Cambridge A Level Mathematics 9709 — 2018 May/June Paper 4 · Variant 2

1 A man has mass 80 kg. He runs along a horizontal road against a constant resistance force of magnitude P N. The total work done by the man in increasing his speed from 4 m s−1 to 5.5 m s−1 while running a distance of 60 metres is 1200 J. Find the value of P. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 KE gain = 1 2 × 80 × (5.52 – 42) [= 570] B1 Either initial or final KE correct WD against Res = 60P B1 [ 1 2 × 80 × (5.52 – 42) + 60P = 1200] M1 Four term work-energy equation P = 10.5 A1 4

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2 A train of mass 240 000 kg travels up a slope inclined at an angle of 4Å to the horizontal. There is a constant resistance of magnitude 18 000 N acting on the train. At an instant when the speed of the train is 15 m s−1 its deceleration is 0.2 m s−2. Find the power of the engine of the train. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Driving force DF = 15 P B1 Correct use of P = Fv ( ) DF 240 000 sin4 18 000 240 000 0.2 g  − − = × −    M1 A four-term Newton 2nd law equation A1 Correct equation Power is 2 060 000 (W) A1 Allow 2060 kW or 2.06 MW 4

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3 3 N 2 N 60Å 1Å P N The three coplanar forces shown in the diagram have magnitudes 3 N, 2 N and P N. Given that the three forces are in equilibrium, find the values of 1 and P. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 M1 Attempt to resolve forces horizontally (2 terms) θ = 41.4 A1 [P = 3 sin 60 + 2 sin θ] M1 Attempt to resolve forces vertically (3 terms) P = 3.92 A1 4 First alternative method for Q3 ( ) ( ) 2 3 sin 120 sin150 sin 90 θ θ = = − + P M1 Attempt two terms of Lami’s equation which can be used to find θ θ = 41.4 A1 M1 Attempt an equation which can be used to find P P = 3.92 A1 Second alternative method for Q3 [Triangle with sides 2, 3, P and angles opposite of 30, 90 – θ, 60 + θ] ( ) ( ) 2 3 sin 60 sin30 sin 90 θ θ = = + − P M1 Attempt two terms from the triangle of forces which can be used to find θ θ = 41.4 A1 M1 Attempt an equation which can be used to find P P = 3.92 A1

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4 A particle P moves in a straight line ABCD with constant acceleration. The distances AB and BC are 100 m and 148 m respectively. The particle takes 4 s to travel from A to B and also takes 4 s to travel from B to C. (i) Show that the acceleration of P is 3 m s−2 and find the speed of P at A. 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(ii) P reaches D with a speed of 61 m s−1. Find the distance CD. 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Mark scheme: 4(i) For example 100 = 4u + 8a or 100 = 1 2 (u + v) × 4 or 148 = 4v + 8a or any equation in two of the variables u, v, w, a M1 two of the variables below a is acceleration u is speed at A v is speed at B w is speed at C A1 One correct equation For example 248 = 8u + 32a or two further correct equations in 3 unknowns such as 148 = 4v + 8a and v = u + 4a or 148 = 1 2 (v + w) × 4 and 248 = 1 2 (u + w) × 8 A1 A second correct equation in the same two variables or two further correct equations leading to three equations in three of the unknowns u, v, w, a M1 Attempt to solve for a or u This must reach a = ... or u = ... a = 3 A1 AG u = 19 B1 6 Question Answer Marks Guidance 4(ii) 2 2 61 19 2 3 = + × × s M1 Attempt equation for s = AD [ 560 = s → CD = 560 – 248] M1 Attempt to find CD Distance CD is 312 A1 3 Alternative method for 4(ii) Speed at C is 19 + 8 × 3 [= 43] M1 Attempt to find speed at C 2 2 61 43 2 3   = + × ×   CD M1 Attempt to find CD Distance CD is 312 A1

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5 A particle of mass 20 kg is on a rough plane inclined at an angle of 60Å to the horizontal. Equilibrium is maintained by a force of magnitude P N acting on the particle, in a direction parallel to a line of greatest slope of the plane. The greatest possible value of P is twice the least possible value of P. Find the value of the coefficient of friction between the particle and the plane. 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Mark scheme: 5 R = 20g cos 60 [= 100] B1 F = µ × 20g cos 60 [= 100µ] M1 Use F = µR M1 Resolve along plane in either case (Pmax =) 20g sin 60 + F A1 One correct equation (Pmin =) 20g sin 60 – F A1 Second correct equation 20g sin 60 + F = 2(20g sin 60 – F) M1 Use of Pmax = 2Pmin to give four term equation in F or µ or P µ = 3 0.577 3 = A1 7 lternative solution for final 3 marks if Pmin is taken as acting down the plane Pmin = F – 20g sin 60 A1 20g sin 60 + F = 2(F – 20g sin 60) M1 µ = 3 3 5.196 = A1

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6 A particle P moves in a straight line passing through a point O. At time t s, the acceleration, a m s−2, of P is given by a = 6 −0.24t. The particle comes to instantaneous rest at time t = 20. (i) Find the value of t at which the particle is again at instantaneous rest. 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(ii) Find the distance the particle travels between the times of instantaneous rest. 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Mark scheme: 6(i) M1 Attempt to integrate a v = 6t – 0.12t2 (+ c) A1 0 = 6 × 20 – 0.12 × 202 + c DM1 Substitute v = 0, t = 20 in an equation with arbitrary constant 0.12t2 – 6t + 72 = 0 DM1 Substitute v = 0 and attempt to solve a 3-term quadratic t = 30 A1 5 6(ii) s = 3t2 – 0.04t3 – 72t (+ k) M1 Attempt to integrate v s(30) – s(20) = –540 – (–560) DM1 Use of limits 20 and their 30 Distance travelled = 20 A1 3

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7 A 1.6 kg 2.5 m P B 2.4 kg 1 m 30Å As shown in the diagram, a particle A of mass 1.6 kg lies on a horizontal plane and a particle B of mass 2.4 kg lies on a plane inclined at an angle of 30Å 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 inclined plane. The distance AP is 2.5 m and the distance of B from the bottom of the inclined plane is 1 m. There is a barrier at the bottom of the inclined plane preventing any further motion of B. The part BP of the string is parallel to a line of greatest slope of the inclined plane. 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. 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(ii) It is given instead that the horizontal plane is rough and that the coefficient of friction between A and the horizontal plane is 0.2. The inclined plane is smooth. Find the total distance travelled by A. 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Mark scheme: 7(i) [T = 1.6a, 2.4g sin 30 – T = 2.4a] System is 2.4g sin 30 = 4a M1 A1 Two correct equations M1 Solve for a or T a = 3 A1 T = 4.8 A1 5 7(ii) Friction force on A is F = 0.2 × 1.6g [= 3.2] B1 From F = µR T – F = 1.6a 2.4g sin 30 – T = 2.4a System is 2.4g sin 30 – F = 4a M1 Attempt Newton’s 2nd law for both particles or for the system A1 Correct equations for A and B or correct system equation M1 Attempt to solve for a a = 2.2 A1 v2 = 2 × 2.2 × 1 M1 Attempt to find v or v2 when B reaches the barrier Subsequent acceleration of A is –2 B1 4.4 = 2 × 2 × s M1 Attempt to find distance A travels while decelerating to v = 0 Total distance travelled is 2.1 m A1 9 Question Answer Marks Guidance 7(ii) Alternative method for Q7 [Work-Energy applied to A and B] F = 0.2 × 1.6g [= 3.2] B1 From F = µR = 0.2 × 1.6g = 3.2 M1 Attempt PE loss as B reaches the barrier PE loss = 2.4g sin 30 [= 12] A1 M1 Attempt KE gain for both A and B KE gain = 1 2 (1.6 + 2.4)v2 [= 2v2] A1 [2.4g sin 30 = 1 2 × 4 × v2 + 3.2 × 1] [v2 = 4.4] M1 Apply work-energy equation for the motion until B reaches the barrier (Three relevant terms) KE loss = 1 2 × 1.6 × 4.4 B1 Find KE loss as A comes to rest after B has stopped [ 1 2 × 1.6 × 4.4 = 3.2d] [d = 1.1] M1 Apply work-energy equation where d is the extra distance travelled by A leading to a positive value for d Total distance = 2.1 m A1 Distance = d + 1 Question Answer Marks Guidance 7(ii) Alternative scheme for first 6 marks of 7(ii) [Work-energy applied to A] Friction = 0.2 × 1.6g [= 3.2] B1 [2.4g sin 30 – T = 2.4a T – F = 1.6a] M1 Apply Newton’s 2nd law to A and B and solve for T T = 6.72 A1 [ 1 2 × 1.6 × v2] M1 Attempt KE for A only A1 Correct KE for A [6.72 × 1 = 1 2 × 1.6 × v2 + 3.2 × 1] M1 Use work/energy equation for A Alternative scheme for first 6 marks of 7(ii) [Work-energy applied to B] Friction = 0.2 × 1.6g [= 3.2] B1 [2.4g sin 30 – T = 2.4a T – F = 1.6a] M1 Apply Newton’s 2nd law to A and B and solve for T T = 6.72 A1 M1 Find energy loss/gain for B Allow either term ±( 1 2 × 2.4 × v2 – 2.4g sin 30) A1 2.4g sin 30 = 1 2 × 2.4 × v2 + 6.72 × 1 M1 Use work/energy equation for B

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