Cambridge A Level Mathematics 9709 — 2018 May/June Paper 4 · Variant 1

1 A particle P is projected vertically upwards with speed 24 m s−1 from a point 5 m above ground level. Find the time from projection until P reaches the ground. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 –5 = 24t – 5t 2 M1 1 Use s = ut + at 2 2 5t 2 – 24t – 5 = 0 M1 Solve relevant 3 term quadratic t = 5 A1 3 Alternative scheme for Question 1 0 = 24 – 10t1 → t1 = 2.4 M1 Attempt to find the time taken to reach the highest point 0 = 242 + 2 × (–10) × h → h = 28.8 M1 Find total height h reached and attempt to 1 2 find time taken from highest point to And 33.8 = gt2 → t2 = 2.6 ground level 2 t = t1 + t2 = 5 A1

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2 6 N 8 N O 10 N The diagram shows three coplanar forces acting at the point O. The magnitudes of the forces are 6 N, 8 N and 10 N. The angle between the 6 N force and the 8 N force is 90Å. The forces are in equilibrium. Find the other angles between the forces. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 [10 cos α = 8 or 10 cos β = 6] M1 Introduce α or β, an angle between the 10N force and the vertical or horizontal and attempt to resolve forces α = 36.9 or β = 53.1 A1 Angle between 6N and 10N is 126.9 B1 Angle between 8N and 10N is 143.1 B1 4 Alternative scheme for Question 2 10 6 8 M1 Attempt to use Lami’s theorem = = sin90 sin γ sin δ γ (8 and 10), δ (6 and 10) All correct A1 Angle between 8N and 10N is γ =143.1 B1 Angle between 6N and 10N is δ =126.9 B1

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3 100 N P 1Å 30Å A particle P of mass 8 kg is on a smooth plane inclined at an angle of 30Å to the horizontal. A force of magnitude 100 N, making an angle of 1Å with a line of greatest slope and lying in the vertical plane containing the line of greatest slope, acts on P (see diagram). (i) Given that P is in equilibrium, show that 1 = 66.4, correct to 1 decimal place, and find the normal reaction between the plane and P. 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(ii) Given instead that 1 = 30, find the acceleration of P. 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Mark scheme: 3(i) M1 Attempt to resolve forces along the plane (2 terms) 100 cos θ = 8 g sin 30 → θ = 66.4 A1 [R = 8 g cos 30 + 100 sin θ] M1 Resolve forces perpendicular to the plane (3 terms) R = 161 A1 4 3(ii) 100 cos 30 – 8g sin 30 = 8a M1 Apply Newton’s 2nd law parallel to the plane (3 terms) a = 5.83 A1 2

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4 A particle P moves in a straight line starting from a point O. At time t s after leaving O, the displacement s m from O is given by s = t3 −4t2 + 4t and the velocity is v m s−1. (i) Find an expression for v in terms of t. 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(ii) Find the two values of t for which P is at instantaneous rest. 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(iii) Find the minimum velocity of P. 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Mark scheme: 4(i) M1 Attempt differentiation v = 3t 2 – 8t + 4 A1 2 4(ii) 3t 2 – 8t + 4 = 0 M1 Set v = 0 and attempt to solve a relevant 3 term quadratic 2 A1 t = and t = 2 3 2 4(iii) [6t – 8 = 0] M1 Differentiate v and equate to 0 4 4 4 M1 Solve for t and attempt v [t = , v =3( )2 – 8( ) + 4] 3 3 3 4 A1 v = – 3 3 Alternative scheme for Question 4(iii) 8 4 M1 Attempt to complete the square for v [v = 3(t 2 – t) + 4 = 3(t – )2 +......] 3 3 4 4 4 M1 Find value of t for minimum v and attempt [t = , v = 3(t – )2 – ] to find v 3 3 3 4 A1 v = – 3

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5 A sprinter runs a race of 200 m. His total time for running the race is 20 s. He starts from rest and accelerates uniformly for 6 s, reaching a speed of 12 m s−1. He maintains this speed for the next 10 s, before decelerating uniformly to cross the finishing line with speed V m s−1. (i) Find the distance travelled by the sprinter in the first 16 s of the race. Hence sketch a displacement- time graph for the 20 s of the sprinter’s race. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ displacement (m) 200 0 time (s) 0 20 (ii) Find the value of V. 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Mark scheme: 5(i) 1 M1 Use constant acceleration equations or [s1 = (0 + 12) × 6] find area in (t,v) graph to find the distance 2 s1 travelled in the first 6 seconds [s2 = 10 × 12] M1 Use constant acceleration equations or find area in (t,v) graph to find s2 the distance travelled between 6s and 16s Distance for first 16 s is A1 36 + 10 × 12 = 156 Curve concave up for 0 < t < 6 B1 Co-ordinates refer to (t,s) in a starting at (0 , 0) ending at (6 , 36) displacement-time graph Line, positive gradient, 6 < t < 16 starts at B1 (6 , 36) ends at (16 , 156) Curve concave down, 16 < t < 20 from B1 (16 , 156) to (20 , 200) 6 5(ii) 1 M1 Use relevant constant acceleration [44 = (12 + V) × 4] equations or the area property of a v–t 2 graph V = 10 A1 2

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6 A car has mass 1250 kg. (i) The car is moving along a straight level road at a constant speed of 36 m s−1 and is subject to a constant resistance of magnitude 850 N. Find, in kW, the rate at which the engine of the car is working. 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(ii) The car travels at a constant speed up a hill and is subject to the same resistance as in part (i). The hill is inclined at an angle of 1Å to the horizontal, where sin 1Å = 0.1, and the engine is working at 63 kW. Find the speed of the car. 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(iii) The car descends the same hill with the engine of the car working at a constant rate of 20 kW. The resistance is not constant. The initial speed of the car is 20 m s−1. Eight seconds later the car has speed 24 m s−1 and has moved 176 m down the hill. Use an energy method to find the total work done against the resistance during the eight seconds. 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Mark scheme: 6(i) [P = DF × v = 850 × 36] M1 Apply P = DF × v with DF = Resistance force Power = rate of working = 30.6 kW A1 2 6(ii) [DF = 1250 g × 0.1 + 850] M1 Driving force comprising of resistance plus a weight component 63000 M1 P DF = DF = v v v = 30 so speed of car is 30 ms–1 A1 3 6(iii) 1 B1 [= 110 000] Gain in KE = × 1250 × (242 – 202) 2 Loss in PE = 1250 g × 176 × 0.1 B1 [= 220 000] WD by car’s engine = 20 000 × 8 B1 [= 160 000] [160 000 + 220 000 = M1 4 term work energy equation WD against resistance + 110 000] WD = 270 000 J = 270 kJ A1 5

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7 P A B 0.8 kg 1.2 kg 45Å 30Å The diagram shows a triangular block with sloping faces inclined to the horizontal at 45Å and 30Å. Particle A of mass 0.8 kg lies on the face inclined at 45Å and particle B of mass 1.2 kg lies on the face inclined at 30Å. The particles are connected by a light inextensible string which passes over a small smooth pulley P fixed at the top of the faces. The parts AP and BP of the string are parallel to lines of greatest slope of the respective faces. The particles are released from rest with both parts of the string taut. In the subsequent motion neither particle reaches the pulley and neither particle reaches the bottom of a face. (i) Given that both faces are smooth, find the speed of A after each particle has travelled a distance of 0.4 m. 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(ii) It is given instead that both faces are rough. The coefficient of friction between each particle and a face of the block is -. Find the value of - for which the system is in limiting equilibrium. 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Mark scheme: 7(i) A T – 0.8 g sin 45 = 0.8a M1 Apply Newton 2nd law to either A or to B B 1.2g sin 30 – T = 1.2a or to the system System 1.2 g sin 30 – 0.8 g sin 45 = 2a A1 One correct equation A1 A second correct equation a = 0.171 M1 Solve for a v2 = 2 × a × 0.4 M1 Use v2 = u2 + 2as with u = 0 v = 0.370 so speed of A is 0.370 ms–1 A1 6 Alternative scheme for Question 7(i) M1 Attempt KE gain or PE loss 1 1 A1 v is the required speed of A KE gain = × 0.8 × v2 + × 1.2 × v2 2 2 PE loss = A1 1.2 g × 0.4 sin 30 – 0.8 g × 0.4 sin 45 1 1 M1 4 term energy equation × 0.8 × v2 + × 1.2 × v2 = 2 2 1.2 g × 0.4 sin 30 – 0.8 g × 0.4 sin 45 M1 Solving for v v = 0.370 so speed of A is 0.370 ms–1 A1 7(ii) RA = 0.8 g cos45 = 4 2 B1 For either RA or RB RB = 1.2 g cos30 = 6 3 FA = 4 2 µ and FB = 6 3 µ M1 Either FA or FB used A 0.8 g sin 45 + FA = T M1 Resolve parallel to the plane either for B 1.2 g sin 30 – FB = T both particles A and B or for the system or system equation: equation 12 sin 30 – 8 sin 45 = FA + FB Correct equation(s) A1 M1 Eliminate T and solve for µ A1 6 − 4 2 ( ) µ = 6 3 + 4 √ 2 ( ) = 0.0214 6

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