Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 5 · Variant 2

1 A particle P of mass 0.2 kg is released from rest at a point O on a smooth horizontal surface. A horizontal force of magnitude te−v N directed away from O acts on P, where v m s−1 is the velocity of P at time t s after release. Find the velocity of P when t = 2. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 M1 Uses Newton's Second Law to set 0.2dv/dt = te−v up a differential equation. Allow a for dv/dt. ∫ e v dv = 5 ∫t dt leading to ev = 5 2t /2 ( + c) M1 Separates the variables and integrates. ev – 1 = 2.5 2t A1 Substitutes t = 0, v = 0. v(2) = ln11 = 2.4 A1 4

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2 0.6 m 0.2 m A uniform solid cone has height 0.6 m and base radius 0.2 m. A uniform hollow cylinder, open at both ends, has the same dimensions. An object is made by putting the cone inside the cylinder so that the base of the cone coincides with one end of the cylinder (see diagram, which shows a cross-section). The total weight of the object is 60 N and its centre of mass is 0.25 m from the base of the cone. Calculate the weight of the cone. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 0.15W + 0.3(60 – W) = 0.25 × 60 M1A1 Attempts to take moments about the base of the cone. W = weight of the cone. W = 20 N A1 3

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3 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. P moves down the line of greatest slope through O. The velocity of P is v m s−1 when its displacement from O is x m. A retarding force of magnitude 0.2v2 N acts on P in the direction PO. (i) Show that vdv = 5 −0.5v2. 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(ii) Express v in terms of x. 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Mark scheme: 3(i) 0.4vdv/dx = 0.4gsin30 – 0.2 v 2 M1 Uses Newton's Second Law down the plane. Allow a for vdv/dx. vdv/dx = 5 – 0.5 2v A1 AG 2 3(ii) ∫ v / (5 − 0.5v 2 )dv = ∫x dx M1 Separates the variables and attempts to integrate. –ln(5 – 0.5 v 2 ) = x ( + c ) A1 c = –ln5 [5 – 0.5 v 2 = 5 e−x ] M1 Puts x = 0, v =0 to find c and attempts to solve for v. − x A1 v = (10 − 10e ) 4

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4 2.4 m A B 0.5 m P A light elastic string has natural length 2 m and modulus of elasticity 39 N. The ends of the string are attached to fixed points A and B which are at the same horizontal level and 2.4 m apart. A particle P of mass m kg is attached to the mid-point of the string and hangs in equilibrium at a point 0.5 m below AB (see diagram). (i) Show that m = 0.9. 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P is projected vertically downwards from the equilibrium position, and comes to instantaneous rest at a point 1.6 m below AB. (ii) Calculate the speed of projection of P. 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Mark scheme: 4(i) 2 2 B1 e = (0.5 + 1.2 ) – 1 = 0.3 T = 39 × 0.3/1 M1 Uses T = λx/L. mg = 2 × (39 × 0.3/1) × 0.5/1.3 M1 Resolves vertically. m = 0.9 A1 AG 4 4(ii) 2 2 B1 E = extension when the particle E = (1.6 + 1.2 ) – 1 = 1 m comes to instantaneous rest. EE = 39 × 21 /(2 × 1) or 39 × 0.32 /(2 × 1) B1 0.9 v 2 /2 + 0.9g(1.6 – 0.5) M1A1 Set up a 4 term energy equation 2 involving EE, KE and PE. = 2[39 × 21 /(2 × 1) – 39 × 0.3 /(2 × 1)] v = 7.54 m −s1 A1 5

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5 O 0.8 m G A 12 N B OAB is a uniform lamina in the shape of a quadrant of a circle with centre O and radius 0.8 m which has its centre of mass at G. The lamina is smoothly hinged at A to a fixed point and is free to rotate in a vertical plane. A horizontal force of magnitude 12 N acting in the plane of the lamina is applied to the lamina at B. The lamina is in equilibrium with AG horizontal (see diagram). (i) Calculate the length AG. 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(ii) Find the weight of the lamina. 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Mark scheme: 5(i) OG = 2 × 0.8sin(π/4)/(3π/4) ( 0.48016…m) B1 AG 2 = (0.8sin45)2 + (0.8cos45 – OG 2) M1 Uses Pythagoras's Theorem 2 2 2 OR the cosine formula. OR AG = 0.8 + OG – 2 × 0.8 × OGcos45 AG = 0.572(11...) m A1 3 5(ii) tanBAG = (0.8cos45 – OG)/(0.8sin45) M1 Uses trigonometry to find angle BAG. BAG = 8.5965° =8.6(0)° A1 W × AG = 12 × 2 × 0.8sin45 × sinBAG M1 Takes moments about A. 0.572W = 12 × 2 × 0.8sin45 × sin8.6 A1FT W = 3.55 N A1 5

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6 One end of a light elastic string of natural length 0.4 m and modulus of elasticity 8 N is attached to a fixed point O on a smooth horizontal plane. The other end of the string is attached to a particle P of mass 0.2 kg which moves on the plane in a circular path with centre O. The speed of P is v m s−1 and the extension of the string is x m. (i) Given that v = 2.5, find x. 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It is given instead that the kinetic energy of P is twice the elastic potential energy stored in the string. (ii) Form two simultaneous equations and hence find x and v. 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Mark scheme: 6(i) T = 0.2 × 2. 5 2 /(0.4 + e) B1 Uses Newton's Second Law towards the centre of the circle. T = 8e/0.4 B1 Uses T = λx/L. 1.25/(0.4 + e) = 20e→20 2e + 8e – 1.25 = 0 M1 Eliminates T to find e. e = 0.12(0) m A1 4 6(ii) 0.2 v 2 /2 = 2[8 x 2 /(2 × 0.4)] B1 Uses KE = 2EE. 0.2 v 2 /(0.4 + x) = 8x/0.4 B1 Uses T = λx/L and T = m v 2 /r. M1 Attempts to solve the 2 equations to find v or x. x = 0.4 and v = 5.66 or 4 2 A1A1 5

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7 A small ball B is projected from a point O which is h m above a horizontal plane. At time 2 s after projection B has speed 18 m s−1 and is moving in the direction 30Å above the horizontal. (i) Find the initial speed and the angle of projection of B. 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B has speed 38 m s−1 immediately before it strikes the plane. (ii) Calculate h. 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B bounces when it strikes the plane, and leaves the plane with speed 20 m s−1 but with its horizontal component of velocity unchanged. (iii) Find the total time which elapses between the initial projection of B and the instant when it strikes the plane for the second time. 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Mark scheme: 7(i) U H =18cos30 and U V =18sin30 + 2g(=29) B1 U= [(18cos30) 2 + 29 2 ] or tanθ=29/(18cos30) M1 Uses Pythagoras's Theorem and trigonometry. U = 32.9(24..) m −s1 A1 θ = 61.7° A1 4 7(ii) v 2 = 38 2 – (18cos30)2= (+/–29)2 + 2gh M1 Uses 2 ways to find v, the vertical velocity at the ground and equates. h = 18 A1 OR mgh + m × 32.924 2 /2 = m × 38 2 /2 M1 h = 18 A1 2 7(iii) – [( 38 2 − (18cos3 0) 2 ] = 29–gt M1 Uses v = u + at for first part of flight. t = 6.36(6) A1 v = [ 2 0 2 − (18cos3 0) 2 ] = 12.5(3) M1 Uses v = u + at for second part of flight. –12.5(3)= 12.5(3) – g ′t ′t = 2.50(6) A1 T ( = 6.366 + 2.506) = 8.87 A1 5

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