Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 5 · Variant 2

1 ABC is a uniform semicircular arc with diameter AC = 0.5 m. The arc rotates about a fixed axis through A and C with angular speed 2.4 rad s−1. Calculate the speed of the centre of mass of the arc. [3]

Mark scheme: 1 OG = 0.25 sin (π / 2)/(π / 2) B1 0.159 (15..) v = 0.159 × 2.4 M1 v = 0.382 ms–1 A1 [3] 2.4 × cv (OG)

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2 A 45° 0.8 m B 60° A uniform rod AB has weight 6 N and length 0.8 m. The rod rests in limiting equilibrium with B in contact with a rough horizontal surface and AB inclined at 60◦to the horizontal. Equilibrium is maintained by a force, in the vertical plane containing AB, acting at A at an angle of 45◦to AB (see diagram). Calculate (i) the magnitude of the force applied at A, [3] (ii) the least possible value of the coefficient of friction at B. [4]

Mark scheme: 2 (i) M1 Takes moments about B 6 × 0.4cos60 = 0.8 Pcos45 A1 P is the force at A P = 2.12N A1 [3] (ii) F = Psin75 (F is friction force at B) B1 Must use correct angle (cos15) R = 6 + Pcos75 (R is normal reaction at B) B1 Must use correct angle (sin15) µ = (2.12sin75) / (6 + 2.12cos75) M1 µ = 0.313 A1 [4]

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3 A particle P of mass 0.2 kg is released from rest and falls vertically. At time t s after release P has speed v m s−1. A resisting force of magnitude 0.8v N acts on P. (i) Show that the acceleration of P is (10 −4v) m s−2. [2] (ii) Find the value of v when t = 0.6. [5]

Mark scheme: 3 (i) 0.2 dv / dt = 0.2g – 0.8v M1 Use Newton’s Second Law, – sign essential a = (dv / dt =)10 – 4v AG A1 [2] (ii) ∫ 1 / (10 – 4v) dv = ∫dt M1 Separates variables and attempts to integrate −ln1 (10 – 4v) = t (+ c) 4 A1 [c = −ln1 10] M1 Attempts to find the constant or uses the 4 correct limits −ln 1 (10 – 4v) = 0.6 – 1 ln4 A1 4 4 v = 2.27 A1 [5]

More questions on Kinematics of motion in a straight line

4 0.67 m P 45° A particle P is moving inside a smooth hollow cone which has its vertex downwards and its axis vertical, and whose semi-vertical angle is 45◦. A light inextensible string parallel to the surface of the cone connects P to the vertex. P moves with constant angular speed in a horizontal circle of radius 0.67 m (see diagram). The tension in the string is equal to the weight of P. Calculate the angular speed of P. [6]

Mark scheme: 4 Rcos45 – Tcos45 = mg M1 Resolves vertically for P Rcos45 = mg + mg cos45 A1 May be implied for later work Rsin45 + Tsin45 = mω2 × 0.67 M1 Uses Newton’s Second Law horizontally for P M1 Obtaining an equation in m (and g) mg + mg cos45 + mg sin45 = mω2 × 0.67 A1 ω = 6(.00) rads–1 A1 [6] GCE A LEVEL – October/November 2012 9709 52 OR 4 M1 Resolves radial acceleration parallel to the slope for P Acceleration = ω2 × 0.67cos45 A1 May be implied by later work mω2 × 0.67cos45 = T + mg cos45 M1 Uses Newton’s Second Law parallel to the slope for P M1 Obtaining an equation in m (and g) mω2 × 0.67cos45 = mg + mg cos45 A1 ω = 6(.00) rads–1 A1 2 2 2

More questions on Kinematics of motion in a straight line

5 A particle P is projected with speed 30 m s−1 at an angle of 60◦above the horizontal from a point O on horizontal ground. For the instant when the speed of P is 17 m s−1 and increasing, (i) show that the vertical component of the velocity of P is 8 m s−1 downwards, [2] (ii) calculate the distance of P from O. [5]

Mark scheme: 5 (i) v2 = 172 – (30 cos60)2 M1 Finds vertical speed v = –8 A1 [2] – may be implied by later work (ii) –8 = 30 sin60 – gt M1 Finds relevant time t = 3.4 A1 3.398 y = [(30 sin60)2 – 82] / (2g) (= 30.55) B1 Or y = (30 sin60) × 3.4 – g 3.42/2 (= 30.53) OP2 = (30 cos60 × 3.4)2 + 30.552 M1 Use of Pythagoras OP = 59.4 m A1 [5] Accept 59.5

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6 B A 0.6 m O C D A uniform lamina OABCD consists of a semicircle BCD with centre O and radius 0.6 m and an isosceles triangle OAB, joined along OB (see diagram). The triangle has area 0.36 m2 and AB = AO. (i) Show that the centre of mass of the lamina lies on OB. [4] (ii) Calculate the distance of the centre of mass of the lamina from O. [4]

Mark scheme: 6 (i) Height of triangle = 0.36 / 0.3(= 1.2 m) B1 Semi-circle C of M = 2 × 0.6 / (3π / 2) B1 Centre of mass lamina from BOD 0.36 × (1.2 / 3) = π × 0.62 / 2 × 2 × 0.6 / (3π / 2) M1 Equating moments idea 0.144 = 0.144 A1 [4] Evidence of checking equality OR 0.36 × (1.2 / 3) – π × 0.62 / 2 ×2 ×0.6 /(3π/2) = distance × total area M1 Table of moments idea Distance = 0 A1 (ii) 0.36 × 0.3 A1 Correct sum of parts = (0.36 + π 0.62 / 2) × OG A1 Correct moment of whole OG = 0.117 m A1 [4] GCE A LEVEL – October/November 2012 9709 52

More questions on Energy, work and power

7 A light elastic string has natural length 3 m and modulus of elasticity 45 N. A particle P of weight 6 N is attached to the mid-point of the string. The ends of the string are attached to fixed points A and B which lie in the same vertical line with A above B and AB = 4 m. The particle P is released from rest at the point 1.5 m vertically below A. (i) Calculate the distance P moves after its release before first coming to instantaneous rest at a point vertically above B. (You may assume that at this point the part of the string joining P to B is slack.) [4] (ii) Show that the greatest speed of P occurs when it is 2.1 m below A, and calculate this greatest speed. [5] (iii) Calculate the greatest magnitude of the acceleration of P. [3]

Mark scheme: 7 (i) M1 Energy conservation, no KE, 2 EE terms 45 × 12 / (2 × 1.5) + 0.6 gh = 45 h2 / (2 × 1.5) A1 5h2 – 2h – 5 = 0 M1 Simplifies, tries to solve a 3 term quadratic equation h = 1.22 m A1 [4] (ii) 45e / 1.5 = 45(1 – e) / 1.5 + 6 M1 Finds equilibrium position (e = 0.6) AP = (1.5 + 0.6) = 2.1 AG A1 0.6 v2 / 2 = 0.6 g × 0.6 + 45 (1)2 / (2 × 1.5) M1 Energy conservation with KE/PE/EE – 4.5(0.6)2 / (2 × 1.5) – 45(0.4)2 / (2 × 1.5) A1 terms v = 6 ms–1 A1 [5] (iii) 0.6 a = ± (0.6g + 45 × 1 / 1.5) M1* Top a = ± 60 ms–2 0.6 a = ± (0.6g – 45 × 1.22 / 1.5) M1* Bottom a = ± 51 ms–2 | a | = 60 ms–2 A**1 [3] Needs acceleration at both extreme positions considered.

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