Cambridge A Level Mathematics 9709 — 2011 Oct/Nov Paper 5 · Variant 3

1 A particle is projected with speed 17 m s−1 at an angle of 50◦above the horizontal from a point on horizontal ground. Calculate the speed of the particle 2 s after the instant of projection. [3]

Mark scheme: 1 17sin50 – 2g B1 Vertical component of velocity v2 = (17sin50 – 2g)2 + (17cos50)2 M1 Pythagoras with 2 perpendicular components v = 13(.0) ms–1 A1 [3]

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2 An object is made from two identical uniform rods AB and BC each of length 0.6 m and weight 7 N. The rods are rigidly joined to each other at B and angle ABC = 90◦. (i) Calculate the distance of the centre of mass of the object from B. [1] The object is freely suspended at A and a force of magnitude F N is applied to the rod BC at C. The object is in equilibrium with AB inclined at 45◦to the horizontal. (ii) (a) A 45° 0.6 m B 0.6 m F N C Fig. 1 Calculate F given that the force acts horizontally as shown in Fig. 1. [2] (b) A 45° 0.6 m B 0.6 m C F N Fig. 2 Calculate F given instead that the force acts perpendicular to the rod as shown in Fig. 2. [2]

Mark scheme: 2 (i) 0.212 B1 [1] From (0.6/2)cos45 (ii) (a) 0.3cos45 × (2 × 7) = (2 × 06sin45) × F M1 Moments about A F = 3.5 A1 [2] (ii) (b) 0.3cos45 × (2 × 7) = 0.6F M1 Or Ans (i)/cos45 F = 4.95 A1 [2]

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3 A particle P is projected with speed 25 m s−1 at an angle of 45◦above the horizontal from a point O on horizontal ground. At time t s after projection the horizontal and vertically upward displacements of P from O are x m and y m respectively. (i) Express x and y in terms of t and hence show that the equation of the path of P is y = x −0.016x2. [4] (ii) Calculate the horizontal distance between the two positions at which P is 2.4 m above the ground. [2]

Mark scheme: 3 (i) x = (25cos45)t B1 y = (25sin45)t – gt2/2 B1 y = x(25sin45)/(25cos45) – g[x/(25cos45)2]/2 M1 Eliminates t between 2 simultaneous equations y = x – 0.016x2 A1 [4] (ii) 2.4 = x – 0.016x2 M1 Creates and attempts to solve a quadratic equation (x = 2.5, 60) Distance = 57.5 m A1 [2] 2

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4 A particle P of mass 0.4 kg is projected horizontally with velocity 8 m s−1 from a point O on a smooth horizontal surface. The motion of P is opposed by a resisting force of magnitude 0.2v2 N, where v m s−1 is the velocity of P at time t s after projection. 8 (i) Show that v = . [4] 1 + 4t (ii) Calculate the distance OP when t = 1.5. [4]

Mark scheme: 4 (i) 0.4δv/δt = 0.2v2 M1 Newton’s Second Law with a = δv/δt ∫ −2 A1 v δv = −5.0 ∫ δt –v–1 = –0.5t (+ c) t = 0, v = 8, hence c = –0.125 M1 v = 1/(0.125 + 0.5t) = 8/(1 + 4t) AG A1 [4] (ii) δx/δt = 8/(1 + 4t) M1* x = 8 ∫ δt / 1( + 4t ) x = 84 ln(1 + 4t) (+ c) A1 Accept c = 0 assumed t = 1.5, x = 84 *ln(1 + 4 × 1.5) D* Or limits used 84 [ln(1 + 4t] 5.10 M1 OP = 3.89 m A1 4 GCE AS/A LEVEL – October/November 2011 9709 53 2

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5 P Q R O 0.4 m 0.4 m 0.4 m w rad s–1 One end of a light inextensible string of length 1.2 m is attached to a fixed point O on a smooth horizontal surface. Particles P, Q and R are attached to the string so that OP = PQ = QR = 0.4 m. The particles rotate in horizontal circles about O with constant angular speed ω rad s−1 and with O, P, Q and R in a straight line (see diagram). R has mass 0.2 kg, and the tensions in the parts of the string attached to Q are 6 N and 10 N. (i) Show that ω = 5. [2] (ii) Calculate the mass of Q. [3] (iii) Given that the kinetic energy of P is equal to the kinetic energy of R, calculate the tension in the part of the string attached to O. [4] [Questions 6 and 7 are printed on the next page.]

Mark scheme: 5 (i) 0.2ω2 × 1.2 = 6 M1 Uses radial acceleration on R, 1 force ω = 5 A1 [2] (ii) mω2 × 2 × 0.4 = 10 – 6 M1 Uses radial acceleration on Q, 2 forces A1 m = 0.2 kg A1 [3] (iii) 0.2 × (5 × 1.2)2/2 = M(5 × 0.4)2/2 M1 M = 1.8 kg A1 1.8 × 52 × 0.4 = T – 10 DM1 T = 28 N A1 [4]

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6 A uniform solid consists of a hemisphere with centre O and radius 0.6 m joined to a cylinder of radius 0.6 m and height 0.6 m. The plane face of the hemisphere coincides with one of the plane faces of the cylinder. (i) Calculate the distance of the centre of mass of the solid from O. [4] [The volume of a hemisphere of radius r is 3πr3.]2 (ii) 0.6 m O 0.6 m 0.48 m A cylindrical hole, of length 0.48 m, starting at the plane face of the solid, is made along the axis of symmetry (see diagram). The resulting solid has its centre of mass at O. Show that the area of the cross-section of the hole is 16π3 m2. [4] (iii) It is possible to increase the length of the cylindrical hole so that the solid still has its centre of mass at O. State the increase in the length of the hole. [1]

Mark scheme: 6 (i) M1 Table of moments idea π0.62 × 0.6 × 0.3 – 2π 0.63/3 × 3 × 0.6/8 A1 Correct elements = (π0.63 + 2π0.63/3)d A1 Correct composite d = 0.09 m A1 [4] (ii) M1 Table of moments idea (about O) A1 Correct elements 2 3 π0.63 × 83 × 0.6 – π × 0.63 × 0.3 A1 + 0.48A × 0.36 = 0 A = 3π/16 m2 A1 [4] OR M1 Table of moments idea ( about O) A1 Correct elements [ 23 π × 0.63 + π × 0.63] × 0.09 = 0.48A × 0.36 A1 A = 3π/16 A1 (iii) Increase in length [= 2 × (0.6 – 0.48)] = 0.24m B1 [1] Remove cylinder with centre of mass at O GCE AS/A LEVEL – October/November 2011 9709 53

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7 One end of a light elastic string of natural length 0.4 m and modulus of elasticity 20 N is attached to a particle P of mass 0.8 kg. The other end of the string is attached to a fixed point O at the top of a smooth plane inclined at 30◦to the horizontal. The particle rests in equilibrium on the plane. (i) Calculate the extension of the string. [2] P is projected from its equilibrium position up the plane along a line of greatest slope. In the subsequent motion P just reaches O, and later just reaches the foot of the plane. Calculate (ii) the speed of projection of P, [4] (iii) the length of the line of greatest slope of the plane. [4]

Mark scheme: 7 (i) 0.8gsin30 = 20e/0.4 M1 e = 0.08 m A1 [2] (ii) M1 Conservation of KE, PE, EE 0.8v2/2 + 20 × 0.082/(2 × 0.4) A1 Correct start terms, signs accurate = 0.8g(0.4 + 0.08)sin30 A1 Correct final term, sign accurate v = 2.1(0) ms–1 A1 [4] (iii) M1* 0.8gdsin30 = 20(d – 0.4)2/(2 × 0.4) A1 4d = 25(d – 0.4)2 25d2 – 24d + 4 = 0 D* Obtains and solves a 3 term quadratic M1 equation. d = 0.745 m A1 [4]

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