Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 5 · Variant 2

1 A small ball B is projected with speed 38 m s−1 at an angle of 30° to the horizontal from a point on horizontal ground. Find the speed of B when the path of B makes an angle of 20° above the horizontal. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 M1 Equate initial horizontal velocity to final horizontal velocity vcos20 = 38cos30 A1 v = 35(.0) m s–1 A1 3

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2 A uniform solid object is made by attaching a cone to a cylinder so that the circumferences of the base of the cone and a plane face of the cylinder coincide. The cone and the cylinder each have radius 0.3 m and height 0.4 m. (i) Calculate the distance of the centre of mass of the object from the vertex of the cone. [4] [The volume of a cone is 130r2h.] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The object has weight W N and is placed with its plane circular face on a rough horizontal surface. A force of magnitude kW N acting at 30° to the upward vertical is applied to the vertex of the cone. The object does not slip. (ii) Find the greatest possible value of k for which the object does not topple. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) M1 Attempt to take moments about the vertex of the cone (π × 0.32 × 0.4/3) × (3 × 0.4/4) + (π × 0.32 × 0.4 × (0.4 + 0.2)) A1 = (π × 0.32 × 0.4/3 + π × 0.32 × 0.4) x A1 x = 0.525 m A1 4 2(ii) M1 Attempt to take moments about a point on the circumference of the base of the cone kWcos30 × 0.3 + kWsin30 × 0.8 = 0.3W A1 k = 0.455 A1 3

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3 A particle P of mass 0.4 kg is projected horizontally along a smooth horizontal plane from a point O. After projection the velocity of P is v m s−1 and its displacement from O is x m. A force of magnitude 8x N directed away from O acts on P and a force of magnitude 2e−x + 4 N opposes the motion of P. One end of a light elastic string of natural length 0.5 m is attached to O and the other end of the string is attached to P. (i) Show that vdv = 20x −10 −5e−x before the elastic string becomes taut. 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(ii) Given that the initial velocity of P is 6 m s−1, find v when the string first becomes taut. 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When the string is taut, the acceleration of P is proportional to e−x. (iii) Find the modulus of elasticity of the string. 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Mark scheme: 3(i) M1 Use Newton's Second Law horizontally vdv/dx = 20x – 10 – 5e–x A1 2 3(ii) ∫vdv = ) (20 10 5 − ∫ − − x x e dx M1 Attempt to integrate the equation from part (i) 0.5 2 2 6 0 / 2 10 10 5 −     = − +     v x v x x e M1 Use correct limits in the integration v = 5.2(0) A1 3 3(iii) 8x – 4 = T = λ(x – 0.5)/0.5 ( = λx/0.5 – λ) M1 Use T = λx/L λ = 4 N A1 2

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5 A particle P of mass 0.7 kg is attached to a fixed point O by a light elastic string of natural length 0.6 m and modulus of elasticity 15 N. The particle P is projected vertically downwards from the point A, 0.8 m vertically below O. The initial speed of P is 2 m s−1. (i) Find the distance below A of the point at which P comes to instantaneous rest. 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(ii) Find the greatest speed of P in the motion. 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Mark scheme: 5(i) B1 M1 Set up a 4 term energy equation 15(0.2 + x)2/(2 × 0.6) = 15 × 0.22/(2 × 0.6) + 0.7gx + (0.7 × 22)/2 A1 0.5 + 5x + 12.5x2 = 0.5 + 7x + 1.4 x = 0.424 m A1 4 Question Answer Marks Guidance 5(ii) 0.7g = 15e/0.6 M1 Use T = λx/L e = 0.28 m A1 (0.7 × 22)/2 + 0.7g(0.28 – 0.2) + (15 × 0.22)/(2 × 0.6) = (0.7 × v2)/2 + (15 × 0.282)/ (2 × 0.6) M1 Set up a 5 term energy equation v = 2.06 m s–1 A1 4

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6 B E x m D C r m r m O F G H A The diagram shows a uniform lamina ABCDEFGH. The lamina consists of a quarter-circle OAB of radius r m, a rectangle DEFG and two isosceles right-angled triangles COD and GOH. The rectangle has DG = EF = r m and DE = FG = x m. (i) Given that the centre of mass of the lamina is at O, express x in terms of r. 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(ii) Given instead that the rectangle DEFG is a square with edges of length r m, state with a reason whether the centre of mass of the lamina lies within the square or the quarter-circle. 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Mark scheme: 6(i) Centre of mass of triangles below O = r/6 B1 Centre of mass of quadrant below O = (2rsinπ/4)/(3π/4) B1 M1 Attempt to take moments about O (rx)(x/2) = (r2/4)(r/6)+ (π r2/4)(2rsinπ/4)/(3π/4) A1 ( ) 2 2 2 2 / 24 2 / 3 = + x r r M1 Attempt to express x in terms of r x = 1.01r A1 6 Question Answer Marks Guidance 6(ii) Within quadrant as the square will be smaller than the rectangle Or if x ˂ r in part (i), within the square as the square will be larger than the rectangle B1ft 1

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7 0.45 m A B R 0.3 m 0.3 m 60Å 6 rad s−1 P A rough horizontal rod AB of length 0.45 m rotates with constant angular velocity 6 rad s−1 about a vertical axis through A. A small ring R of mass 0.2 kg can slide on the rod. A particle P of mass 0.1 kg is attached to the mid-point of a light inextensible string of length 0.6 m. One end of the string is attached to R and the other end of the string is attached to B, with angle RPB = 60Å (see diagram). R and P move in horizontal circles as the system rotates. R is in limiting equilibrium. (i) Show that the tension in the portion PR of the string is 1.66 N, correct to 3 significant figures. 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(ii) Find the coefficient of friction between the ring and the rod. 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Mark scheme: 7(i) Tcos30 + Ucos30 = 0.1g (= 1) B1 Resolve vertically for P Note T and U are the tensions in PR and PB respectively Tcos60 – Ucos60 = 0.1 × 62 × 0.3 ( = 1.08) M1A1 Use Newton's Second Law horizontally 2Tcos30cos60 = 1.08cos30 + 1cos60 M1 Attempt to eliminate U T ( = 1.65735) = 1.66 N AG A1 5 7(ii) F – Tcos60 = 0.2 × 62 × 0.15 or R = 0.2g + Tcos30 M1 Use Newton's Second Law horizontally or resolve vertically F ( = 1.9086) = 1.91 N A1 R ( = 3.4353) = 3.44 N A1 µ = 1.9086/3.4353 M1 Use F = µR µ = 0.556 A1 Accept µ = 0.56 5

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