Cambridge A Level Mathematics 9709 — 2020 May/June Paper 4 · Variant 2

1 A tram starts from rest and moves with uniform acceleration for 20 s. The tram then travels at a constant speed, V m s−1, for 170 s before being brought to rest with a uniform deceleration of magnitude twice that of the acceleration. The total distance travelled by the tram is 2.775 km. (a) Sketch a velocity-time graph for the motion, stating the total time for which the tram is moving. [2] ........................................................................................................................................................ ........................................................................................................................................................ (b) Find V. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the magnitude of the acceleration. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) Trapezium, deceleration steeper than acceleration B1 Time from 0 to 200 B1 2 1(b) 0.5(170 200) 2775 v + = M1 15 v = A1 2 1(c) 15 20 a = ÷ M1 0.75 a = A1 2

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2 4P N 3P N 30Å 30Å 1Å P N 20 N Coplanar forces of magnitudes 20 N, P N, 3P N and 4P N act at a point in the directions shown in the diagram. The system is in equilibrium. Find P and 1. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Resolving forces in either direction M1 20cos 4 cos30 P θ = A1 4 2 sin30 20sin P P θ + = A1 3 cos 10 sin 4 P P θ θ = = 2 2 3 1 1 100 16 P P + = M1 3.29 P = A1 55.3 θ = A1 6

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3 T N 60Å 2.5 kg 20Å A particle of mass 2.5 kg is held in equilibrium on a rough plane inclined at 20Å to the horizontal by a force of magnitude T N making an angle of 60Å with a line of greatest slope of the plane (see diagram). The coefficient of friction between the particle and the plane is 0.3. Find the greatest and least possible values of T. 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Mark scheme: 3 + = B1 Attempt at resolving in any direction M1 T cos 60 = F + 25 sin 20 A1 T cos 60 + F = 25 sin 20 A1 Use of F R μ = M1 cos60 25sin 20 0.3(25cos20 sin60) T T = ± − 25sin20 0.3 25cos20 cos60 0.3sin60 T ± × = ± M1 6.26 T = A1 20.5 T = A1 8

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4 Small smooth spheres A and B, of equal radii and of masses 4 kg and 2 kg respectively, lie on a smooth horizontal plane. Initially B is at rest and A is moving towards B with speed 10 m s−1. After the spheres collide A continues to move in the same direction but with half the speed of B. (a) Find the speed of B after the collision. 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A third small smooth sphere C, of mass 1 kg and with the same radius as A and B, is at rest on the plane. B now collides directly with C. After this collision B continues to move in the same direction but with one third the speed of C. (b) Show that there is another collision between A and B. 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(c) A and B coalesce during this collision. Find the total loss of kinetic energy in the system due to the three collisions. 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Mark scheme: 4(a) × + = × + M1 5 and 10 A B v v = = A1 2 4(b) Conservation of momentum B, C 2 10 [ 0] 2 3 v v × + = × + M1 4 v = A1 A B v v > , hence another collision A1 3 4(c) Conservation of momentum A, B M1 1 14 4 5 2 4 4 2 (ms ) 3 their their v v v − × + × = + = A1 KE initial = 2 1 4 10 2 × × M1 KE final = 2 2 1 14 1 6 ( ) 1 12 2 3 2 their their × × + × × A1 Loss of KE = 412 188 200 3 3 − = A1 5

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6 A particle P moves in a straight line. The velocity v m s−1 at time t s is given by v = 2t + 1 for 0 ≤t ≤5, v = 36 −t2 for 5 ≤t ≤7, v = 2t −27 for 7 ≤t ≤13.5. (a) Sketch the velocity-time graph for 0 ≤t ≤13.5. [3] (b) Find the acceleration at the instant when t = 6. 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(c) Find the total distance travelled by P in the interval 0 ≤t ≤13.5. 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Mark scheme: 6(a) Correct for 0 5 t ≤≤ B1 Correct for 5 7 t ≤≤ B1 Correct for 7 13.5 t ≤≤ B1 3 6(b) 2 a t = − by differentiating M1 12 a = − A1 2 6(c) 5 6 7 13.5 2 2 0 5 6 7 (2 1)d (36 )d (36 )d (2 27)d s t t t t t t t t = + + − + − + −     M1 5 6 7 13.5 2 2 0 5 6 7 (2 1)d (36 )d (36 )d (2 27)d s t t t t t t t t = + + − + − + −     A1 3 2 2 [ ] [36 ] 27 3 t s t t t t t = + + − + − M1 All correct A1 84.25 s = A1 5

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