Cambridge A Level Mathematics 9709 — 2008 Oct/Nov Paper 5 · Variant 1

1 One end of a light elastic rope of natural length 2.5 m and modulus of elasticity 80 N is attached to a fixed point A. A stone S of mass 8 kg is attached to the other end of the rope. S is held at a point 6 m vertically below A and then released. Find the initial acceleration of S. [4]

Mark scheme: 1 [T = 80x3.5/2.5 (= 112)] M1 For using T = λ x / L M1 For using Newton’s second law 8a = T – 8g A1 Acceleration is 4 ms −2 A1 4 [4]

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2 h cm 24 cm r cm r cm A uniform solid cylinder has height 24 cm and radius r cm. A uniform solid cone has base radius r cm and height h cm. The cylinder and the cone are both placed with their axes vertical on a rough horizontal plane (see diagram, which shows cross-sections of the solids). The plane is slowly tilted and both solids remain in equilibrium until the angle of inclination of the plane reaches α◦, when both solids topple simultaneously. (i) Find the value of h. [2] (ii) Given that r = 10, find the value of α. [2]

Mark scheme: 2 (i) [r /( h / 4) = r /( 24 / 2]) M1 For using r / y cone = r / y cylinder h = 48 A1 2 (ii) tanα = 10/12 M1 For using tanα o = r/ y α = 39.8 A1 2 [4]

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3 A particle P of mass 0.5 kg moves along the x-axis on a horizontal surface. When the displacement of P from the origin O is x m the velocity of P is v m s−1 in the positive x-direction. Two horizontal forces act on P; one force has magnitude (1 + 0.3x2) N and acts in the positive x-direction, and the other force has magnitude 8e−x N and acts in the negative x-direction. dv (i) Show that v = 2 + 0.6x2 −16e−x. [2] dx (ii) The velocity of P as it passes through O is 6 m s−1. Find the velocity of P when x = 3. [5]

Mark scheme: 3 (i) [0.5a = 1 + 0.3x 2 – 8e −]x M1 For using Newton’s second law v(dv/dx) = 2 + 0.6x 2 – 16e − x A1 2 (ii) M1 For separating variables and integrating v 2 /2 = 2x + 0.2x 3 + 16e − x (+c) A1 M1 For using v(0) = 6 v 2 /2 = 2x + 0.2x 3 + 16e − x + 2 A1 Velocity is 5.33 ms −1 A1 5 [7]

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4 (i) A 0.35 m Fig. 1 A small sphere A of mass 0.15 kg is moving inside a fixed smooth hollow cylinder whose axis is vertical. A moves with constant speed 1.2 m s−1 in a horizontal circle of radius 0.35 m, and is continuously in contact with both the plane base and the curved surface of the cylinder. Fig. 1 shows a vertical cross-section of the cylinder through its axis. Find the magnitude of the force exerted on A by (a) the base of the cylinder, (b) the curved surface of the cylinder. [3] (ii) 0.2 m A B Fig. 2 Sphere A is now attached to one end of a light inextensible string. The string passes through a small smooth hole in the middle of the base of the cylinder. Another small sphere B, of mass 0.25 kg, is attached to the other end of the string. B hangs in equilibrium below the hole while A is moving in a horizontal circle of radius 0.2 m (see Fig. 2). Find the angular speed of A. [4] [Questions 5, 6 and 7 are printed on the next page.]

Mark scheme: 4 (i) (a) Magnitude is 1.5 N B1 From R = mg (b) [S = mv 2 /r] M1 For using Newton’s second law Magnitude is 0.617 N A1 3 (ii) Tension is 2.5 N B1 [T = mrω 2 ] M1 For using Newton’s second law 2.5 = 0.15x0.2 ω 2 A1ft Angular speed is 9.13 rads −1 A1 4 [7]

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5 B 1.2 m A 0.8 m C E 20° D ABCD is a central cross-section of a uniform rectangular block of mass 35 kg. The lengths of AB and BC are 1.2 m and 0.8 m respectively. The block is held in equilibrium by a rope, one end of which is attached to the point E of a rough horizontal floor. The other end of the rope is attached to the block at A. The rope is in the same vertical plane as ABCD, and EAB is a straight line making an angle of 20◦with the horizontal (see diagram). (i) Show that the tension in the rope is 187 N, correct to the nearest whole number. [5] (ii) The block is on the point of slipping. Find the coefficient of friction between the block and the floor. [4]

Mark scheme: 5 (i) Moment of W about D = W(0.4 2 + 0.6 2 ) 1 / 2 cos (20 o + tan −1 23 ) or W(0.6cos20 o – 0.4sin20 o ) = (0.427W) B2 M1 For taking moments about D 0.8T = 350x0.427 A1ft Tension is 187 N A1 5 (ii) R = 350 + Tsin20 o B1 F = Tcos20 o B1 [µ = 176 / 414 ] M1 For using µ = F/R Coefficient is 0.424 A1 4 [9] GCE A/AS LEVEL – October/November 2008 9709 05 FIRST ALTERNATIVE 5 Moment of W about E =  o 2 2 12 o −1 2  W  8.0 / sin 20 + ( 4.0 + 6.0 ) cos( 20 + tan )  3   B2 M1 For taking moments about E 2.34R = 2.766x350 A1ft R = 350 + Tsin20 o B1 (i) Tension is 187 N A1 (ii) F = Tcos20 o B1 [µ = 176 / 414 ] M1 For using µ = F/R Coefficient is 0.424 A1 9 [9] SECOND ALTERNATIVE 5 Distance of line of action of R from G = 1 (0.4 2 + 0.6 2 ) 2 cos(20 o + tan −1 23 ) and distance of line of action of F from G = 1 (0.4 2 + 0.6 2 ) 2 sin(20 o + tan −1 23 ) B2 M1 For taking moments about G ,the centre of mass of the block 0.4T + 0.581F = 0.427R A1ft R = 350 + Tsin20 o B1 F = Tcos20 o B1 (i) Tension is 187 N A1 (ii) [µ = 176 / 414 ] M1 For using µ = F/ R Coefficient is 0.424 A1 9 [9]

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6 A light elastic string has natural length 4 m and modulus of elasticity 2 N. One end of the string is attached to a fixed point O of a smooth plane which is inclined at 30◦to the horizontal. The other end of the string is attached to a particle P of mass 0.1 kg. P is held at rest at O and then released. The speed of P is v m s−1 when the extension of the string is x m. (i) Show that v2 = 45 −5(x −1)2. [5] Hence find (ii) the distance of P from O when P is at its lowest point, [2] (iii) the maximum speed of P. [2]

Mark scheme: 6 (i) EE gain = 2x 2 /(2x4) B1 PE loss = 0.1g(4 + x)sin30 o B1 [ 12 0.1v 2 = 2 + 0.5x – 0.25x 2 ] M1 For using KE gain=PE loss-EE gain v 2 = 40 + 10x – 5x 2 = 45 – (5x 2 – 10 x + 5) M1 For attempting to express v 2 in the required form v 2 = 45 – 5(x – 1) 2 A1 5 AG (ii) 5(x – 1) 2 = 45 M1 For substituting v = 0 Distance is 8 m A1 2 (iii) M1 For using x = 1 for maximum v Maximum speed is 6.71 ms −1 A1 2 [9] GCE A/AS LEVEL – October/November 2008 9709 05

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7 A particle P is projected from a point O on horizontal ground with speed V m s−1 and direction 60◦ upwards from the horizontal. At time t s later the horizontal and vertical displacements of P from O are x m and y m respectively. (i) Write down expressions for x and y in terms of V and t and hence show that the equation of the trajectory of P is y = (√3)x −20x2 . [5] V2 P passes through the point A at which x = 70 and y = 10. Find (ii) the value of V, [2] (iii) the direction of motion of P at the instant it passes through A. [3]

Mark scheme: 7 (i) x = Vtcos60 o B1 y = Vtsin60 o – 12 gt 2 B1 M1 For eliminating t y = xsin60 o /cos60 – 12 gx 2 /(V 2 cos 2 60 o ) A1 For any correct form 2 2 A1 5 AG y = 3 x – 20x /V (ii) 10 = 3 x70 – 20x70 2 /V 2 M1 For substituting x = 70, y = 10 and attempting to solve for V V = 29.7 A1 2 (iii) [dy/dx = 3 – 40x/V 2 ] M1 For differentiating y(x) Gradient is –1.44… at A A1ft Direction is 55.3 o downwards from the horizontal A1 3 [10] ALTERNATIVE FOR (iii) (iii) . . M1 For attempting to find either x or y at A . . [ x = 148...., y = −21.42... ] A1ft Direction is 55.3 o downwards from the horizontal A1 3

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