Cambridge A Level Mathematics 9709 — 2009 Oct/Nov Paper 4 · Variant 1

1 A car of mass 1000 kg moves along a horizontal straight road, passing through points A and B. The power of its engine is constant and equal to 15 000 W. The driving force exerted by the engine is 750 N at A and 500 N at B. Find the speed of the car at A and at B, and hence find the increase in the car’s kinetic energy as it moves from A to B. [4]

Mark scheme: 1 [15000 = 750vA, 15000 = 500vB] M1 For using P = Fv Speeds are 20ms–1 and 30ms–1 A1 [KE gain = ½ 1000(302 – 202)] M1 For using KE = ½ mv2 Increase is 250 000J (or 250kJ) A1ft 4 ft 500(vB2 – vA2)

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2 A 6 m s–1 B D E 0.65 m C A smooth narrow tube AE has two straight parts, AB and DE, and a curved part BCD. The part AB is vertical with A above B, and DE is horizontal. C is the lowest point of the tube and is 0.65 m below the level of DE. A particle is released from rest at A and travels through the tube, leaving it at E with speed 6 m s−1 (see diagram). Find (i) the height of A above the level of DE, [2] (ii) the maximum speed of the particle. [2]

Mark scheme: 2 (i) [mgh = ½ m62] M1 For using PE loss = KE gain Height is 1.8m A1 2 (ii) ½ mv2 = mg(1.8 + 0.65) or ½ mv2 – ½ m62 = mg × 0.65 B1ft Maximum speed is 7ms–1 B1 2

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3 P N 12 N 40° 80° Q N Two forces have magnitudes P N and Q N. The resultant of the two forces has magnitude 12 N and acts in a direction 40◦clockwise from the force of magnitude P N and 80◦anticlockwise from the force of magnitude Q N (see diagram). Find the value of Q. [4]

Mark scheme: 3 For resolving forces parallel to or perpendicular to:– the force of magnitude QN, or M1 the resultant Q – Pcos60o = 12cos80o and Psin60o = 12sin80o Qcos80o + Pcos40o = 12 and Psin40o = Qsin80o A1 [Q – 12sin80ocos60o/sin60o = 12cos80o For eliminating P Qcos80o + Qsin80ocos40o/sin40o = 12] M1 Q = 8.91 A1 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (First alternative) M2 For resolving forces perp. to P Qcos30o = 12cos50o A1 Q = 8.91 A1 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (Second alternative) For triangle of forces with sides P, Q and 12, and values of any 2 angles shown or implied (Note P, Q and –R are M1 in equil.) Angles opposite Q and 12 are 40o and 60o respectively A1 Q/sin40o = 12/sin60o M1 For using the sine rule Q = 8.91 A1 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (Third alternative) For force diagram showing P, Q and –R, M1 and values of any 2 angles at ‘O’ shown. Angle between P and Q is 120o and between P and –R is 140o A1 [Q/sin140o = 12/sin120o] M1 For using Lami’s theorem Q = 8.91 A1 4 GCE A/AS LEVEL – October/November 2009 9709 41

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4 A 50 cm S 30 cm 40 cm P A particle P of weight 5 N is attached to one end of each of two light inextensible strings of lengths 30 cm and 40 cm. The other end of the shorter string is attached to a fixed point A of a rough rod which is fixed horizontally. A small ring S of weight W N is attached to the other end of the longer string and is threaded on to the rod. The system is in equilibrium with the strings taut and AS = 50 cm (see diagram). (i) By resolving the forces acting on P in the direction of PS, or otherwise, find the tension in the longer string. [3] (ii) Find the magnitude of the frictional force acting on S. [2] (iii) Given that the coefficient of friction between S and the rod is 0.75, and that S is in limiting equilibrium, find the value of W. [3]

Mark scheme: 4 (i) For angle between AP and vertical = 36.9o (or sin–10.6) or for angle between PS and vertical = 53.1o (or sin–10.8) B1 May be implied For resolving forces on P in the direction [TPS + (TPAcos90o) = 5sin36.9o] M1 of PS (2 non–zero terms required) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (First alternative) For the angle between PA and the horizontal through P is 53.1o and the angle between PS and the horizontal through P is 36.9o B1 May be implied [0.6TPA = 0.8TPS and 0.8TPA + 0.6TPS = 5 For resolving forces on P vertically and {0.8(0.8/0.6) + 0.6}TPS = 5] M1 horizontally and eliminating TPA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (Second alternative) For ∆ of forces with sides TPA, TPS and 5, with angles opposite TPS and 5 shown as 36.9o and 90o B1 May be implied [TPS = 5sin36.9o] M1 For using trig. in ∆ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (Third alternative) For force diag. showing TPA, TPS and 5, with angles between TPS and TPA, and between 5 and TPA being shown as 90o and 143.1o B1 May be implied [TPS /sin143.1o = 5/sin90o] M1 For using Lami’s rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Tension is 3N A1 3 Accept 3.00 (ii) [F = T cos(sin–10.6)] M1 For resolving forces on S horizontally Frictional force is 2.4N A1 2 Accept 2.40 (iii) R = 2.4/0.75 B1ft [W + T sin(sin–10.6) = R] M1 For resolving forces on S vertically W = 1.4 A1ft 3 ft W = 7T/15 or W = 4F/3 – 1.8

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5 A particle P of mass 0.6 kg moves upwards along a line of greatest slope of a plane inclined at 18◦to the horizontal. The deceleration of P is 4 m s−2. (i) Find the frictional and normal components of the force exerted on P by the plane. Hence find the coefficient of friction between P and the plane, correct to 2 significant figures. [6] After P comes to instantaneous rest it starts to move down the plane with acceleration a m s−2. (ii) Find the value of a. [2] [Questions 6 and 7 are printed on the next page.]

Mark scheme: 5 (i) M1 For using Newton’s second law – F – 0.6gsin 18o = 0.6(–4) A1 Frictional component is 0.546N A1 [R = 0.6gcos18o] M1 For resolving forces normal to the plane Normal component is 5.71N A1 Coefficient is 0.096 B1ft 6 (ii) 0.6gsin18o – 0.546 = 0.6a or 2(0.6gsin18o) = 0.6(a + 4) B1ft a = 2.18 B1 2 SR For candidates who use ‘a’ for the upwards acceleration , instead of as defined in the question. –0.6gsin18o + 0.546 = 0.6a a = –2.18 B1 a = 2.18 accompanied by satisfactory explanation for dropping the minus sign. B1 GCE A/AS LEVEL – October/November 2009 9709 41

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6 P Q 5 m Particles P and Q, of masses 0.55 kg and 0.45 kg respectively, are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. The particles are held at rest with the string taut and its straight parts vertical. Both particles are at a height of 5 m above the ground (see diagram). The system is released. (i) Find the acceleration with which P starts to move. [3] The string breaks after 2 s and in the subsequent motion P and Q move vertically under gravity. (ii) At the instant that the string breaks, find (a) the height above the ground of P and of Q, [2] (b) the speed of the particles. [1] (iii) Show that Q reaches the ground 0.8 s later than P. [4]

Mark scheme: 6 (i) For using Newton’s second law to P or M − m M1 Q, or for using a = g M + m 0.55g – T = 0.55a and T – 0.45g = 0.45a or a = [(0.55 – 0.45)/(0.55+ 0.45)]g A1 Acceleration is 1ms–2 A1 3 (ii) (a) For using s = 5 – ½ a22 for P or M1 s = 5 + ½ a22 for Q Height of P is 3m and height of Q is 7m A1ft 2 ft 5 – 2a and 5 + 2a (b) Speed is 2ms–1 B1ft 1 ft 2a (iii) For using s = ut + ½ gt2 for P or for Q [3 = 2tP + 5tP2, 7 = –2tQ + 5tQ2] M1 (NB a = g) tP = 0.6 A1 Accept tQ = 0.2 + 1.2 following consideration of upward and downward tQ = 1.4 A1 motion under gravity of Q separately Q is 0.8s later than P A1 4 AG

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7 A particle P starts from rest at the point A at time t = 0, where t is in seconds, and moves in a straight line with constant acceleration a m s−2 for 10 s. For 10 ≤t ≤20, P continues to move along the line 800 with velocity v m s−1, where v = −2. Find t2 (i) the speed of P when t = 10, and the value of a, [2] (ii) the value of t for which the acceleration of P is −a m s−2, [4] (iii) the displacement of P from A when t = 20. [6]

Mark scheme: 7 (i) Speed is 6ms–1 B1 a = 0.6 B1ft 2 ft v/10 (ii) M1 For differentiating v(t) a(t) = –1600/t3 (second stage) A1 [0.6 = 1600/t3 t3 = (1600/0.6)] M1 For attempting to solve a(t) = –0.6 t = 13.9 A1 4 SR in parts (i) and (ii) (treated as a single entity) for candidates who assume there is necessarily continuity of acceleration at t = 10 (max 5/6) Speed is 6ms–1 B1 For differentiating v(t) M1 a(t) = –1600/t3 A1 a = a(10) = –1.6 B1 − 1600 For t = 3 B1 6.1 (iii) s1 = 30m B1 M1 For integrating v(t) s = (800t–1)/(–1) – 2t (+ C) A1 For using limits 10, 20 or for using s(10) = s1 to find c (= 130) and M1 evaluating s(20). s2 = (–40 – 40) – (–80 – 20) or s = –40 – 40 + 130 A1 Any correct form Displacement from A is 50m A1 6

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