Cambridge A Level Mathematics 9709 — 2014 Oct/Nov Paper 4 · Variant 2

1 A particle P is projected vertically upwards with speed 11 m s−1 from a point on horizontal ground. At the same instant a particle Q is released from rest at a point h m above the ground. P and Q hit the ground at the same instant, when Q has speed V m s−1. (i) Find the time after projection at which P hits the ground. [2] (ii) Hence find the values of h and V. [2]

Mark scheme: 1 (i) [–11 = 11 – 10t] M1 For using v = u – gt (or equivalent method) to find the duration of motion Time after projection is 2.2 seconds A1 2 (ii) h = 0 + ½ g × 2.22 = 24.2 B1 V = 0 + g × 2.2 = 22 B1 2

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2 y 30 N 25 N " ! x 20 N Three coplanar forces act at a point. The magnitudes of the forces are 20 N, 25 N and 30 N, and the directions in which the forces act are as shown in the diagram, where sin ! = 0.28 and cos ! = 0.96, and sin " = 0.6 and cos " = 0.8. (i) Show that the resultant of the three forces has a zero component in the x-direction. [2] (ii) Find the magnitude and direction of the resultant of the three forces. [2] (iii) The force of magnitude 20 N is replaced by another force. The effect is that the resultant force is unchanged in magnitude but reversed in direction. State the magnitude and direction of the replacement force. [1]

Mark scheme: 2 (i) [X = 25 × 0.96 – 30 × 0.8 = 0] M1 For resolving forces in the x direction Component in x-direction is zero A1 2 AG (ii) [Y = 25 × 0.28 – 20 + 30 × 0.6 = 5] M1 For resolving forces in the y direction Resultant has magnitude 5 N and acts in the positive y direction A1 2 (iii) Replacement has magnitude 30 N and acts in the –ve y direction B1 1

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3 A train of mass 200 000 kg moves on a horizontal straight track. It passes through a point A with speed 28 m s−1 and later it passes through a point B. The power of the train’s engine at B is 1.2 times the power of the train’s engine at A. The driving force of the train’s engine at B is 0.96 times the driving force of the train’s engine at A. (i) Show that the speed of the train at B is 35 m s−1. [2] (ii) For the motion from A to B, find the work done by the train’s engine given that the work done against the resistance to the train’s motion is 2.3 × 106 J. [3]

Mark scheme: 3 (i) [vB = 1.2 × 28 ÷ 0.96] M1 For using P = Fv and the factors 1.2 and 0.96 and an equation in vB only Speed of the train at B is 35 ms–1 A1 2 AG (ii) KE increase = 100 000(352 – 282) B1 WD by engine For using WD by engine = KE = 44.1 × 106 + 2.3 × 106 J M1 increase + WD against resistance Work done is 46 400 kJ or 46.4 × 106 J A1 3 or 46 400 000 J

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4 40 N X N A B 30Å C Forces of magnitude X N and 40 N act on a block B of mass 15 kg, which is in equilibrium in contact with a horizontal surface between points A and C on the surface. The forces act in the same vertical plane and in the directions shown in the diagram. (i) Given that the surface is smooth, find the value of X. [2] (ii) It is given instead that the surface is rough and that the block is in limiting equilibrium. The frictional force acting on the block has magnitude 10 N in the direction towards A. Find the coefficient of friction between the block and the surface. [5]

Mark scheme: 4 (i) [Xcos30o = 40cos60o] M1 For resolving forces horizontally X = 23.1 (= 40 / √3) A1 2 (ii) [Xcos30o – 10 = 40cos60o] M1 For resolving forces horizontally X = 60 ÷ √3 or 34.6 A1 [R + Xsin30o + 40sin60o = 15g] M1 For resolving forces vertically (R = 98.038) [µ = 10 ÷ (150 – 30/√3 – 20√3)] M1 For using F = µR Coefficient is 0.102 A1 5

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5 B A Particles A and B, each of mass 0.3 kg, are connected by a light inextensible string. The string passes over a small smooth pulley fixed at the edge of a rough horizontal surface. Particle A hangs freely and particle B is held at rest in contact with the surface (see diagram). The coefficient of friction between B and the surface is 0.7. Particle B is released and moves on the surface without reaching the pulley. (i) Find, for the first 0.9 m of B’s motion, (a) the work done against the frictional force acting on B, [2] (b) the loss of potential energy of the system, [1] (c) the gain in kinetic energy of the system. [2] At the instant when B has moved 0.9 m the string breaks. A is at a height of 0.54 m above a horizontal floor at this instant. (ii) Find the speed with which A reaches the floor. [3] [Questions 6 and 7 are printed on the next page.]

Mark scheme: 5 (i) (a) [F = 0.7 × 3, WD = 2.1 × 0.9] M1 For using F = µR and WD = Fs Work done is 1.89 J A1 2 (b) Loss of PE = 3 × 0.9 = 2.7 J B1 1 (c) [KE gain = 2.7 – 1.89] M1 For ‘gain in KE = loss in PE – WD by friction’ Gain in KE = 0.81 J A1 2 (ii) ½(0.3 + 0.3)vat break2 = 0.81] M1 For using ½ (mA + mB)v2 = gain in KE 2 2 vfloor = vat break + 2g × 0.54 M1 For using v2 = u2 + 2gs Speed at the floor is 3.67 ms–1 A1 3 Alternative method for (i) (c) and (ii) (c) [T – 2.1 = 0.3a and 3 – T = 0.3a M1 For applying Newton’s 2nd law to → a = 1.5] both particles and finding a and [v2 = 2 × 1.5 × 0.9 = 2.7] using v2 = 0 + 2as and attempting KE KE = 0.5 × (0.3 + 0.3) × 2.7 = 0.81 J A1 2 (ii) [vat break 2 = 2.7] M1 For using their v2 in (i)(c) as 2 vat break 2 2 vfloor = vat break + 2g × 0.54 M1 For using v2 = u2 + 2gs Speed at floor = 3.67 ms–1 (= 1.5√6) A1 3 Alternative method for (ii) (ii) [0.3 × g × 0.54] or [½ × 0.3 × (v2 – 2.7)] M1 For attempting PE loss or KE gain for the falling particle only [1.62 = ½ × 0.3 × (v2 – 2.7)] M1 For using PE loss = KE gain of this particle Speed at floor = 3.67 ms–1 (= 1.5√6) A1 3

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6 ABC is a line of greatest slope of a plane inclined at angle ! to the horizontal, where sin ! = 0.28 and cos ! = 0.96. The point A is at the top of the plane, the point C is at the bottom of the plane and the length of AC is 5 m. The part of the plane above the level of B is smooth and the part below the level of B is rough. A particle P is released from rest at A and reaches C with a speed of 2 m s−1. The coefficient of friction between P and the part of the plane below B is 0.5. Find (i) the acceleration of P while moving (a) from A to B, (b) from B to C, [3] (ii) the distance AB, [3] (iii) the time taken for P to move from A to C. [3]

Mark scheme: 6 (i) (a) (a) Acceleration is 2.8 ms–2 B1 Using acceleration = g sin α (b) [mg × 0.28 – 0.5mg × 0.96 = ma] M1 For using Newton’s 2nd law Acceleration is – 2 ms–2 A1 3 (ii) For using v2 = u2 + 2as for AB and M1 for BC and using AB + BC = 5 2 vB = 2 × 2.8(AB) and 22 = 5.6(AB) – 2 × 2(5 – AB) A1 ft incorrect answers in (i) Distance is 2.5 m A1 3 Alternative method for (ii) [mg × 5 × 0.28 = ½ m 22 + M1 For using Loss in PE = Gain in KE µ × mg × 0.96 × BC] + WD against Friction for the motion from A to C 14 = 2 + 4.8 × BC A1 Correct equation BC = 12/4.8 = 2.5 m A1 3 (iii) M1 For using t = 2s ÷ (u + v) for AB and BC T = 2 × 2.5 ÷ (0 + √14) + 2 × 2.5 ÷ (√14+ 2) A1 Time taken is 2.21 s A1 3

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7 v m s−1 O t s 3 5 15 The diagram shows the velocity-time graph for the motion of a particle P which moves on a straight line BAC. It starts at A and travels to B taking 5 s. It then reverses direction and travels from B to C taking 10 s. For the first 3 s of P’s motion its acceleration is constant. For the remaining 12 s the velocity of P is v m s−1 at time t s after leaving A, where v = −0.2t2 + 4t −15 for 3 ≤t ≤15. (i) Find the value of v when t = 3 and the magnitude of the acceleration of P for the first 3 s of its motion. [3] (ii) Find the maximum velocity of P while it is moving from B to C. [3] (iii) Find the average speed of P, (a) while moving from A to B, (b) for the whole journey. [6]

Mark scheme: 7 (i) v = – 4.8 B1 [± 4.8 = 3a] M1 For using v = 0 + at Magnitude of acceleration is 1.6 ms–2 A1 3 (ii) [–0.4t + 4 (= 0 when t = 10)] M1 For finding the value of t when dv/dt = 0 M1 For evaluating v(10) as vmax (the graph excludes the possibility of v(10) as vmin) vmax = –0.2 × 100 + 4 × 10 – 15 → Maximum velocity is 5 ms–1 A1 3 (iii) (a) Distance 0 to 3 s = ½ × 3 × 4.8 ( = 7.2) B1 d t M1 Attempt to integrate and use limits 2.0t 2 + 4t − 15 ) Distance 3 to 5s = − ∫35(− Distance = ± 4.5333…m A1 Average speed = (7.2 + 4.533) ÷ 5 = 2.35 ms–1 B1 (b) Distance BC M1 ft for errors in coefficients in cubic  2.0t 3 2  15 expression =  − + 2t − 15t  3 5   and Av speed = (AB + BC) ÷ 15 Av speed = (45.066 ÷ 15) = 3.00 ms–1 A1 6

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