Cambridge A Level Mathematics 9709 — 2016 Oct/Nov Paper 5 · Variant 2

1 A stone S is thrown horizontally from the top T of a high tower. At the instant 1.6 s after S is thrown, the line ST makes an angle of 30Å below the horizontal. Find the speed with which S is thrown. [3]

Mark scheme: 1 Y = g1.62 /2 B1 12.8 m 12.8/ (1.6V) = tan30 M1 1.6V = X = 22.17 m V = 13.9 m s–1 A1 [3]

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2 A particle P of mass 0.5 kg is attached to one end of a light elastic string with modulus of elasticity 24 N and natural length 0.6 m. The other end of the string is attached to a fixed point A. The particle P hangs in equilibrium vertically below A. (i) Find the distance AP. [2] The particle P is raised to A and released from rest. (ii) Calculate the greatest speed of P in the subsequent motion. [3]

Mark scheme: 2 (i) 5 = 24e /0.6 M1 Hence e = 0.125 AP = 0.725 m A1 [2] (ii) 24 x 0.1252/2 x 0.6 B1 EE at eqm (= 0.3125) 0.5g x 0.725 = M1 KE/EE/PE conservation 24 x 0.1252/ 2 x 0.6 + 0.5v2 /2 v = 3.64 m s–1 A1 [3]

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3 10 N 60Å A 30Å F N 0.5 m 45Å B A non-uniform rod AB of length 0.5 m is freely hinged to a fixed point at A. The rod is in equilibrium at an angle of 30Å with the horizontal with B below the level of A. Equilibrium is maintained by a force of magnitude F N applied at B acting at 45Å above the horizontal in the vertical plane containing AB. The force exerted by the hinge on the rod has magnitude 10 N and acts at an angle of 60Å above the horizontal (see diagram). (i) By resolving horizontally and vertically, calculate F and the weight of the rod. [4] (ii) Find the distance of the centre of mass of the rod from A. [3]

Mark scheme: 3 (i) Fcos45 = 10cos60 M1 Resolving horizontally F = 7.07 A1 7.071 ..= 5√2 Fsin45 + 10sin60 = W M1 Resolving vertically W = 13.7 A1 13.660.. = 5(√2+√3) [4] (ii) M1 Moments about A Wdcos30 = (Fsin75)0.5 A1 d = 0.289 m A1 [3]

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4 A particle P is projected with speed 20 m s−1 at an angle of 30Å above the horizontal from a point O on horizontal ground. P subsequently bounces when it first strikes the ground at the point A. (i) Find the time after projection when P first strikes the ground, and the distance OA. [3] When P bounces at A the horizontal component of the velocity of P is unchanged. The vertical component of velocity is 8 m s−1 immediately after bouncing. P strikes the ground for the second time at B where it remains at rest. (ii) Calculate the first and last times after projection at which the speed of P is 18 m s−1. [5]

Mark scheme: 4 (i) –20sin30 = 20sin30 – gT M1 T = 2 s A1 OA = 34.6 m B1 [3] (ii) Vv 2 = 182 – (20cos30)2 M1 VV = (±) 4.899 A1 4.899 = 20sin30 – gt M1 t = 0.51(0) s A1 –4.899 = 8 – gt t = 1.29 T = 3.29 s A1 [5]

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5 A particle P of mass 0.4 kg is released from rest at a point O on a smooth plane inclined at 30Å to the horizontal. A force of magnitude 3e−t N directed up a line of greatest slope acts on P, where t s is the time after release. dv (i) Show that = 7.5e−t −5, where v m s−1 is the velocity of P up the plane at time t s. [2] dt (ii) Express v in terms of t. [3] (iii) Find the distance of P from O when v has its maximum value. [3]

Mark scheme: 5 (i) 0.4dv/dt = 3e–t – 0.4gsin30 M1 dv/dt = 7.5e–t – 5 AG A1 [2] (ii) M1 Integrates accn v t – t M1 Limits or finds integration constant 7.5e – 5 ) dt ∫0 d v = ∫0 ( v = 7.5 –7.5e–t – 5t A1 [3] (iii) Solves dv/dt = 0 M1 t = 0.405(46…) x 0.405 7.5 – 7.5e – 5t ) d t ∫0 – t M1 Integrates expression for v and uses d x = ∫0 ( A1 t = 0.405 x = 0.13(0) m [3]

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6 C 0.4 m D r m E 1.8 m F B A The diagram shows the cross-section ABCDEF through the centre of mass of a uniform prism which rests with AB on rough horizontal ground. ABCD is a rectangle with AB = CD = 0.4 m and BC = AD = 1.8 m. The other part of the cross-section is a semicircle with diameter DF and radius r m. (i) Given that the prism is on the point of toppling, show that r = 0.6. [3] A force of magnitude P N is applied to the prism, acting at 60Å to the upwards vertical along a tangent to the semicircle at a point between D and E. The prism has weight 15 N and is in equilibrium on the point of toppling about B. (ii) Show that P = 3.26, correct to 3 significant figures. [4] (iii) Find the smallest possible value of the coefficient of friction between the prism and the ground. [2] [Question 7 is printed on the next page.]

Mark scheme: 6 (i) CoM semi-circle from DF = 4r/3π B1 (0.4 x 1.8) x 0.2 = (πr2 /2) x (4r/3π) M1 Moments about A r = 0.6 AG A1 [3] (ii) Pcos60(0.4 + 0.6cos60) B1 Moment of vertical component Pcos30(1.8 – 0.6 + 0.6sin60) B1 Moment of horiz component 15 x 0.4 = M1 Pcos60(0.4 + 0.6cos60) + Pcos30(1.8 –0.6 + 0.6sin60) P = 3.26 N AG A1 3.2622... [4] (iii) µ =3.262sin60/(15 – 3.262cos60) M1 µ = 0.211 A1 [2]

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7 0.4 m O B 60Å A small ball B of mass 0.5 kg moves in a horizontal circle with centre O and radius 0.4 m on the smooth inner surface of a hollow cone fixed with its vertex down. The axis of the cone is vertical and the semi-vertical angle is 60Å (see diagram). (i) Show that the magnitude of the force exerted by the cone on B is 5.77 N, correct to 3 significant figures, and calculate the angular speed of B. [4] One end of a light elastic string of natural length 0.45 m and modulus of elasticity 36 N is attached to B. The other end of the string is attached to the point on the axis 0.3 m above O. The ball B again moves on the surface of the cone in the same horizontal circle as before. (ii) Calculate the speed of B. [6]

Mark scheme: 7 (i) Rcos30 = 0.5g M1 R = 5.77(35...) AG A1 Rsin30 = 0.5ω2 x 0.4 M1 ω = 3.8(0) rad s–1 A1 [4] (ii) T = 36(0.5 – 0.45) /0.45 M1 4 N Vert cmpt = 4 x 0.3/0.5 = 2.4 A1 Horiz cmpt = 4 x 0.4/0.5 = 3.2 A1 Rcos30 +2.4 = 0.5g M1 R = 3(.00…) N 0.5v2 /0.4 =3.2 + Rsin30 M1 v = 1.94 m s–1 A1 [6]

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