Cambridge A Level Physics 9702 — 2011 Oct/Nov Paper 2 · Variant 1

1 (a) Define density. .......................................................................................................................................... ..................................................................................................................................... [1] (b) Explain how the difference in the densities of solids, liquids and gases may be related to the spacing of their molecules. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (c) A paving slab has a mass of 68 kg and dimensions 50 mm × 600 mm × 900 mm. (i) Calculate the density, in kg m–3, of the material from which the paving slab is made. density = ...................................... kg m–3 [2] (ii) Calculate the maximum pressure a slab could exert on the ground when resting on one of its surfaces. pressure = ............................................ Pa [3]

Mark scheme: 1 (a) density = mass / volume B1 [1] (b) density of liquids and solids same order as spacing similar / to about 2× B1 density of gases much less as spacing much more or density of gases much lower hence spacing much more B1 [2] (c) (i) density = 68 / [50 × 600 × 900 × 10–9] C1 = 2520 (allow 2500) kg m–3 A1 [2] (ii) P = F / A C1 = 68 × 9.81 / [50 × 600 × 10–6] C1 = 2.2 × 104 Pa A1 [3]

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2 (a) Define the torque of a couple. For Examiner’s .......................................................................................................................................... Use ..................................................................................................................................... [2] (b) A uniform rod of length 1.5 m and weight 2.4 N is shown in Fig. 2.1. 1.5 m rope A 8.0 N pin rod weight 2.4 N rope B 8.0 N Fig. 2.1 The rod is supported on a pin passing through a hole in its centre. Ropes A and B provide equal and opposite forces of 8.0 N. (i) Calculate the torque on the rod produced by ropes A and B. torque = .......................................... N m [1] (ii) Discuss, briefly, whether the rod is in equilibrium. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [2] (c) The rod in (b) is removed from the pin and supported by ropes A and B, as shown in For Fig. 2.2. Examiner’s Use 1.5 m rope A rope B 0.30 m P weight 2.4 N Fig. 2.2 Rope A is now at point P 0.30 m from one end of the rod and rope B is at the other end. (i) Calculate the tension in rope B. tension in B = ............................................. N [2] (ii) Calculate the tension in rope A. tension in A = ............................................. N [1]

Mark scheme: 2 (a) torque is the product of one of the forces and the distance between forces M1 the perpendicular distance between the forces A1 [2] (b) (i) torque = 8 × 1.5 = 12 N m A1 [1] (ii) there is a resultant torque / sum of the moments is not zero M1 (the rod rotates) and is not in equilibrium A1 [2] (c) (i) B × 1.2 = 2.4 × 0.45 C1 B = 0.9(0) N A1 [2] (ii) A = 2.4 – 0.9 = 1.5 N / moments calculation A1 [1] 1

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3 A ball is thrown against a vertical wall. The path of the ball is shown in Fig. 3.1. For Examiner’s Use P 15.0 m s–1 wall 60.0° S F 6.15 m 9.95 m Fig. 3.1 (not to scale) The ball is thrown from S with an initial velocity of 15.0 m s–1 at 60.0° to the horizontal. Assume that air resistance is negligible. (a) For the ball at S, calculate (i) its horizontal component of velocity, horizontal component of velocity = ........................................ m s–1 [1] (ii) its vertical component of velocity. vertical component of velocity = ........................................ m s–1 [1] (b) The horizontal distance from S to the wall is 9.95 m. The ball hits the wall at P with a velocity that is at right angles to the wall. The ball rebounds to a point F that is 6.15 m from the wall. Using your answers in (a), (i) calculate the vertical height gained by the ball when it travels from S to P, height = ............................................. m [1] (ii) show that the time taken for the ball to travel from S to P is 1.33 s, For Examiner’s Use [1] (iii) show that the velocity of the ball immediately after rebounding from the wall is about 4.6 m s–1. [1] (c) The mass of the ball is 60 × 10–3 kg. (i) Calculate the change in momentum of the ball as it rebounds from the wall. change in momentum = ........................................... N s [2] (ii) State and explain whether the collision is elastic or inelastic. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [1]

Mark scheme: 3 (a) (i) horizontal velocity = 15 cos 60° = 7.5 m s–1 A1 [1] (ii) vertical velocity = 15 sin 60° = 13 m s–1 A1 [1] (b) (i) v2 = u2 + 2as s = (13)2 / (2 × 9.81) = 8.6(1) m A1 [1] using g = 10 then max. 1 (ii) t = 13 / 9.81 = 1.326 s or t = 9.95 / 7.5 = 1.327 s A1 [1] (iii) velocity = 6.15 / 1.33 M1 = 4.6 m s–1 A0 [1] (c) (i) change in momentum = 60 × 10–3 [–4.6 – 7.5] C1 = (–)0.73 N s A1 [2] (ii) final velocity / kinetic energy is less after the collision or relative speed of separation < relative speed of approach M1 hence inelastic A0 [1] GCE AS/A LEVEL – October/November 2011 9702 21

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4 (a) Distinguish between gravitational potential energy and electric potential energy. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... ..................................................................................................................................... [2] (b) A body of mass m moves vertically through a distance h near the Earth’s surface. Use the defining equation for work done to derive an expression for the gravitational potential energy change of the body. [2] (c) Water flows down a stream from a reservoir and then causes a water wheel to rotate, as shown in Fig. 4.1. reservoir stream 120 m water wheel Fig. 4.1 As the water falls through a vertical height of 120 m, gravitational potential energy is converted to different forms of energy, including kinetic energy of the water. At the water wheel, the kinetic energy of the water is only 10% of its gravitational potential energy at the reservoir. (i) Show that the speed of the water as it reaches the wheel is 15 m s–1. [2] (ii) The rotating water wheel is used to produce 110 kW of electrical power. Calculate For the mass of water flowing per second through the wheel, assuming that the Examiner’s production of electric energy from the kinetic energy of the water is 25% efficient. Use mass of water per second = ....................................... kg s–1 [3]

Mark scheme: 4 (a) electrical potential energy (stored) when charge moved and gravitational potential energy (stored) when mass moved B1 due to work done in electric field and work done in gravitational field B1 [2] (b) work done = force × distance moved (in direction of force) and force = mg M1 mg × h or mg × ∆h A1 [2] (c) (i) 0.1 × mgh = ½ mv2 B1 0.1 × m × 9.81 × 120 = 0.5 × m × v2 B1 v = 15.3 m s–1 A0 [2] (ii) P = 0.5 m v2 / t C1 m / t = 110 × 103 / [0.25 × 0.5 × (15.3)2] C1 = 3740 kg s–1 A1 [3]

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5 (a) Define the ohm. For Examiner’s ..................................................................................................................................... [1] Use (b) Determine the SI base units of resistivity. base units of resistivity = ................................................. [3] (c) A cell of e.m.f. 2.0 V and negligible internal resistance is connected to a variable resistor R and a metal wire, as shown in Fig. 5.1. 2.0 V R metal wire 900 mm Fig. 5.1 The wire is 900 mm long and has an area of cross-section of 1.3 × 10–7 m2. The resistance of the wire is 3.4 Ω. (i) Calculate the resistivity of the metal wire. resistivity = ................................................. [2] (ii) The resistance of R may be varied between 0 and 1500 Ω. For Calculate the maximum potential difference (p.d.) and minimum p.d. possible across Examiner’s the wire. Use maximum p.d. = ................................................... V minimum p.d. = ....................................................V [2] (iii) Calculate the power transformed in the wire when the potential difference across the wire is 2.0 V. power = ............................................. W [2] (d) Resistance R in (c) is now replaced with a different variable resistor Q. State the power transformed in Q, for Q having (i) zero resistance, power = ............................................. W [1] (ii) infinite resistance. power = ............................................. W [1]

Mark scheme: 5 (a) ohm = volt / ampere B1 [1] (b) ρ = RA / l or unit is Ω m C1 units: V A–1 m2 m–1 = N m C–1 A–1 m2 m–1 C1 = kg m2 s–2 A–1 s–1 A–1 m2 m–1 = kg m3 s–3 A–2 A1 [3] (c) (i) ρ = [3.4 × 1.3 × 10–7] / 0.9 C1 = 4.9 × 10–7 (Ω m) A1 [2] (ii) max = 2.(0) V A1 min = 2 × (3.4 /1503.4) = 4.5 × 10–3 V A1 [2] (iii) P = V2 / R or P = VI and V = IR C1 = (2)2 / 3.4 = 1.18 (allow 1.2) W A1 [2] (d) (i) power in Q is zero when R = 0 B1 [1] (ii) power in Q = 0 / tends to zero as R = infinity B1 [1] GCE AS/A LEVEL – October/November 2011 9702 21

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6 (a) State Hooke’s law. For Examiner’s .......................................................................................................................................... Use ..................................................................................................................................... [1] (b) The variation with extension x of the force F for a spring A is shown in Fig. 6.1. 8.0 L 6.0 F / N 4.0 2.0 0 0 2 4 6 8 10 x / 10–2 m Fig. 6.1 The point L on the graph is the elastic limit of the spring. (i) Describe the meaning of elastic limit. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [1] (ii) Calculate the spring constant kA for spring A. kA = ....................................... N m–1 [1] (iii) Calculate the work done in extending the spring with a force of 6.4 N. For Examiner’s Use work done = .............................................. J [2] (c) A second spring B of spring constant 2kA is now joined to spring A, as shown in Fig. 6.2. spring A spring B 6.4 N Fig. 6.2 A force of 6.4 N extends the combination of springs. For the combination of springs, calculate (i) the total extension, extension = ............................................. m [1] (ii) the spring constant. spring constant = ....................................... N m–1 [1]

Mark scheme: 6 (a) extension is proportional to force (for small extensions) B1 [1] (b) (i) point beyond which (the spring) does not return to its original length when the load is removed B1 [1] (ii) gradient of graph = 80 N m–1 A1 [1] (iii) work done is area under graph / ½ Fx / ½ kx2 C1 = 0.5 × 6.4 × 0.08 = 0.256 (allow 0.26) J A1 [2] (c) (i) extension = 0.08 + 0.04 = 0.12 m A1 [1] (ii) spring constant = 6.4 / 0.12 = 53.3 N m–1 A1 [1]

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7 (a) Two isotopes of the element uranium are 23592U and 23892U. For Examiner’s Explain the term isotope. Use .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (b) (i) In a nuclear reaction, proton number and neutron number are conserved. Other than proton number and neutron number, state a quantity that is conserved in a nuclear reaction. ............................................................................................................................. [1] (ii) When a nucleus of uranium-235 absorbs a neutron, the following reaction may take place. 23592U + ab n 141x Ba + 36y Kr + 3 ab n State the values of a, b, x and y. a = ................. b = ................. x = ................. y = ................. [3] (c) When the nucleus of 23892U absorbs a neutron, the nucleus decays, emitting an α-particle. State the proton number and nucleon number of the nucleus that is formed as a result of the emission of the α-particle. proton number = ...................................................... nucleon number = ...................................................... [2]

Mark scheme: 7 (a) nuclei with the same number of protons B1 and a different number of neutrons B1 [2] (b) (i) (mass + energy) (taken together) is conserved (B1) momentum is conserved (B1) one point required max. 1 B1 [1] (ii) a = 1 and b = 0 B1 x = 56 B1 y = 92 B1 [3] (c) proton number = 90 B1 nucleon number = 235 B1 [2]

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