Cambridge A Level Mathematics 9709 — 2011 May/June Paper 5 · Variant 2

1 A 4 N B A uniform rod AB of weight 16 N is freely hinged at A to a fixed point. A force of magnitude 4 N acting perpendicular to the rod is applied at B (see diagram). Given that the rod is in equilibrium, (i) calculate the angle the rod makes with the horizontal, [2] (ii) find the magnitude and direction of the force exerted on the rod at A. [4]

Mark scheme: 1 (i) 16Lcosθ = 4 × 2L M1 Moments about A, accept L = 1 θ = 60 o or π /3 c or 1.05 c A1 [2] (ii) X = 4sin60 o and Y = 16 – 4cos60 o B1 = √[(4sin60 o ) 2 + (16 – 4cos60 o ) 2 ] M1 tanα = (16 – 4cos60 o )/(4sin60 o ) = 14.4 N A1ft ft cv(X,Y). α = 76.1 o α = 76.1 o B1 R = 14.4 N [4]

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2 A uniform lamina ABCD consists of a semicircle BCD with centre O and diameter 0.4 m, and an isosceles triangle ABD with base BD 0.4 m and perpendicular height h m. The centre of mass of = the lamina is at O. (i) Find the value of h. [4] (ii) D X A m h 0.4 O m C B The lamina is suspended from a vertical string attached to a point X on the side AD of the triangle (see diagram). Given the lamina is in equilibrium with AD horizontal, calculate XD. [3]

Mark scheme: 2 (i) C of M semi-circle = 4 × 0.2/(3π ) B1 (0.08488…) 2 M1 Moments about a relevant point. π 2.0 2.0 4.0 h h × 4 × = × A1 2 3π 2 3 = 0.283 A1 [4] (ii) tanθ = 0.283/0.2 M1 tanADO = h/0.2 , ADO = 54.75 o cosθ = XD/0.2 ( = 0.5774) M1 For candidates ADO XD = 0.115 m A1 OR tanα = 0.2/0.283 M1 tanDAO = 0.2/h, DAO = 35.25 sinα = XD/0.2 ( = 0.5774) M1 For candidate’s DAO XD = 0.115 m A1 [3]

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3 V 60° P 0.1 m A particle P of mass 0.5 kg is attached to the vertex V of a fixed solid cone by a light inextensible string. P lies on the smooth curved surface of the cone and moves in a horizontal circle of radius 0.1 m with centre on the axis of the cone. The cone has semi-vertical angle 60◦(see diagram). (i) Calculate the speed of P, given that the tension in the string and the contact force between the cone and P have the same magnitude. [4] (ii) Calculate the greatest angular speed at which P can move on the surface of the cone. [4]

Mark scheme: 3 (i) Rcos30 o + Tcos60 o = 0.5g M1 or with R = T = F F = 0.5g/(cos30 o + cos60 o ) A1 F = 3.660… = R = T Tsin60 o – R sin30 o = 0.5v 2 /0.1 M1 Newton’s Second Law with radial acceleration v = 0.518 ms −1 A1 [4] (ii) R = 0 B1 Could be implied Tcos60 o = 0.5g M1 T = 10 N Tsin60 o = 0.5 × ω 2 × 0.1 M1 Newton’s Second Law with radial acceleration ω = 13.2 rads −1 A1 OR R = 0 B1 Could be implied mv 2 sin30 o /r or mrω 2 sin30 o M1 = mgcos30 o M1 ω = 13.2 rad s −1 A1 [4] GCE AS/A LEVEL – May/June 2011 9709 52

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4 One end of a light elastic string of natural length 0.5 m and modulus of elasticity 12 N is attached to a fixed point O. The other end of the string is attached to a particle P of mass 0.24 kg. P is projected vertically upwards with speed 3 m s−1 from a position 0.8 m vertically below O. (i) Calculate the speed of the particle when it is moving upwards with zero acceleration. [5] (ii) Show that the particle moves 0.6 m while it is moving upwards with constant acceleration. [4]

Mark scheme: 4 (i) 0.24g = 12(x)/0.5 M1 Finds position for equilibrium x = 0.1 A1 EITHER 1 2 × 0.24 × 32 + 12 × (0.8 – 0.5) 2 /(2 × 0.5) = M1 Energy balance, initial to equilibrium 2 2 positions 0.24v /2 + 12 × 0.1 /(2 × 0.5) + 0.24g(0.8 – 0.5 – 0.1) A1 v = 3.61 ms −1 A1 OR 0.24vdv/dx = mg – 12x/0.5 M1 Using Newton’s Second Law 0.24v 2 /2 = 2.4x – 12x 2 ( + c) A1 v = 3, x = 0.3, c = 1.44 x = 0.1, v = 3.61 ms −1 A1 Or uses limits [5] (ii) 0.24 × 3 2 /2 + 12 × (0.8 – 0.5) 2 /(2 × 0.5) = M1 Initial KE + initial EE = Final PE 0.24g(0.8 + x) A1 x = 0.1m A1 s = (0.5 + 0.1) = 0.6 m A1 OR 1 2 × 12 × 0.3 2 /0.5 + 12 × 0.24 × 3 2 M1 Initial EE + Initial KE = (KE + PE) at 1 2 equilibrium position = 2 × 0.24v + 0.24 × 10 × 0.3 v = 12 A1 Either 0 = 12 – 2 × 10s M1 Using v 2 = u 2 + 2as s = 0.6 A1 Or 12 × 0.24 × 12 = 0.24 × 10s M1 Using KE at equilibrium position = Final PE A1 s = 0.6 OR 1 2 × 12 × 0.3 2 /0.5 + 12 × 0.24 × 3 2 M1 Initial EE + Initial KE = Final PE where y is the distance above the start = 0.24 × 10y A1 y = 0.9 A1 s = 0.9 – 0.3 = 0.6 A1 [4] GCE AS/A LEVEL – May/June 2011 9709 52

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5 A particle P of mass 0.4 kg moves in a straight line on a horizontal surface and has velocity v m s−1 at time t s. A horizontal force of magnitude k√v N opposes the motion of P. When t 0, v 9 and = = when t 2, v 4. = = dv (i) Express in terms of k and v, and hence show that v 1 [5] dt = 4(t −6)2. (ii) Find the distance travelled by P in the first 3 seconds of its motion. [4]

Mark scheme: 5 (i) dv/dt = –2.5k v B1 0.4dv/dt = –k v ∫ v −5.0 dv = –2.5k∫ dt M1 v 5.0 /0.5 = –2.5kt (+ c) A1 LHS = 0.8 v t = 0, v = 9 hence c = 6 and M1 v = (6 – t)/2 t = 2, v = 4 hence k = 0.4 Uses correct limits v = (6 – t) 2 /4 = (t – 6) 2 /4 AG A1 [5] (ii) x = ∫ (t – 6) 2 /4dt M1 ∫(6 – t) 2 /4dt x = (t – 6) 3 /(3 × 4) (+ c) A1 –(6 – t) 3 /(3 × 4) (+ c) t = 0, x = 0 hence c = 18 M1 Or uses limits 0, 3 x(3) = 18 – (3 – 6) 3 /12 x(3) = 15.75 A1 Accept 15.7 or 15.8 OR 1 ∫ v 2 dv = ∫ –dx M1 From mvdv/dx = –k v 3 2 3 v 2 = – x ( + c) A1 3 x = 18 – 23 v 2 M1 Using v = 9, x = 0 so c = 18 x = 15.75 A1 Put t = 3 to find v = 2.25 [4]

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6 A particle P is projected with speed 26 m s−1 at an angle of 30◦below the horizontal, from a point O which is 80 m above horizontal ground. (i) Calculate the distance from O of the particle 2.3 s after projection. [4] (ii) Find the horizontal distance travelled by P before it reaches the ground. [3] (iii) Calculate the speed and direction of motion of P immediately before it reaches the ground. [4]

Mark scheme: 6 (i) x = (26cos30 o ) × 2.3 B1 = 51.788.. y = (26sin30 o ) × 2.3 + g × 2.3 2 /2 B1 = 56.35 d 2 = 51.8 2 + 56.35 2 M1 d = 76.5 m A1 [4] (ii) 80 = (26sin30 o )t + 10t 2 /2 M1 or v 2 =(26sin30 o ) 2 +2 × 10 × 80 with v = 42.06 t = 2.91s [or (42.06–13)/10] A1 = 26sin30 o + 10t solved for t x = (2.906 × 26cos30 o ) = 65.4 m A1 OR 80 = xtan30 o + 10x 2 /(2 × 26 2 × cos 2 30 o ) M1 Uses trajectory equation M1 Attempts to solve the quadratic equation x = 65.4 A1 [3] (iii) v 2 = (26sin30 o ) 2 + 2g × 80 B1 v = 42.06. Accept v = 26sin30 o + 10 × 2.91 or award correct method to find α 2 o 2 o 2 V = (26sin30 ) + 2g × 80 + (26cos30 ) M1 V = 47.7 ms −1 A1 α = tan −[(42.06)/(26cos301 o )] = 61.8 o A1 Below horizontal (1.08) [4]

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