Cambridge A Level Mathematics 9709 — 2020 May/June Paper 4 · Variant 3

1 Particles P of mass m kg and Q of mass 0.2 kg are free to move on a smooth horizontal plane. P is projected at a speed of 2 m s−1 towards Q which is stationary. After the collision P and Q move in opposite directions with speeds of 0.5 m s−1 and 1 m s−1 respectively. Find m. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Use of conservation of momentum M1 m × 2 + 0 = m × (–0.5) + 0.2 × 1 A1 m = 0.08 A1 3

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2 A minibus of mass 4000 kg is travelling along a straight horizontal road. The resistance to motion is 900 N. (a) Find the driving force when the acceleration of the minibus is 0.5 m s−2. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the power required for the minibus to maintain a constant speed of 25 m s−1. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) F – 900 = 4000 × 0.5 (M1 for use of Newton’s second law, 3 terms) M1 F = 2900 N A1 2(b) 900 × 25 (M1 for use of P = Fv with F = resistance only) M1 22 500 W or 22.5 kW A1

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3 40 N F N !Å 60Å 45Å 30Å 20 N 50 N Four coplanar forces of magnitudes 40 N, 20 N, 50 N and F N act at a point in the directions shown in the diagram. The four forces are in equilibrium. Find F and !. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Attempt to resolve, either direction with correct number of terms M1 Fcosα = 40sin30 + 20sin60 – 50sin45 (= 1.965...) A1 Fsinα = 50cos45 + 20cos60 – 40cos30 (= 10.714...) A1 Method for either F or α M1 ( ) ( ) ( ) ( ) 2 2 1.965... 10.714... 10.9 10.893 F = + = A1 α = tan–1(10.714... / 1.965...) = 79.6 (79.606...) A1 6

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4 A car starts from rest and moves in a straight line with constant acceleration a m s−2 for a distance of 50 m. The car then travels with constant velocity for 500 m for a period of 25 s, before decelerating to rest. The magnitude of this deceleration is 2a m s−2. (a) Sketch the velocity-time graph for the motion of the car. [1] v (m s−1) t (s) (b) Find the value of a. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the total time for which the car is in motion. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Trapezium shape with gradient of right-hand side approximately 2 times left side B1 1 4(b) Constant velocity = 500/25 = 20 ms–1 B1 202 = 0 + 2a × 50 M1 a = 4 A1 3 4(c) Time to accelerate = 20/4 = 5 s B1 Deceleration time = 2.5 s B1 So total time = 5 + 25 + 2.5 = 32.5 s B1 3

More questions on Kinematics of motion in a straight line

5 A block B of mass 4 kg is pushed up a line of greatest slope of a smooth plane inclined at 30Å to the horizontal by a force applied to B, acting in the direction of motion of B. The block passes through points P and Q with speeds 12 m s−1 and 8 m s−1 respectively. P and Q are 10 m apart with P below the level of Q. (a) Find the decrease in kinetic energy of the block as it moves from P to Q. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find the work done by the force pushing the block up the slope as the block moves from P to Q. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) At the instant the block reaches Q, the force pushing the block up the slope is removed. Find the time taken, after this instant, for the block to return to P. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Decrease in KE = 1 2 × 4 × (122 – 82) 160 J A1 2 5(b) PE gained = 4g × 10sin30 (= 200) B1 Total work done = 200 – 160 M1 Total work done = 40 J A1 FT 3 5(c) –4gsin30 = 4a M1 a = –5 A1 –10 = 8t – 1 2 × 5t2 M1 t = 4.16 s A1 4

More questions on Kinematics of motion in a straight line

6 A particle travels in a straight line PQ. The velocity of the particle t s after leaving P is v m s−1, where v = 4.5 + 4t −0.5t2. (a) Find the velocity of the particle at the instant when its acceleration is zero. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The particle comes to instantaneous rest at Q. (b) Find the distance PQ. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) (M1 for differentiation) M1 When a = 0, t = 4 A1 At t = 4, v = 12.5 A1 3 6(b) Velocity = 0 when 4.5 + 4t – 0.5t2 = 0 M1 t = 9 (reject t = –1) A1 2 (4.5 4 0.5 ) t t dt + −  M1 2 3 1 4.5 2 6 t t t + − [+ c] A1 Apply limits (0 and 9) M1 Distance = 81 m A1 6

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7 A 3m kg 1 2m kg B 0.8 m Two particles A and B, of masses 3m kg and 2m 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 edge of a plane. The plane is inclined at an angle 1 to the horizontal. A lies on the plane and B hangs vertically, 0.8 m above the floor, which is horizontal. The string between A and the pulley is parallel to a line of greatest slope of the plane (see diagram). Initially A and B are at rest. (a) Given that the plane is smooth, find the value of 1 for which A remains at rest. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ It is given instead that the plane is rough, 1 = 30Å and the acceleration of A up the plane is 0.1 m s−2. (b) Show that the coefficient of friction between A and the plane is 1 3. [5] 10 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) When B reaches the floor it comes to rest. Find the length of time after B reaches the floor for which A is moving up the plane. [You may assume that A does not reach the pulley.] [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) B1 3mg sin θ – T = 0 (M1 for resolving forces parallel to the plane and solving for θ) M1 θ = 41.8 (41.810...) A1 3 7(b) R = 3mgcos30 B1 Use of F = μR M1 2mg – T = 0.1 × 2m OR T – 3mg sin30 –μ × 3mg cos30 = 0.1 × 3m M1 2mg – 0.2m – 3mg sin30 – μ × 3mg cos30 = 0.1 × 3m M1 3 10 μ = A1 5 7(c) v2 = 0 + 2 × 0.1 × 0.8 (v = 0.4) M1 –3mg sin30 – μ × 3mg cos30 = 3ma (a = –6.5) M1 0 = –0.4 – 6.5t M1 t = 0.4/6.5 = 0.0615 s A1 4

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