Cambridge A Level Mathematics 9709 — 2020 May/June Paper 4 · Variant 1
9709/41/M/J/20 · 7 questions · 50 marks · ≈56 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper16 pages
















Mark scheme12 pages
Answers below. Sit the paper first if you are practising.












Questions as text
Q1 · 50 N A !
1 50 N A ! 100 N ! 50 N Three coplanar forces of magnitudes 100 N, 50 N and 50 N act at a point A, as shown in the diagram. The value of cos ! is 5.4 Find the magnitude of the resultant of the three forces and state its direction. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................
Mark scheme: 1 Resultant = 100 – 2 × 50cos α M1 20 N A1 Direction is to the left (or equivalent) B1 3
Q2 · A car of mass 1800 kg is towing a trailer of mass 400 kg along a straight horizontal road
2 A car of mass 1800 kg is towing a trailer of mass 400 kg along a straight horizontal road. The car and trailer are connected by a light rigid tow-bar. The car is accelerating at 1.5 m s−2. There are constant resistance forces of 250 N on the car and 100 N on the trailer. (a) Find the tension in the tow-bar. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the power of the engine of the car at the instant when the speed is 20 m s−1. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Mark scheme: 2(a) [T – 100 = 400 × 1.5] M1 T = 700 N A1 2 2(b) F – 250 – 100 = 2200 × 1.5 (F = 3650 N) (M1 for using Newton’s second law for the system or for the car using the result from 2(a)) M1 For use of power =Fv M1 73 000 W or 73 kW A1 3
Q3 · A particle P is projected vertically upwards with speed 5 m s−1 from a point A which is…
3 A particle P is projected vertically upwards with speed 5 m s−1 from a point A which is 2.8 m above horizontal ground. (a) Find the greatest height above the ground reached by P. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the length of time for which P is at a height of more than 3.6 m above the ground. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Mark scheme: 3(a) s = 1.25 A1 [Height above ground =] 4.05 m A1 3 3(b) Use of s = ut + ½ at2 M1 0.8 = 5t – 5t2 A1 t = 0.2 or 0.8 M1 Length of time = 0.6 s A1 4
Q4 · 30Å T N The diagram shows a ring of mass 0.1 kg threaded on a fixed horizontal rod
4 30Å T N The diagram shows a ring of mass 0.1 kg threaded on a fixed horizontal rod. The rod is rough and the coefficient of friction between the ring and the rod is 0.8. A force of magnitude T N acts on the ring in a direction at 30Å to the rod, downwards in the vertical plane containing the rod. Initially the ring is at rest. (a) Find the greatest value of T for which the ring remains at rest. 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(b) Find the acceleration of the ring when T = 3. 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Mark scheme: 4(a) Resolving forces in either direction M1 R = T sin30 + 0.1g, F = T cos30 A1 T cos30 = 0.8 (T sin30 + 0.1g) M1 T = 1.72 (1.7166...) A1 4 4(b) R = 3sin30 + 0.1g B1 3 cos30 – 0.8(3sin30 + 0.1g) = 0.1a M1 a = 5.98 ms–2 (5.9807...) A1 3
Q5 · 45Å 4 m P A child of mass 35 kg is swinging on a rope
5 45Å 4 m P A child of mass 35 kg is swinging on a rope. The child is modelled as a particle P and the rope is modelled as a light inextensible string of length 4 m. Initially P is held at an angle of 45Å to the vertical (see diagram). (a) Given that there is no resistance force, find the speed of P when it has travelled half way along the circular arc from its initial position to its lowest point. 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(b) It is given instead that there is a resistance force. The work done against the resistance force as P travels from its initial position to its lowest point is X J. The speed of P at its lowest point is 4 m s−1. Find X. 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Mark scheme: 5(a) Attempt at finding PE lost M1 PE lost = 35g (4cos22.5 – 4cos45) A1 1 2 × 35v2 = 35g (4cos22.5 – 4cos45) M1 Speed = 4.16 ms–1 (4.1643...) A1 4 5(b) Use of the work-energy equation in the form: PE lost = KE gain + WD against resistance M1 1 2 × 35 × 42 = 35g (4 – 4cos45) – X A1 X = 130 (130.05...) A1 3
Q6 · A particle moves in a straight line AB
6 A particle moves in a straight line AB. The velocity v m s−1 of the particle t s after leaving A is given by v = k t2 −10t + 21 , where k is a constant. The displacement of the particle from A, in the direction towards B, is 2.85 m when t = 3 and is 2.4 m when t = 6. (a) Find the value of k. Hence find an expression, in terms of t, for the displacement of the particle from A. 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(b) Find the displacement of the particle from A when its velocity is a minimum. 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Mark scheme: 6(a) ( ) 2 10 21 d k t t t − + M1 3 2 1 5 21 C 3 s k t t t = + + + A1 3 2 1 2.85 3 5 3 21 3 C 3 k = × −× + × + or 3 2 1 2.4 6 5 6 21 6 C 3 k = × −× + × + M1 2.85 = 27k + C, 2.4 = 18k + C (A1 for both) A1 Solving for k M1 k = 0.05 A1 3 2 1 0.05 5 21 1.5 3 s t t t = − + + A1 7 6(b) Differentiating v or completing the square for v M1 a = 0.05(2t –10) A1 Min value of v is at t = 5. M1 Displacement at t = 5 is 2.58 m (2.5833...) A1 4
Q7 · 0.3 kg P 2.5 m Q 0.2 kg 30Å 1.5 m A particle P of mass 0.3 kg, lying on a smooth plane…
7 0.3 kg P 2.5 m Q 0.2 kg 30Å 1.5 m A particle P of mass 0.3 kg, lying on a smooth plane inclined at 30Å to the horizontal, is released from rest. P slides down the plane for a distance of 2.5 m and then reaches a horizontal plane. There is no change in speed when P reaches the horizontal plane. A particle Q of mass 0.2 kg lies at rest on the horizontal plane 1.5 m from the end of the inclined plane (see diagram). P collides directly with Q. (a) It is given that the horizontal plane is smooth and that, after the collision, P continues moving in the same direction, with speed 2 m s−1. Find the speed of Q after the collision. 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(b) It is given instead that the horizontal plane is rough and that when P and Q collide, they coalesce and move with speed 1.2 m s−1. Find the coefficient of friction between P and the horizontal plane. 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Mark scheme: 7(a) (M1 for applying Newton’s second law parallel to the plane) M1 v2 = 0 + 2 × 2.5 × a M1 v = 5 A1 0.3 × 5 + 0 = 0.3 × 2 + 0.2 w M1 Velocity of Q = 4.5 ms–1 A1 5 Question Answer Marks 7(b) 0.3 × z + 0 = 0.5 × 1.2 M1 Velocity of P before collision z = 2 A1 Friction force on P after reaches horizontal plane F = μ × 0.3 g B1 μ × 0.3g × 1.5 = 1 2 × 0.3 × 52 – 1 2 × 0.3 × 22 M1 Coefficient μ = 0.7 A1 Alternative method for question 7(b) 0.3 × z + 0 = 0.5 × 1.2 M1 Velocity of P before collision z = 2 A1 Friction force on P after reaches horizontal plane F = μ × 0.3 g B1 a = (52 – 22) / (2 × 1.5) = 7, F = 0.3 × 7 M1 Coefficient μ = 0.7 A1 5
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