Cambridge A Level Mathematics 9709 — 2010 Oct/Nov Paper 4 · Variant 3

1 A particle P is released from rest at a point on a smooth plane inclined at 30◦to the horizontal. Find the speed of P (i) when it has travelled 0.9 m, (ii) 0.8 s after it is released. [4]

Mark scheme: 1 a = gsin30o B1 2 [(i) v1 = 2(gsin30o)0.9 M1 For using v2 = 2as 2 or ½ mv1 = mg(0.9sin30o) or 1/2 mv2 = mgh or (ii) v2 = (gsin30o)0.8] or v = at (i) Speed is 3 ms–1 or (ii) Speed is 4ms–1 A1 (ii) Speed is 4ms–1 or (i) Speed is 3 ms–1 B1 [4] 2 2

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2 A 1.8 m B C The diagram shows the vertical cross-section ABC of a fixed surface. AB is a curve and BC is a horizontal straight line. The part of the surface containing AB is smooth and the part containing BC is rough. A is at a height of 1.8 m above BC. A particle of mass 0.5 kg is released from rest at A and travels along the surface to C. (i) Find the speed of the particle at B. [2] (ii) Given that the particle reaches C with a speed of 5 m s−1, find the work done against the resistance to motion as the particle moves from B to C. [2]

Mark scheme: 2 (i) [½ v2 = 10x1.8] M1 For using ½ mv2 = mgh Speed is 6 ms–1 A1 [2] (ii) [WD = ½x0.5(62 – 52) or M1 For using WD = loss of KE 0.5x10x1.8 = ½x0.5x52] or KEA + PEA – WD = KEC + PEC Work done is 2.75 J A1 [2]

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3 A P Q 3 Ö3 N B 30° C A small smooth pulley is fixed at the highest point A of a cross-section ABC of a triangular prism. Angle ABC 90◦and angle BCA 30◦. The prism is fixed with the face containing BC in contact = = with a horizontal surface. Particles P and Q are attached to opposite ends of a light inextensible string, which passes over the pulley. The particles are in equilibrium with P hanging vertically below the pulley and Q in contact with AC. The resultant force exerted on the pulley by the string is 3 √3 N (see diagram). (i) Show that the tension in the string is 3 N. [2] The coefficient of friction between Q and the prism is 0.75. (ii) Given that Q is in limiting equilibrium and on the point of moving upwards, find its mass. [5]

Mark scheme: 3 (i) [2T cos30o = 3 3 M1 For expressing resultant in terms of T and equating with value or T/sin30o = 3 3 /sin120o or for using sine rule or T2 = T2 + (3 3 )2 – 2T(3 3 )cos30o or for using cosine rule or √{(Tcos30o)2 + (T + Tcos60o)2} = 3 3 ] or for finding Rx and Ry and equating resultant to 3 3 Tension is 3 N A1 [2] AG (ii) [T = F + mg sin30] M1 For resolving forces on Q parallel to AC R = mg cos30 B1 M1 For using F = µ R 3 = 0.75(10cos30o) m + 10 m sin 30o A1 Mass is 0.261 kg A1 [5]

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4 A particle starts from rest at a point X and moves in a straight line until, 60 seconds later, it reaches a point Y. At time t s after leaving X, the acceleration of the particle is 0.75 m s−2 for 0 t 4, < < 0 m s−2 for 4 t 54, < < −0.5 m s−2 for 54 < t < 60. (i) Find the velocity of the particle when t 4 and when t 60, and sketch the velocity-time graph. = = [5] (ii) Find the distance XY. [2]

Mark scheme: 4 (i) v(4) = 0.75x4 B1 v(54) = v(4) and v(60) = v(54) – 0.5(60 – 54) B1 Velocity is 3 ms–1 when t = 4 and 0 when B1 t = 60 M1 Graph consists of 3 straight line segments with 1st and 3rd having +ve and –ve slopes respectively; v is single valued and continuous throughout, and v(0) = 0. 2nd segment has zero slope; end points of ft incorrect value(s) for v(4) and v(60) segments are seen to be correct{(0,0), (4,3), (54,3), (60,0)} A1ft [5] (ii) [XY = ½ (60 + 50)x3 M1 For using area property for distance 2 or or s1 = ½ a1t12, s2 = u2t2, s3 = ½ a3t3 XY = ½ x0.75x42 + 3x50 – ½ x0.5x62] and XY= s1 + s2 – s3 Distance is 165 m A1 [2] GCE AS/A LEVEL – October/November 2010 9709 43 2 2 2 2 2 2

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5 A force of magnitude F N acts in a horizontal plane and has components 27.5 N and −24 N in the x-direction and the y-direction respectively. The force acts at an angle of α◦below the x-axis. (i) Find the values of F and α. [4] A second force, of magnitude 87.6 N, acts in the same plane at 90◦anticlockwise from the force of magnitude F N. The resultant of the two forces has magnitude R N and makes an angle of θ◦with the positive x-axis. (ii) Find the values of R and θ. [3]

Mark scheme: 5 (i) [F2 = 27.52 + (–24)2] M1 For using F2 = X2 + Y2 (may be scored in (ii)) F = 36.5 A1 o = –(–24/27.5)] M1 For using tanα o = –Y/X [tanα α = 41.1 A1 [4] (ii) R = 94.9 B1 o o + θ = tan–1(87.6/36.5); M1 For using tan(α o + θ o) = 87.6/F [α o o or (α + θ o) = cos–1 (36.5/94.9) or cos(α + θ o) = F/R o o or θ = tan–1(87.6sin48.9o – 24)/(27.5 + or tanθ = Y/X 87.6cos48.9o)] θ = 26.3 A1ft [3] ft 67.4 – incorrect α i ( ) & ( )

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6 A particle travels along a straight line. It starts from rest at a point A on the line and comes to rest again, 10 seconds later, at another point B on the line. The velocity t seconds after leaving A is 0.72t2 −0.096t3 for 0 ≤t ≤5, 2.4t −0.24t2 for 5 ≤t ≤10. (i) Show that there is no instantaneous change in the acceleration of the particle when t 5. [4] = (ii) Find the distance AB. [4] [Question 7 is printed on the next page.]

Mark scheme: 6 (i) M1 For using a(t) = v& (t) a1(t) = 1.44t – 0.288t2, a2(t) = 2.4 – 0.48t A1 [a1 = 1.44x5 – 0.288x25, a2 = 2.4 – 0.48x5] M1 For evaluating a1(5) and a2(5) a1 = a2 (= 0) no instantaneous change A1 [4] (ii) M1 For using s = ∫vdt s1 = 0.24t3 – 0.024t4, s2 = 1.2t2 – 0.08t3 A1 [{(0.24x53 – 0.024x54) – (0 – 0)} + M1 For using limits 0 to 5 and 5 to 10 or {(1.2x102 – 0.08x103) – (1.2x52 – 0.08x53)}] equivalent Distance is 35 m A1 [4]

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7 A car of mass 1250 kg travels along a horizontal straight road. The power of the car’s engine is constant and equal to 24 kW and the resistance to the car’s motion is constant and equal to R N. The car passes through the point A on the road with speed 20 m s−1 and acceleration 0.32 m s−2. (i) Find the value of R. [3] The car continues with increasing speed, passing through the point B on the road with speed 29.9 m s−1. The car subsequently passes through the point C. (ii) Find the acceleration of the car at B, giving the answer in m s−2 correct to 3 decimal places. [2] (iii) Show that, while the car’s speed is increasing, it cannot reach 30 m s−1. [2] (iv) Explain why the speed of the car is approximately constant between B and C. [1] (v) State a value of the approximately constant speed, and the maximum possible error in this value at any point between B and C. [1] The work done by the car’s engine during the motion from B to C is 1200 kJ. (vi) By assuming the speed of the car is constant from B to C, find, in either order, (a) the approximate time taken for the car to travel from B to C, (b) an approximation for the distance BC. [4]

Mark scheme: 7 (i) DF = 24000/20 B1 [DF – R = 1250x0.32] M1 For using Newton’s second law (3 terms) R = 800 A1 [3] (ii) 24000/29.9 – 800 = 1250a B1 Acceleration is 0.002 ms–2 B1 [2] (iii) [a = (24000/30 – 800)/1250 M1 For finding a when v = 30 or for using 24000/v – 800 > 0 v < 30] a > 0 to obtain an inequality for v Car not accelerating when v = 30 or Speed cannot reach 30 ms–1 A1 [2] AG (iv) 29.9 ≤ v < 30 speed approximately constant B1 [1] (v) 30 ms–1 (max error 0.1) or 29.95 ms–1 (max error 0.05) or 29.9 ms–1 (max error 0.1) B1 [1] (vi) (a) [24 = 1200/T] M1 For using P = ∆WD/∆t Time taken is 50 s A1 (b) [s = 30x50 or 29.95x50 or 29.9x50] M1 For using s = vt Distance BC is 1500 m or 1500 m or 1495 m A1 [4] GCE AS/A LEVEL – October/November 2010 9709 43 ALTERNATIVE FOR PART (vi) (b) [1200 000 = 800d] M1 For using ‘no change in KE’ WD by car’s engine = WD against resistance’ (may be implied) Distance BC is 1500 m A1 (a) [t = 1500/30 or 1500/29.95 or 1500/29.9] M1 For using t = s/v Time taken is 50 s or 50.1 s or 50.2 s A1

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