Cambridge A Level Mathematics 9709 — 2010 Oct/Nov Paper 5 · Variant 1

1 A horizontal circular disc rotates with constant angular speed 9 rad s−1 about its centre O. A particle of mass 0.05 kg is placed on the disc at a distance 0.4 m from O. The particle moves with the disc and no sliding takes place. Calculate the magnitude of the resultant force exerted on the particle by the disc. [3]

Mark scheme: 1 0.05 x 92 x 0.4 = H M1 H = 1.62 N A1 .1 62 2 + 5.0 2 R = 1.70 N A1 [3]

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2 A 0.8 m O 23 p rad B A bow consists of a uniform curved portion AB of mass 1.4 kg, and a uniform taut string of mass m kg which joins A and B. The curved portion AB is an arc of a circle centre O and radius 0.8 m. Angle AOB is 23π radians (see diagram). The centre of mass of the bow (including the string) is 0.65 m from O. Calculate m. [6]

Mark scheme: 2 OG = 0.8sin(π /3)/(π /3) B1 0.66159 OM = 0.8cos(π /3) B1 0.4 M1 For taking moments about O 0.65(m + 1.4) = 0.4m + 0.66159x1.4 A1 0.25m = 0.01159 x 1.4 M1 For collecting like terms m = 0.0649 A1 OR OG = 0.8sin(π /3)/(π /3) B1 0.66159 OM = 0.8cos(π /3) B1 0.4 M1 Taking moments about M (1.4 + m) × 0.25 = 1.4 × 0.26159 A1 0.25m = 1.4x0.01159 M1 For collecting like terms m = 0.0649 A1 [6] 2

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3 A 0.2 m 30° P One end of a light inextensible string of length 0.2 m is attached to a fixed point A which is above a smooth horizontal surface. A particle P of mass 0.6 kg is attached to the other end of the string. P moves in a circle on the surface with constant speed v m s−1, with the string taut and making an angle of 30◦to the horizontal (see diagram). (i) Given that v = 1.5, calculate the magnitude of the force that the surface exerts on P. [4] (ii) Given instead that P moves with its greatest possible speed while remaining in contact with the surface, find v. [3]

Mark scheme: 3 (i) 0.6x1.52/(0.2cos30°) = Tcos30° M1 Uses N2L horizontally with component of tension T = 9 N A1 R = 0.6g – 9sin30° M1 Resolves vertically, 3 terms R = 1.5 N A1 [4] (ii) Tsin30° = 0.6g M1 Resolves vertically, 2 terms 0.6v2/(0.2cos30°) = 12cos30° M1 v2 = 3, v =1.73 A1 [3] GCE A LEVEL – October/November 2010 9709 51

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4 B 1.7 m 220 N A 0.8 m 70 N A uniform beam AB has length 2 m and weight 70 N. The beam is hinged at A to a fixed point on a vertical wall, and is held in equilibrium by a light inextensible rope. One end of the rope is attached to the wall at a point 1.7 m vertically above the hinge. The other end of the rope is attached to the beam at a point 0.8 m from A. The rope is at right angles to AB. The beam carries a load of weight 220 N at B (see diagram). (i) Find the tension in the rope. [3] (ii) Find the direction of the force exerted on the beam at A. [4]

Mark scheme: 4 (i) T x 0.8 = 70x1sinα + 220x2sinα M1 Moments about A (3 terms) sinα = 1.5/1.7 A1 cosα = 0.8/1.7 α = 61.9° A1 T = 562.5 N [3] (ii) H = 562.5cosα = 265 N B1 H = 264.70 N V = 562.5sinα – 70 – 220 M1 V = 206.3 N tanα = 265/206.3 M1 α = 52.1° (with vertical) A1 Or 37.9 (with horizontal) OR X = (70+220)cosα = 136.6 B1 Resolving along the rod AB Y = 562.5 – (70+220)sinα = 306.7 M1 Resolving perpendicular to AB tanθ = 306.7/136.6 M1 θ = 65.99° or 66.0° (with beam) A1 [4]

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5 A particle P of mass 0.28 kg is attached to the mid-point of a light elastic string of natural length 4 m. The ends of the string are attached to fixed points A and B which are at the same horizontal level and 4.8 m apart. P is released from rest at the mid-point of AB. In the subsequent motion, the acceleration of P is zero when P is at a distance 0.7 m below AB. (i) Show that the modulus of elasticity of the string is 20 N. [4] (ii) Calculate the maximum speed of P. [3]

Mark scheme: 5 (i) 2Tcosθ = 0.28g M1 Tension component = weight 2T x 0.7/2.5 = 2.8, T = 5 A1 5 = λ x 0.5/2 M1 Hookes Law λ = 20 N A1 [4] (ii) 0.28v2/2 + 2x20x0.52 /(2x2) = M1 PE/EE/KE conservation with 4 terms 0.28gx0.7 +2x20x0.42/(2x2) A1 v = 2.75 ms–1 A1 [3] GCE A LEVEL – October/November 2010 9709 51

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7 10 m s–1 A 45° 30° O A particle P is projected from a point O with initial speed 10 m s−1 at an angle of 45◦above the horizontal. P subsequently passes through the point A which is at an angle of elevation of 30◦from O (see diagram). At time t s after projection the horizontal and vertically upward displacements of P from O are x m and y m respectively. (i) Write down expressions for x and y in terms of t, and hence obtain the equation of the trajectory of P. [3] (ii) Calculate the value of x when P is at A. [3] (iii) Find the angle the trajectory makes with the horizontal when P is at A. [4]

Mark scheme: 7 (i) x = (10cos45°)t and y = (10sin45°)t – gt2 /2 B1 y = (10sin45° / 10cos45°)x – 10(x/10cos45°)2/2 M1 y = x – x2/10 A1 [3] (ii) y/x = tan30° M1 1 – x/10 = tan30° A1 x = 4.23 A1 4.2264… [3] (iii) dy/dx = 1 – 2x/10 M1 4.2264 = (10cos45°)t tanθ = dy/dx B1 t = 0.5977 10 sin 45° − 10 x .05977 tanθ =1 – 2x4.23/10(= 0.15472..) M1 tanθ = 10 cos 45° θ = 8.79° A1 [4]

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