Cambridge A Level Mathematics 9709 — 2012 May/June Paper 4 · Variant 3
9709/43/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 · 6 N 24° 0.5 m s–1 A ring is threaded on a fixed horizontal bar
1 6 N 24° 0.5 m s–1 A ring is threaded on a fixed horizontal bar. The ring is attached to one end of a light inextensible string which is used to pull the ring along the bar at a constant speed of 0.5 m s−1. The string makes a constant angle of 24◦with the bar and the tension in the string is 6 N (see diagram). Find the work done by the tension in a period of 8 s. [3]
Mark scheme: 1 M1 For using WD = Fdcosα WD = 6 × (0.5 × 8)cos24o A1 Work done is 21.9 J A1 [3]
Q2 · A q° R 11.2 N B A smooth ring R of mass 0.16 kg is threaded on a light inextensible string
2 A q° R 11.2 N B A smooth ring R of mass 0.16 kg is threaded on a light inextensible string. The ends of the string are attached to fixed points A and B. A horizontal force of magnitude 11.2 N acts on R, in the same vertical plane as A and B. The ring is in equilibrium. The string is taut with angle ARB = 90◦, and the part AR of the string makes an angle of θ◦with the horizontal (see diagram). The tension in the string is T N. (i) Find two simultaneous equations involving T sin θ and T cos θ. [3] (ii) Hence find T and θ. [3]
Mark scheme: 2 (i) M1 For resolving forces horizontally or vertically Tcosθ + Tsinθ = 11.2 A1 (or – Tcosθ + Tsinθ = 0.16g) – Tcosθ + Tsinθ = 0.16g (or Tcosθ + Tsinθ = 11.2) A1 [3] (ii) [Tcosθ = 4.8 and Tsinθ = 6.4 and For finding Tcosθ and Tsinθ and hence T2 = 4.82 + 6.42 or tanθ = 6.4/4.8] finding T or θ , OR [4T2(cos2θ + sin2θ ) = for finding the value of (11.2 – 1.6)2 + (11.2 + 1.6)2 4T2(cos2θ + sin2θ ) or of or 2Tsinθ ÷ 2Tcosθ = 2Tsinθ ÷ 2Tcosθ or of (11.2 + 1.6) ÷ (11.2 – 1.6) (Tcosθ + Tsinθ ) ÷ (– Tcosθ + Tsinθ ) or (Tcosθ + Tsinθ ) ÷ (– Tcosθ + Tsinθ ) = 11.2 ÷ 1.6] M1 T = 8 (or θ = 53.1) A1 θ = 53.1 or T = 8 A1 [3] F i ∫ d
Q3 · A particle P travels from a point O along a straight line and comes to instantaneous rest…
3 A particle P travels from a point O along a straight line and comes to instantaneous rest at a point A. The velocity of P at time t s after leaving O is v m s−1, where v = 0.027(10t2 −t3). Find (i) the distance OA, [4] (ii) the maximum velocity of P while moving from O to A. [3]
Mark scheme: 3 (i) M1 For using s = v∫dt s = 0.027(10t3/3 – t4/4) (+C) A1 s = 0.027[10 000/3 – 10000/4] DM1 For finding the value of t at A and using limits or equivalent Distance is 22.5 m A1 [4] (ii) [0.027(20t – 3t2) = 0 t = 20/3]] M1 For using dv/dt = 0 vmax = 0.027(4000/9 – 8000/27) A1ft ft incorrect t in 0.027(10t2 – t3) Maximum speed is 4 ms–1 A1 [3] GCE AS/A LEVEL – May/June 2012 9709 43
Q4 · A car of mass 1230 kg increases its speed from 4 m s−1 to 21 m s−1 in 24.5 s
4 A car of mass 1230 kg increases its speed from 4 m s−1 to 21 m s−1 in 24.5 s. The table below shows corresponding values of time t s and speed v m s−1. t 0 0.5 16.3 24.5 v 4 6 19 21 (i) Using the values in the table, find the average acceleration of the car for 0 < t < 0.5 and for 16.3 < t < 24.5. [2] While the car is increasing its speed the power output of its engine is constant and equal to P W, and the resistance to the car’s motion is constant and equal to R N. (ii) Assuming that the values obtained in part (i) are approximately equal to the accelerations at v = 5 and at v = 20, find approximations for P and R. [5]
Mark scheme: 4 (i) [When 4 < v < 6, aave = (6 – 4)/(0.5 – 0); ∆v For using a ≈ when 19 < v <21 ∆t aave = (21 – 19)/(24.5 – 16.3)] M1 Average accelerations are 4 ms–2 and 0.244 ms–2 A1 [2] (ii) DF(5) = P/5 and DF(20) = P/20 B1 [DF – R = ma] M1 For using Newton’s 2nd law P/5 – R = 1230 × 4 and A1ft ft incorrect average a values P/20 – R = 1230 × 0.244 P = 30800 (or R = 1240) B1 R = 1240 (or P = 30800) B1ft [5] ft P/5 – 1230a1 or P/20 – 1230a2 or 5(1230a1 + R) or 20(1230a2 + R)
Q5 · A lorry of mass 16 000 kg moves on a straight hill inclined at angle α◦to the horizontal
5 A lorry of mass 16 000 kg moves on a straight hill inclined at angle α◦to the horizontal. The length of the hill is 500 m. (i) While the lorry moves from the bottom to the top of the hill at constant speed, the resisting force acting on the lorry is 800 N and the work done by the driving force is 2800 kJ. Find the value of α. [4] (ii) On the return journey the speed of the lorry is 20 m s−1 at the top of the hill. While the lorry travels down the hill, the work done by the driving force is 2400 kJ and the work done against the resistance to motion is 800 kJ. Find the speed of the lorry at the bottom of the hill. [4]
Mark scheme: 5 (i) WD against resistance = 800 × 500 B1 [2 800 000 = PE gain + 400 000] M1 For using WD by the driving force = PE gain + WD against resistance [2 400 000 = 16000g × 500sinα ] M1 For using PE gain = mgLsinα α = 1.7 A1 [4] (ii) [KE gain = 2 400 000 + 2 400 000 – M1 For using KE gain = WD by the driving 800 000] force + PE loss – WD against resistance 4000 000 J A1ft ft PE gain [ ½ 16000(v2 – 202) = 4 000 000] M1 For KE gain = ½ m(v2 – 202) and attempting to solve for v Speed is 30 ms–1 A1 [4] SR (max 2/4) for candidates who assume constant driving force and constant resistance without justification Uses Newton’s Second Law and v2 = u2 + 2as [4800 + 16000gsin α – 1600 = 16000a, v2 = 202 + 2a × 500)] M1 Speed is 30 ms–1 A1 Alternative Method for Part (i) (i) Driving force = 2800 000 ÷ 500 B1 [DF – mgsinα – R = m × 0] M1 For using Newton’s second law [16000 × 10sinα = 5600 – 800] DM1 For solving the resultant equation for α A1 α = 1.7 [4] GCE AS/A LEVEL – May/June 2012 9709 43
Q6 · 5.9 N 5.9 N a a Fig
6 5.9 N 5.9 N a a Fig. 1 Fig. 2 A block of weight 6.1 N is at rest on a plane inclined at angle α to the horizontal, where tan α = 1160. The coefficient of friction between the block and the plane is µ. A force of magnitude 5.9 N acting parallel to a line of greatest slope is applied to the block. (i) When the force acts up the plane (see Fig. 1) the block remains at rest. Show that µ ≥45. [5] (ii) When the force acts down the plane (see Fig. 2) the block slides downwards. Show that µ < 76. [2] (iii) Given that the acceleration of the block is 1.7 m s−2 when the force acts down the plane, find the value of µ. [2] [Question 7 is printed on the next page.]
Mark scheme: 6 (i) M1 For resolving forces parallel to the plane F = 5.9 – 6.1 sinα A1 R = 6.1cosα B1 [5.9 – 6.1 sinα ≤ µ (6.1cosα ) ] M1 For using F ≤ µR 4 µ > A1 [5] AG 5 (ii) [6.1 × (11/61) + 5.9 – µ6.1 × (60/61) > 0] M1 For using F = µR and ‘net downward force > 0’ 7 µ < A1 [2] AG 6 (iii) [6.1 × (11/61) + 5.9 – µ 6.1 × (60/61) = For using Newton’s 2nd law and F = µR 0.61 × 1.7] M1 µ = 0.994 A1 [2]
Q7 · A B 0.65 m Two particles A and B have masses 0.12 kg and 0.38 kg respectively
7 A B 0.65 m Two particles A and B have masses 0.12 kg and 0.38 kg respectively. The particles are attached to the ends of a light inextensible string which passes over a fixed smooth pulley. A is held at rest with the string taut and both straight parts of the string vertical. A and B are each at a height of 0.65 m above horizontal ground (see diagram). A is released and B moves downwards. Find (i) the acceleration of B while it is moving downwards, [2] (ii) the speed with which B reaches the ground and the time taken for it to reach the ground. [3] B remains on the ground while A continues to move with the string slack, without reaching the pulley. The string remains slack until A is at a height of 1.3 m above the ground for a second time. At this instant A has been in motion for a total time of T s. (iii) Find the value of T and sketch the velocity-time graph for A for the first T s of its motion. [3] (iv) Find the total distance travelled by A in the first T s of its motion. [2]
Mark scheme: 7 (i) For using Newton’s second law [T – 0.12g = 0.12a & 0.38g – T = 0.38a; M − m for A and B or for using a = g .038 − .012 M + m a = g ] M1 .038 + .012 Acceleration is 5.2 ms–2 A1 [2] (ii) [v2 = 2 × 5.2 × 0.65; 0.65 = ½ 5.2TB2] M1 For using v2 = 2ah or s = ½ at2 Speed of B is 2.6ms–1 or TB = 0.5 A1ft ft incorrect a TB = 0.5 or Speed of B is 2.6ms–1 B1 [3] (iii) [– 2.6 = 2.6 – 10(T – 0.5)] M1 For using –V = V – g(T – TB) or equivalent T = 1.02 A1ft ft incorrect V and/or TB Correct graph for 0 < t < 1.02 B1ft [3] ft incorrect values of V, T and TB 0.5 1.02 (iv) [0.65 + 0.5(1.02 – 0.5)2.6] M1 For using ‘total distance T A − T B = ½ (VTB) + 2 x ½ V 2 Total distance is 1.326 m (accept 1.33) A1 [2]
What was in this paper
The subtopics covered by these 7 questions, and how many questions each got. Open one in a new tab to see every Cambridge question on it.
What you needed in this session
Cambridge’s own grade thresholds for 2012 May/June, Paper 4 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.