Cambridge A Level Mathematics 9709 — 2018 Feb/March Paper 5 · Variant 2

1 A uniform rectangular block has a square base ABCD with AB = BC = 0.4 m. The height of the block is h m. The block is placed with its base on a rough plane inclined at 30° to the horizontal. The block does not slide. It is given that the block is on the point of toppling when the diagonal AC lies along a line of greatest slope. Calculate h. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 d = 2 (0.2 + 0.22) (= 0.2828) OR AC = 2 (0.4 +0.42) (= 0.56568..) B1 Note d = 1 2 AC tan30 = 0.2828 / (h / 2) M1 h = 0.98(0) A1 2 6 / 5 3

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2 An object is projected with speed 15 m s−1 at an angle of 35° above the horizontal from a point on horizontal ground. Find the speed and direction of motion of the object at time 2 s after the instant of projection. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 u = 15cos35 ( = 12.287) B1 Use horizontal motion v = 15sin35 – 2g ( = –11.396) B1 Use vertical motion V = 2 (12.287 + 11.3962) OR tanθ = ±11.396 / 12.287 M1 V = 16.8 m s–1 A1 θ = 42.8° below the horizontal A1 5

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3 A small ball B is connected to one end of a light elastic string of natural length 0.4 m and modulus of elasticity 12 N. The other end of the string is attached to a fixed point A. The ball is projected with speed 1 m s−1 vertically downwards from a position 0.4 m vertically below A, and reaches its greatest speed at the point 0.7 m below A. (i) Show that the mass of B is 0.9 kg. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Calculate the greatest speed of B. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) mg = 12(0.7 – 0.4) / 0.4 M1 Use T = λx / L m = 0.9 kg AG A1 2 Question Answer Marks Guidance 3(ii) EPE = 12(0.7 – 0.4)2 / (2 × 0.4) B1 Correct EPE term 0.9v2 / 2 = 0.9g(0.7 – 0.4) + 0.9 × 12 / 2 – 12(0.7 – 0.4)2 / (2 × 0.4) M1 Attempts a 4 term energy equation A1 Correct equation v = 2 m s–1 A1 4

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4 A particle P is projected from a point O on horizontal ground. At the instant t s after projection, the horizontal and vertically upwards displacements of P from O are x m and y m respectively. The equation of the trajectory of P is y = 3x −0.05x2. (i) Find the angle of projection and the initial speed of P. 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(ii) Find the coordinates of P at the instant when OP makes an angle of 45° with the horizontal. 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(iii) For the instant when P is at its greatest height above the ground, calculate this height and the corresponding value of t. 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Mark scheme: 4(i) (tanθ = 3) θ = 71.6° B1 Use the formula sheet for the trajectory equation 0.05 = g / (2V 2cos271.6) M1 V = 10 10 = 31.6 m s–1 A1 3 4(ii) x = 3x – 0.05x2 M1 Use y = x x = 40 and y = 40 A1 2 Question Answer Marks Guidance 4(iii) dy / dx = 3 – 0.1x = 0 M1 Use the fact that the gradient is zero at the highest point x = 30, y = (3 × 30 – 0.05 × 302) = 45 A1 30 = (31.6cos71.6)t M1 Use horizontal motion t = 3.01 A1 t = 3 if exact arithmetic used 4

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5 0.4 m P 30Å 0.4 m One end of a light inextensible string of length 0.4 m is attached to the lowest point of a hemisphere of radius 0.4 m fixed with its axis vertical. A particle P of mass 0.3 kg is attached to the other end of the string. The string is straight and makes an angle of 30° with the horizontal. P moves on the smooth inner surface of the hemisphere in a horizontal circle (see diagram). (i) Calculate the smallest possible angular speed of P. 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(ii) Given that the greatest possible tension in the string is 5 N, calculate the greatest possible speed of P. 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Mark scheme: 5(i) EITHER: Rcos60 = 0.3g (M1 Resolve vertically R = 6 N A1 6cos30 = 0.3ω 2 × 0.4cos30 M1 Use Newton's Second Law horizontally ω = 5 2 = 7.07 rad s–1 A1) OR: 0.3gcos30 = 0.3 × (0.4cos30)ω 2cos60 (M1 Resolve along the tangent A1 Correct equation ω = 5 2 = 7.07 rad s–1 M1 Attempt to solve for ω A1) 4 Question Answer Marks Guidance 5(ii) Rcos60 = 0.3g + 5sin30 M1 Resolve vertically R = 11 N A1 11cos30 + 5cos30 = 0.3v2 / (0.4cos30) M1 Resolve horizontally v = 4 m s–1 A1 4

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6 1 v m s−1 P N O A small object of mass 0.2 kg rests at a point O on a rough horizontal surface. The coefficient of friction between the object and the surface is 0.5. A force of magnitude P N acting at an angle 1 below the horizontal is applied to the object. The velocity of the object is v m s−1 away from O at time t s after the force begins to act (see diagram). It is given that tan 1 = 3 and that P = 0.4t for 0 ≤t ≤8. 4 (i) Find the value of t when the object starts to move. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ dv (ii) Show that, when the force is acting and the object is in motion, = t −5. 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When t = 8 the force of magnitude P N ceases to act. (iii) Find the distance travelled by the object after t = 8 before it comes to rest. 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Mark scheme: 6(i) R = 0.2g + 0.4tsinθ ( = 2 + 0.24t) F = 0.5(2 + 0.24t) = 1 + 0.12t M1 Resolve vertically and use F = µR 0.4tcosθ = 1 + 0.12t M1 Resolve horizontally t = 5 A1 3 6(ii) 0.2dv/dt = 0.4t × 0.8 – (1 + 0.12t) M1 Use Newton's Second Law horizontally dv / dt = t – 5 AG A1 2 Question Answer Marks Guidance 6(iii) d ∫v = ( ) 5 d ∫ − t t v = t2 / 2 – 5t + c M1 Attempt to integrate the equation from part(ii) v = 0 when t = 5 hence c = 12.5 A1 Finds the constant of integration, c v = 82 / 2 – 5 × 8 + 12.5 = 4.5 A1 Find v when t = 8 a = −0.5 × 0.2g / 0.2 = –5 m s–1 and s = 4.52 / (2 × 5) M1 Finds a and uses 2 v = 2 u + 2as s = 2.025 m A1 5

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7 A 48Å 0.6 m D r m B 0.25 m 0.3 m 0.3 m C Fig. 1 ABCD is a uniform square lamina with sides of length 0.6 m. A circular hole of radius r m is made in the lamina. The centre of the hole is 0.3 m from AB and 0.25 m from AD. The lamina is freely suspended at A and hangs with the axis of symmetry making an angle of 48° with the horizontal (see Fig. 1). (i) Show that r = 0.214, correct to 3 significant figures. 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........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 15 N 60Å A D B C Fig. 2 The lamina is held in equilibrium with AD horizontal by a force of magnitude 15 N acting in the plane of the lamina applied at D. The line of action of this force makes an angle of 60° with the vertical (see Fig. 2). (ii) Find the weight of the original square lamina, before the hole was made. 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Mark scheme: 7(i) tan48 = x /0.3 M1 x is the distance of the centre of mass from AD x = 0.3332 A1 0.62 × 0.3 = πr 2 × 0.25 + (0.62 – πr 2) x OR 0.62 × 0.3 = πr 2 × 0.35 + (0.62 – πr 2) × (0.6 – x ) M1 Take moments about AD Take moments about BC πr 2 × (0.3332 – 0.25) = 0.62 × (0.3332 – 0.3) OR πr 2(0.6 – 0.3332 – 0.35) = 0.62 (0.6 – 0.3332 – 0.3) A1 r = 0.214 AG A1 5 Question Answer Marks Guidance 7(ii) 0.3W = 0.6 × 15cos60 M1 Take moments about C.(W = weight of the lamina) W = 15 A1 Square = 15 × 0.62 / (0.62 – π × 0.2142) M1 Recognise that the ratio of weights = ratio of areas Square = 25(.0) N A1 4

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