Cambridge A Level Mathematics - Further 9231 — 2020 Oct/Nov Paper 3 · Variant 2

1 1 6 ag A a a O A fixed smooth solid sphere has centre O and radius a. A particle of mass m is projected downwards a with the with speed 16 ag from the point A on the surface of the sphere, where OA makes an angle upward vertical through O (see diagram). The particle moves in part of a vertical circle on the surface of the sphere. It loses contact with the sphere at the point B, where OB makes an angle b with the upward vertical through O. Given that cos a = 23 , find the value of cosb. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 At B, 2 cos mv mg a β = : ( 2 cos ) v ag β = ( ) 2 2 1 1 cos cos 2 2 mv mu mga α β = + − M1A1 Energy equation with 4 terms and correct dimensions 2 3 ( 2 cos ) 2 ga v ag β = − Substitute for u, cosα and v: 2 cos 2 cos 6 3 ag ag ag β β   = + −     M1 Eliminate to find cosβ 1 cos 2 β = A1 5

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2 A B a 2m m u Two uniform smooth spheres A and B of equal radii have masses 2m and m respectively. Sphere B is at rest on a smooth horizontal surface. Sphere A is moving on the surface with speed u and collides with B. Immediately before the collision, the direction of motion of A makes an angle a with the line of centres of the spheres, where tan a = 43 (see diagram). The coefficient of restitution between the spheres is 1.3 Find the speed of A after the collision. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Speeds v and w after collision 2 2 cos mv mw mu α + = M1 Momentum equation with m. Correct masses, allow sin instead of cos cos w v eu α −= M1 Restitution, with consistent signs ( ) 1 1 3 1 1 cos 2 . 2 3 3 5 3 3 v u e u u α   = − = − =     A1 Square of speed of A = ( ) 2 2 1 sin 3u u α  +     M1 Uses correct speed perpendicular to motion 2 2 1 4 3 5 u u     = +         Speed = 13 15u (= 0.867u) A1 5

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3 An object consists of a uniform solid circular cone, of vertical height 4r and radius 3r, and a uniform solid cylinder, of height 4r and radius 3r. The circular base of the cone and one of the circular faces of the cylinder are joined together so that they coincide. The cone and the cylinder are made of the same material. (a) Find the distance of the centre of mass of the object from the end of the cylinder that is not attached to the cone. 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(b) Show that the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface. 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Mark scheme: 3(a) Volume Centre of mass from base Cone ( ) 2 1 3 . 4 3 r r π 4r r + Cylinder ( ) 2 3 . 4 r r π 2r Combined ( ) 2 4 3 . 4 3 r r π x B1 Distances correct Taking moments about base of cylinder: ( ) ( ) ( ) 2 2 2 4 1 . 3 .4 3 .4 .5 3 .4 .2 3 3 x r r r r r r r r π π π = + M1 A1 Moments equation 11 4 x r = A1 4 3(b) Condition: cos OG OA θ < (where O is vertex of cone and OA is slant height of cone) B1 Correct condition for equilibrium 5 4 4 5 4 5 r r r   + × <     M1 Expression in terms of r 21 < 25 True A1 Correct conclusion, with correct working 3

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4 A particle P of mass m is moving in a horizontal circle with angular speed ~ on the smooth inner surface of a hemispherical shell of radius r. The angle between the vertical and the normal reaction of the surface on P is i. g (a) Show that cos i = 2 . 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The plane of the circular motion is at a height x above the lowest point of the shell. When the angular speed is doubled, the plane of the motion is at a height 4x above the lowest point of the shell. (b) Find x in terms of r. 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Mark scheme: 4(a) cos N mg θ ↑ = B1 2 sin sin N mr θ θω ← = B1 cos mg N θ = so 2 cos g r θ ω = B1 AG 3 4(b) cos r x r θ − = = 2 g r ω B1 Using trig of situation: must involve x In new situation: 2 4 4 g r x r r ω − = × M1 Using new situation with 4x and 2ω seen ( ) 4 4 r x r x − = − M1 Combining 1 5 x r = A1 4

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5 A particle P is projected with speed u m s-1 at an angle of i above the horizontal from a point O on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of P from O at a subsequent time t s are denoted by x m and y m respectively. (a) Starting from the equation of the trajectory given in the List of formulae (MF19), show that gx 2 2 i) . 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When i = tan -1 2 , P passes through the point with coordinates (10, 16). (b) Show that there is no value of i for which P can pass through the point with coordinates (18, 30). 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Mark scheme: 5(a) θ θ = 2 2 tan (1 2 gx y x u θ = − + tan2 ) θ Allow derived from first principles AG 1 5(b) ( ) 2 10 100 16 20 1 4 2u × = − + M1 Substitute into result (a) 2 625, ( 25 u u = = ) A1 Use equation again: ( ) 2 10 324 30 18tan 1 (tan ) 2 625 θ θ × = − + × M1 2 2.592(tan ) ) 18tan 32.592 0 θ θ − + = A1 3 term quadratic. Alternatives include: 2 54 375 679 0, t t − + = 2 324 2250 4074 0 t t − + = Discriminant = 324 4 2.592 32.592 − × × = −13.91 M1 Discriminant for alternatives: −6039 and −217404 As this is less than 0, no real solutions for θ A1 CWO 6

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6 One end of a light elastic string, of natural length a and modulus of elasticity k, is attached to a particle P of mass m. The other end of the string is attached to a fixed point Q. The particle P is projected vertically upwards from Q. When P is moving upwards and at a distance 43 a directly above Q, it has a speed 2ga . At this point, its acceleration is 73 g downwards. Show that k = 4mg and find in terms of a the greatest height above Q reached by P. 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Mark scheme: 6 7 .3 T mg m g + = M1 With 3 a T k a = giving 4 k mg = A1 AG Let greatest height above Q be 4 3 a x + Gain in GPE = mgx and Loss in KE = ½ m.2ga B1 The length being found may be expressed as the total extension of the string or the greatest height above Q. GPE and KE Gain in EPE = 2 2 1 4 . 2 3 3 mg a a x a       + −               B1 EPE Note: initial EPE = 2 9 mga 2 2 2 4 2 2 3 9 9 mg ax a a x mgx mga a   + + − + =     M1 A1 Energy equation, correct number of terms 2 2 7 2 0 3 ax x a + − = M1 Simplify to quadratic 1 3 x a = so greatest height is 5 3 a A1

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7 A particle P of mass m kg moves in a horizontal straight line against a resistive force of magnitude mkv2 N , where v ms -1 is the speed of P after it has moved a distance x m and k is a positive constant. The initial speed of P is u ms -1 . 1 1 (a) Show that x = ln 2 when v = 2 u . 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Beginning at the instant when the speed of P is 12 u , an additional force acts on P. This force has 5mmagnitude N and acts in the direction of increasing x. v (b) Show that when the speed of P has increased again to u ms -1 , the total distance travelled by P is given by an expression of the form 1 A - ku 3 ln 3 3k e B - ku o, stating the values of the constants A and B. 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Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 7(a) 2 dv mv kmv dx = − B1 N2L, with m lnv kx c = − + M1 Separate variables and integrate 0, : ln x v u c u = = = M1 Use initial condition 1 1 : ln , 2 2 v u kx = = − 1 ln 2 x k = A1 AG 4 7(b) 2 d 5 d v m mv mkv x v = − + B1 N2L (allow missing m in this part) 2 3 d d 5 v v x kv = − ( ) 3 1 ln 5 ( ) 3 kv x d k − − = + M1A1 Separate variables and integrate Using (a) ( ) ( ) 3 3 1 1 1 ln 5 ln 5 ln2 3 3 kv x ku k k k − − = − − − M1M1 Use condition. M0 if 1 , 0 2 v u x = = used unless 1 ln2 k is added on later Rearrange dependent on ln solution 3 3 1 40 ln 3 5 ku x k ku   − =   −   M1A1 Use v u = 7

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