8.2 × 10–4 or 1.476 × 108 C1 = 0.15 (0.148) GPa A1 [2] (b) (i) wavelength = 3 × 108 / 12 × 1012 C1 = 25 µm A1 [2] (ii) infra-red / IR B1 [1] (c) (i) arrow drawn up to the left of 7.5 N force approximately 5° to 40° to west of north A1 [1] (ii) 1. correct vector triangle or working to show magnitude of resultant force = 6.6 N allow 6.5 to 6.7 N if scale diagram M1 [1] 2. magnitude of acceleration = 6.6 / 0.75 [scale diagram: (6.5 to 6.7) / 0.75] C1 = 8.8 m s–2 [scale diagram: 8.7 – 8.9 m s–2] A1 [2] (iii) 19° [use of scale diagram allow 17° to 21° (a diagram must be seen)] B1 [1]

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2 A ball is thrown from A to B as shown in Fig. 2.1. V 60° A B Fig. 2.1 The ball is thrown with an initial velocity V at 60° to the horizontal. The variation with time t of the vertical component Vv of the velocity of the ball from t = 0 to t = 0.60 s is shown in Fig. 2.2. 6.0 Vv 4.0 2.0 velocity / m s–1 0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 t / s –2.0 –4.0 –6.0 Fig. 2.2 Assume air resistance is negligible. (a) (i) Complete Fig. 2.2 for the time until the ball reaches B. [2] (ii) Calculate the maximum height reached by the ball. height = .......................................................m [2] (iii) Calculate the horizontal component Vh of the velocity of the ball at time t = 0. Vh = ................................................. m s−1 [2] (iv) On Fig. 2.2, sketch the variation with t of Vh. Label this sketch Vh. [1] (b) The ball has mass 0.65 kg. Calculate, for the ball, (i) the maximum kinetic energy, maximum kinetic energy = ........................................................J [3] (ii) the maximum potential energy above the ground. maximum potential energy = ........................................................J [2]

Mark scheme: 2 (a) (i) straight line from t = 0.60 s to t = 1.2 s and |Vv| = 5.9 at t = 1.2 s M1 Vv = – 5.9 at t = 1.2 s i.e. line is for negative values of Vv A1 [2] (ii) s = 0 + ½ × 9.81 × (0.6)2 or area of graph = (5.9 × 0.6) / 2 C1 = 1.8 (1.77) m = 1.8 (1.77) m A1 [2] (iii) Vh = V cos 60° and Vv = V sin 60° or Vh = 5.9 / tan 60° or Vh = 5.9 tan 30° C1 Vh = 3.4 m s−1 A1 [2] (iv) horizontal line at 3.4 from t = 0 to t = 1.2 s [to half a small square] B1 [1] (b) (i) KE = ½ mv2 C1 = ½ × 0.65 × (6.81)2 [allow if valid method to find v] C1 = 15 (15.1) J A1 [3] (ii) PE = 0.65 × 9.81 × 1.77 C1 = 11(11.3) J A1 [2]

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3 (a) Define electric field strength. ................................................................................................................................................... ...............................................................................................................................................[1] (b) A sphere S has radius 1.2 × 10–6 m and density 930 kg m−3. Show that the weight of S is 6.6 × 10−14 N. [2] (c) Two horizontal metal plates are 14 mm apart in a vacuum. A potential difference (p.d.) of 1.9 kV is applied across the plates, as shown in Fig. 3.1. metal plate +1.9 kV 14 mm metal plate sphere S Fig. 3.1 A uniform electric field is produced between the plates. The sphere S in (b) is charged and is held stationary between the plates by the electric field. (i) Calculate the electric field strength between the plates. electric field strength = .................................................V m−1 [2] (ii) Calculate the magnitude of the charge on S. charge = .......................................................C [2] (iii) The magnitude of the p.d. applied to the plates is increased. Explain why S accelerates towards the top plate. ........................................................................................................................................... .......................................................................................................................................[2]

Mark scheme: 3 (a) electric field strength is force per unit positive charge B1 [1] (b) mass = volume × density (any subject, allow usual symbols or defined symbols) C1 = 4/3 × π × (1.2 × 10–6)3 × 930 (= 6.73 × 10–15) weight = 4/3 × π × (1.2 × 10–6)3 × 930 × 9.81 = 6.6 × 10–14 N M1 [2] (c) (i) E = 1.9 × 103 / 14 × 10–3 C1 = 1.4 (1.36) × 105 V m–1 A1 [2] (ii) F = QE Q = 6.6 × 10–14 / 1.36 × 105 C1 = 4.9 (4.86) × 10–19 C [allow 4.7 × 10–19 C if 1.4 × 105 used] A1 [2] (iii) electric force increases / is greater (than weight) B1 charge (on S) is negative to give resultant / net / sum / total force up B1 [2]

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4 (a) Compare the molecular motion of a liquid with (i) a solid, ........................................................................................................................................... .......................................................................................................................................[2] (ii) a gas. ........................................................................................................................................... .......................................................................................................................................[1] (b) (i) A ductile material in the form of a wire is stretched up to its breaking point. On Fig. 4.1, sketch the variation with extension x of the stretching force F. ductile material F 0 0 x Fig. 4.1 [1] (ii) On Fig. 4.2, sketch the variation with extension x of the stretching force F for a brittle material up to its breaking point. brittle material F 0 0 x Fig. 4.2 [1] (c) Describe a similarity and a difference between ductile and brittle materials. similarity: ................................................................................................................................... ................................................................................................................................................... difference: ................................................................................................................................. ................................................................................................................................................... [2]

Mark scheme: 4 (a) (i) solid: (molecules) vibrate B1 no translational motion / fixed position, liquid: translational motion B1 [2] (ii) gas: molecules have random (and translational) motion B1 [1] (b) (i) ductile: straight line through origin then curving towards x-axis B1 [1] (ii) brittle: straight line through origin with no or negligible curved region B1 [1] (c) similarity: obey Hooke’s law / F ∝ x or have elastic regions B1 difference: brittle no or (very) little plastic region ductile has (large(r)) plastic region B1 [2]

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5 A battery of electromotive force (e.m.f.) 12 V and internal resistance r is connected in series to two resistors, each of constant resistance X, as shown in Fig. 5.1. 12 V r I1 X X Fig. 5.1 The current Ι1 supplied by the battery is 1.2 A. The same battery is now connected to the same two resistors in parallel, as shown in Fig. 5.2. 12 V r I2 X X Fig. 5.2 The current Ι2 supplied by the battery is 3.0 A. (a) (i) Show that the combined resistance of the two resistors, each of resistance X, is four times greater in Fig. 5.1 than in Fig. 5.2. [2] (ii) Explain why Ι2 is not four times greater than Ι1. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (iii) Using Kirchhoff’s second law, state equations, in terms of e.m.f., current, X and r, for 1. the circuit of Fig. 5.1, ........................................................................................................................................... 2. the circuit of Fig. 5.2. ........................................................................................................................................... [2] (iv) Use the equations in (iii) to calculate the resistance X. X = .......................................................Ω [1] (b) Calculate the ratio power transformed in one resistor of resistance X in Fig. 5.1 . power transformed in one resistor of resistance X in Fig. 5.2 ratio = ...........................................................[2] (c) The resistors in Fig. 5.1 and Fig. 5.2 are replaced by identical 12 V filament lamps. Explain why the resistance of each lamp, when connected in series, is not the same as the resistance of each lamp when connected in parallel. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2]

Mark scheme: 5 (a) (i) in series 2X or in parallel X / 2 M1 other relationship given and 4× greater in series (than in parallel) A1 [2] (ii) due to the internal resistance B1 total resistance for series circuit is not four times greater than resistance for parallel circuit B1 [2] (iii) 1. E = I1(2X + r) or 12 = 1.2(2X + r) A1 2. E = I2(X/2 + r) or 12 = 3.0(X/2 + r) A1 [2] (iv) 2X + r = 10 and X/2 + r = 4 X = 4.0 Ω A1 [1] (b) P = I2R or V2 / R or VI C1 ratio = [(1.2)2 × 4] / [(1.5)2 × 4] = 0.64 A1 [2] (c) the resistance (of a lamp) changes with V or I B1 V or I is greater in parallel circuit or circuit 2 or V or I is less in series circuit or circuit 1 B1 [2]

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6 (a) State one difference and one similarity between longitudinal and transverse waves. difference: ................................................................................................................................. ................................................................................................................................................... similarity: ................................................................................................................................... ................................................................................................................................................... [2] (b) A laser is placed in front of two slits as shown in Fig. 6.1. slits laser 0.35 mm 2.5 m screen Fig. 6.1 (not to scale) The laser emits light of wavelength 6.3 × 10–7 m. The distance from the slits to the screen is 2.5 m. The separation of the slits is 0.35 mm. An interference pattern of maxima and minima is observed on the screen. (i) Explain why an interference pattern is observed on the screen. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) Calculate the distance between adjacent maxima. distance = .......................................................m [2] (c) State and explain the effect, if any, on the distance between adjacent maxima when the laser is replaced by another laser emitting ultra-violet radiation. ................................................................................................................................................... ...............................................................................................................................................[1]

Mark scheme: 6 (a) difference: vibration / oscillation (of particles) / displacement of particles is parallel to energy transfer / wavefronts in longitudinal and perpendicular for transverse B1 or transverse can be polarised, longitudinal cannot be polarised similarity: both transfer / propagate energy B1 [2] (b) (i) waves from slits are coherent / constant phase relationship (B1) waves overlap (at screen) with a phase difference or have a path difference (B1) maxima where phase difference is integer ×360° (or ×2π rad) or path difference is integer ×λ or equivalent explanation of minima e.g. (n+½)×360° (B1) max. 2 [2] (ii) maxima spacing = λD / a C1 = (6.3 × 10–7 × 2.5) / 0.35 × 10–3 = 4.5 × 10–3 m A1 [2] (c) (ultra-violet has) shorter wavelength, hence smaller separation / distance A1 [1]

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210 7 In the decay of a nucleus of 84 Po, an α-particle is emitted with energy 5.3 MeV. The emission is represented by the nuclear equation 210 A 84 Po B X + α + energy (a) (i) On Fig. 7.1, complete the number and name of the particle, or particles, represented by A and B in the nuclear equation. number name of particle or particles A B Fig. 7.1 [1] 210 (ii) State the form of energy given to the α-particle in the decay of 84 Po. .......................................................................................................................................[1] 210 (b) A sample of polonium 84 Po emits 7.1 × 1018 α-particles in one day. Calculate the mean power output from the energy of the α-particles. power = ...................................................... W [2]

Mark scheme: 7 (a) (i) A: 206, nucleon(s) or neutron(s) and proton(s) } B: 82, proton(s) } all correct A1 [1] (ii) kinetic / EK / KE B1 [1] (b) energy = 5.3 × 1.6 × 10–13 (J) [= 8.48 × 10–3 (J)] C1 power = (7.1 × 1018 × 5.3 × 1.6 × 10–13) / (3600 × 24) = 70 (69.7) W A1 [2]

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