Cambridge A Level Mathematics 9709 — 2012 May/June Paper 4 · Variant 2

1 A block is pulled in a straight line along horizontal ground by a force of constant magnitude acting at an angle of 60◦above the horizontal. The work done by the force in moving the block a distance of 5 m is 75 J. Find the magnitude of the force. [3]

Mark scheme: 1 M1 For using WD = Fdcosα F × 5cos60o = 75 A1 Magnitude of the force is 30 N A1 [3]

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2 12 N F N P a° 15 N Three coplanar forces of magnitudes F N, 12 N and 15 N are in equilibrium acting at a point P in the directions shown in the diagram. Find α and F. [4]

Mark scheme: 2 [12 = 15sinα ] M1 For resolving forces in the direction of the force of magnitude 12 N α = 53.1 A1 [F = 15cosα ] M1 For resolving forces in the direction of the force of magnitude F N F = 9 N A1 [4] 2 ALTERNATIVE 1 [Fsin α = 12cosα and Fcosα + 12sinα M1 For resolving forces in the x and y = 15 sinα ÷ cosα = directions and eliminating F from the 12cosα ÷ 15 – 12sinα resultant equations 15sinα – 12 sin2α = 12cos2α 15sinα A1 = 12 α = 53.1 M1 For substituting into Fsin α = 12cosα or Fcosα +12sinα =15 F = 9 N A1 [4] 2 ALTERNATIVE 2 [sin α =12/15] M1 For using correct triangle of forces to find α α = 53.1 A1 [F2 = 152 – 122] M1 For using correct triangle of forces to find F F = 9 N A1 [4] 2 ALTERNATIVE 3 [12 ÷ sin(180 – α ) = 15 ÷ sin90 M1 For using Lami’s rule and 12 = 15sinα ] sin (180o – α) = sinα α = 53.1 A1 [F ÷ sin 143.1 = 15 ÷ sin90] M1 For using Lami’s rule and value of α to find F F = 9 N A1 [4] SR (max 2/4) For candidates who have sin and cos interchanged. Allow B1 for α = 36.9 and allow B1 for F = 9 following correct work relative to the cos/sin interchange error. GCE AS/A LEVEL – May/June 2012 9709 42

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3 A particle P moves in a straight line, starting from the point O with velocity 2 m s−1. The acceleration 2 of P at time t s after leaving O is 2t 3 m s−2. (i) Show that t 5 3 = 5 when the velocity of P is 3 m s−1. [4] 6 (ii) Find the distance of P from O when the velocity of P is 3 m s−1. [3]

Mark scheme: 3 (i) M1 For an attempt to find v(t) using integration of a(t) v = 1.2t5/3 + 2 A1 DM1 For attempting to solve v(t) = 3 for t5/3 or For confirming v = 3 by substituting t5/3 = 5/6 into the expression found for v(t) t5/3 = 5/6 A1 [4] AG (ii) M1 For integrating and using s(0) = 0 (may be implied by absence of +C) to find s(t) s = 0.45t8/3 + 2t A1 Distance is 2.13 m A1 [3]

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4 25° T N A ring of mass 4 kg is attached to one end of a light string. The ring is threaded on a fixed horizontal rod and the string is pulled at an angle of 25◦below the horizontal (see diagram). With a tension in the string of T N the ring is in equilibrium. (i) Find, in terms of T, the horizontal and vertical components of the force exerted on the ring by the rod. [4] The coefficient of friction between the ring and the rod is 0.4. (ii) Given that the equilibrium is limiting, find the value of T. [3]

Mark scheme: 4 (i) M1 For resolving forces horizontally Horizontal component is Tcos25o (0.906T) A1 M1 For resolving forces vertically Vertical component is 4g + Tsin 25o (40 + 0.423T) A1 [4] (ii) M1 For using F = 0.4R 0.906T = 16 + 0.169T A1ft May be implied by correct answer for T T = 21.7 N A1 [3]

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5 O S2 B 2 kg S1 A 3 kg A block A of mass 3 kg is attached to one end of a light inextensible string S1. Another block B of mass 2 kg is attached to the other end of S1, and is also attached to one end of another light inextensible string S2. The other end of S2 is attached to a fixed point O and the blocks hang in equilibrium below O (see diagram). (i) Find the tension in S1 and the tension in S2. [2] The string S2 breaks and the particles fall. The air resistance on A is 1.6 N and the air resistance on B is 4 N. (ii) Find the acceleration of the particles and the tension in S1. [5]

Mark scheme: 5 (i) Tension in S1 is 30 N B1 Tension in S2 is 50 N B1 [2] (ii) M1 For applying Newton’s second law to A or to B 3g – T – 1.6 = 3a (or 2g + T – 4 = 2a) A1 2g + T – 4 = 2a (or 3g – T – 1.6 = 3a) or (3g + 2g) – (1.6 + 4) = (3 + 2)a B1 Acceleration is 8.88 ms–2 B1 Tension is 1.76 N A1 [5] SR (max. 1 / 2) for candidates who do not give numerical answers in (i). Allow B1 for Tension in S1 is 3g and Tension in S2 is 5g GCE AS/A LEVEL – May/June 2012 9709 42

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6 A car of mass 1250 kg travels from the bottom to the top of a straight hill which has length 400 m and is inclined to the horizontal at an angle of α, where sin α = 0.125. The resistance to the car’s motion is 800 N. Find the work done by the car’s engine in each of the following cases. (i) The car’s speed is constant. [4] (ii) The car’s initial speed is 6 m s−1, the car’s driving force is 3 times greater at the top of the hill than it is at the bottom, and the car’s power output is 5 times greater at the top of the hill than it is at the bottom. [5]

Mark scheme: 6 (i) PE gain = 1250 × 10 x 400 × 0.125 B1 WD against resistance is 800 × 400 J B1 M1 For using WD by car’s engine = Gain in PE + WD against resistance WD by car’s engine is 945 000 J (945 kJ) A1 [4] (ii) For using P = Fv v 2 P2 F 1 = × [v2/6 = 5 × (1/3)] M1 v 1 P1 F 2 v2 = 10 A1 KE gain = ½ 1250(102 – 62) B1ft [WD by car’s engine = 945 000 + 40 000] M1 For using WD by car’s engine = (Gain in PE + WD against resistance) + KE gain WD by car’s engine is 985 000 J (985 kJ) A1ft [5] ft incorrect ans(i) Alternative scheme for part (i) (i) M1 For using Newton’s second law with a = 0 DF = 1250g × 0.125 + 800 A1 M1 For using WD = DF × 400 WD by car’s engine is 945 00 J (945 kJ) A1 [4] GCE AS/A LEVEL – May/June 2012 9709 42 d

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7 3 m s–1 X Z Y The frictional force acting on a small block of mass 0.15 kg, while it is moving on a horizontal surface, has magnitude 0.12 N. The block is set in motion from a point X on the surface, with speed 3 m s−1. It hits a vertical wall at a point Y on the surface 2 s later. The block rebounds from the wall and moves directly towards X before coming to rest at the point Z (see diagram). At the instant that the block hits the wall it loses 0.072 J of its kinetic energy. The velocity of the block, in the direction from X to Y, is v m s−1 at time t s after it leaves X. (i) Find the values of v when the block arrives at Y and when it leaves Y, and find also the value of t when the block comes to rest at Z. Sketch the velocity-time graph. [9] (ii) The displacement of the block from X, in the direction from X to Y, is s m at time t s. Sketch the displacement-time graph. Show on your graph the values of s and t when the block is at Y and when it comes to rest at Z. [4]

Mark scheme: 7 (i) [– 0.12 = 0.15a] M1 For using Newton’s 2nd law a = –0.8 ms–2 A1 [v = 3 – 0.8 × 2] M1 For using v = u + at to find speed of approach vapproach = 1.4 A1 2 2 [½ 0.15(1.42 – vr )] M1 For using KE loss = ½ m(va – vr2) vreturn = – 1 A1 M1 For using 0 = vreturn + a(t – 2) t = 3.25 s when block comes to rest A1 Alternative for the M1 A1 immediately above. tYZ = 1.25 B1 t =3.25s when block is at rest B1ft ft incorrect values of v and t (although For correct sketch B1ft [9] vreturn must be negative) (ii) [XY = ½ (3 + 1.4) × 2, YZ = ½ 1.25 × 1] M1 For using area property (or equivalent) to find distances XY and YZ s = 4.4 at Y and 3.775 at Z, stated or on A1 (accept 3.77 or 3.78) graph Curve starts at origin, s increases, slope B1ft ft incorrect value for s(2) decreases (convex upwards) for 0 < t < 2, value of s(2) shown Curve starts at (2, 4.4), s decreases, magnitude of slope decreases to zero at (3.25, 3.775) B1ft [4] ft incorrect values of s and t

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