Cambridge A Level Mathematics 9709 — 2020 Feb/March Paper 4 · Variant 2

1 A lorry of mass 16 000 kg is travelling along a straight horizontal road. The engine of the lorry is working at constant power. The work done by the driving force in 10 s is 750 000 J. (a) Find the power of the lorry’s engine. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) There is a constant resistance force acting on the lorry of magnitude 2400 N. Find the acceleration of the lorry at an instant when its speed is 25 m s−1. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) Power = 750000/10 = 75000 W or 75 kW B1 Power = WD/Time 1 1(b) Driving force DF = 75000/25 B1FT Using P = DF × v [DF – 2400 = 16000a] M1 Using Newton’s 2nd law a = 0.0375 ms–2 A1 Allow a = 3 80 3

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2 A particle P of mass 0.4 kg is on a rough horizontal floor. The coefficient of friction between P and the floor is -. A force of magnitude 3 N is applied to P upwards at an angle ! above the horizontal, where tan ! = 34. The particle is initially at rest and accelerates at 2 m s−2. (a) Find the time it takes for P to travel a distance of 1.44 m from its starting point. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find -. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) M1 For using a complete method which would lead to an equation for finding a value of t such as s = ut + ½ at2 with u = 0, s = 1.44 and a = 2 t = 1.2 s A1 2 2(b) R = 0.4g – 3 × 3 5 = 0.4g – 3 sin 36.9 [= 2.2] B1 [3 × 4 5 – F = 3 cos 36.9 – F = 0.4 × 2] [F = 1.6] M1 Use Newton’s 2nd law, 3 terms, to find F. 4 5 3 5 3 0.4 2 1.6 0.4 3 2.2 g μ   × − × = =   −×     M1 Use of F R μ = μ = 0.727 A1 Allow μ = 8 11 4

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3 B h m A 0.5 m C The diagram shows the vertical cross-section of a surface. A, B and C are three points on the cross- section. The level of B is h m above the level of A. The level of C is 0.5 m below the level of A. A particle of mass 0.2 kg is projected up the slope from A with initial speed 5 m s−1. The particle remains in contact with the surface as it travels from A to C. (a) Given that the particle reaches B with a speed of 3 m s−1 and that there is no resistance force, find h. 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(b) It is given instead that there is a resistance force and that the particle does 3.1 J of work against the resistance force as it travels from A to C. Find the speed of the particle when it reaches C. 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Mark scheme: 3(a) or Final KE = ½ × 0.2 × 32 B1 ½ × 0.2 × 52 = 0.2gh + ½ × 0.2 × 32 M1 Use conservation of energy h = 0.8 A1 3 3(b) Apply work-energy equation from A to C M1 ½ × 0.2 × 52 – 3.1 + 0.2g × 0.5 = ½ × 0.2v2 A1 Correct work-energy equation Speed = 2 ms–1 A1 3

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4 A cyclist travels along a straight road with constant acceleration. He passes through points A, B and C. The cyclist takes 2 seconds to travel along each of the sections AB and BC and passes through B with speed 4.5 m s−1. The distance AB is 4 of the distance BC. 5 (a) Find the acceleration of the cyclist. 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(b) Find AC. 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Mark scheme: 4(a) Use the constant acceleration equations to obtain an expression for either sAB or sBC in terms of a M1 sAB = 2 × 4.5 – ½ × a × 22 A1 or sAB = ½(vA + vB) × 2 = 9 – 2a sBC = 2 × 4.5 + ½ × a × 22 A1 or sBC = ½(vB + vC) × 2 = 9 + 2a [2 × 4.5 – ½a × 22 = 4 5 (2 × 4.5 + ½a × 22)] M1 Use the given information to find a valid equation for a a = 0.5 ms–2 A1 Alternative method for question 4(a) [4.5 = u + 2a, sAC = 4u + 8a, sAB = 2u + 2a] M1 Any two relevant equations in u, a, sAB and sAC where u is the velocity at A Two correct equations A1 Three correct equations A1 [2(4.5 – 2a) + 6a = 5 4 {2(4.5 – 2a) + 2a}] M1 Use the given information that BC = 5/4AB to find a valid equation such as the one shown OE involving a only a = 0.5 ms–2 A1 Alternative method for question 4(a) [AC = 4.5 × 4] M1 Using AC = vB × 4 since vB is the average velocity over AC BC = 5/9 × AC or AB = 4/9 × AC M1 BC = 10 or AB = 8 A1 [10 = 4.5 × 2 + 2a or 8 = 4.5 × 2 – 2a] M1 Using s = ut + ½ at2 for BC or s = vt – ½ at2 for AB a = 0.5 ms–2 A1 Question Answer Marks Guidance 5 4(b) sAB = 2 × 4.5 – ½ × 0.5 × 22 = 8 OR sBC = 2 × 4.5 + ½ × 0.5 × 22 = 10 M1 Attempt to find the value of sAB or sBC OR attempt to find sAB directly as sAC = 3.5 × 4 + ½ × a × 42 or ½ (4.5 – 2a + 4.5 + 2a) × 4 or add the 2 expressions found in 4(a) for sAB and sBC sAC = 8 + 5/4 × 8 = 18 m OR sAC = 10 + 4/5 × 10 = 18 m A1 2

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5 F N 4 N !Å 30Å 3 N P 6 N Coplanar forces, of magnitudes F N, 3 N, 6 N and 4 N, act at a point P, as shown in the diagram. (a) Given that ! = 60, and that the resultant of the four forces is in the direction of the 3 N force, find F. 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(b) Given instead that the four forces are in equilibrium, find the values of F and !. 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Mark scheme: 5(a) [4 sin 30 + F sin 60 – 6 = 0] M1 Resolve forces vertically and equate to zero Correct equation A1 F = 4.62 A1 Allow F = 8 3 or F = 8 3 3 3 Question Answer Marks Guidance 5(b) Resolve forces either vertically or horizontally M1 F sin α + 4 sin 30 – 6 = 0 and F cos α + 3 – 4 cos 30 = 0 A1 Both equations correct [F sin α = 4] [F cos α = 0.464102...] [F2 = 42 + 0.4642] or 4 0.464 sin83.4 cos83.4 F   = =     M1 Attempt to solve for F using Pythagoras or from a value found for α 1 4 tan 0.464 α −     =         or 1 1 4 0.464 sin cos 4.03 4.03 α − −       = =             M1 Attempt to solve for α using trigonometry or from a value found for F F = 4.03 and α = 83.4 A1 Both correct as shown [F = 4.0268…, α = 83.382…] 5

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6 On a straight horizontal test track, driverless vehicles (with no passengers) are being tested. A car of mass 1600 kg is towing a trailer of mass 700 kg along the track. The brakes are applied, resulting in a deceleration of 12 m s−2. The braking force acts on the car only. In addition to the braking force there are constant resistance forces of 600 N on the car and of 200 N on the trailer. (a) Find the magnitude of the force in the tow-bar. 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(b) Find the braking force. 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(c) At the instant when the brakes are applied, the car has speed 22 m s−1. At this instant the car is 17.5 m away from a stationary van, which is directly in front of the car. Show that the car hits the van at a speed of 8 m s−1. 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(d) After the collision, the van starts to move with speed 5 m s−1 and the car and trailer continue moving in the same direction with speed 2 m s−1. Find the mass of the van. 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Mark scheme: 6(a) [T – 200 = 700 × –12] Car: –T – 600 – F = 1600 × –12 System: –600 – 200 – F = 2300 × –12 the car and to the system and eliminate the braking force, F. Magnitude of T = 8200 N A1 2 6(b) Car [T – F – 600 = 1600 × –12] or System [–600 – 200 – F = 2300 × –12] M1 Apply Newton’s second law either to the car or to the system with braking force = F and use of their T from 6(a) Braking force F = 26800 N A1 2 6(c) [v2 = 222 + 2 × –12 × 17.5] M1 A complete method using constant acceleration equations which would lead to an equation for finding v, using u = 22, s = 17.5 and a = –12 v = 8 ms–1 A1 AG 2 6(d) [2300 × 8 + m × 0 = 2300 × 2 + m × 5] M1 For applying the conservation of momentum equation to the system of car, trailer and van, where m = mass of the van A1 Correct equation m = 2760 kg A1 3

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7 A particle moves in a straight line through the point O. The displacement of the particle from O at time t s is s m, where s = t2 −3t + 2 for 0 ≤t ≤6, 24 s = −t2 + 25 for t ≥6. t 4 (a) Find the value of t when the particle is instantaneously at rest during the first 6 seconds of its motion. 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At t = 6, the particle hits a barrier at a point P and rebounds. (b) Find the velocity with which the particle arrives at P and also the velocity with which the particle leaves P. 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(c) Find the total distance travelled by the particle in the first 10 seconds of its motion. 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Mark scheme: 7(a) [v = 2t – 3] t = 1.5 A1 2 7(b) Velocity at arrival = 9 ms–1 B1 t = 6 used in v 2 24 0.5 v t t = − − M1 For differentiation of s for t ⩾ 6 Velocity when leaves = –3.67 ms–1 A1 Allow v = –11/3 3 7(c) At t = 0, s = 2 or at t = 6, s = 20 B1 SOI At t = 1.5, s = –0.25 B1 SOI At t = 10, s = 2.4 B1 SOI [Total distance = 2 + 0.25 + 0.25 + 20 + (20 – 2.4)] M1 Evidence of distance rather than displacement involving all three sections, (0, 1.5), (1.5, 6) and (6, 10) So total distance travelled = 40.1 m A1 5

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