Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 4 · Variant 1

1 A block of mass 3 kg is initially at rest on a smooth horizontal floor. A force of 12 N, acting at an angle of 25Å above the horizontal, is applied to the block. Find the distance travelled by the block in the first 5 seconds of its motion. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 [12 cos 25 = 3a] M1 For use of Newton’s second law a = 4 cos 25 = 3.625 A1 [s = ½ × 4cos 25 × 52] M1 For use of s = ut + ½at2 OE Distance = 45.3 m A1 4

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2 A tractor of mass 3700 kg is travelling along a straight horizontal road at a constant speed of 12 m s−1. The total resistance to motion is 1150 N. (i) Find the power output of the tractor’s engine. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The tractor comes to a hill inclined at 4Å above the horizontal. The power output is increased to 25 kW and the resistance to motion is unchanged. (ii) Find the deceleration of the tractor at the instant it begins to climb the hill. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the constant speed that the tractor could maintain on the hill when working at this power. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) Power = 1150 × 12 = 13 800W B1 For use of P = F × v Allow 13.8 kW 1 2(ii) 25000 B1 P Driving force = Using F = 12 v 25000 M1 For applying Newton’s 2nd law up the – 1150 – 3700g sin 4 = 3700a slope, 4 terms 12 a = –0.445 m s–2 A1 3 2(iii) 25000 M1 For stating the equation for constant v, – 1150 – 3700gsin 4 = 0 with 3 terms, and solving for v v v = 6.70 m s-1 A1 2

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3 A roller-coaster car (including passengers) has a mass of 840 kg. The roller-coaster ride includes a section where the car climbs a straight ramp of length 8 m inclined at 30Å above the horizontal. The car then immediately descends another ramp of length 10 m inclined at 20Å below the horizontal. The resistance to motion acting on the car is 640 N throughout the motion. (i) Find the total work done against the resistance force as the car ascends the first ramp and descends the second ramp. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The speed of the car at the bottom of the first ramp is 14 m s−1. Use an energy method to find the speed of the car when it reaches the bottom of the second ramp. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) 640 × 18 M1 For use of work done = F × d Work done = 11 520 J A1 2 3(ii) KE at start B1 = ½ × 840 × 142 = 82 320 J PE gained = 840g × 8sin 30 B1 – 840g × 10sin 20 = 4870 J ½ × 840 × v2 = 82 320 – 11 520 – 4870 M1 For using work – energy equation with 4 terms and solving for v v = 12.5 m s–1 A1 4

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4 v (m s−1) V 10 t (s) 0 1.5 3.5 6 10 T −15 The diagram shows the velocity-time graph of a particle which moves in a straight line. The graph consists of 5 straight line segments. The particle starts from rest at a point A at time t = 0, and initially travels towards point B on the line. (i) Show that the acceleration of the particle between t = 3.5 and t = 6 is −10 m s−2. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The acceleration of the particle between t = 6 and t = 10 is 7.5 m s−2. When t = 10 the velocity of the particle is V m s−1. Find the value of V. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) The particle comes to rest at B at time T s. Given that the total distance travelled by the particle between t = 0 and t = T is 100 m, find the value of T. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) ( − 25) B1 AG Acceleration = = –10 m s–2 2.5 1 4(ii) V = –15 + 7.5 × 4 M1 Using v–t graph OE V = 15 m s–1 A1 2 4(iii) Using v = 0 at t = 4.5 and t = 8 B1 M1 Attempting to use area to find total distance travelled ½ × (4.5 + 2) × 10 M1 For setting up an equation for total + ½ × (8 – 4.5) × 15 distance travelled and solving for T + ½ × (T – 8) × 15 = 100 T = 13.5 A1 4

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5 A particle starts from a point O and moves in a straight line. The velocity of the particle at time t s after leaving O is v m s−1, where v = 1.5 + 0.4t for 0 ≤t ≤5, 100 v = −0.1t for t ≥5. t2 (i) Find the acceleration of the particle during the first 5 seconds of motion. 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(ii) Find the value of t when the particle is instantaneously at rest. 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(iii) Find the total distance travelled by the particle in the first 10 seconds of motion. 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Mark scheme: 5(i) Acceleration = 0.4 m s–2 B1 1 5(ii) 100 M1 For setting v = 0 and solving for t – 0.1t = 0 t 2 t = 10 s A1 2 5(iii) Distance t = 0 to t = 5 is B1 Trapezium rule or integration ½ (1.5 + 3.5) × 5 = 12.5  100  M1 For integration 0.1t  dt s ( t ) = ∫  2 −  t  100 2 A1 Correct integration = − − 0.05t ( + C ) t s (10 ) − s ( 5 ) M1 Use limits 5 and 10 used or find + C Total distance = 12.5 + 6.25 = 18.75 m A1 5

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6 50 N ! G N P 50Å F N 3F N Coplanar forces, of magnitudes F N, 3F N, G N and 50 N, act at a point P, as shown in the diagram. (i) Given that F = 0, G = 75 and ! = 60Å, find the magnitude and direction of the resultant force. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Given instead that G = 0 and the forces are in equilibrium, find the values of F and !. 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Mark scheme: 6(i) M1 For resolving forces (either direction) X = 75 + 50 cos 60 (= 100) A1 For both equations, unevaluated Y = 50 sin 60 (= 43.3) Resultant = √(1002 + 43.32) = 109 N B1  43.3  B1 Must state anticlockwise from the positive Angle = arctan   = 23.4º x-axis or show in a diagram  100  4 6(ii) 50 cos α – F cos 50 = 0 B1 Resolving forces horizontally 50 sin α – 3F – F sin 50 = 0 B1 Resolving forces vertically ( 3 F + F sin50 ) M1 For division to find θ or for using tan α = Pythagoras to find F ( F cos50 ) α = 80.3 A1 F = 13.1 A1 5

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7 B A 2.5 N 1Å 25Å Two particles A and B of masses 0.9 kg and 0.4 kg respectively are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley which is attached to the top of two inclined planes. The particles are initially at rest with A on a smooth plane inclined at angle 1Å to the horizontal and B on a plane inclined at angle 25Å to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes. A force of magnitude 2.5 N is applied to B acting down the plane (see diagram). (i) For the case where 1 = 15 and the plane on which B rests is smooth, find the acceleration of B. 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(ii) For a different value of 1, the plane on which B rests is rough with coefficient of friction between the plane and B of 0.8. The system is in limiting equilibrium with B on the point of moving in the direction of the 2.5 N force. Find the value of 1. 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Mark scheme: 7(i) M1 For applying Newton’s 2nd law to either particle (correct number of terms) T – 0.9 g sin 15 = 0.9a A1 2.5 + 0.4 g sin 25 – T = 0.4a A1 1.3a = 1.86… M1 Solving simultaneously for a a = 1.43 m s–2 A1 5 7(ii) F = 0.8 × 0.4g cos 25 B1 2.5 + 0.4 g sin 25 – T – F = 0 M1 For using equilibrium of forces acting on particle B with 4 terms T – 0.9 g sin θ = 0 M1 For using equilibrium of forces acting on particle A with 2 terms M1 For solving for θ θ = 8.2º A1 5

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