Cambridge A Level Mathematics 9709 — 2013 May/June Paper 4 · Variant 3

1 A straight ice track of length 50 m is inclined at 14Å to the horizontal. A man starts at the top of the track, on a sledge, with speed 8 m s−1. He travels on the sledge to the bottom of the track. The coefficient of friction between the sledge and the track is 0.02. Find the speed of the sledge and the man when they reach the bottom of the track. [4]

Mark scheme: 1 [(W / g) a = W sin α – 0.02 W cos α] M1 For using Newton’s second law a = (sin 14o – 0.02 cos 14o) g (= 2.225 … ) A1 [v2 = 82 + 2 × 2.225 … × 50] M1 For using v2 = u2 + 2 a s Speed is 16.9 m s-1 A1 [4] Alternative Scheme 1 WD against friction = 0.02 W cos α × 50 B1 PE loss = W × 50 sin α B1 For using Gain in KE = Loss in PE M1 – WD against friction Speed is 16.9 m s-1 A1 [4]

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2 B 3.24 m q A Particle A of mass 1.6 kg and particle B of mass 2 kg are attached to opposite ends of a light inextensible string. The string passes over a small smooth pulley fixed at the top of a smooth plane, which is inclined at angle 1, where sin 1 = 0.8. Particle A is held at rest at the bottom of the plane and B hangs at a height of 3.24 m above the level of the bottom of the plane (see diagram). A is released from rest and the particles start to move. (i) Show that the loss of potential energy of the system, when B reaches the level of the bottom of the plane, is 23.328 J. [3] (ii) Hence find the speed of the particles when B reaches the level of the bottom of the plane. [2]

Mark scheme: 2 (i) M1 PE loss = B’s loss – A’s gain Loss of PE = 2g × 3.24 – 1.6 g (3.24 × 0.8) A1 Loss is 23.328 J. A1 [3] AG (ii) ½ (1.6 + 2) v2 = 23.328 B1 Speed is 3.6 m s-1 B1 [2] SR (max 1/2) for using Newton’s second law and v2 = u2 + 2 a s 2 g – T = 2 a and T – 1.6g × 0.8 = 1.6a a = 2 v2 = 2 × 2 × 3.24 v = 3.6 B1

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3 A car has mass 800 kg. The engine of the car generates constant power P kW as the car moves along a straight horizontal road. The resistance to motion is constant and equal to R N. When the car’s speed is 14 m s−1 its acceleration is 1.4 m s−2, and when the car’s speed is 25 m s−1 its acceleration is 0.33 m s−2. Find the values of P and R. [6]

Mark scheme: 3 M1 For using DF = P / v For using Newton’s 2nd law for M1 both speeds / accelerations 1000 P / 14 – R = 800 x 1.4 and 1000 P / 25 – R = 800 x 0.33 A1 M1 For solving for P P = 27.2 A1 R = 825 B1 [6] Accept 825.5 GCE AS/A LEVEL – May/June 2013 9709 43

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4 An aeroplane moves along a straight horizontal runway before taking off. It starts from rest at O and has speed 90 m s−1 at the instant it takes off. While the aeroplane is on the runway at time t seconds after leaving O, its acceleration is 1.5 + 0.012t m s−2. Find (i) the value of t at the instant the aeroplane takes off, [4] (ii) the distance travelled by the aeroplane on the runway. [3]

Mark scheme: 4 (i) M1 For integrating a (t) to obtain v (t) V (t) = 1.5 t + 0.006 t2 A1 Constant of integration zero or absent [0.006 t2 + 1.5 t - 90 = 0 t2 + 250t – 15000 = 0] For using v (t) = 90 and solving for (t – 50) (t + 300) = 0] DM1 t (dependent on integration) Leaves the ground when t = 50 A1 [4] (ii) M1 For integrating v (t) and using limits 0 to candidate’s answer for part (i) s = 0.75 t2 + 0.002 t3 A1ft ft if there is a non-zero constant of integration C in part (i) s = 0.75 t2 + 0.002 t3 + C t Distance is 2125 m A1ft [3] Accept 2120 or 2130 ft t from part (i) in 0.75 t2 + 0.002 t3

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5 A particle P is projected vertically upwards from a point on the ground with speed 17 m s−1. Another particle Q is projected vertically upwards from the same point with speed 7 m s−1. Particle Q is projected T seconds later than particle P. (i) Given that the particles reach the ground at the same instant, find the value of T. [2] (ii) At a certain instant when both P and Q are in motion, P is 5 m higher than Q. Find the magnitude and direction of the velocity of each of the particles at this instant. [6]

Mark scheme: 5 (i) T = 2 x time to max. height for P – [T = 2 x 1.7 – 2 x 0.7] 2 x time to max. height for Q [for P 17 t – 5 t2 = 0 or For using T = time for P to and return to ground – time for Q to for Q 7 t = 5 t2 = 0] M1 return to ground T = 2 A1 [2] SR (max 1/2) for candidates who find difference in time to maximum height T = 1.7 – 0.7 = 1 B1 (ii) M1 For using hP – hQ = 5 and s = u t – 5 t2 for both P and Q 17(t + 2) – 5(t + 2)2 – (7t – 5t2) = 5 or 17t – 5t2 – 7(t - 2) + 5(t – 2)2 = 5 A1 ft ft T from part (i) t = 0.9 or t = 2.9 A1 M1 For using v = u – 10 t for P and Q GCE AS/A LEVEL – May/June 2013 9709 43 vP = 17 – 10 (0.9 + 2), vQ = 7 – 10 × 0.9 Magnitudes are 12 m s-1 & 2 m s-1 A1 ft ft using tP and tP – T or using tQ and tQ + T The direction for both is vertically downwards A1 [6]

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6 y F N 100 N a° 30° x 60° 120 N A small box of mass 40 kg is moved along a rough horizontal floor by three men. Two of the men apply horizontal forces of magnitudes 100 N and 120 N, making angles of 30Å and 60Å respectively with the positive x-direction. The third man applies a horizontal force of magnitude F N making an angle of !Å with the negative x-direction (see diagram). The resultant of the three horizontal forces acting on the box is in the positive x-direction and has magnitude 136 N. (i) Find the values of F and !. [6] (ii) Given that the box is moving with constant speed, state the magnitude of the frictional force acting on the box and hence find the coefficient of friction between the box and the floor. [3]

Mark scheme: 6 (i) For resolving the applied forces on the box in the x-direction or the y- M1 direction. 100 cos 30o + 120 cos 60o – F cosα = 136 (F cos α = 10.6025 … ) or 100 sin 30o – 120 sin 60o + F sin α =0 (F sin α =53.9230 … ) A1 100 sin 30o – 120 sin 60o + F sin α = 0 (F sin α =53.9230 … ) or 100 cos 30o + 120 cos 60o – F cos α = 136 (F cos α = 10.6025 … ) B1 M1 for using F2 = (F cos α)2 + (F sin α) or tan α = F sin α ÷ F cos α F = 55.0 or α = 78.9 A1 α = 78.9 or F = 55.0 B1 [6] (ii) Magnitude is 136 N B1 R = 40 g B1 Coefficient is 0.34 B1 [3] GCE AS/A LEVEL – May/June 2013 9709 43 d

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7 A 0.48 m P B 0.45 m Particle A of mass 1.26 kg and particle B of mass 0.9 kg are attached to the ends of a light inextensible string. The string passes over a small smooth pulley P which is fixed at the edge of a rough horizontal table. A is held at rest at a point 0.48 m from P, and B hangs vertically below P, at a height of 0.45 m above the floor (see diagram). The coefficient of friction between A and the table is 7.2 A is released and the particles start to move. (i) Show that the magnitude of the acceleration of the particles is 2.5 m s−2 and find the tension in the string. [5] (ii) Find the speed with which B reaches the floor. [2] (iii) Find the speed with which A reaches the pulley. [4]

Mark scheme: 7 (i) M1 For applying Newton’s 2nd law to A or to B T – (2 / 7) 1.26 g = 1.26 a or A1 0.9 g - T = 0.9 a 0.9g – T = 0.9 a or T – (2 / 7) 1.26 g = 1.26 a or 0.9 g – (2 / 7) 1.26 g = (0.9 + 1.26) a B1 Acceleration is 2.5 m s-2 B1 AG Tension is 6.75 N A1 [5] (ii) [v2 = 2 × (2.5) × 0.45] M1 For using v2 = 2 a h Speed is 1.5 m s-1 A1 [2] (iii) [– (2 / 7) 1.26 g = 1.26 a] M1 For applying Newton’s 2nd law to A a = – 20 / 7 A1 [v2 = 2.25 + 2 (–20 / 7) (0.03)] M1 For using v2 = vB2 + 2 a s Speed is 1.44 m s-1 A1 [4]

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