Cambridge A Level Mathematics 9709 — 2022 Feb/March Paper 4 · Variant 2

1 A crane is used to raise a block of mass 600kg vertically upwards at a constant speed through a height of 15m. There is a resistance to the motion of the block, which the crane does 10000J of work to overcome. (a) Find the total work done by the crane. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Given that the average power exerted by the crane is 12.5kW, find the total time for which the block is in motion. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) M1 Attempt potential energy. Total work done by crane = [90 000 + 10 000 =] 100 000 J A1 2 1(b) 100 000 = 12 500 × t M1 Use of work done = power × time to set up an equation from which t can be found. Time = 8 s A1 FT FT on their work done = 100 000 Alternative scheme for question 1(b) Average force Total WD 15 F = Average velocity 15 = = s v t t Total WD 15 12500 15 P Fv t = → = × M1 A complete method, using = P Fv , for setting up an equation from which t can be found. Time = 8 s A1 FT 2

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2 A particle P is projected vertically upwards from horizontal ground with speed ums−1. P reaches a maximum height of 20m above the ground. (a) Find the value of u. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the total time for which P is at least 15m above the ground. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) OR 0 = u – 10t and 2 1 2 20 10 vt t = + ´ ´ or 2 1 2 20 10 ut t = - ´ ´ or 0 20 2 u t + = ´ M1 Complete method to set up an equation in u only. Use of v2 = u2 + 2as or finding time to reach maximum height (t = 2) and using this value to set up another equation in u only. u = 20 A1 2 2(b) 2 1 15 20 10 2 = − × × t t M1 Use of 2 1 2 = + s ut at and attempt to set up an equation from which a relevant t value can be found. Must be using their u and 10 = − a t = 1 or t = 3 A1 Total time = 2 s A1 CWO Alternative method for question 2(b) 2 1 5 10 2 = × ×t M1 Use of 2 1 2 = + s ut at and attempt to set up an equation from which a relevant t value can be found. Must be using 0 = u and 10 = a t = 1 A1 Total time = 2 s A1 CWO 3

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3 A car of mass mkg is towing a trailer of mass 300kg down a straight hill inclined at 3Å to the horizontal at a constant speed. There are resistance forces on the car and on the trailer, and the total work done against the resistance forces in a distance of 50m is 40000J. The engine of the car is doing no work and the tow-bar is light and rigid. (a) Find the value of m. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The resistance force on the trailer is 200N. (b) Find the tension in the tow-bar between the car and the trailer. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) PE lost in 50 m = (m + 300) g × 50 sin 3 B1 (m + 300) g × 50 sin 3 – 40 000 = 0 M1 Use of the work-energy equation. m = 1230 to 3 sf A1 m = 1228.6 Alternative method for question 3(a) Resistance force R = 40000 50 [= 800 N] B1 ( ) 300 sin3 0 + − = m g R M1 Apply Newton’s second law to the system, 3 terms. m = 1230 to 3 sf A1 m = 1228.6 3 3(b) T + 300 g sin 3 – 200 = 0 (Trailer) or mg sin 3 = T + 600 (Car) M1 Apply Newton’s 2nd law either to the trailer or to the car using a = 0, three terms in either case. T = 43[.0] N to 3 sf A1 2

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4 The total mass of a cyclist and her bicycle is 70kg. The cyclist is riding with constant power of 180W up a straight hill inclined at an angle ! to the horizontal, where sin ! = 0.05. At an instant when the cyclist’s speed is 6ms−1, her acceleration is −0.2ms−2. There is a constant resistance to motion of magnitude F N. (a) Find the value of F. 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(b) Find the steady speed that the cyclist could maintain up the hill when working at this power. 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Mark scheme: 4(a) Forward force exerted by cyclist driving force = 180 6 [= 30 N] B1 DF – F – 70g sin α = 70 × ‒ 0.2 M1 Attempt Newton’s second law, 4 terms required. A value must be used for sin α. 30 – F – 70g × 0.05 = 70 × ‒ 0.2 A1 Correct equation F = 9 A1 From exact working only 4 4(b) 180 v – F – 70g × sin α = 0 M1 Apply Newton’s second law up the hill with a = 0. Must have 3 relevant terms using their F from 4(a). A value for sin α must be used. v = 4.09 m s−1 A1 Allow 45 11 2

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5 Four coplanar forces act at a point. The magnitudes of the forces are 10N, F N, GN and 2F N. The directions of the forces are as shown in the diagram. (a) Given that the forces are in equilibrium, find the values of F and G. 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(b) Given instead that F = 3, find the value of G for which the resultant of the forces is perpendicular to the 10N force. 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Mark scheme: 5(a) Attempt to resolve vertically or horizontally M1 Correct number of terms. G sin 60° + 2F sin 40° − 10 = 0 A1 Correct resolution vertically. F + G cos 60° − 2F cos 40° = 0 A1 Correct resolution horizontally. Attempt to solve simultaneously for F or G M1 From equations with 3 relevant terms in each F = 4.53, G = 4.82 A1 For both correct. 5 5(b) G sin 60° + 2 × 3 sin 40° − 10 = 0 M1 Resolve forces parallel to the 10 N force and equate this expression to zero, 3 terms. G = 7.09 to 3 sf A1 2

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6 A cyclist starts from rest at a fixed point O and moves in a straight line, before coming to rest k seconds 1 2 −3 later. The acceleration of the cyclist at time t s after leaving O is ams−2, where a = 2t−1 5t 2 for 0 < t ≤k. (a) Find the value of k. 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(b) Find the maximum speed of the cyclist. 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(c) Find an expression for the displacement from O in terms of t. Hence find the total distance travelled by the cyclist from the time at which she reaches her maximum speed until she comes to rest. 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Mark scheme: 6(a) For an attempt at integration *M1 Power of at least one term increased by 1 and the coefficient changed. [ ] 1 3 2 2 2 4 5 = − + v t t C A1 Correct v 1 3 2 2 2 4 0 5 − = t t DM1 Equating their 2-term v to zero and attempt to solve for t or k. k = 10 A1 Final answer t = 10 is A0 4 6(b) Max speed when 1 1 2 2 3 2 0 5 −− = t t M1 Attempt to solve a = 0 and find a value of t. t = 10 3 A1 Maximum speed = 4.87 ms−1 to 3 sf B1 Allow maximum speed as 8 30 9 3 Question Answer Marks Guidance 6(c) For an attempt at integration of their v *M1 Power of at least one term increased by 1 and the coefficient changed. [ ] 3 5 2 2 8 4 3 25 = − + s t t C A1 Correct s Substitute their t = 10 3 and t = 10 DM1 Use their 10 = t and their ( ) 10 0 3 = ≠ t correctly. Distance = 20.7 m A1 Distance = 20.7479... 4

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7 A bead, A, of mass 0.1kg is threaded on a long straight rigid wire which is inclined at sin−1 7 to 25 the horizontal. A is released from rest and moves down the wire. The coefficient of friction between A and the wire is -. When A has travelled 0.45m down the wire, its speed is 0.6ms−1. (a) Show that - = 0.25. 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Another bead, B, of mass 0.5kg is also threaded on the wire. At the point where A has travelled 0.45m down the wire, it hits B which is instantaneously at rest on the wire. A is brought to instantaneous rest in the collision. The coefficient of friction between B and the wire is 0.275. (b) Find the time from when the collision occurs until A collides with B again. 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Mark scheme: 7(a) a = 0.4 A1 R = 0.1g × cos α = 0.1g × 24 25 = 0.1g × cos 16.3° 24 0.96 25 R é ù ê ú = = ê ú ë û B1 Must use a value for cos α. 0.1g × 7 25 – F = 0.1 × 0.4 [0.28 – F = 0.04 → F = 0.24] M1 Newton’s second law, 3 terms. F = µ × 0.1g × 24 25 24 0.96 25 F μ μ é ù ê ú = = ê ú ë û M1 Use of F = µR, where R is a component of 0.1g µ = 0.25 A1 AG Must be from exact working µ = 0.25 only Question Answer Marks Guidance 7(a) Alternative scheme for question 7(a) Attempt PE loss or KE gain M1 Use of either PE = mgh or KE = ½ mv2 PE loss 7 63 0.1 0.45sin16.3 0.1 0.45 0.126 25 500   = × × = × × × = =     g g KE gain 2 1 9 0.1 0.6 0.018 2 500   = × × = =     A1 Both correct. R = 0.1g × cosα = 0.1g × 24 25 = 0.1g × cos 16.3° 24 0.96 25 R é ù ê ú = = ê ú ë û B1 Must use a value for cosα. 2 7 1 0.1 0.45 0.1 0.6 0.45 25 2 × × × = × × + × g F 63 9 54 500 500 125 μ   = + ×     or [ ] 0.126 0.018 0.432 μ = + × M1 Use of work-energy equation as PE loss = KE gain + WD against friction F = µ × 0.1g × 24 25 24 0.96 25 F μ μ é ù ê ú = = ê ú ë û M1 Use of F = µR, where R is a component of 0.1g µ = 0.25 A1 AG Must be from exact working µ = 0.25 only 6 Question Answer Marks Guidance 7(b) 0.1 × 0.6 = 0.5v M1 Use of conservation of momentum, 2 terms. v = 0.12 A1 For B 0.5g × 7 25 – 0.275 × 0.5g × 24 25 = 0.5a [leading to a = 0.16] B1 Apply Newton’s second law for particle B, 3 terms. Allow correct unsimplified expression in a only. sA = 0 + 2 1 0.4 2 × t sB = 2 1 0.12 0.16 2 + × t t *M1 Attempt an expression for either sA or sB. Must see uA = 0 and uB ≠ 0 but uB must have been found from a momentum equation. For both sA and sB and attempt to solve sA = sB to find t DM1 Must be from 3 terms leading to a 2-term quadratic. If energy used in 7(a) then must find a = 0.4 for A. Their working must be leading to a positive t value. Required time is t = 1 s A1 6

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