Cambridge A Level Mathematics 9709 — 2017 Feb/March Paper 5 · Variant 2

1 A small ball is projected with speed 15 m s−1 at an angle of 60Å above the horizontal. Find the distance from the point of projection of the ball at the instant when it is travelling horizontally. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 0 = 15sin60 – gt (t = 3 3 /4 = 1.30) M1 Uses v = u + at vertically x = 15cos60 x 1.3 (= 9.7428..) A1 y = (15sin60)2/(2g) (= 8.4375..) B1 Uses v2 = u2 + 2as vertically 2 M1 Applies Pythagoras's theorem D = (9.74 + 8.442) D = 12.9 m A1 Total: 5

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2 A cylindrical container is open at the top. The curved surface and the circular base of the container are both made from the same thin uniform material. The container has radius 0.2 m and height 0.9 m. (i) Show that the centre of mass of the container is 0.405 m from the base. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The container is placed with its base on a rough inclined plane. The container is in equilibrium on the point of slipping down the plane and also on the point of toppling. (ii) Find the coefficient of friction between the container and the plane. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) M = 2π x 0.2 x 0.9 + π x 0.22 B1 M = total mass of the container (2π x 0.2 x 0.9 + π x 0.22) Lx M1 Takes moments about the base = 2π x 0.2 x 0.9 x 0.9/2 Lx = 0.405 m AG A1 Total: 3 2(ii) tanθ= 0.2/0.405 M1 θis the angle of slope of the plane µ= tanθ B1 µ= 0.494 A1 Total: 3

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3 A particle P is projected with speed 20 m s−1 at an angle of 60Å below the horizontal, from a point O which is 30 m above horizontal ground. (i) Calculate the time taken by P to reach the ground. 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(ii) Calculate the speed and direction of motion of P immediately before it reaches the ground. 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Mark scheme: 3(i) 1 M1 1 30 = (20sin60)t + gt2 Uses s = ut + at2 vertically 2 2 5t2 + 10 3 t – 30 = 0 M1 Sets up a quadratic equation and attempts to solve it t = 1.27 A1 Total: 3 3(ii) v2 = (20sin60)2 + 2g x 30 (hence v = 30) B1 Uses v2 = u2 + 2as vertically V = 30 2 + (20cos60)2 or M1 tanθ= 30/(20cos60) V = 31.6 ms –1 A1 θ= 71.6° with the horizontal A1 Or 18.4° with the downward vertical Total: 4

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4 A 0.6 m D 0.75 m B 0.9 m C The diagram shows a uniform lamina ABCD with AB = 0.75 m, AD = 0.6 m and BC = 0.9 m. Angle BAD = angle ABC = 90Å. (i) Show that the distance of the centre of mass of the lamina from AB is 0.38 m, and find the distance of the centre of mass from BC. 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The lamina is freely suspended at B and hangs in equilibrium. (ii) Find the angle between BC and the vertical. 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Mark scheme: 4(i) A = 0.6 x 0.75 + 0.3 x 0.75/2 (= 0.5625) B1 A = total area of the lamina 0.5625 Lx = 0.75 x 0.6 x 0.3 + 1 0.3 x 0.75 x M1 Takes moments about AB 2 (0.6 + 0.3/3) Lx = 0.38 m (from AB) AG A1 0.5625 Ly = 0.75 x 0.6 x 0.375 + 1 0.3 x 0.75 M1 Takes moments about BC 2 x 0.25 Ly = 0.35 m (from BC) A1 Total: 5 4(ii) tanθ= 0.35/0.38 M1 tanθ= Ly / Lx where θ is the required angle θ= 42.6° A1 Total: 2

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5 A 0.5 m 60Å P B Q 7 rad s−1 Two particles P and Q have masses 0.4 kg and m kg respectively. P is attached to a fixed point A by a light inextensible string of length 0.5 m which is inclined at an angle of 60Å to the vertical. P and Q are joined to each other by a light inextensible vertical string. Q is attached to a fixed point B, which is vertically below A, by a light inextensible string. The string BQ is taut and horizontal. The particles rotate in horizontal circles about an axis through A and B with constant angular speed 7 rad s−1 (see diagram). The tension in the string joining P and Q is 1.5 N. (i) Find the tension in the string AP and the value of 7. 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(ii) Find m and the tension in the string BQ. 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Mark scheme: 5(i) Tcos60 = 1.5 + 0.4g M1 Resolve vertically for P T = 11 N A1 2 M1 Uses Newton's Second Law horizontally Tsin60 = 0.4 ω x0.5sin60 for P ω = 55 = 7.42 A1 Total: 4 5(ii) m = 0.15 (from mg = 1.5) B1 Resolves vertically for Q T * = 0.15 x 7.422 x 0.5sin60 M1 Uses Newton's Second Law horizontally for Q T * = 3.57 N A1 Total: 3

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6 O and A are fixed points on a rough horizontal surface, with OA = 1 m. A particle P of mass 0.4 kg is projected horizontally with speed U m s−1 from A in the direction OA and moves in a straight line. After projection, when the displacement of P from O is x m, the velocity of P is v m s−1. The 0.8 coefficient of friction between the surface and P is 0.4. A force of magnitude N acts on P in the x direction PO. (i) Show that, while the particle is in motion, vdv = −4 −2 [3] dx x. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ It is given that P comes to instantaneous rest between x = 2.0 and x = 2.1. (ii) Find the set of possible values of U. 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Mark scheme: 6(i) Friction = 0.4 x 0.4g B1 Uses F = µR 0.4vdv/dx = –0.4 x 0.4g – 0.8/x M1 Uses Newton's Second Law and a = vdv/dx vdv/dx = –4 – 2/x A1 Total: 3 6(ii)  2  M1 Separates the variables and attempts to ∫vdv = ∫−− 4  d x integrate  x  v2/2 = –4x – 2lnx ( + c) A1 v = U when x = 1 hence c = U2 /2 + 4 M1 Attempts to find c 0= –4 x 2 – 2ln2 + U2/2 + 4 [U2 = 10.7(725)] M1 Put v = 0 and x = 2 0= –4 x 2.1 – 2ln2.1 + U2/2 + 4 Put v = 0 and x = 2.1 [U2 =11.7(677)] 3.28 ˂ U ˂ 3.43 A1 Total: 5

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7 One end of a light elastic string of natural length 0.6 m and modulus of elasticity 24 N is attached to a fixed point O. The other end of the string is attached to a particle P of mass 0.4 kg which hangs in equilibrium vertically below O. (i) Calculate the extension of the string. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ P is projected vertically downwards from the equilibrium position with speed 5 m s−1. (ii) Calculate the distance P travels before it is first at instantaneous rest. 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When P is first at instantaneous rest a stationary particle of mass 0.4 kg becomes attached to P. (iii) Find the greatest speed of the combined particle in the subsequent motion. 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Mark scheme: 7(i) 0.4g = 24e/0.6 M1 Uses T = λx/L e = 0.1 m A1 Total: 2 7(ii) Initial EE = 24 x 0.12/(2 x 0.6) (= 0.2 J) B1 Uses EE = λx2/2L 0.4 x 52/2 + 0.4gd=24(0.1 + d)2/(2 x 0.6) –24 M1 A1 Set up a 4 term energy equation involving x 0.12/(2 x 0.6) EE, PE and KE d = 0.5 m A1 Total: 4 7(iii) e = 0.2 B1 0.8v2/2=24 x 0.62/(2 x 0.6)– 24 x 0.22/(2 x M1 A1 Set up a 4 term energy equation in EE, PE 0.6) – 0.8g x 0.4 and KE v = 2 2 = 2.83 ms–1 A1 Total: 4

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