Cambridge A Level Mathematics 9709 — 2010 Oct/Nov Paper 4 · Variant 1

1 v (m s–1 ) V P t (s) O 2 4 Q Two particles P and Q move vertically under gravity. The graphs show the upward velocity v m s−1 of the particles at time t s, for 0 ≤t ≤4. P starts with velocity V m s−1 and Q starts from rest. (i) Find the value of V. [2] Given that Q reaches the horizontal ground when t = 4, find (ii) the speed with which Q reaches the ground, [1] (iii) the height of Q above the ground when t = 0. [2]

Mark scheme: 1 (i) M1 For using –g = (0 – V)/(2 – 0) or 0 = V – gt V = 20 A1 [2] (ii) Speed is 40 ms–1 B1 [1] (iii) M1 For using h = ½ 4 × 40 or h = ½g × 42 or 402 = 2gh Height is 80 m A1 [2]

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2 A car of mass 600 kg travels along a horizontal straight road, with its engine working at a rate of 40 kW. The resistance to motion of the car is constant and equal to 800 N. The car passes through the point A on the road with speed 25 m s−1. The car’s acceleration at the point B on the road is half its acceleration at A. Find the speed of the car at B. [5]

Mark scheme: 2 [F – R = ma] M1 For using Newton’s second law (3 terms) FA – 800 = 600aA A1 FA = 40000/25 (1600) B1 40000/vB – 800 = 600 (400/600) A1 Speed is 33.3 ms–1 A1 [5]

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3 P1 P2 X A C B The diagram shows three particles A, B and C hanging freely in equilibrium, each being attached to the end of a string. The other ends of the three strings are tied together and are at the point X. The strings carrying A and C pass over smooth fixed horizontal pegs P1 and P2 respectively. The weights of A, B and C are 5.5 N, 7.3 N and W N respectively, and the angle P1XP2 is a right angle. Find the angle AP1X and the value of W. [5]

Mark scheme: 3 M1 For using triangle of forces or for resolving in dirn XP1 or for using Lami’s theorem or for resolving forces at X vertically and horizontally (equations must contain not more than one unknown angle) For correct ∆ or resolve XP1 A1 and cosα = 5.5/7.3; or 5.5/sin(90° + α) = 7.3/sin90° (Lami); or 5.5cosα + Wsinα = 7.3 and 5.5sinα = Wcosα. Angle AP1X = 41.1° or 0.718c A1 For correct triangle and W2 = 7.32 – 5.52; A1ft ft incorrect α or W/sin(180° – 41.1°) = 7.3/sin90°; or Wsin41.1° = 7.3 – 5.5cos41.1° or Wcos41.1° = 5.5sin41.1° W = 4.8 A1 [5]

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4 A particle P starts from a fixed point O at time t = 0, where t is in seconds, and moves with constant acceleration in a straight line. The initial velocity of P is 1.5 m s−1 and its velocity when t = 10 is 3.5 m s−1. (i) Find the displacement of P from O when t = 10. [2] Another particle Q also starts from O when t = 0 and moves along the same straight line as P. The acceleration of Q at time t is 0.03t m s−2. (ii) Given that Q has the same velocity as P when t = 10, show that it also has the same displacement from O as P when t = 10. [5]

Mark scheme: 4 (i) (1.5 + 3.5)/2 = s/10 B1 For using (u + v)/2 = s/t Displacement is 25 m B1 [2] (ii) M1 For using v = ∫ adt v = 0.015t2 (+ C) A1 [3.5 = 0.015 × 100 + C → C = 2] B1 [s = 0.005t3 + 2t + (0)] M1 For using s = ∫ vdt Displacement is 25 m, same as P. A1 [5] GCE AS/A LEVEL – October/November 2010 9709 41 2 2 2 2

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5 A particle of mass 0.8 kg slides down a rough inclined plane along a line of greatest slope AB. The distance AB is 8 m. The particle starts at A with speed 3 m s−1 and moves with constant acceleration 2.5 m s−2. (i) Find the speed of the particle at the instant it reaches B. [2] (ii) Given that the work done against the frictional force as the particle moves from A to B is 7 J, find the angle of inclination of the plane. [4] When the particle is at the point X its speed is the same as the average speed for the motion from A to B. (iii) Find the work done by the frictional force for the particle’s motion from A to X. [3]

Mark scheme: 5 (i) [v2 = 32 + 2 × 2.5 × 8] M1 For using v2 = u2 + 2as Speed is 7 ms–1 A1 [2] (ii) KE gain = ½ 0.8(72 – 32) (= 16) B1ft ft incorrect speed PE loss = 16 + 7 B1ft ft incorrect expression for KE [0.8 × 10 × 8sinα = 23] M1 For using PE loss = mgLsinα Angle is 21.1° or 0.368c A1 [4] (ii) ALTERNATIVELY F = 7/8 B1 [0.8 × 10sinα – F = 0.8 × 2.5] M1 For using Newton’s second law 0.8 × 10sinα – 0.875 = 0.8 × 2.5 A1 Angle is 21.1° or 0.368c A1 (iii) 52 = 32 + 2 × 2.5s (s = 3.2) B1 [WD/7 = 3.2/8 M1 For using WD proport’l to dist. or WD = 0.875 × 3.2 or WD = F(AX) or WD = 8 × 3.2 × (23/64) or WD = PE loss – KE gain – ½ 0.8(52 – 32)] Work done is 2.8 J A1 [3]

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6 A X B 3 m H m h m A smooth slide AB is fixed so that its highest point A is 3 m above horizontal ground. B is h m above the ground. A particle P of mass 0.2 kg is released from rest at a point on the slide. The particle moves down the slide and, after passing B, continues moving until it hits the ground (see diagram). The speed of P at B is vB and the speed at which P hits the ground is vG. (i) In the case that P is released at A, it is given that the kinetic energy of P at B is 1.6 J. Find (a) the value of h, [3] (b) the kinetic energy of the particle immediately before it reaches the ground, [1] (c) the ratio vG : vB. [2] (ii) In the case that P is released at the point X of the slide, which is H m above the ground (see diagram), it is given that vG : vB = 2.55. Find the value of H correct to 2 significant figures. [3]

Mark scheme: 6 (i) (a) PE loss = 0.2g(3 – h) B1 [0.2g(3 – h) = 1.6] M1 For using PE loss = KE gain h = 2.2 A1 [3] (b) KE is 6 J B1 [1] (c) [vG / vB = (3/(3 – 2.2))½ M1 For using v2 ∝(3 – ht) or (vG / vB)2 = Ans. (i)(b) ÷ 1.6 or vG / vB = 6 / 6.1 ] Ratio is 1.94 A1 [2] Accept 60 ÷ 4 or 15 ÷ 2 (ii) M1 For using v2 ∝(H – ht) or using ½ m(2.55vB)2 = mgH and ½ mvB2 = mg(H – 2.2) and eliminating vB2 H/(H – 2.2) = 2.552 A1 H = 2.6 A1 [3] GCE AS/A LEVEL – October/November 2010 9709 41

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7 3.2 N Q 30° P Particles P and Q, of masses 0.2 kg and 0.5 kg respectively, are connected by a light inextensible string. The string passes over a smooth pulley at the edge of a rough horizontal table. P hangs freely and Q is in contact with the table. A force of magnitude 3.2 N acts on Q, upwards and away from the pulley, at an angle of 30◦to the horizontal (see diagram). (i) The system is in limiting equilibrium with P about to move upwards. Find the coefficient of friction between Q and the table. [6] The force of magnitude 3.2 N is now removed and P starts to move downwards. (ii) Find the acceleration of the particles and the tension in the string. [4]

Mark scheme: 7 (i) M1 For resolving forces on Q vertically R + 3.2sin30° = 0.5g A1 M1 For resolving forces on Q horizontally and using T = WP F + 0.2g = 3.2cos30° A1 [µ = (3.2cos30° – 2)/(5 – 3.2sin30°)] M1 For using F = µR Coefficient is 0.227 A1 [6] (ii) 2 – T = 0.2a B1 T – 0.227 × 5 = 0.5a B1ft Allow B1ft for 2 – 0.227 × 5 = (0.2 + 0.5)a instead of one of the above equations M1 For solving for a or T Acceleration is 1.24 ms–2 and tension is A1 Allow a = 1.25 1.75 N [4]

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