Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 4 · Variant 3

1 A crate of mass 500 kg is being pulled along rough horizontal ground by a horizontal rope attached to a winch. The winch produces a constant pulling force of 2500 N and the crate is moving at constant speed. Find the coefficient of friction between the crate and the ground. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 B1 Use of F=µR [2500=µ×500g] M1 Resolving horizontally µ=0.5 A1 3

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2 A train of mass 150 000 kg ascends a straight slope inclined at !Å to the horizontal with a constant driving force of 16 000 N. At a point A on the slope the speed of the train is 45 m s−1. Point B on the slope is 500 m beyond A. At B the speed of the train is 42 m s−1. There is a resistance force acting on the train and the train does 4 × 106 J of work against this resistance force between A and B. Find the value of !. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 B1 2 2 1 1 2 2 150000 45 150000 42 × × − × × (=19575000) B1 Correct expression for KE loss M1 For 5 term work energy equation (or 4 terms if using loss in KE as 1 term) 150000g × 500sinα=19575000+16000×500 – 4×106 A1 α =1.8 A1 5

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3 50 N 20Å 60 N !Å 30Å R N 100 N Three coplanar forces of magnitudes 50 N, 60 N and 100 N act at a point. The resultant of the forces has magnitude R N. The directions of these forces are shown in the diagram. Find the values of R and !. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Resolving horizontally or vertically M1 50cos20 + 60 – 100sin30 (=56.984…) A1 100cos30 – 50sin20 (= 69.501…) A1 2 2 (56.984... 69.501... ) R = + or 1 56.984... tan 69.501... α −  =     M1 Method to find either R or α R=89.9 (89.876…) A1 α=39.3 (39.348…) A1 6

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4 A car travels along a straight road with constant acceleration. It passes through points P, Q, R and S. The times taken for the car to travel from P to Q, Q to R and R to S are each equal to 10 s. The distance QR is 1.5 times the distance PQ. At point Q the speed of the car is 20 m s−1. (i) Show that the acceleration of the car is 0.8 m s−2. 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(ii) Find the distance QS and hence find the average speed of the car between Q and S. 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Mark scheme: 4(i) 2 20 10 0.5 10 PQ s a = × − × or 2 20 10 0.5 10 QR s a = × + × M1 For use of 2 1 2 = − s vt at or 2 1 2 = + s ut at OE suvat to find PQ or QR s = 200–50a and 1.5s = 200 + 50a A1 OE 1.5(200 – 50a) = 200 + 50a Æ 100 = 125a Æ a = 0.8 ms–2 B1 AG 3 4(ii) Distance 2 1 20 20 0.8 20 2 = × + × × QS M1 Using 2 1 2 = + s ut at Distance=560 m A1 Average speed between Q and 1 560 28ms 20 − = = S B1 3

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5 A cyclist is travelling along a straight horizontal road. The total mass of the cyclist and his bicycle is 80 kg. His power output is a constant 240 W. His acceleration when he is travelling at 6 m s−1 is 0.3 m s−2. (i) Show that the resistance to the cyclist’s motion is 16 N. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the steady speed that the cyclist can maintain if his power output and the resistance force are both unchanged. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) The cyclist later ascends a straight hill inclined at 3Å to the horizontal. His power output and the resistance force are still both unchanged. Find his acceleration when he is travelling at 4 m s−1. 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Mark scheme: 5(i) Driving force 240 6 = (= 40 N) B1 [40 – R = 80 × 0.3] M1 Use of Newton’s Second Law (3 terms) Resistance is 16 N A1 AG 3 5(ii) 240 16   =     v M1 Use of P=Fv with DF=resistance Steady speed is 15 ms–1 A1 2 5(iii) Use of Newton’s Second Law M1 (4 terms) 240 16 80 sin3 80 4 − − = g a A1 Acceleration is 0.0266 ms–2 A1 3

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7 A B m kg km kg 0.81 m Two particles A and B have masses m kg and km kg respectively, where k > 1. The particles are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley and the particles hang vertically below it. Both particles are at a height of 0.81 m above horizontal ground (see diagram). The system is released from rest and particle B reaches the ground 0.9 s later. The particle A does not reach the pulley in its subsequent motion. (i) Find the value of k and show that the tension in the string before B reaches the ground is equal to 12m N. [7] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ At the instant when B reaches the ground, the string breaks. (ii) Show that the speed of A when it reaches the ground is 5.97 m s−1, correct to 3 significant figures, and find the time taken, after the string breaks, for A to reach the ground. 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(iii) Sketch a velocity-time graph for the motion of particle A from the instant when the system is released until A reaches the ground. [2]

Mark scheme: 7(i) 2 1 0.81 0 0.9 2   = + × ×     a M1 For use of 2 1 2 = + s ut at a = 2 A1 T – mg = ma or kmg – T = kma M1 Use of Newton’s Second Law for A or B or use of ( ) ( ) − = + B A B A m m g a m m T – mg = ma and kmg – T = kma or ( ) ( )   − =   +     km m g a km m A1 ( ) ( ) 2 ... 1 kg g a k k − = = → = + M1 Solves to find k k = 1.5 A1 T = 10m + 2m = 12m N B1 AG 7 7(ii) Velocity of A when string breaks = 2 × 0.9 (=1.8 ms–1 upwards) B1FT For use of v=u+at ft a from (i) v2=1.82+2g×1.62 → v=... M1 For use of suvat to find vA at ground Speed is 5.97 ms–1 A1 AG Time taken ( ) 1.8 5.97 0.777 + = = s g (0.7769...) B1 4 Question Answer Marks Guidance 7(iii) Straight line from (0, 0) to (0.9, 1.8) B1 Straight line from (0.9, 1.8) to approx. (1.7, –6) B1FT FT 0.9 + t from (ii) for 1.7 2

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