Cambridge A Level Physics 9702 — 2012 Oct/Nov Paper 2 · Variant 2

1 (a) The drag force D on an object of cross-sectional area A, moving with a speed v through Use a fluid of density ρ, is given by 1 2 D = CρAv 2 where C is a constant. Show that C has no unit. [2] (b) A raindrop falls vertically from rest. Assume that air resistance is negligible. (i) On Fig. 1.1, sketch a graph to show the variation with time t of the velocity v of the raindrop for the first 1.0 s of the motion. 10.0 8.0 6.0 v / m s–1 4.0 2.0 0 0 1.0 2.0 3.0 4.0 5.0 t / s Fig. 1.1 [1] (ii) Calculate the velocity of the raindrop after falling 1000 m. velocity = ........................................ m s–1 [2] (c) In practice, air resistance on raindrops is not negligible because there is a drag force. For This drag force is given by the expression in (a). Examiner’s Use (i) State an equation relating the forces acting on the raindrop when it is falling at terminal velocity. [1] (ii) The raindrop has mass 1.4 × 10–5 kg and cross-sectional area 7.1 × 10–6 m2. The density of the air is 1.2 kg m–3 and the initial velocity of the raindrop is zero. The value of C is 0.60. 1. Show that the terminal velocity of the raindrop is about 7 m s–1. [2] 2. The raindrop reaches terminal velocity after falling approximately 10 m. On Fig. 1.1, sketch the variation with time t of velocity v for the raindrop. The sketch should include the first 5 s of the motion. [2]

Mark scheme: 1 (a) units for D identified as kg m s–2 M1 all other units shown: units for A: m2 units for v2: m2 s–2 units for ρ: kg m–3 kg m s −2 C = with cancelling / simplification to give C no units A1 [2] kg m − 3 m 2 m 2 s − 2 (b) (i) straight line from (0,0) to (1,9.8) ± half a square B1 [1] (ii) ½ mv2 = mgh or using v2 = 2 as C1 v = (2 × 9.81 × 1000)1/2 = 140 m s–1 A1 [2] (c) (i) weight = drag (D) ( + upthrust) B1 [1] Allow mg or W for weight and D or expression for D for drag (ii) 1. mg = 1.4 ×10–5 × 9.81 C1 1.4 × 10–5 × 9.81 = 0.5 × 0.6 × 1.2 × 7.1 × 10–6 × v2 M1 v = 7.33 m s–1 A0 [2] 2. line from (0,0) correct curvature to a horizontal line at velocity of 7 m s–1 M1 line reaches 7 m s–1 between 1.5 s and 3.5 s A1 [2]

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2 (a) State Newton’s second law. For Examiner’s .......................................................................................................................................... Use ......................................................................................................................................[1] (b) A ball of mass 65 g hits a wall with a velocity of 5.2 m s–1 perpendicular to the wall. The ball rebounds perpendicularly from the wall with a speed of 3.7 m s–1. The contact time of the ball with the wall is 7.5 ms. Calculate, for the ball hitting the wall, (i) the change in momentum, change in momentum = ........................................... N s [2] (ii) the magnitude of the average force. force = ............................................. N [1] (c) (i) For the collision in (b) between the ball and the wall, state how the following apply: 1. Newton’s third law, .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] 2. the law of conservation of momentum. .................................................................................................................................. ..............................................................................................................................[1] (ii) State, with a reason, whether the collision is elastic or inelastic. .................................................................................................................................. ..............................................................................................................................[1]

Mark scheme: 2 (a) (resultant) force = rate of change of momentum / allow proportional to or change in momentum / time (taken) B1 [1] (b) (i) ∆p = (–) 65 × 10–3 (5.2 + 3.7) C1 = (–) 0.58 N s A1 [2] (ii) F = 0.58 / 7.5 × 10–3 = 77(.3) N A1 [1] (c) (i) 1. force on the wall from the ball is equal to the force on ball from the wall M1 but in the opposite direction A1 [2] (statement of Newton’s third law can score one mark) 2. momentum change of ball is equal and opposite to momentum change of the wall / change of momentum of ball and wall is zero B1 [1] (ii) kinetic energy (of ball and wall) is reduced / not conserved so inelastic B1 [1] (Allow relative speed of approach does not equal relative speed of separation.) GCE AS/A LEVEL – October/November 2012 9702 22

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3 (a) With reference to the arrangement of atoms, distinguish between metals, polymers and For amorphous solids. Examiner’s Use metals: ............................................................................................................................. .......................................................................................................................................... polymers: ......................................................................................................................... .......................................................................................................................................... amorphous solids: ............................................................................................................ .......................................................................................................................................... [3] (b) On Fig. 3.1, sketch the variation with extension x of force F to distinguish between a metal and a polymer. F F 0 0 0 x 0 x metal polymer Fig. 3.1 [2]

Mark scheme: 3 (a) metal: regular / repeated / ordered arrangement / pattern / lattice or long range order (of atoms / molecules / ions) B1 polymer: tangled chains (of atoms / molecules) or long chains (of atoms / molecules / ions) B1 amorphous: disordered / irregular arrangement or short range order (of atoms / molecules / ions) B1 [3] (b) metal: straight line or straight line then curving with less positive gradient B1 polymer: curve with decreasing gradient with steep increasing gradient at end B1 [2]

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4 Fig. 4.1 shows an arrangement for producing stationary waves in a tube that is closed at one For end. Examiner’s Use signal generator loudspeaker tube Fig. 4.1 (a) Explain how waves from the loudspeaker produce stationary waves in the tube. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[3] (b) One of the stationary waves that may be formed in the tube is represented in Fig. 4.2. P S Fig. 4.2 (i) Describe the motion of the air particles in the tube at 1. point P, ..............................................................................................................................[1] 2. point S. ..............................................................................................................................[1] (ii) The speed of sound in the tube is 330 m s–1 and the frequency of the waves from the loudspeaker is 880 Hz. Calculate the length of the tube. length = ............................................. m [3]

Mark scheme: 4 (a) waves (travels along tube) reflect at closed end / end of tube B1 incident and reflected waves or these two waves are in opposite directions M1 interfere or stationary wave formed if tube length equivalent to λ / 4, 3λ / 4, etc. A1 [3] (b) (i) 1. no motion (as node) / zero amplitude B1 [1] 2. vibration backwards and forwards / maximum amplitude along length B1 [1] (ii) λ = 330 / 880 (= 0.375 m) C1 L = 3λ / 4 C1 L = 3 / 4 × (0.375) = 0.28 (0.281) m A1 [3]

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5 Fig. 5.1 shows a 12 V power supply with negligible internal resistance connected to a uniform For metal wire AB. The wire has length 1.00 m and resistance 10 Ω. Two resistors of resistance Examiner’s 4.0 Ω and 2.0 Ω are connected in series across the wire. Use 12 V I1 I2 C metal wire A B 40 cm I3 4.0 1 2.0 1 D Fig. 5.1 Currents І1, І2 and І3 in the circuit are as shown in Fig. 5.1. (a) (i) Use Kirchhoff’s first law to state a relationship between І1, І2 and І3. ..............................................................................................................................[1] (ii) Calculate І1. І1 = .............................................. A [3] (iii) Calculate the ratio x, where power in metal wire x = . power in series resistors x = ................................................. [3] (b) Calculate the potential difference (p.d.) between the points C and D, as shown in Fig. 5.1. The distance AC is 40 cm and D is the point between the two series resistors. p.d. = .............................................. V [3]

Mark scheme: 5 (a) (i) І1 = І2 + І3 B1 [1] (ii) І = V / R or І2 = 12 / 10 (= 1.2 A) C1 R = [1/6 + 1 / 10]–1 [total R = 3.75 Ω] or І3 = 12 / 6 (= 2.0 A) C1 І1 = 12 / 3.75 = 3.2 A or І1 = 1.2 + 2.0 = 3.2 A A1 [3] (iii) power = VІ or І2R or V2 / R C1 power in wire I 22 R w V I 2 V 2 / R w x = = or or C1 power in series resistors I 32 R s V I 3 V 2 / R s x = 12 × 1.2 / 12 × 2.0 = 0.6(0) allow 3 / 5 or 3:5 A1 [3] (b) p.d. BC: 12 – 12 × 0.4 = 7.2 (V) / p.d. AC = 4.8 (V) C1 p.d. BD: 12 – 12 × 4 / 6 = 4.0 (V) / p.d. AD = 8.0 (V) C1 p.d. = 3.2 V A1 [3]

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6 (a) State Hooke’s law. For Examiner’s .......................................................................................................................................... Use ......................................................................................................................................[1] (b) A spring is attached to a support and hangs vertically, as shown in Fig. 6.1. An object M of mass 0.41 kg is attached to the lower end of the spring. The spring extends until M is at rest at R. spring M R S Fig. 6.1 The spring constant of the spring is 25 N m–1. Show that the extension of the spring is about 0.16 m. [2] (c) The object M in Fig. 6.1 is pulled down a further 0.060 m to S and is then released. For M, just as it is released, (i) state the forces acting on M, ..............................................................................................................................[1] (ii) calculate the acceleration of M. acceleration = ........................................ m s–2 [3] (d) Describe and explain the energy changes from the time the object M in Fig. 6.1 is For released to the time it first returns to R. Examiner’s Use .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] Please turn over for Question 7.

Mark scheme: 6 (a) extension is proportional to force / load B1 [1] (b) F = mg C1 x = (mg / k ) = 0.41 × 9.81 / 25 = (4.02 / 25) M1 x = 0.16 m A0 [2] GCE AS/A LEVEL – October/November 2012 9702 22 (c) (i) weight and (reaction) force from spring (which is equal to tension in spring) B1 [1] (ii) F – weight or 0.06 × 25 = ma C1 F = 0.2209 × 25 = 5.52 (N) or 0.22 × 25 = 5.5 a = (5.52 – 0.41× 9.81) / 0.41 or 1.5 / 0.41 and (5.5 – 4.02) C1 a = 3.7 (3.66) m s–2 gives 3.6 m s–2 A1 [3] (d) elastic potential energy / strain energy to kinetic energy and gravitational potential energy B1 stretching / extension reduces and velocity increases / height increases B1 [2] 3 3 4 1

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7 A nuclear reaction between two helium nuclei produces a second isotope of helium, two For protons and 13.8 MeV of energy. The reaction is represented by the following equation. Examiner’s Use ......... ......... 3 3 2He + 2He .........He + 2 .........p + 13.8 MeV (a) Complete the nuclear equation. [2] (b) By reference to this reaction, explain the meaning of the term isotope. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (c) State the quantities that are conserved in this nuclear reaction. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (d) Radiation is produced in this nuclear reaction. State (i) a possible type of radiation that may be produced, ..............................................................................................................................[1] (ii) why the energy of this radiation is less than the 13.8 MeV given in the equation. ..............................................................................................................................[1] (e) Calculate the minimum number of these reactions needed per second to produce power of 60 W. number = ........................................... s–1 [2]

Mark scheme: 7 (a) 32 He + 32 He → 42 He + 2 11 p + Q A numbers correct (4 and 1) B1 Z numbers correct (2 and 1) B1 [2] (b) both nuclei have 2 protons B1 the two isotopes have 1 neutron and two neutrons B1 [2] [allow 1 for ‘same number of protons but different number of neutrons’] (c) proton number and neutron number B1 energy – mass B1 momentum B1 [2] (d) (i) γ radiation B1 [1] (ii) product(s) must have kinetic energy B1 [1] (e) 13.8 MeV = 13.8 × 1.6 × 10–19 × 106 (= 2.208 × 10–12) C1 60 = n × 13.8 × 1.6 × 10–13 n = 2.7(2) × 1013 s–1 A1 [2]

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