Cambridge A Level Mathematics 9709 — 2016 Feb/March Paper 5 · Variant 2

1 A particle is projected from a point on horizontal ground. At the instant 2 s after projection, the particle has travelled a horizontal distance of 30 m and is at its greatest height above the ground. Find the initial speed and the angle of projection of the particle. [5]

Mark scheme: 1 Vsinθ = 2g ( = 20) B1 Using vertical motion to greatest height Vcosθ = 30/2 ( = 15) B1 Using horizontal motion V 2 = 152 + 20 2 or tanθ = 20/15 M1 Using Pythagoras or trigonometry V = 25 ms–1 A1 θ = 53.1° A1 5

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2 0.8 m P N 1 A uniform solid hemisphere of weight 60 N and radius 0.8 m rests in limiting equilibrium with its curved surface on a rough horizontal plane. The axis of symmetry of the hemisphere is inclined at an angle of 1 to the horizontal, where cos 1 = 0.28. Equilibrium is maintained by a horizontal force of magnitude P N applied to the lowest point of the circular rim of the hemisphere (see diagram). (i) Show that P = 8.75. [3] (ii) Find the coefficient of friction between the hemisphere and the plane. [2]

Mark scheme: 2 (i) 60(3 × 0.8/8) × 0.28 = P(0.8 − 0.8 × 0.28) M1 An attempt at taking moments A1 P = 8.75 AG A1 3 (ii) µ = 8.75/60 M1 µ = 0.146 A1 2

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3 A stone is thrown with speed 9 m s−1 at an angle of 60Å above the horizontal from a point on horizontal ground. Find the distance between the two points at which the path of the stone makes an angle of 45Å with the horizontal. [5]

Mark scheme: 3 Vh = 9cos60 B1 Vv = (±)4.5 B1 Or 9cos60 − 4.5 = 4.5 – gt M1 t = 0.9 A1 Distance = 4.05 A1 5 From 0.9 × 9cos60

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4 0.56 m C D E F 2 m 1.2 m B G A A uniform lamina is made by joining a rectangle ABCD, in which AB = CD = 0.56 m and BC = AD = 2 m, and a square EFGA of side 1.2 m. The vertex E of the square lies on the edge AD of the rectangle (see diagram). The centre of mass of the lamina is a distance h m from BC and a distance v m from BAG. (i) Find the value of h and show that v = h. [4] The lamina is freely suspended at the point B and hangs in equilibrium. (ii) State the angle which the edge BC makes with the horizontal. [1] Instead, the lamina is now freely suspended at the point F and hangs in equilibrium. (iii) Calculate the angle between FG and the vertical. [2]

Mark scheme: 4 (i) 2 × 0.56 × 0.28 + 1.2 2 (0.56 + 1.2/2) = M1 Moments about BC h(2 × 0.56 + 1.2 2 ) h = 0.775 A1 2 × 0.56 × 1 + 1.2 2 (1.2/2) = v(2 × 0.56 + M1 Moments about BAG 1.2 2 ) v = 0.775 A1 4 (ii) 45° B1 1 (iii) tanθ=(0.56 + 1.2 – 0.775) / (1.2 – 0.775) M1 θ =66.7° A1 2

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5 A particle P of mass 0.6 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 24 N. The other end of the string is attached to a fixed point A, and P hangs in equilibrium. (i) Calculate the extension of the string. [2] P is projected vertically downwards from the equilibrium position with speed 4.5 m s−1. (ii) Find the distance AP when the speed of P is 3.5 m s−1 and P is below the equilibrium position. [4] (iii) Calculate the speed of P when it is 0.5 m above the equilibrium position. [3]

Mark scheme: 5 (i) 24e/0.8 = 0.2g M1 e = 0.2 A1 2 (ii) 24 × 0.2 2 / (2 × 0.8) (= 0.6) B1 ft(cv0.2) Initial EE 0.6 × 4.5 2 / 2 + 0.6gd + 24 × 0.2 2 / (2 × 0.8) M1 PE/EE/KE balance attempt = 0.6 × 3.5 2 / 2 + 24 × (0.2 + d 2) / (2 × 0.8) A1 d = distance particle falls d = 0.4 so AP ( = 0.8 + 0.2 + 0.4) = 1.4m A1 4 (iii) 24 × 0.2 2 / (2 × 0.8) + 0.6 × 4.5 2 / 2 = M1 PE/EE/KE balance, 4 terms. Award A1 B1ft for initial KE if not already 0.6 v 2 /2 + 0.6g × 0.5 seen in part ii v = 3.5 m −s1 A1 3

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