Cambridge A Level Mathematics 9709 — 2011 May/June Paper 4 · Variant 2

1 A load is pulled along horizontal ground for a distance of 76 m, using a rope. The rope is inclined at 5◦above the horizontal and the tension in the rope is 65 N. (i) Find the work done by the tension. [2] At an instant during the motion the velocity of the load is 1.5 m s−1. (ii) Find the rate of working of the tension at this instant. [2]

Mark scheme: 1 (i) [WD = 65 × 76cos5o] M1 For using WD = Tdcosα Work done is 4920 J A1 [2] (ii) [P = 65 cos5o × 1.5] M1 For using P = Tvcosα Rate of working is 97.1 W A1ft [2] ft for the value of ans(i) × 1.5÷76 SR for candidates who assume without justification that the speed is constant (max 1/2) t = 76 ÷ 1.5 = 50.6…s rate = WD/t = 4960÷50.6.. = 97.1W B1

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2 An object of mass 8 kg slides down a line of greatest slope of an inclined plane. Its initial speed at the top of the plane is 3 m s−1 and its speed at the bottom of the plane is 8 m s−1. The work done against the resistance to motion of the object is 120 J. Find the height of the top of the plane above the level of the bottom. [4]

Mark scheme: 2 M1 For using ‘loss of PE = gain in KE + WD against resistance’ PE loss = ½ 8(82 – 32) + 120 (= 340 J) A1 [340 = 8gh] DM1 For using PE = mgh Height is 4.25 m A1 [4] SR for candidates who assume without justification that the resistance to motion is constant, usually implicitly by using constant acceleration formulae (max 3/4) For using Newton’s second law with 3 terms, v2 – u2 = 2as and h = s sinα M1 For attempting to eliminate α, a and s from the equations (80sinα – 120/s = 8a 64 – 9 = 2as, h = s sinα) M1 80s sinα – 120 = 4(64 – 9) → 80h – 120 = 220 → h = 4.25 A1

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3 v (m s–1 ) 50 8 0 t (s) 0 5 12 102 The velocity-time graph shown models the motion of a parachutist falling vertically. There are four stages in the motion: • falling freely with the parachute closed, • decelerating at a constant rate with the parachute open, • falling with constant speed with the parachute open, • coming to rest instantaneously on hitting the ground. (i) Show that the total distance fallen is 1048 m. [2] The weight of the parachutist is 850 N. (ii) Find the upward force on the parachutist due to the parachute, during the second stage. [5]

Mark scheme: 3 (i) [ ½ 5 × 50 + ½ 7(8 + 50) + 90 × 8] M1 For using the area property for distance or s = ½ (u + v)t Distance is 1048 m A1 [2] AG (ii) M1 For use of the gradient property for acceleration (deceleration) a = (8 – 50)/(12 – 5) or d = (50 – 8)/(12 – 5) A1 M1 For using Newton’s second law (3 terms) 850 – F = 85a (or –85d) A1 Upward force is 1360 N A1 [5] GCE AS/A LEVEL – May/June 2011 9709 42

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4 12 N F N P 30° q° 10 N The three coplanar forces shown in the diagram act at a point P and are in equilibrium. (i) Find the values of F and θ. [6] (ii) State the magnitude and direction of the resultant force at P when the force of magnitude 12 N is removed. [2]

Mark scheme: 4 (i) M1 For resolving forces in the i and j directions Fcosθ = 12cos30° (= 10.932) A1 Fsinθ = 10 – 12sin30° (= 4) A1 M1 For using F2 = X2 + Y2 or tanθ = Y/X F = 11.1 or θ = 21.1 (accept 21.0) A1 θ = 21.1 (accept 21.0) or F = 11.1 B1 [6] SR for candidates who consistently have cos for sin and vice versa (max 4/6) M1 as above (resolving) A1 for Fsinθ = 12sin30° and Fcosθ = 10 – 12cos30° M1 as above F2 = .... & tanθ =.... A1 for F = 6.01 and θ = 93.7 (ii) Magnitude is 12N B1 Direction is 30o clockwise from +ve ‘x’ axis B1 [2] alternative for 4(i) For triangle of forces with sides 12, F and 10 and at least one of the angles (90° – θ ) or 60° or (θ + 30°) B1 M1 For use of cosine rule (with θ absent) or use of sine rule (with F absent) and use of sin(A ± B) = sinAcosB ± sinBcosA F2 = 122 + 102 – 2 × 12 × 10cos60o or (12cos30°)sinθ = (10 – 12sin30°)cosθ A1 F = 11.1 or θ = 21.1 (accept 21.0) A1 M1 For correct method for θ or F θ = 21.1 (accept 21.0) or F = 11.1 A1 [6] second alternative for 4(i) For using Lami’s theorem with 12 N and 10 N M1 12/sin(90 + θ ) = 10/sin(150 – θ ) A1 12/cosθ = 20 ÷ (cosθ + 3 ½ sinθ ) → 12 × 3 ½ sinθ = 8cosθ → tanθ = 2 ÷ (3 × 3 ½ ) A1 → θ = 21.1 For using Lami’s theorem with F N and (12 N or 10 N) M1 F/sin120o = 12/sin111.1o (or 10/sin128.9o) A1 F = 11.1 A1 [6] GCE AS/A LEVEL – May/June 2011 9709 42 Alternative for 4(ii) For X = 11.1cos21.1o and Y = 11.1sin21.1o – 10, M1 R2 = X2 + Y2 and tanФ = Y/X Magnitude 12 N and direction 30o clockwise from +ve x-axis A1 [2]

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5 Two particles P and Q are projected vertically upwards from horizontal ground at the same instant. The speeds of projection of P and Q are 12 m s−1 and 7 m s−1 respectively and the heights of P and Q above the ground, t seconds after projection, are hP m and hQ m respectively. Each particle comes to rest on returning to the ground. (i) Find the set of values of t for which the particles are travelling in opposite directions. [3] (ii) At a certain instant, P and Q are above the ground and 3hP = 8hQ. Find the velocities of P and Q at this instant. [5]

Mark scheme: 5 (i) M1 For using 0 = u – gt to find times at maximum heights. Times to max. height are 1.2s and 0.7s A1 Range of values is 0.7 < t < 1.2 A1 [3] (ii) M1 For using h = ut – ½ gt2 and attempting to solve 3hA = 8hB for t 36t – 1.5gt2 = 56t – 4gt2 A1 t = 8/g A1 M1 For using v = u – gt Velocities are 4m–1 and –1ms–1 A1 [5] Alternative for part 5(ii) For using 3hP = 8hQ → 3(vP2 – 144) ÷ (–20) = 8(vQ2 – 49)÷(–20) → 3vP2 – 8vQ2 B1 = 40 For using vP = 12 – 10t and vQ = 7 – 10t → vP – vQ = 5 B1 For eliminating vQ (or vP) and solving for vP (or vQ). M1 vP2 – 16vP + 48 = 0 → vP = 4 (or 4, 12) A1 Upward velocities are 4 ms–1 and –1 ms–1 A1 [5]

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6 A 80 cm B 50 cm 50 cm R A small smooth ring R, of mass 0.6 kg, is threaded on a light inextensible string of length 100 cm. One end of the string is attached to a fixed point A. A small bead B of mass 0.4 kg is attached to the other end of the string, and is threaded on a fixed rough horizontal rod which passes through A. The system is in equilibrium with B at a distance of 80 cm from A (see diagram). (i) Find the tension in the string. [3] (ii) Find the frictional and normal components of the contact force acting on B. [4] (iii) Given that the equilibrium is limiting, find the coefficient of friction between the bead and the rod. [2]

Mark scheme: 6 (i) M1 For resolving forces on R vertically 2T cosα = 0.6g A1 Where α = ½ angle ARB Tension is 5N A1 [3] (ii) [F = T sinα ] M1 For resolving forces on B horizontally Frictional component is 4N A1 [N = 0.4g + T cosα ] M1 For resolving forces on B vertically Normal component is 7 N A1 [4] (iii) M1 For using µ = F/N Coefficient is 4/7 or 0.571 A1ft [2] ft conditional on both M1 marks scored in (ii); ft F and/or N GCE AS/A LEVEL – May/June 2011 9709 42 Alternative for Q6(i)/(ii) (i) For finding the relevant angles and using M1 Lami’s theorem 6/sin106.26° = T/sin126.87° A1 Tension is 5N A1 [3] (ii) F/sin126.87° = 5/sin90° B1 Frictional component is 4N B1 (R – 4)/sin143.13° = 5/sin90° B1 Normal component is 7 N B1 [4]

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7 A walker travels along a straight road passing through the points A and B on the road with speeds 0.9 m s−1 and 1.3 m s−1 respectively. The walker’s acceleration between A and B is constant and equal to 0.004 m s−2. (i) Find the time taken by the walker to travel from A to B, and find the distance AB. [3] A cyclist leaves A at the same instant as the walker. She starts from rest and travels along the straight road, passing through B at the same instant as the walker. At time t s after leaving A the cyclist’s speed is kt3 m s−1, where k is a constant. (ii) Show that when t = 64.05 the speed of the walker and the speed of the cyclist are the same, correct to 3 significant figures. [5] (ii) Find the cyclist’s acceleration at the instant she passes through B. [2]

Mark scheme: 7 (i) [1.3 = 0.9 + 0.004T, M1 For using v = u + at or v2 = u2 + 2as 1.32 = 0.92 + 2 × 0.004S] Time is 100 s (or distance is 110 m) A1 Distance is 110 m (or time is 100 s) B1 [3] = ¼ kt4 B1 (ii) ∫ kt 3 dt [k( ¼ 1004 – 0) = 110] M1 For using limits 0 to T and equating definite integral to S k = 4.4 × 10–6 A1 [vW = 0.9 + 0.004 × 64.05, M1 For attempting to find the speed of the vC = 4.4 × 10–6 × 64.053] walker and of the cyclist. Both are equal to 1.16 ms–1 correct to 3 sf. A1 [5] (iii) Acceleration = 3kt2 B1 Acceleration at B is 0.132 ms–2 B1 [2]

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