Cambridge A Level Mathematics - Further 9231 — 2023 May/June Paper 3 · Variant 1

1 One end of a light elastic string, of natural length a and modulus of elasticity 3mg, is attached to a fixed point O. The other end of the string is attached to a particle P of mass m. The string hangs with P vertically below O. The particle P is pulled vertically downwards so that the extension of the string is 2a. The particle P is then released from rest. (a) Find the speed of P when it is at a distance 34 a below O. [3] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... (b) Find the initial acceleration of P when it is released from rest. [2] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

Mark scheme: 1(a)   2 3 2 2 mg a a   2 2 1 3 3 3 2 2 4 2 mg mv mg a a a a           M1 Dimensionally correct energy equation. Must have one KE, one EPE term and at least one GPE. Allow sign errors. 15 2 v ag  2.74 ag     A1 AEF 3 1(b) T mg mA   and 3 2 mg T a a   M1 N2L and Hooke’s law Acceleration = 5g [upwards] A1 Allow 50  or 5g  2

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2 P u a i A particle P of mass m is moving with speed u on a fixed smooth horizontal surface. It collides at an angle a with a fixed smooth vertical barrier. After the collision, P moves at an angle i with the barrier, where tan i = 12 (see diagram). The coefficient of restitution between P and the barrier is e. The particle P loses 20% of its kinetic energy as a result of the collision. Find the value of e. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Parallel to wall cos cos v u Perpendicular to wall sin sin v eu    M1 Both Dividing, 1 2tan e   A1 AEF KE reduced by 20%, so   1 4 1 2 2 2 2 2 2 5 2 cos sin mu e mu      M1 Dimensionally correct equation in u or v , but not both. Must have either  or , but not both. Must see 4 5 on the correct side of the equation. Eliminate e: 4 cos 5  A1 2 3 e  A1 Question Answer Marks Guidance 2 Alternative method for question 2 Parallel to wall cos cos v u    Perpendicular to wall sin sin v eu    M1 Both     5 2 5 5 2 5 sin , cos sin , cos 5 5 5 5 v v u u e               A1   2 2 2 2 2 2 2 2 4 5 5 v v u u cos u sin e           A1 AEF, e.g. 2 2 1 4 5 v e        2 2 2 2 1 4 1 2 1 4 2 5 2 5 5 v mv mu m e                 M1 Dimensionally correct equation in v . Must have either  or , but not both. Must see 4 5 or 2 5 on the correct side of the equation. 2 3 e  A1 5

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3 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held at the point A, where OA makes an angle i with the downward vertical through O, and with the string taut. The particle P is projected perpendicular to OA in an upwards direction with speed u. It then starts to move along a circular path in a vertical plane. The string goes slack when P is at B, where angle AOB is 90° and the speed of P is 45 ag . (a) Find the value of sini. 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(b) Find, in terms of m and g, the tension in the string when P is at A. 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Mark scheme: 3(a) At B, 4 sin 5 m ag mg a  until tension 0  used. Mass must be seen. No sign error. 4 sin 5  A1 2 3(b) At A, 2 cos mu T mg a    B1 Energy   2 1 1 4 cos sin 2 2 5 ag mu m mga       M1 A1 Energy equation with 4 terms, dimensionally correct. Mass must be present, allow sign errors. Must see 1 2 in the KE terms. Solve to find T M1 Complete method leading to an expression in mg for T . 21 5 T mg  A1 CWO 5

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4 y a d 2a O C x An object is formed from a solid hemisphere, of radius 2a, and a solid cylinder, of radius a and height d. The hemisphere and the cylinder are made of the same material. The cylinder is attached to the plane face of the hemisphere. The line OC forms a diameter of the base of the cylinder, where C is the centre of the plane face of the hemisphere and O is common to both circumferences (see diagram). Relative to axes through O, parallel and perpendicular to OC as shown, the centre of mass of the object is ( x , y ). 32 a 2 + 3 ad (a) Show that x = and find an expression, in terms of a and d, for .y [5] 16a + 3 d ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... The object is placed on a rough plane which is inclined to the horizontal at an angle i where sin i = 16 . The object is in equilibrium with CO horizontal, where CO lies in a vertical plane through a line of greatest slope. (b) Find d in terms of a. 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Mark scheme: 4(a) [Mass is proportional to volume] Volume Distance of centre of mass from vertical axis Distance of centre of mass from OC Hemisphere   3 2 2 3 a  2a 3 2 8 a   Cylinder 2 a d  a 1 2 d Object   3 2 2 2 3 a a d    x y   3 2 3 2 2 16 2 2 3 3 a a d x a a a d a                M1 A1 Moments equation, dimensionally correct, correct number of terms. Allow sign errors. Simplify to x 2 32 3 16 3 a ad a d    A1 AG. At least one line of intermediate working.   3 2 3 2 2 16 3 1 2 π π 2 π 3 3 8 2 a a d y a a a d d                    M1 Moments equation, dimensionally correct, correct number of terms. Allow sign errors. y     2 2 3 8 2 16 3 d a a d    A1 AEF 5 Question Answer Marks Guidance 4(b) 2 sin 2 a x a    B1 2 1 32 3 2 2 6 16 3 a ad a a a d          5 16 3 32 3 3 a d a d    M1 Remove fractions 8 3 d a  A1 3

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5 A light elastic string of natural length a and modulus of elasticity mmg has one end attached to a fixed point O on a smooth horizontal surface. When a particle of mass m is attached to the free end of the string, it moves with speed v in a horizontal circle with centre O and radius x. When, instead, a particle of mass 2m is attached to the free end of the string, this particle moves with speed 12 v in a horizontal circle with centre O and radius 34 x . (a) Find x in terms of a. 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(b) Given that v = 12ag , find the value of m. 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Mark scheme: 5(a) Hooke’s law,   1 mg T x a a    or 2 3 4 mg T x a a          Also, 1T = 2 mv x and equate   2 mv mg x a x a    M1   2 gx x a v a    Dimensionally correct terms. Similarly: 2 3 1 2 4 2 3 4 x mg a m v a x                M1 2 3 3 2 4 gx v x a a          Must have 1 2 v and 3 4 x on the RHS. Their dimensionally correct 2. T Equate expressions for 2 v and solve for x in terms of a . M1 4 x a  A1 WWW 5 SC B3 for answer of 4a using  instead of mg  . 5(b)   2 a v xg x a   or 2 2 3 3 4 a v xg x a         and substitute 4 x a  , 12 v ag  M1 FT their expression for x. 1 A1 CAO 2

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6 A particle P moving in a straight line has displacement x m from a fixed point O on the line and velocity v m s -1 at time t s. The acceleration of P, in ms -2 , is given by 6v v + 9 . When t = 0, x = 2 and v = 72. (a) Find an expression for v in terms of x. 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(b) Find an expression for x in terms of t. 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Mark scheme: 6(a) 6 9 dv v v v dx   and attempt to separate variables and integrate 2 9 6 v x A    A1 2, 72 6 x v A    M1 Use initial condition to find constant.   2 9 1 9 v x    A1 Correct, AEF. 4 6(b)     2 9 2 , 9 2 dx dx x x dt dt x x             1 1 1 9 2 2 dx dt x x          M1 Separate variables and write in the form a b dx dt x x c          1 ln 9 2 2 x t B x         A1 Integrate, any correct form. 1 1 0, 2 ln 2 2 t x B    M1 Use initial condition to find constant. 2 18 ln 2 x t x         18 2 2 t x e x   M1 Take logarithms 18 18 2 2 t t e x e   or 18 2 2 1 t x e   A1 Any correct form 5

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7 At time t s, a particle P is projected with speed 40 m s -1 at an angle i above the horizontal from a point O on a horizontal plane and moves freely under gravity. The greatest height achieved by P during its flight is H m and the corresponding time is T s. (a) Obtain expressions for H and T in terms of i. [2] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... During the time between t = T and t = 3, P descends a distance 14 H . (b) Find the value of i. 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(c) Find the speed of P when t = 3. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 7(a) 2 80sin H   or 2 800sin g  4sin T   or 40sin g  B1 2 7(b) Between t T  and 3 t    2 1 1 10 3 4 2 H T      M1 A1 No extra terms. Use results from part (a)   2 2 1 80sin 5 3 4sin 4     2 4sin 8sin 3 0      M1 Substitute their expressions for H and T from part (a) and obtain a quadratic equation in sin with no more than three terms. 1 sin 2  , 30   A1 Single answer. NFWW. Alternative method for question 7 part (b)  2 3 1 3 40 3 sin 10 3 4 2 H y        M1 A1 120 sin 45  Use results from (a): 2 380sin 120 sin 45 4     2 4sin 8sin 3 0      M1 Substitute their expressions for H and T from part (a) and obtain a quadratic equation in sin with no more than three terms. 1 sin 2  , 30   A1 Single answer. NFWW. 4 Question Answer Marks Guidance 7(c) When 3 t  speeds 40cos  and 40sin 10 3     B1 Square and add to find square of speed:     2 2 2 20 3 10 v   M1 Must be numerical. 2 1300, 10 13 v v   [= 36.1] A1 3

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