Cambridge A Level Mathematics 9709 — 2014 May/June Paper 5 · Variant 2

1 A particle is projected with speed 12 m s−1 from a point on horizontal ground. The particle strikes the ground 1.6 s after the instant of projection. Calculate the angle of projection. [2]

Mark scheme: 1 –12sinθ = 12sinθ – 1.6g M1 sin θ = (1.6g / 2)12 θ = 41.8° A1 2 OR 0 = 12sin θ × 1.6 – 12 × 10 × 1.62 M1 Uses s = ut + 12 at 2 θ = 41.8° A1 OR 0 = 12sinθ – 10 × 0.8 M1 Uses v = u + at at the highest point θ = 41.8° A1

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2 10 N A 0.8 m 30 0.4 m B 6 N A non-uniform rod AB of weight 6 N rests in limiting equilibrium with the end A in contact with a rough vertical wall. AB = 1.2 m, the centre of mass of the rod is 0.8 m from A, and the angle between AB and the downward vertical is . A force of magnitude 10 N acting at an angle of 30 to the upwards vertical is applied to the rod at B (see diagram). The rod and the line of action of the 10 N force lie in a vertical plane perpendicular to the wall. Calculate (i) the value of , [4] (ii) the coefficient of friction between the rod and the wall. [2]

Mark scheme: 2 (i) 10cos30 × 1.2sinθ – 10sin30 × 1.2cosθ M1 = 6 × 0.8sinθ A1 5.5923..sinθ = 6cosθ M1 Creating a 3 term solvable equation in θ = 47(.0) A1 4 sinθ and cosθ OR 10 × 1.2sin(θ – 30) = 6 × 0.8sinθ or M1 10 × 1.2cos(120 – θ) = 6 × 0.8sinθ A1 5.5923..sinθ = 6cosθ M1 Creating a 3 term solvable equation in θ = 47(.0) A1 sinθ and cosθ (ii) µ = (10cos30 – 6)/(10sin30) M1 For using F = µR with a reasonable µ = 0.532 A1 2 attempt to find F and R

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3 A light elastic string has natural length 0.8 m and modulus of elasticity 16 N. One end of the string is attached to a fixed point O, and a particle P of mass 0.4 kg is attached to the other end of the string. The particle P hangs in equilibrium vertically below O. (i) Show that the extension of the string is 0.2 m. [2] P is projected vertically downwards from the equilibrium position. P first comes to instantaneous rest at the point where OP = 1.4 m. (ii) Calculate the speed at which P is projected. [3] (iii) Find the speed of P at the first instant when the string subsequently becomes slack. [2]

Mark scheme: 3 (i) 0 .4 g = 16 e/ 0 .8 M1 Uses mg = 16 ext / 8.0 e = 0.2 AG A1 2 (ii) EE at C = 16 × 6.0 2 / (2 × 8.0 ) B1 4.0u 2 / 2 + 16 × 2.0 2 / (2 × 8.0 ) M1 KE/EE/PE balance attempted with 4 2 terms. + 4.0 g (4.1 − 0.1 ) = 16 × 6.0 / ( 2 × 8.0 ) u = 2.83 ms–1 A1 3 8 not allowed (iii) 16 × 6.0 2 / (2 × 8.0 ) M1 KE/EE/PE balance attempted with 3 2 terms. = 4.0v / 2 + 4.0 g (4.1 − 8.0 ) v =2.45 ms–1 A1 2

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4 A particle P is projected with speed 20 m s−1 at an angle of 40 above the horizontal from a point O on horizontal ground. (i) Find the height of P above the ground when P has speed 18 m s−1. [2] (ii) Calculate the length of time for which the speed of P is less than 18 m s−1, and find the horizontal distance travelled by P during this time. [6]

Mark scheme: 4 (i) 18 2 − ( 20 cos 40 ) 2 = 20 2 − ( 20 cos 40 ) 2 M1 Uses vertical motion with –2gh v 2 = u 2 − 2 gs h = 3.8 m A1 2 OR 2 2 Uses energy equation m × 20 / 2 − m × 18 = mgh M1 h = 3.8 m A1 GCE A LEVEL – May/June 2014 9709 52 (ii) V 2 = 18 2 − ( 20 cos 40 ) 2 M1 V is the vertical component of the V = 9.4483 A1 velocity of P when P's speed is 18 9.4483 = –9.4483 + gt M1 M1 for using their V t = 1.89 s A1 x = 1.89 × 20cos40 M1 M1 scored if their time is used x = 29(.0) m A1 6 OR 2 1 2 8.3 = ( 20 sin 40 )T − gT / 2 M1 Uses s = ut + at 2 T = 2.23(0), 0.34(1) A1 t =2.23(0) – 0.34(1) M1 t = 1.89 A1

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5 B A 1.8 m O 1.8 m C A uniform metal frame OABC is made from a semicircular arc ABC of radius 1.8 m, and a straight rod AOC with AO = OC = 1.8 m (see diagram). (i) Calculate the distance of the centre of mass of the frame from O. [3] A uniform semicircular lamina of radius 1.8 m has weight 27.5 N. A non-uniform object is formed by attaching the frame OABC around the perimeter of the lamina. The object is freely suspended from a fixed point at A and hangs in equilibrium. The diameter AOC of the object makes an angle of 22 with the vertical. (ii) Calculate the weight of the frame. [5]

Mark scheme: 5 (i) OG(arc) = 1.8sin( π /2)/( π /2) B1 1.1459.. or 3.6/π OX 8.1( × 2 + π ×8.1 ) = .11459 × π ×8.1 M1 OX = 0.7(00) m A1 3 0.70017.. (ii) OY = 1.8tan22 B1 C of M solid = 0.727247..m from O OG(lamina) = 2×8.1 sin( π / 2 ) /( 3 π / 2 ) B1 C of M lamina = 0.763943.. or 2.4/π 1.8tan22 × (W + 27.5) = M1 275.  4 ×8.1 /  3 π  − .0727247  =     0.7W + 0.763943 × 27.5 A1 W = 37(.3)N A1 W(0.727247 – 0.70017) 5 Accept to 2sf as sensitive to rounding error

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6 A particle P of mass 0.6 kg is released from rest at a point above ground level and falls vertically. The motion of P is opposed by a force of magnitude 3v N, where v m s−1 is the speed of P. Immediately before P reaches the ground, v = 1.95. (i) Calculate the time after its release when P reaches the ground. [5] P is now projected horizontally with speed 1.95 m s−1 across a smooth horizontal surface. The motion of P is again opposed by a force of magnitude 3v N, where v m s−1 is the speed of P. (ii) Calculate the distance P travels after projection before coming to rest. [3] [Question 7 is printed on the next page.]

Mark scheme: 6 (i) 0.6dv / dt =0.6g – 3v B1 Newton's Second Law ∫ 1 / (10 − 5v )d v = ∫ d t M1 6.0 ∫ 1 / (6.0g − 3v )dv = ∫ d t − 15 ln (10 − 5v ) = t ( + c ) A1 6.0−3 ln (6.0 g − 3v ) = t ( + c ) Finds c or uses limits twice M1 t = 0.738 s A1 5 (ii) 0.6vdv / dx = –3v B1 Newton's Second Law ∫ 2.0dv = − ∫ dx M1 Integration with use of limits or finding c x = 0.39 m A1 3

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