Cambridge A Level Mathematics 9709 — 2019 May/June Paper 5 · Variant 1

1 A 0.8 m 0.15 m O P v m s−1 A particle P of mass 0.3 kg is attached to a fixed point A by a light inextensible string of length 0.8 m. The fixed point O is 0.15 m vertically below A. The particle P moves with constant speed v m s−1 in a horizontal circle with centre O (see diagram). (i) Show that the tension in the string is 16 N. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the value of v. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) 0.15 cos 0.3 0.8 T T g θ   = × =     M1 the vertical T = 16 N AG A1 2 1(ii) 2 2 2 0.8 0.15 r = − B1 r = 0.78581... 2 0.78581... 0.3 16sin 16 0.8 0.78581... v θ   = × =     M1 Use Newton’s Second Law horizontally v = 6.416 A1 3

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2 A particle is projected with speed V m s−1 at an angle of 1Å above the horizontal. At the instant 4 s after projection the speed of the particle is 16 m s−1 and its direction of motion is 30Å above the horizontal. Find V and 1. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 cos 16cos30 8 3 13.856... V θ = = = B1 ( ) sin 16sin30 4 48 V g θ = + = B1 Use vertical motion ( ) ( ) 2 2 2 16cos30 16sin30 4 V g = + + OR ( ) 16sin30 4 tan 16cos30 g θ + = M1 Use Pythagoras’s theorem or trigonometry of a right angled triangle V = 50(.0) A1 θ = 73.9° A1 5

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3 0.2 m A 0.2 m 0.7 m The diagram shows the cross-section through the centre of mass of a uniform solid object. The object is a cylinder of radius 0.2 m and length 0.7 m, from which a hemisphere of radius 0.2 m has been removed at one end. The point A is the centre of the plane face at the other end of the object. Find the distance of the centre of mass of the object from A. [5] [The volume of a hemisphere is 230r3.] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Volume of hemisphere = ( ) 3 0.2 2π 0.0053333π 3 × = B1 Distance of centre of mass from object base ( ) 0.2 0.7 3 0.625 8 = −× = B1 3 3 2 0.2 0.2 0.2 π 0.2 0.7 2π 0.7 3 2π 0.35 0.028π 3 8 3 x    × × − × + −× × × = ×         M1A1 Take moments about the plane face x = 0.285 m A1 5

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4 A small ball is projected with speed 25 m s−1 at an angle of 30Å above the horizontal from a point O on horizontal ground. At time t s after projection the horizontal and vertically upwards displacements of the ball from O are x m and y m respectively. (i) Express x and y in terms of t and hence find the equation of the trajectory of the ball. 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(ii) Find x for the position of the ball when its path makes an angle of 15Å below the horizontal. 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Mark scheme: 4(i) B1 2 25sin30 2 gt y t = − B1 Use vertical motion 2 25cos30 25sin30 25cos30 2 x g x y         = −     M1 Eliminate t 2 4 375 3 x x y = − or 2 0.577 0.0107 y x x = − A1 4 Question Answer Marks Guidance 4(ii) d 1 8 d 375 3 y x x = − or d 0.577 0.0214 d y x x = − M1A1 Differentiate the equation from part (i) to find the gradient 1 8 tan15 375 3 x − = − or tan15 0.577 0.0214x − = − M1 Attempt to solve x = 39.6 or x = 39.5 A1 4 Alternative method for question 4(ii) tan15 12.5 3 y y x v v v = = M1 ( ) 12.5 3tan15 5.8 yv = = downwards A1 –5.8 = 12.5 – 10t leading to t = 1.83 M1 Vertical motion using v = u + at 25 3 1.83 39.6 2 X = × = A1 4

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5 A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length 0.5 m and modulus of elasticity 6 N. The other end of the string is attached to a fixed point O. The particle P is released from rest at the point 0.5 + x m vertically below O. The particle P comes to instantaneous rest at O. (i) Find x. 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(ii) Find the greatest speed of P. 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Mark scheme: 5(i) ( ) ( ) 2 6 0.4 0.5 2 0.5 x g x + = × M1 Set up an energy equation 6x2 – 4x – 2 = 0 or 3x2 – 2x – 1 = 0 M1 Attempt to solve a 3 term quadratic equation x = 1 (ignore 1 3 − if seen) A1 3 5(ii) 6 0.4 0.5 e g = M1 Use x T l λ = to find the extension at the equilibrium position 1 3 e = A1 PE change = 1 0.4 0.5 3 g  +     B1ft Ft for candidate’s e ( ) 2 2 1 6 0.4 1 3 0.4 0.5 2 3 2 0.5 V g         = + −   ×   M1 Set up a three term energy equation V = 3.65 ms–1 A1 5

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6 A 0.3 m B C 0.6 m ABC is a uniform lamina in the form of a triangle with AB = 0.3 m, BC = 0.6 m and a right angle at B (see diagram). (i) State the distances of the centre of mass of the lamina from AB and from BC. [2] Distance from AB........................................................................................................................... ........................................................................................................................................................ ........................................................................................................................................................ Distance from BC .......................................................................................................................... ........................................................................................................................................................ ........................................................................................................................................................ The lamina is freely suspended at B and hangs in equilibrium. (ii) Find the angle between AB and the horizontal. 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A force of magnitude 12 N is applied along the edge AC of the lamina in the direction from A towards C. The lamina, still suspended at B, is now in equilibrium with AB vertical. (iii) Calculate the weight of the lamina. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) B1 From BC = 0.1 B1 2 6(ii) 0.1 tan 0.2 θ = M1 θ is the angle between AB and the horizontal θ = 26.6° A1 2 6(iii) 12cos26.6 × 0.3 = W × 0.2 M1A1 Take moments about B. (W is the weight of the lamina) W = 16.1 N A1 3

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7 A particle P of mass 0.5 kg is attached to a fixed point O by a light elastic string of natural length 1 m and modulus of elasticity 16 N. The particle P is projected vertically upwards from O with speed 6 m s−1. A resisting force of magnitude 0.1x2 N acts on P when P has displacement x m above O. After projection the upwards velocity of P is v m s−1. (i) Show that, before the string becomes taut, vdv = −10 −0.2x2. [2] dx ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the velocity of P at the instant the string becomes taut. 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(iii) Find an expression for the acceleration of P while it is moving upwards after the string becomes taut. 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(iv) Verify that P comes to instantaneous rest before the extension of the string is 0.5 m. 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Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................

Mark scheme: 7(i) 2 d 0.5 0.5 0.1 d v v g x x = − − M1 2 d 10 0.2 d v v x x = − − AG A1 2 Question Answer Marks Guidance 7(ii) ( ) 2 d 10 0.2 d v v x x = − − ∫ ∫ M1 Attempt to integrate the expression in part (i) 2 3 0.2 10 2 3 v x c − − × + A1 2 0.2 10 18 2 3 v   = − − +     M1 Either use limits or find c and put x = 1 v = 3.98 (329…) ms–1 7(iii) ( ) 2 16 1 d 0.5 0.5 0.1 d 1 x v v g x x − = − − − M1 Use Newton’s Second Law vertically when string becomes taut 2 2 d 10 0.2 32 32 22 32 0.2 d v v x x x x x = − − − + = − − A1 2 7(iv) ( ) 2 d 22 32 0.2 d v v x x x = − − ∫ ∫ M1 Attempt to integrate after the string becomes taut 2 2 3 32 0.2 22 2 2 3 v x x x k = − − + A1 x = 1, v= 3.98 (329…) hence k = 2. Now put x = 1.5 2 3 1.5 1.5 22 1.5 32 0.2 2 1.225 2 3 × − × − × + = − M1 Either use limits or find k and put x = 1.5 As 2 2 v cannot be negative, P comes to rest before the extension of the string is 0.5. A1 4

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