Cambridge A Level Mathematics 9709 — 2011 May/June Paper 5 · Variant 3

1 A 60° P 0.2 m A particle P of mass 0.4 kg is attached to a fixed point A by a light inextensible string. The string is inclined at 60◦to the vertical. P moves with constant speed in a horizontal circle of radius 0.2 m. The centre of the circle is vertically below A (see diagram). (i) Show that the tension in the string is 8 N. [2] (ii) Calculate the speed of the particle. [2]

Mark scheme: 1 (i) Tsin30° = 0.4g M1 Resolves vertically T = 8N A1 [2] (ii) Tcos30° = 0.4v 2 / 0.2 ( = 0.4ω 2 × 0.2) M1 Newton’s Second Law radially v = 1.86 ms −1 A1ft ft only on T from part (i) [2] 2 2 2

More questions on Forces and equilibrium

2 A stone is thrown with speed 15 m s−1 horizontally from the top of a vertical cliff 20 m above the sea. Calculate (i) the distance from the foot of the cliff to the point where the stone enters the sea, [3] (ii) the speed of the stone when it enters the sea. [3]

Mark scheme: 2 (i) 20 = gt 2 / 2 (t = 2) M1 y = –gx 2 /(2 × 15 2 ) use of trajectory equation x = 15 × 2 DM1 –20 = –10x 2 / (2 × 15 2 ) x = 30 A1 [3] (ii) v = (g × 2) = 20 B1 v = √(152 + 202) M1 v = 25 A1 [3]

More questions on Kinematics of motion in a straight line

3 F N O 0.4 m 20° A smooth hemispherical shell, with centre O, weight 12 N and radius 0.4 m, rests on a horizontal plane. A particle of weight W N lies at rest on the inner surface of the hemisphere vertically below O. A force of magnitude F N acting vertically upwards is applied to the highest point of the hemisphere, which is in equilibrium with its axis of symmetry inclined at 20◦to the horizontal (see diagram). (i) Show, by taking moments about O, that F 16.48 correct to 4 significant figures. [3] = (ii) Find the normal contact force exerted by the plane on the hemisphere in terms of W. Hence find the least possible value of W. [3]

Mark scheme: 3 (i) M1 Moments about O F × 0.4sin20° = 12 × (0.4 / 2)cos20° A1 F = 16.48 AG A1 [3] (ii) R = –16.48 + 12 + W B1 Equates forces vertically –16.48 + 12 + W = 0 M1 Works with R = 0 W = 4.48 A1 [3] √ 2 2

More questions on Forces and equilibrium

4 The ends of a light elastic string of natural length 0.8 m and modulus of elasticity λ N are attached to fixed points A and B which are 1.2 m apart at the same horizontal level. A particle of mass 0.3 kg is attached to the centre of the string, and released from rest at the mid-point of AB. The particle descends 0.32 m vertically before coming to instantaneous rest. (i) Calculate λ. [4] (ii) Calculate the speed of the particle when it is 0.25 m below AB. [4]

Mark scheme: 4 (i) e = √(0.62 + 0.322) – 0.4 ( = 0.28) B1 Extension of half string = 0.28 m 0.3g × 0.32 = 2[λ (0.28 2 – 0.2 2 )] / (2 × 0.4) M1, A1 PE loss = EE gain λ = 10 A1 [4] (ii) e = √(0.62 + 0.252) – 0.4 B1 Extension of half string = 0.25 m 0.3g × 0.25 = 0.3v 2 / 2 + M1 PE loss = KE gain + EE gain 2[10(0.25 2 – 0.2 2 ) / (2 × 0.4)] A1ft N.B. 0.25 is extension of half string v = 1.12 A1 ft on candidates λ only [4]

More questions on Energy, work and power

5 One end of a light elastic string of natural length 0.3 m and modulus of elasticity 6 N is attached to a fixed point O on a smooth horizontal plane. The other end of the string is attached to a particle P of mass 0.2 kg, which moves on the plane in a circular path with centre O. The angular speed of P is ω rad s−1. (i) For the case ω = 5, calculate the extension of the string. [4] (ii) Express the extension of the string in terms of ω, and hence find the set of possible value of ω. [4]

Mark scheme: 5 (i) T = 6e / 0.3 B1 0.2 × 5 2 (0.3 + e) = 6e / 0.3 M1, A1 Newton’s Second Law radially e = 0.1 A1 [4] (ii) 0.2ω 2 (0.3 + e) = 6e / 0.3 M1 Newton’s Second Law radially e = 0.06ω 2/(20 – 0.2ω 2) A1 Other forms acceptable 20 – 0.2ω 2 > 0 M1 Uses denominator > 0 (0 <) ω < 10 A1 Disregard lower limit [4] GCE AS/A LEVEL – May/June 2011 9709 53 2

More questions on Newton’s laws of motion

6 3 m s–1 P O 0.5 m A O and A are fixed points on a horizontal surface, with OA 0.5 m. A particle P of mass 0.2 kg is = projected horizontally with speed 3 m s−1 from A in the direction OA and moves in a straight line (see diagram). At time t s after projection, the velocity of P is v m s−1 and its displacement from O is x m. 0.4 The coefficient of friction between the surface and P is 0.5, and a force of magnitude N acts on x2 P in the direction PO. (i) Show that, while the particle is in motion, vdv 5 2 . [2] dx = − + x2 (ii) Calculate the distance travelled by P before it comes to rest, and show that P does not subsequently move. [7] [Question 7 is printed on the next page.]

Mark scheme: 6 (i) 0.2a = –0.2g0.5 – 0.4/x 2 M1 Uses Newton’s Second law vdv/dx = –(5 + 2x −)2 AG A1 [2] M1 Separates variables and integrates (ii) ∫vdv = –∫ (5 + 2 x −2 ) d x v 2 /2 = –5x + 2/x ( + c) A1 3 2 /2 = –5 × 0.5 +2/0.5 + c M1 Hence c = 3, or [v 2 / 2] 30 = [–5x + 2/x] 5.0x x = 1 A1 From 0 = –5x + 2/x = 3 Travels ( = 1 – 0.5) = 0.5m A1 F towards O ( 0.4) less than maximum M1, A1 Compares 0.5 × 0.2g and 0.4/1 2 friction ( = 1) [7] 2

More questions on Kinematics of motion in a straight line

7 B a m C a m A 1 m O E D ABCDE is the cross-section through the centre of mass of a uniform prism resting in equilibrium with DE on a horizontal surface. The cross-section has the shape of a square OBCD with sides of length a m, from which a quadrant OAE of a circle of radius 1 m has been removed (see diagram). (i) Find the distance of the centre of mass of the prism from O, giving the answer in terms of a, π and √2. [5] (ii) Hence show that 3a2(2 −a) < 32π −2, and verify that this inequality is satisfied by a 1.68 but not by a 1.67. [4] = =

Mark scheme: 7 (i) OG quadrant = 2sin(π /4) / (3π /4) B1 8 / (3 2 π ) a 2 (a 2 /2) = π /4[2sin(π /4) / (3π /4)] M1 –1 each error, min zero +(a 2 – π /4)x A2 There must be 3 moment terms x = 2 2 (3a 3 – 2) / (12a 2 – 3π ) A1 Other forms acceptable [5] (ii) xcos45° > 1 B1 (6a 3 – 4) / (12a 2 – 3π ) > 1 M1 3a 2 (2 – a) < 3π /2 – 2 AG A1 True when a = 1.68, not when a = 1.67 AG B1 RHS = 2.712.. compared with [4] LHS = 2.709.. and 2.76.. respectively

More questions on Coordinate geometry