Cambridge A Level Mathematics 9709 — 2017 May/June Paper 5 · Variant 2

1 A particle P of mass 0.2 kg moves with speed 4 m s−1 and angular speed 5 rad s−1 in a horizontal circle on a smooth surface. P is attached to one end of a light elastic string of natural length 0.6 m. The other end of the string is attached to the point on the surface which is the centre of the circular motion of P. (i) Find the radius of this circle. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the modulus of elasticity of the string. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) Total: 1 1(ii) T = 0.2 × 2 5 × 0.8 M1 Uses Newton’s Second Law horizontally T = 4 N A1 FT FT with their radius from part (i) 4 = λ(0.8 – 0.6) / 0.6 M1 Uses T = λx / L λ = 12 A1 Total: 4

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2 A 0.7 m 60Å 6 N P 4 N 60Å 0.7 m B The ends of two light inextensible strings of length 0.7 m are attached to a particle P. The other ends of the strings are attached to two fixed points A and B which lie in the same vertical line with A above B. The particle P moves in a horizontal circle which has its centre at the mid-point of AB. Both strings are inclined at 60Å to the vertical. The tension in the string attached to A is 6 N and the tension in the string attached to B is 4 N (see diagram). (i) Find the mass of P. 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(ii) Calculate the speed of P. 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Mark scheme: 2(i) 6cos60 = 4cos60 + mg M1 Resolve vertically m = 0.1 kg A1 Total: 2 2(ii) radius = 0.7sin60 B1 6sin60 + 4sin60 = 0.1 2v / (0.7sin60) M1 Uses Newton's Second Law horizontally with 3 terms v = 7.25 m 1 s− A1 Total: 3 Height of C of M of each vertical face above the base = 0.1 m

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3 An open box in the shape of a cube with edges of length 0.2 m is placed with its base horizontal and its four sides vertical. The four sides and base are uniform laminas, each with weight 3 N. (i) Calculate the height of the centre of mass of the box above its base. 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The box is now fitted with a thin uniform square lid of weight 3 N and with edges of length 0.2 m. The lid is attached to the box by a hinge of length 0.2 m and weight 2 N. The lid of the box is held partly open. (ii) Find the angle which the lid makes with the horizontal when the centre of mass of the box (including the lid and hinge) is 0.12 m above the base of the box. 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Mark scheme: 3(i) B1 5 × 3y = 4 × 3 × 0.1 M1 Takes moments about the base. y is the height of the C of M above the base y = 0.08 m A1 Total: 3 Question Answer Marks Notes 3(ii) Moment of lid about the base = 3 × (0.2 + 0.1sinθ) B1 θ is the angle the lid makes with the horizontal (6 × 3 + 2) × 0.12 = 5 × 3 × 0.08 + 2 × 0.2 + 3 × (0.2 + 0.1sinθ) M1 Take moments about the base A1 θ = 41.8° A1 Total: 4 M1 Uses Newton’s Second Law vertically

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4 A small object of mass 0.4 kg is released from rest at a point 8 m above the ground. The object descends vertically and when its downwards displacement from its initial position is x m the object has velocity v m s−1. While the object is moving, a force of magnitude 0.2v2 N opposes the motion. (i) Show that vdv = 10 −0.5v2. 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(ii) Express v in terms of x. 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(iii) Find the increase in the value of v during the final 4 m of the descent of the object. 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Mark scheme: 4(i) 0.4a = 0.4g – 0.2 2v vdv / dx = 10 – 0.5 2v A1 AG Total: 2 4(ii) 2 d / (10 0. ) 5 v v v ∫ − = dx ∫ M1 Separates the variables and attempts to integrate – ln(10 – 0.5 2v ) = x ( + c ) A1 x = 0, v = 0 hence c = –ln10 M1 Attempts to find c using x = 0, v = 0 v = (20 20 ) e x − − A1 10–0.5 2v = 10 e x ln −+ = 10 e x − Total: 4 4(iii) Increase= 8 (20 20e− − − 4 (20 20e− − M1 M1 if x values are substituted into their value for part (ii) Increase = 0.0404 m 1 s− A1 Allow 0.04 Total: 2

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5 A particle of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 6 N. The other end of the string is attached to a fixed point O. The particle is projected vertically downwards from O with initial speed 2 m s−1. (i) Calculate the greatest speed of the particle during its descent. 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(ii) Find the greatest distance of the particle below O. 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Mark scheme: 5(i) e = 0.4 m A1 EE = 6 × 2 0.4 /(2 × 0.8) B1 FT FT for their e 0.3 2v / 2 – 0.3 × 2 2 / 2 = 0.3 g(0.8 + 0.4) – 6 × 2 0.4 / (2 × 0.8) M1 Sets up a 4 term energy equation involving EE, KE and PE v = 4.9(0) m 1 s− or 2 6 A1 Total: 5 5(ii) 0.3 × 2 2 /2 + 0.3 gL = 6(L− 0.8 2) / (2 × 0.8) M1 Sets up a 3 term energy equation involving EE, KE and PE A1 L = 2.18 m A1 Ignore answers less than 0.8 Total: 3 M1 Takes moments about A

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6 3 N 30Å B 0.6 m 60Å A The end A of a non-uniform rod AB of length 0.6 m and weight 8 N rests on a rough horizontal plane, with AB inclined at 60Å to the horizontal. Equilibrium is maintained by a force of magnitude 3 N applied to the rod at B. This force acts at 30Å above the horizontal in the vertical plane containing the rod (see diagram). (i) Find the distance of the centre of mass of the rod from A. 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The 3 N force is removed, and the rod is held in equilibrium by a force of magnitude P N applied at B, acting in the vertical plane containing the rod, at an angle of 30Å below the horizontal. (ii) Calculate P. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ In one of the two situations described, the rod AB is in limiting equilibrium. (iii) Find the coefficient of friction at A. 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Mark scheme: 6(i) 3 × 0.6 = 8cos60 x x = 0.45 m A1 Total: 2 6(ii) Pcos60 × 0.6 = 8 × 0.45cos60 M1 Takes moments about A P = 6 N A1 Total: 2 Question Answer Marks Notes 6(iii) µ = 3cos30 / (8 – 3sin30) M1 Uses F = µR used µ = 6cos30 / (8 + 6sin30) M1 µ = 0.4 or 0.472 A1 µ = 0.472 accept 0.47 A1 Total: 4 tanθ = 2 B1 Note θ = 63.4349..°

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7 A particle P is projected from a point O with speed V m s−1. At time 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 = 2x −25x2 . V2 (i) Write down the value of tan 1, where 1 is the angle of projection of P. [1] ........................................................................................................................................................ ........................................................................................................................................................ When t = 4, P passes through the point A where x = y = a. (ii) Calculate V and a. 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(iii) Find the direction of motion of P when it passes through A. 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Mark scheme: 7(i) Total: 1 Question Answer Marks Notes 7(ii) EITHER: a = 2a – 25 2 a / 2 V (25a = 2 V ) (B1 Substitutes x = y = a into the trajectory equation a = Vcos63.4349.. × 4 B1 Horizontal motion 2 V = 25 × 4 × Vcos63.4349.. M1 Attempts to eliminate a V = 44.7(213..) or 20 5 A1 a = 80 A1) OR: a = Vsin63.4349..× 4 – g 2 4 /2 (B1 Uses s = ut + a 2t / 2 vertically a = Vcos63.4349..× 4 B1 Horizontal motion Vsin63.4349..× 4 − g 2 4 / 2 = Vcos63.4349..× 4 M1 Attempts to solve the 2 equations V = 44.7(213..)or 20 5 A1 a = 80 A1) Total: 5 7(iii) vv = 44.7213..sin63.4349.. – 4g ( = 0) M1 vv = vertical component of the velocity α = 1 tan−+/− 0 / (44.7213..cos63.4349..) M1 tanα = vv / hv where hv = horizontal velocity α = 0° A1 Total: 3

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