Cambridge A Level Mathematics 9709 — 2016 May/June Paper 4 · Variant 1
9709/41/M/J/16 · 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 paper4 pages
Mark scheme7 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Q1 · A lift moves upwards from rest and accelerates at 0.9 m s−2 for 3 s
1 A lift moves upwards from rest and accelerates at 0.9 m s−2 for 3 s. The lift then travels for 6 s at constant speed and finally slows down, with a constant deceleration, stopping in a further 4 s. (i) Sketch a velocity-time graph for the motion. [3] (ii) Find the total distance travelled by the lift. [2]
Mark scheme: Part Qu Answer Marks Guidance Mark 1 (i) Trapezium seen B1 v–t graph with three straight lines, with positive, zero and negative gradients, continuous 0, 3, 9, 13 shown on the t axis B1 v = 2.7 soi in either part B1 [3] (ii) [0.5 × (6 + 13) × 2.7] M1 Using area of trapezium Total distance = 25.65 m A1 [2] Allow Distance = 513/20 m Alternative method for 1(ii) (ii) Stage 1 M1 Complete method to find the total s1 = 0.5 × 0.9 × 32 = 4.05 distance travelled by the lift using Stage 2 constant acceleration equations s2 = 2.7 × 6 = 16.2 for all three stages Stage 3 s3 = 0.5 × (2.7 + 0) × 4 = 5.4 Total distance = 25.65 m A1 [2]
Q2 · A box of mass 25 kg is pulled, at a constant speed, a distance of 36 m up a rough plane…
2 A box of mass 25 kg is pulled, at a constant speed, a distance of 36 m up a rough plane inclined at an angle of 20Å to the horizontal. The box moves up a line of greatest slope against a constant frictional force of 40 N. The force pulling the box is parallel to the line of greatest slope. Find (i) the work done against friction, [1] (ii) the change in gravitational potential energy of the box, [2] (iii) the work done by the pulling force. [2]
Mark scheme: 2 (i) WD = 40 × 36 = 1440 J B1 [1] (ii) M1 Using PE = mgh PE = 25 × g × 36 sin 20 = 3080 J A1 [2] [PE = 3078.18] (iii) WD by pulling force = M1 For using (i) + (ii) WD by pulling force = Gain in PE + WD against F WD = 4520 J A1 [2] [WD = 4518.18] Alternative for (iii) (iii) [(25g sin 20+ 40) × 36] M1 For attempting to find the pulling force and multiply it by 36 to find the work done WD = 4520 J A1 [2] [WD = 4518.18]
Q3 · A car of mass 1000 kg is moving along a straight horizontal road against resistances of…
3 A car of mass 1000 kg is moving along a straight horizontal road against resistances of total magnitude 300 N. (i) Find, in kW, the rate at which the engine of the car is working when the car has a constant speed of 40 m s−1. [3] (ii) Find the acceleration of the car when its speed is 25 m s−1 and the engine is working at 90% of the power found in part (i). [3]
Mark scheme: 3 (i) Driving Force = 300 B1 Using DF = Resistance P = 300 × 40 M1 Using P = Fv P = 12000 W = 12 kW A1 [3] Must give answer in kW (ii) P = 0.9 × 12000 = 10800 B1 ft on 12000 10 800 − 300 = 1000 a M1 Applying Newton’s second law 25 with 3 terms to the car a = 132/1000 = 0.132 ms–2 A1 [3]
Q4 · 50 N 1Å !Å 48 N P N 14 N Coplanar forces of magnitudes 50 N, 48 N, 14 N and P N act at a…
4 50 N 1Å !Å 48 N P N 14 N Coplanar forces of magnitudes 50 N, 48 N, 14 N and P N act at a point in the directions shown in the diagram. The system is in equilibrium. Given that tan ! = 24,7 find the values of P and 1. [6]
Mark scheme: 4 P cos θ = 48 cos α – 14 sin α M1 For resolving forces horizontally and/or and/or vertically P sin θ = 50 – 48 sin α –14 cos α P cos θ= 48(24/25) – 14(7/25) Allow α = 16.3 used throughout = 42.16 A1 P sin θ = 50 – 48(7/25) –14(24/25) = 23.12 A1 M1 For attempting to find P or θ P = 42.16 2 + 23.12 2 = 48.1 A1 Allow P = 34 2 23.12 tan θ = 42.16 θ = 28.7 B1 [6]
Q5 · 10 kg 5 kg !
5 10 kg 5 kg ! Two particles of masses 5 kg and 10 kg are connected by a light inextensible string that passes over a fixed smooth pulley. The 5 kg particle is on a rough fixed slope which is at an angle of ! to the horizontal, where tan ! = 34. The 10 kg particle hangs below the pulley (see diagram). The coefficient of friction between the slope and the 5 kg particle is 12. The particles are released from rest. Find the acceleration of the particles and the tension in the string. [7]
Mark scheme: 5 R = 5g cos α = 4g B1 For finding the normal reaction R F = 0.5 × 4g = 2g acting on the 5 kg particle and using F = µR M1 For applying Newton’s second law to one or both particles or to the system T – 2g – 5gsin α = 5a→ System equation is T – 5g = 5a A1 10g –5g sin α –2g = 5g = 15a 10g – T = 10a A1 [5g = 15a] M1 For eliminating T and solve for a a = g/3 = 3.33 ms–2 A1 T = 10g – 10(g/3) = 20g/3 = 66.7 N B1 [7]
Q6 · A particle P moves in a straight line
6 A particle P moves in a straight line. It starts at a point O on the line and at time t s after leaving O it has a velocity v m s−1, where v = 6t2 −30t + 24. (i) Find the set of values of t for which the acceleration of the particle is negative. [2] (ii) Find the distance between the two positions at which P is at instantaneous rest. [4] (iii) Find the two positive values of t at which P passes through O. [3]
Mark scheme: 6 (i) a = 12t – 30 M1 For differentiating v to find a t< 2.5 A1 [2] (ii) v = 0 at t = 1 and t = 4 B1 Using v = 6(t – 4)(t – 1) s = ∫ ( 6t 2 − 30t + 24 ) dt M1 For using integration to find s 6 3 30 2 = t − t + 24t 3 2 3 2 4 M1 For using limits s = 2t − 15t + 24t 1 Distance = 27 m A1 [4] (iii) 3 2 2t − 15t + 24t = 0 M1 State s = 0 2t 2 − 15t + 24 = 0 M1 Reduce to a quadratic and attempt to solve t = 2.31 and t = 5.19 A1 [3]
Q7 · A particle of mass 30 kg is on a plane inclined at an angle of 20Å to the horizontal
7 A particle of mass 30 kg is on a plane inclined at an angle of 20Å to the horizontal. Starting from rest, the particle is pulled up the plane by a force of magnitude 200 N acting parallel to a line of greatest slope. (i) Given that the plane is smooth, find (a) the acceleration of the particle, [2] (b) the change in kinetic energy after the particle has moved 12 m up the plane. [2] (ii) It is given instead that the plane is rough and the coefficient of friction between the particle and the plane is 0.12. (a) Find the acceleration of the particle. [4] (b) The direction of the force of magnitude 200 N is changed, and the force now acts at an angle of 10Å above the line of greatest slope. Find the acceleration of the particle. [4]
Mark scheme: 7 (i) (a) 200 – 30g sin 20 = 30a M1 For applying Newton’s second law with 3 terms parallel to the plane a = 3.25 ms–2 A1 [2] [a = 3.2465] (b) [v2 = 2 × 3.2465 × 12 = 77.9] M1 For using v2 = u2 + 2as and attempting to find KE change KE change = 0.5 × 30 × 77.9 = 1170 J A1 [2] [KE = 1168.7 J] Alternative method for 7(i)(b) (b) KE change = M1 Using KE gain = 200 × 12 – 30g × 12 sin 20 WD by DF – PE gain KE change = 1170 J A1 [2] (ii) (a) N = 30g cos 20 B1 [N = 281.9] F = 0.12 × 30g cos 20 [= 33.8] M1 Using F = µNa 200 – 30g sin 20 – 33.8 = 30a M1 For using Newton’s second law with 4 terms applied to the particle a = 2.12 ms–2 A1 [4] (b) N + 200 sin 10 = 30g cos 20 M1 For resolving forces [N = 247.2] perpendicular to the plane. Three term equation. F = 0.12 N [= 0.12 × 247.2 = 29.66] M1 N must be from a 3 term equation 200 cos 10 – 29.66 – 30g sin 20 = 30a M1 For using Newton’s second law with 4 terms applied to the particle a = 2.16 ms–2 A1 [4]
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