Cambridge A Level Mathematics 9709 — 2012 May/June Paper 4 · Variant 1
9709/41/M/J/12 · 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 scheme6 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Q1 · A car of mass 880 kg travels along a straight horizontal road with its engine working at…
1 A car of mass 880 kg travels along a straight horizontal road with its engine working at a constant rate of P W. The resistance to motion is 700 N. At an instant when the car’s speed is 16 m s−1 its acceleration is 0.625 m s−2. Find the value of P. [4]
Mark scheme: 1 M1 For using Newton’s 2nd law DF – 700 = 880 × 0.625 A1 [P = 1250 × 16] M1 For using P = (DF)v P = 20 000 A1 [4]
Q2 · 13 N q° 14 N O Forces of magnitudes 13 N and 14 N act at a point O in the directions…
2 13 N q° 14 N O Forces of magnitudes 13 N and 14 N act at a point O in the directions shown in the diagram. The resultant of these forces has magnitude 15 N. Find (i) the value of θ, [3] (ii) the component of the resultant in the direction of the force of magnitude 14 N. [2]
Mark scheme: 2 (i) X = 14 – 13cosθ and Y = 13sinθ or triangle B1 with sides 13, 14, 15 and θ opposite 15 [142 + 132 – 2 × 13 × 14cosθ = 152] M1 For using X2 + Y2 = R2 or cosine rule θ = 67.4 A1 [3] (ii) M1 For evaluating X or 15cos[tan–1(Y/X)] Component is 9 N A1ft [2]
Q3 · 1.25 m s–1 A 160 kg 20 m O A load of mass 160 kg is pulled vertically upwards, from rest…
3 1.25 m s–1 A 160 kg 20 m O A load of mass 160 kg is pulled vertically upwards, from rest at a fixed point O on the ground, using a winding drum. The load passes through a point A, 20 m above O, with a speed of 1.25 m s−1 (see diagram). Find, for the motion from O to A, (i) the gain in the potential energy of the load, [1] (ii) the gain in the kinetic energy of the load. [2] The power output of the winding drum is constant while the load is in motion. (iii) Given that the work done against the resistance to motion from O to A is 20 kJ and that the time taken for the load to travel from O to A is 41.7 s, find the power output of the winding drum. [3]
Mark scheme: 3 (i) PE gain is 32 000 J B1 [1] (ii) [KE gain = ½ 160 × 1.252] M1 For using KE gain = ½ mv2 KE gain is 125 J A1 [2] (iii) WD by drum = 32 000 + 125 + 20 000 B1ft [P = 52 125 ÷ 41.7] M1 For using P = ∆(WD) ÷ ∆ T Power is 1250 W A1 [3] 2
Q4 · A particle P starts at the point O and travels in a straight line
4 A particle P starts at the point O and travels in a straight line. At time t seconds after leaving O the velocity of P is v m s−1, where v = 0.75t2 −0.0625t3. Find (i) the positive value of t for which the acceleration is zero, [3] (ii) the distance travelled by P before it changes its direction of motion. [5]
Mark scheme: 4 (i) [a = 1.5t – 0.1875t2] M1 For using a = dv/dt [0.1875t(8 – t) = 0] DM1 For attempting to solve dv/dt = 0 Acceleration is zero when t = 8 A1 [3] (ii) Changes direction when t = 12 B1 M1 For using s = ∫ vdt s = 0.25t3 – 0.0625t4 ÷ 4 (+ C) A1 [s = 0.25 × 1728 – 0.0625 × 20736 ÷ 4] DM1 For using limits 0 to (12) or equivalent Distance is 108 m A1 [5] GCE AS/A LEVEL – May/June 2012 9709 41
Q5 · O 10 m A 10 m a B The diagram shows the vertical cross-section OAB of a slide
5 O 10 m A 10 m a B The diagram shows the vertical cross-section OAB of a slide. The straight line AB is tangential to the curve OA at A. The line AB is inclined at α to the horizontal, where sin α = 0.28. The point O is 10 m higher than B, and AB has length 10 m (see diagram). The part of the slide containing the curve OA is smooth and the part containing AB is rough. A particle P of mass 2 kg is released from rest at O and moves down the slide. (i) Find the speed of P when it passes through A. [3] The coefficient of friction between P and the part of the slide containing AB is 12.1 Find (ii) the acceleration of P when it is moving from A to B, [3] (iii) the speed of P when it reaches B. [2]
Mark scheme: 5 (i) PE loss = 2g(10 – 10 × 0.28) B1 [ ½ 2v2 = 144] M1 For using ½ mv2 = PE loss Speed is 12 ms–1 A1 [3] (ii) R = 2g x 0.96 B1 [2g × 0.28 – 2g × 0.96 ÷ 12 = 2a] M1 For using Newton’s 2nd law Acceleration is 2 ms–1 A1 [3] (iii) [v2 = 122 + 2 × 2 × 10] M1 For using v2 = u2 + 2as Speed is 13.6 ms–1 A1 [2] d
Q6 · P Q q q Particles P and Q, of masses 0.6 kg and 0.4 kg respectively, are attached to the…
6 P Q q q Particles P and Q, of masses 0.6 kg and 0.4 kg respectively, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of a vertical cross-section of a triangular prism. The base of the prism is fixed on horizontal ground and each of the sloping sides is smooth. Each sloping side makes an angle θ with the ground, where sin θ = 0.8. Initially the particles are held at rest on the sloping sides, with the string taut (see diagram). The particles are released and move along lines of greatest slope. (i) Find the tension in the string and the acceleration of the particles while both are moving. [5] The speed of P when it reaches the ground is 2 m s−1. On reaching the ground P comes to rest and remains at rest. Q continues to move up the slope but does not reach the pulley. (ii) Find the time taken from the instant that the particles are released until Q reaches its greatest height above the ground. [4]
Mark scheme: 6 (i) M1 For using Newton’s 2nd law for P or for Q; or for using (M – m)g × 0.8 = (M + m)a 0.6g × 0.8 – T = 0.6a and T – 0.4g × 0.8 = 0.4a or (0.6 – 0.4)g × 0.8 = (0.6 + 0.4)a A1 M1 For solving for T or for a Tension is 3.84 N or acceleration is 1.6ms–2 A1 Acceleration is 1.6 ms–2 or tension is 3.84 N A1 [5] (ii) 2 = 1.6t1 (t1 = 1.25) B1ft M1 For using 0 + u + at with a = –0.8g 0 = 2 – 0.8gt2 (t2 = 0.25) A1 Time taken in 1.5 s A1ft [4] ft incorrect acceleration in (i) GCE AS/A LEVEL – May/June 2012 9709 41
Q7 · C 2 m 8 N B 1.5 m A A small ring of mass 0.2 kg is threaded on a fixed vertical rod
7 C 2 m 8 N B 1.5 m A A small ring of mass 0.2 kg is threaded on a fixed vertical rod. The end A of a light inextensible string is attached to the ring. The other end C of the string is attached to a fixed point of the rod above A. A horizontal force of magnitude 8 N is applied to the point B of the string, where AB = 1.5 m and BC = 2 m. The system is in equilibrium with the string taut and AB at right angles to BC (see diagram). (i) Find the tension in the part AB of the string and the tension in the part BC of the string. [5] The equilibrium is limiting with the ring on the point of sliding up the rod. (ii) Find the coefficient of friction between the ring and the rod. [5]
Mark scheme: 7 (i) M1 For resolving forces vertically and horizontally at B TC × (2/2.5) – TA × (1.5/2.5) = 0 A1 TC × (1.5/2.5) + TA × (2/2.5) = 8 A1 [0.6 TC + 0.8 (4TC/3) = 8 → (5/3) TC = 8 or For eliminating TA or TC and attempting 0.6(0.75TA) + 0.8TA = 8 → 1.25TA = 8 ] M1 to find TC or TA Tension in AB is 6.4 N; tension in BC is 4.8 N A1 [5] (ii) M1 For resolving forces vertically F + 0.2 g = TA × (1.5/2.5) A1 N = TA × (2/2.5) B1 [ µ = (3.84 – 2 )/5.12] M1 For using µ = F/N with F vertical and N horizontal Coefficient is 0.359 A1 [5] Accept 0.36
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