Cambridge A Level Physics 9702 — 2007 Oct/Nov Paper 2 · Variant 1

1 (a) Distinguish between systematic errors and random errors. systematic errors ............................................................................................................. .......................................................................................................................................... random errors .................................................................................................................. ..................................................................................................................................... [2] (b) A cylinder of length L has a circular cross-section of radius R, as shown in Fig. 1.1. R L Fig. 1.1 The volume V of the cylinder is given by the expression V = πR2L . The volume and length of the cylinder are measured as V = 15.0 ± 0.5 cm3 L = 20.0 ± 0.1 cm. Calculate the radius of the cylinder, with its uncertainty. radius = ........................ ± ........................ cm [5]

Mark scheme: 1 (a) systematic: e.g. constant error (in all readings) cannot be eliminated by averaging error in measuring instrument B1 random: e.g. readings scattered (equally) about true value error due to observer can be eliminated by averaging (only if averaging not included for systematic) B1 [2] (b) 15 = π × R2 × 20 R = 0.4886 cm (accept any number of s.f.) C1 % uncertainty in V = 3.3 % (or 0.5/15) C1 % uncertainty in L = 0.5 % (or 0.1/20) C1 % uncertainty in R = 1.9 % (i.e. one half of the sum) C1 R = 0.489 ± 0.009 cm A1 [5]

More questions on Errors and uncertainties

2 A girl G is riding a bicycle at a constant velocity of 3.5 m s–1. At time t = 0, she passes a boy B sitting on a bicycle that is stationary, as illustrated in Fig. 2.1. B G 3.5 m s–1 t = 0 Fig. 2.1 At time t = 0, the boy sets off to catch up with the girl. He accelerates uniformly from time t = 0 until he reaches a speed of 5.6 m s–1 in a time of 5.0 s. He then continues at a constant speed of 5.6 m s–1. At time t = T, the boy catches up with the girl. T is measured in seconds. (a) State, in terms of T, the distance moved by the girl before the boy catches up with her. distance = ............................. m [1] (b) For the boy, determine (i) the distance moved during his acceleration, distance = ............................. m [2] (ii) the distance moved during the time that he is moving at constant speed. Give your answer in terms of T. distance = ................................. m [1] Examiner’s Use (c) Use your answers in (a) and (b) to determine the time T taken for the boy to catch up with the girl. T = .......................................... s [2] (d) The boy and the bicycle have a combined mass of 67 kg. (i) Calculate the force required to cause the acceleration of the boy. force = .......................................... N [3] (ii) At a speed of 4.5 m s–1, the total resistive force acting on the boy and bicycle is 23 N. Determine the output power of the boy’s legs at this speed. power = ......................................... W [2]

Mark scheme: 2 (a) 3.5 T B1 [1] (b) (i) distance = average speed × time (however expressed) C1 = 14 m A1 [2] (ii) distance = 5.6 × (T – 5) (or 3.5T – 14) A1 [1] (c) 3.5T = 14 + 5.6(T – 5) C1 T = 6.7 s A1 [2] (d) (i) acceleration = (5.6 / 5 =) 1.12 m s–2 C1 force = ma C1 = 75 N A1 [3] (ii) power = (force × speed =) {75 + 23} × 4.5 C1 = 440 W A1 [2] (allow 1/2 for 234 W, 0/2 for 338 W or 104 W)

More questions on Equations of motion

3 (a) (i) Define potential energy. .................................................................................................................................. ............................................................................................................................. [1] (ii) Distinguish between gravitational potential energy and elastic potential energy. gravitational potential energy ................................................................................... .................................................................................................................................. elastic potential energy ............................................................................................ ............................................................................................................................. [2] (b) A small sphere of mass 51 g is suspended by a light inextensible string from a fixed point P. The centre of the sphere is 61 cm vertically below point P, as shown in Fig. 3.1. P 18° 61 cm sphere, mass 51g Fig. 3.1 The sphere is moved to one side, keeping the string taut, so that the string makes an angle of 18° with the vertical. Calculate (i) the gain in gravitational potential energy of the sphere, gain = ……………………….. J [2] Examiner’s Use (ii) the moment of the weight of the sphere about point P. moment = .................................... N m [2]

Mark scheme: 3 (a) (i) potential energy: stored energy available to do work B1 [1] (ii) gravitational: due to height/position of mass OR distance from mass OR moving mass from one point to another B1 elastic: due to deformation/stretching/compressing B1 [2] (b) (i) height raised = (61 – {61 cos18} =) 3.0 cm C1 energy = (mgh = 0.051 × 9.8 × 0.030 =) 1.5 × 10–2 J A1 [2] (ii) moment = force × perpendicular distance = 0.051 × 9.8 × 0.61 × sin18 C1 = 0.094 N m A1 [2] GCE A/AS LEVEL – October/November 2007 9702 02

More questions on Gravitational potential energy and kinetic energy

4 A sample of material in the form of a cylindrical rod has length L and uniform area of cross-section A. The rod undergoes an increasing tensile stress until it breaks. Fig. 4.1 shows the variation with stress of the strain in the rod. 0.02 strain breaking point 0.01 0 0 5 10 stress / 108 Pa Fig. 4.1 (a) State whether the material of the rod is ductile, brittle or polymeric. ..................................................................................................................................... [1] (b) Determine the Young modulus of the material of the rod. Young modulus = ............................................. Pa [2] Examiner’s Use (c) A second cylindrical rod of the same material has a spherical bubble in it, as illustrated in Fig. 4.2. cylindrical rod bubble 1.9 x 103 N 1.9 x 103 N cross-sectional area 3.2 x 10–6 m2 Fig. 4.2 The rod has an area of cross-section of 3.2 × 10–6 m2 and is stretched by forces of magnitude 1.9 × 103 N. By reference to Fig. 4.1, calculate the maximum area of cross-section of the bubble such that the rod does not break. area = ............................................ m2 [3] (d) A straight rod of the same material is bent as shown in Fig. 4.3. Fig. 4.3 Suggest why a thin rod can bend more than a thick rod without breaking. .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2]

Mark scheme: 4 (a) brittle B1 [1] (b) Young modulus = stress / strain C1 = (9.5 × 108) / 0.013 = 7.3 × 1010 Pa (allow ± 0.1 × 1010 Pa) A1 [2] (c) stress = force / area C1 (minimum) area = (1.9 × 103) / (9.5 × 108) = 2.0 × 10–6 m2 C1 (max) area of cross-section = (3.2 – 2.0) × 10–6 = 1.2 × 10–6 m2 A1 [3] (d) when bent, ‘top’ and ‘bottom’ edges have different extensions M1 with thick rod, difference is greater (than with a thin rod) A1 so breaks with less bending A0 [2]

More questions on Stress and strain

5 (a) Fig. 5.1 shows the variation with time t of the displacement y of a wave W as it passes a point P. The wave has intensity I. y 0 t wave W Fig. 5.1 A second wave X of the same frequency as wave W also passes point P. This wave has intensity I. The phase difference between the two waves is 60°. On Fig. 5.1, sketch the variation with time t of the displacement y of wave X. [3] (b) In a double-slit interference experiment using light of wavelength 540 nm, the separation of the slits is 0.700 mm. The fringes are viewed on a screen at a distance of 2.75 m from the double slit, as illustrated in Fig. 5.2 (not to scale). coherent light 0.700 mm wavelength 540 nm screen 2.75 m Fig. 5.2 Examiner’s Use Calculate the separation of the fringes observed on the screen. separation = ................................ mm [3] (c) State the effect, if any, on the appearance of the fringes observed on the screen when the following changes are made, separately, to the double-slit arrangement in (b). (i) The width of each slit is increased but the separation remains constant. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [3] (ii) The separation of the slits is increased. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [2]

Mark scheme: 5 (a) amplitude between 6.5 squares and 7.5 squares on 3 peaks B2 (allow 1 mark if outside this range but between 6.0 and 8.0 squares) correct phase (ignore lead/lag, look at x-axis only and allow ±½ square B1 [3] (b) λ = ax / D C1 540 × 10–9 = (0.700 × 10–3 x) / 2.75 C1 x = 2.12 mm A1 [3] (c) (i) same separation B1 bright areas brighter (1) dark areas, no change (1) (allow ‘contrast greater’ for 1 mark if dark/light areas not discussed) fewer fringes observed (1) any two, 1 each B2 [3] (ii) smaller separation of fringes B1 no change in brightness B1 [2]

More questions on Interference

6 An electric shower unit is to be fitted in a house. The shower is rated as 10.5 kW, 230 V. The shower unit is connected to the 230 V mains supply by a cable of length 16 m, as shown in Fig. 6.1. copper wire cable shower unit 230 V supply 10.5 kW 230 V copper wire 16 m Fig. 6.1 (a) Show that, for normal operation of the shower unit, the current is approximately 46 A. [2] (b) The resistance of the two wires in the cable causes the potential difference across the shower unit to be reduced. The potential difference across the shower unit must not be less than 225 V. The wires in the cable are made of copper of resistivity 1.8 × 10–8 Ω m. Assuming that the current in the wires is 46 A, calculate (i) the maximum resistance of the cable, resistance = ............................... Ω [3] Examiner’s Use (ii) the minimum area of cross-section of each wire in the cable. area = ...................................... m2 [3] (c) Connecting the shower unit to the mains supply by means of a cable having wires with too small a cross-sectional area would significantly reduce the power output of the shower unit. (i) Assuming that the shower is operating at 210 V, rather than 230 V, and that its resistance is unchanged, determine the ratio power dissipated by shower unit at 210 V . power dissipated by shower unit at 230 V ratio = .......................................... [2] (ii) Suggest and explain one further disadvantage of using wires of small cross-sectional area in the cable. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [2]

Mark scheme: 6 (a) power = VI C1 current = 10.5 × 103 / 230 M1 = 45.7 A A0 [2] (b) (i) p.d. across cable = 5.0 V C1 R = 5.0 / 46 C1 = 0.11 Ω A1 [3] (ii) R = ρL / A C1 0.11 = (1.8 × 10–8 × 16 × 2) / A C1 A = 5.3 × 10–6 m2 A1 [3] (wires in parallel, not series, allow max 1/3 marks) GCE A/AS LEVEL – October/November 2007 9702 02 (c) (i) either power = V 2 / R or power ∝ V 2 C1 ratio = (210 / 230)2 = 0.83 A1 [2] (ii) resistance of cable is greater M1 greater power loss/fire hazard/insulation may melt wire may melt/cable gets hot A1 [2]

More questions on Potential difference and power

Use 7 (a) Evidence for the nuclear atom was provided by the α-particle scattering experiment. State the results of this experiment. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..................................................................................................................................... [2] (b) Give estimates for the diameter of (i) an atom, ..............................................................................................................................[1] (ii) a nucleus. ..............................................................................................................................[1]

Mark scheme: 7 (a) most α-particles deviated through small angles B1 (accept ‘undeviated’) few α-particles deviated through angles greater than 90° B1 [2] (b) (i) allow 10–9 m → 10–11 m B1 [1] (ii) allow 10–13 m → 10–15 m B1 [1] (if (i) and (ii) out of range but (ii) = 10–4(i), then allow 1 mark) (if no units or wrong units but (ii) = 10–4(i), then allow 1 mark)

More questions on Atoms, nuclei and radiation