Cambridge A Level Mathematics 9709 — 2007 May/June Paper 4 · Variant 1

1 A particle slides up a line of greatest slope of a smooth plane inclined at an angle α◦to the horizontal. The particle passes through the points A and B with speeds 2.5 m s−1 and 1.5 m s−1 respectively. The distance AB is 4 m (see diagram). Find (i) the deceleration of the particle, [2] (ii) the value of α. [2]

Mark scheme: 1 (i) [1.52 = 2.52 + 2a × 4] M1 For using v2 = u2 + 2as Deceleration is 0.5 ms-2 A1 2 Accept a = -0.5 (ii) M1 For using Newton’s second law or a = (-)gsinα or ½ m(vB2 – vA2 ) = mg(AB)sinα α = 2.9 A1ft 2 ft α = sin-1(-0.1a)

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2 Two forces, each of magnitude 8 N, act at a point in the directions OA and OB. The angle between the forces is θ◦(see diagram). The resultant of the two forces has component 9 N in the direction OA. Find (i) the value of θ, [2] (ii) the magnitude of the resultant of the two forces. [3]

Mark scheme: 2 (i) [8 + 8cosθ = 9] M1 For an equation in θ using component 9N θ = 82.8 A1 2 (ii) For showing θ or (180° – θ ) or B1 This mark may be implied by a θ/2, in a triangle representing the correct equation for R(θ ) in the two forces and the resultant, or for subsequent working using Y = 8sinθ in R2 = X2 + Y2 [R2 = 82 + 82 – 2×8×8cos(180 – θ), M1 For an equation in R or R2 R2 = 82 + 82 + 2×8×8cosθ, cos(θ/2) = (R/2) ÷ 8, Rcos(θ/2) = 9, Rsin(θ/2) = 8sinθ, R2 = 92 + (8sinθ )2, R2 = (8 + 8cosθ )2 + (8sinθ )2] Magnitude is 12 N A1 3

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3 A car travels along a horizontal straight road with increasing speed until it reaches its maximum speed of 30 m s−1. The resistance to motion is constant and equal to R N, and the power provided by the car’s engine is 18 kW. (i) Find the value of R. [3] (ii) Given that the car has mass 1200 kg, find its acceleration at the instant when its speed is 20 m s−1. [3]

Mark scheme: 3 (i) [DF = 18000/30] M1 For using DF = P/v-may be scored in (ii) [R = DF] M1 For using a = 0 (may be implied) R = 600 N A1 3 (ii) M1 For using Newton’s second law (3 terms) 18000/20 – 600 = 1200a A1ft ft wrong R Acceleration is 0.25ms-2 A1 3

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4 Particles P and Q, of masses 0.6 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string which passes over a smooth fixed peg. The particles are held at rest with the string taut. Both particles are at a height of 0.9 m above the ground (see diagram). The system is released and each of the particles moves vertically. Find (i) the acceleration of P and the tension in the string before P reaches the ground, [5] (ii) the time taken for P to reach the ground. [2]

Mark scheme: 4 (i) M1 For applying Newton’s second law to P or to Q (3 terms) 0.6 g – T = 0.6a A1 T – 0.2 g = 0.2a A1 Allow B1 for 0.6 g – 0.2 g = (0.6 + 0.2)a as an alternative for either of the above A marks Acceleration is 5 ms-2 B1 Tension is 3 N A1 5 (ii) [0.9 = ½ 5t2] M1 For using s = ut + ½ at2 Time taken is 0.6 s A1ft 2 ft 1.8/a GCE A/AS LEVEL – May/June 2007 9709 04 5 (i) M1 For using KE = ½ mv2 Increase in KE A1 = ½ 12500(252 – 172) Special case for candidates who assume the acceleration is constant (max 1 mark out of 2) 252 – 172 = 2ad, F = 12500×168/d, KE gain = WD in increasing speed = Fd = 12500×168 B1 [WD = 2100 + 5000] M1 For using WD by DF = KE gain + WD v res Work done by driving force is A1ft 4 ft only when units are consistent and 7100 kJ (or 7100 000 J) both M marks are scored (ii) M1 For an equation with PE gain, WD by DF and WD v res (and KE loss if appropriate) in linear combination PE gain = (7100 + 3300) – A1ft Or equivalent in joules (5000 + 4800×500 ÷ 1000) or PE gain = 3300 + 2100 – 4800×500 ÷ 1000 [3000 000 = 12500×10h] M1 For solving mgh = gain PE found Height is 24m A1 4 Special case for candidates who assume the acceleration is constant (max 3 marks out of 4) 3300000/500 – 4800 – 12500×10sinθ = 12500(-0.336) B1 For using h = 500sinθ M1 Height is 24 m A1 6 (i) dt Q M1 For using s Q = ∫ v sQ = 1.5t2 – 0.1t3 (+ C) A1 M1 For using limits 0 to 10 or equivalent (or 0 to 5 if the candidate states or implies that that vQ is symmetric about t = 5) sQ (10) = 50 (or sQ(5) = 25) A1ft May be implied in subsequent working M1 For using ½ 10vmax = sQ (10) (or ½ 5vmax = sQ (5)) Greatest velocity is 10 ms-1 A1 6 AG Special case for final 2 marks (max 1 mark out of 2) 5v = 50 → v = 10 B1 (ii) aP = 10/5 B1 [3 – 0.6t = 2] M1 For differentiating to find aQ(t) and equating to aP t = 1.67 (or 12/3) A1 3 GCE A/AS LEVEL – May/June 2007 9709 04 7 (i) Tcos60° = 75cos30° → T = 130 B1 Accept 75 3 M1 For resolving forces vertically (4 terms) Tsin60° + 75sin30° + R = 20g A1ft ft consistent sin/cos mix [130sin60° + 75sin30° + R = 200] M1 For substituting for T and solving for R Magnitude is 50 N A1 5 Accept 49.9 (ii) M1 For resolving forces horizontally Tcos60° + 25 = 75cos30° A1ft ft consistent sin/cos mix (T = 14.4) (T = 79.9) [79.9sin60° + 75sin30° + R = 200] M1 For resolving forces vertically (4 terms) and substituting for T R = 93.3 A1 May be implied by final answer [µ= 25/93.3] M1 For using µ = 25/R Coefficient is 0.268 (= 2 – 3 ) A1ft 6 ft for µ = value obtained from 25/candidate’s R, including her/his answer in (i) but excluding R = 20 g

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