Cambridge A Level Mathematics 9709 — 2017 May/June Paper 4 · Variant 1

1 A particle of mass 0.6 kg is dropped from a height of 8 m above the ground. The speed of the particle at the instant before hitting the ground is 10 m s−1. Find the work done against air resistance. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 B1 KE gain = ½ (0.6) 10 2 [= 30] B1 WD against Res = 48 – 30 = 18 J B1 Total: 3

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2 A particle of mass 0.8 kg is projected with a speed of 12 m s−1 up a line of greatest slope of a rough plane inclined at an angle of 10Å to the horizontal. The coefficient of friction between the particle and the plane is 0.4. (i) Find the acceleration of the particle. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the distance the particle moves up the plane before coming to rest. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) R = 0.8g cos 10 [= 7.88] B1 F = 0.4 × 8 cos 10 [= 3.15] M1 Use F = µR –8 sin 10 – 3.2 cos 10 = 0.8a M1 Newton 2 along the plane a = –5.68 ms–2 A1 Total: 4 2(ii) 0 = 12 2 – 2 × 5.68 × s M1 Using v2 = u2 + 2as s = 144/(2 × 5.68) = 12.7 m A1 Total: 2 Question Answer Mark Guidance 3 EITHER: (M1 Resolve horizontally and/or vertically at the 25 N weight A cos 30 + B cos 40 = 25 A1 A sin 30 = B sin 40 A1 M1 Solve for A and/or B A = 17.1 A1 B = 13.3 A1) OR: (M1 Attempt Lami’s theorem 25 sin 70 sin140 sin150 = = A B A1 One correct equation A1 A second correct equation M1 Solve for A and/or B A = 17.1 A1 B = 13.3 A1) Total: 6

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4 A car of mass 800 kg is moving up a hill inclined at 1Å to the horizontal, where sin 1 = 0.15. The initial speed of the car is 8 m s−1. Twelve seconds later the car has travelled 120 m up the hill and has speed 14 m s−1. (i) Find the change in the kinetic energy and the change in gravitational potential energy of the car. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The engine of the car is working at a constant rate of 32 kW. Find the total work done against the resistive forces during the twelve seconds. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) M1 Attempt KE and/or PE with correct dimensions KE gain = ½ × 800 × (142 – 82) = 52800 J A1 PE gain = 800 × 10 × 120 × 0.15 = 144000 J A1 Total: 3 4(ii) WD by engine = 32000 × 12 B1 32000 × 12 = 144000 + 52800 + WD against F M1 Work/Energy equation 4 terms WD against F = 187200 J A1 WD = 187000 to 3sf Total: 3 [12 2 = 20 2 – 2a × AB Use v 2 = u 2 + 2(–a)s

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5 A particle P moves in a straight line ABCD with constant deceleration. The velocities of P at A, B and C are 20 m s−1, 12 m s−1 and 6 m s−1 respectively. (i) Find the ratio of distances AB : BC. 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(ii) The particle comes to rest at D. Given that the distance AD is 80 m, find the distance BC. 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Mark scheme: 5(i) 6 2 = 12 2 – 2a × BC] M1 for AB or BC where a is the deceleration AB = 128/a A1 BC = 54/a A1 AB : BC = 64:27 A1 Allow equivalent unsimplified ratio Total: 4 Question Answer Mark Guidance 5(ii) 0 = 20 2 – 2a × 80 → a = 2.5 M1 Use v 2 = u 2 + 2(–a)AD to find a BC = 54/2.5 M1 Use a to find BC BC = 21.6 m A1 Total: 3

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6 A particle P moves in a straight line passing through a point O. At time t s, the velocity of P, v m s−1, is given by v = qt + rt2, where q and r are constants. The particle has velocity 4 m s−1 when t = 1 and when t = 2. (i) Show that, when t = 0.5, the acceleration of P is 4 m s−2. 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(ii) Find the values of t when P is at instantaneous rest. 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(iii) The particle is at O when t = 3. Find the distance of P from O when t = 0. 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Mark scheme: 6(i) [q + r = 4 and 2q + 4r = 4] M1 Use v = 4 at t = 1 and t = 2 q = 6 and r = –2 so v = 6t – 2t 2 A1 a = 6 – 4t M1 Differentiation used for a At t = 0.5, a = 4 A1 AG Total: 4 6(ii) v = 6t – 2t 2 = 0 M1 Set v = 0 and solve for t t = 0 and t = 3 A1 Total: 2 Question Answer Mark Guidance 6(iii) EITHER: s = ∫(6t – 2t 2) dt (M1 Attempt to integrate v to find s s = 3t 2 – ⅔t 3 + C A1 0 = 3 × 32 – ⅔ × 33 + C M1 Use s = 0 when t = 3 to find C C = –9 so distance = 9 m A1) Valid argument OR: ( ) 3 2 0 6 2 d = − ∫ s t t t (M1 Attempt integration with limits 3 2 3 0 2 3 3   −     t t A1 Correct integration and correct limits but no evaluation [27 – 18 = 9] M1 Evaluation of integral between limits Distance from O at t = 0 is 9 m A1) With explanation Total: 4

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7 P A B 0.8 kg 1.2 kg 60Å 30Å As shown in the diagram, a particle A of mass 0.8 kg lies on a plane inclined at an angle of 30Å to the horizontal and a particle B of mass 1.2 kg lies on a plane inclined at an angle of 60Å to the horizontal. The particles are connected by a light inextensible string which passes over a small smooth pulley P fixed at the top of the planes. The parts AP and BP of the string are parallel to lines of greatest slope of the respective planes. The particles are released from rest with both parts of the string taut. (i) Given that both planes are smooth, find the acceleration of A and the tension in the string. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) It is given instead that both planes are rough, with the same coefficient of friction, -, for both particles. Find the value of - for which the system is in limiting equilibrium. 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Mark scheme: 7(i) [T – 0.8g sin 30 = 0.8a 1.2g sin 60 – T = 1.2a 1.2g sin 60 – 0.8g sin 30 = 2a] M1 For A 4 0.8 − = T a A1 For B 6 3 10.4 1.2 − = − = T T a A1 System equation is 6 3 4 6.4 2 − = = a M1 Solve for a or T 3 3 2 3.20 = − = a ms–2 A1 ( ) 12 1 3 6.56 N 5 T = + = A1 Total: 6 Question Answer Mark Guidance 7(ii) RA = 0.8 cos30 4 3 = g RB = 1.2 cos60 6 = g B1 For either RA or RB FA = 4 3 µ and FB = 6µ M1 Either FA or FB used M1 Resolve parallel to the plane for both particles A and B or system 12 sin 60 – 6µ – T = 0 or T – 8 sin 30 – 4√3 µ = 0 A1 System equation is 12 sin 60 – 8 sin 30 – 6µ – 4√3 µ = 0 M1 Eliminate T and/or find µ ( ) ( ) µ 6 3 4 / 6 4 3 = √− + √ = 0.494 A1 Total: 6

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