Cambridge A Level Mathematics - Further 9231 — 2025 May/June Paper 3 · Variant 4

1 A light spring of natural length a and modulus of elasticity 20mg is placed so that it stands vertically on a horizontal plane. The lower end of the spring is fixed to the plane. A particle of mass m is attached to the upper end of the spring. The particle is pushed vertically downwards until the length of the spring is 3 a . The system is then 5 released from rest. Find the maximum extension of the spring in the subsequent motion. [5] .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... ..........................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Energy equation: M1 Energy equation with two EPE terms and at least one GPE term, 2 2 2 dimensionally correct, allow sign errors. 20 mg ( 5 a ) 3 20 mgx + 5 mga = + mg ( a + x ) Award equivalent marks if any other x is used. 2 a 2 a B1 One correct EPE term seen anywhere. A1 Correct energy equation. 50 x 2 + 5ax − 6a 2 = 0 M1 Dependent on two EPE terms used. Simplify to 3-term quadratic with terms in 2x , ax and a 2 only. x = 103 a A1 Single correct answer. If seen, x = − 52 a must be rejected. Alternative method for question 1 Energy equation: M1 Energy equation with two EPE terms and at least one GPE term, 2 2 2 dimensionally correct, allow sign errors. 20 mg ( 5 a ) 3 20 mg ( h − a ) + 5 mga = + mgh Award equivalent marks if any other h is used. 2 a 2 a B1 One correct EPE term seen anywhere. A1 Correct energy equation. 50h2 − 95ah + 39a 2 = 0 M1 Dependent on two EPE terms used. Simplify to 3-term quadratic with terms in h 2 , ah and a 2 only. h = 1013 a , so the maximum extension is 103 a . A1 Single correct answer. If seen, h = 53 a leading to x = − 52 a must be rejected. 5

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2 A particle P of mass m kg moves along a horizontal straight line against a resistive force of magnitude 2mv 3 N , where v ms -1 is the velocity of P at time t s. When t = 0, v = 1. (a) Find an expression for v in terms of t. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the displacement of P from its initial position when t = 6 . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) dv 3 M1 N2L (m must be seen, possibly as F = ma), no extra terms. m = −2 mv Attempt to integrate by separating variables, allow sign errors. dt −2 2dt  −3 Allow errors in coefficient of v . v dv =  − Allow missing +A . − 12 v −2 = −2t + A A1 Any correct equivalent form. When t = 0 , v = 1 leading to A = − 12 . M1 Use correct initial conditions to find value for A. 1 A1 1 v = AEF, A0 for  . 4t + 1 4t + 1 SC B2 for complete solution except for m not seen. 4 2(b) dx 1 1 M1 FT FT their answer to part 2(a). = so x = 2 4t + 1 [ + B ] Allow missing +B for this mark. dt 4t + 1 When t = 0 , x = 0 leading to B = − 12 . M1 Use correct initial conditions to find value for B. When t = 6 , x = 2 (metres). A1 CAO Alternative method for question 2(b) 6 M1 FT Integration attempt, FT their answer to part 2(a). 6 1 1 x = dt =  4t + 1   2  0 Ignore limits. 0 4t + 1 x = 12 25 − 12 1 M1 Substitute correct limits. x = 2 (metres) A1 CAO 3

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3 A particle P is projected with speed u at an angle a 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 any subsequent time t are denoted by x and y respectively. (a) Derive the equation of the trajectory of P in the form gx 2 2 y = x tan a - sec a . 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It is given that u = 20 2 ms -1 and that the particle P passes through the point where x = 64 m and y = 8 m . (b) Find the possible values of tana. 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Mark scheme: 3(a)  x  B1 Explicitly stated. or t = ( → ) x = (u cos) t    u cos (  ) y = (u sin ) t − 12 gt 2 B1 Explicitly stated. 2 M1 Eliminate t.  x  1  x  y = u sin   − 2 g   Must see this substitution.  u cos  u cos gx 2 2 A1 AG, shown convincingly. y = x tan − sec  2 Dependent on all 3 previous marks being awarded. 2u 4 3(b) 128 2 M1 May be un-simplified, must be an equation in a single 1 + tan  8 = 64tan − 5 ( ) trigonometric function. 16tan 2 − 40tan+ 21 = 0 M1 Obtain 3-term quadratic. AEF, for example 128tan 2 − 320tan+ 168 = 0 . tan = 34 , 74 A1 Both answers required. 3

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4 A h a O B A hollow cone with a smooth inner surface is fixed with its vertex O downwards. The semi-vertical angle of the cone is a, where tan a = 3 . A light inextensible string has a particle A of mass m attached 4 to one end and a particle B of mass m attached to the other end. The string passes through a small hole in the cone at O. Particle B hangs in equilibrium below O. Particle A is on the inner surface of the cone at a height h above the level of O and moves in horizontal circles with constant angular speed ~ (see diagram). Find ~ in terms of g and h. 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Mark scheme: 4 For B: T = mg B1 Must be seen explicitly. = T cos+ mg *M1 Allow sin/cos mix, allow sign error. For A: (  ) R sin Allow this mark with mg instead of T. ( → ) R cos+ T sin  = mr2 *M1 Allow sin/cos mix, allow sign error. Allow this mark with mg instead of T. r = h tan B1 Seen anywhere. Combine equations to find an expression for 2 in terms of g and DM1 mg (1 + cos) May see R = 3mg or R = . h only. sin  g A1 If T = mg is not seen explicitly, maximum possible mark is = 2 h B0 M1 M1 B1 DM1 A0. 6

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5 D h E C 6a A F h B ABCD is a uniform square lamina of side 6a. Points E and F are on DC and AB respectively and are such that DE = FB = h . The quadrilateral BCEF is removed from the square lamina (see diagram). (a) Show that the distance of the centre of mass of the resulting lamina AFED from AD is h 2 - 6 ah + 36a 2 and find a corresponding expression for the distance of the centre of mass from AB. 18 a [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ When the lamina AFED is suspended from the point D, the edge DA makes an angle i with the downward vertical, where tan i = 7 . 15 (b) Find, in terms of a, the two possible values of h. 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Mark scheme: 5(a) Split into rectangle AGED and triangle GFE (DA parallel to EG). M1 All terms present. All terms must be dimensionally correct. Area CoM from AD CoM from AB 1 1 1 18a 2 − 6 ah 2 h 3a Moments about AD: 18a 2 x = 6 ah ( 3 h + 2 a ) Rectangle 6ah 2 h ) + ( )( Triangle 18a 2 − 6ah 13 h + 2 a 2a AFED 18a 2 x y A1 Fully correct equation, does not need to be simplified. h 2 − 6 ah + 36 a 2 A1 AG, shown convincingly. x = 18a 2 2 M1 All terms present. All terms must be dimensionally correct. 18a − 6 ah 2 a ) Moments about AB: 18a y = 6 ah ( 3a ) + ( )( y = 2 a + 13 h A1 Must expand brackets and collect like terms. Alternative method for question 5(a) Split into triangle AED and triangle AEF. M1 All terms present. All terms must be dimensionally correct. Area CoM from AD CoM from AB 1 18a 2 − 3ah 2 a ) AED 3ah 13 h 4a Moments about AD: 18a 2 x = 3ah ( 3 h ) + ( )( AEF 18a 2 − 3ah 2a 2a AFED 18a 2 x y A1 Fully correct equation, does not need to be simplified. h 2 − 6 ah + 36 a 2 A1 AG, shown convincingly. x = 18a 2 2 M1 All terms present. All terms must be dimensionally correct. 18a − 3ah 2 a ) Moments about AB: 18a y = 3ah ( 4 a ) + ( )( y = 2 a + 13 h A1 Must expand brackets and collect like terms. 5(a) Alternative method for question 5(a) Subtract triangle EFH from rectangle AFHD M1 All terms present. All terms must be dimensionally correct. (H is on DC, FH parallel to AD). Area CoM from AD CoM from AB AFHD 36a 2 − 6ah 3a − 12 h 3a Moments about AD: 18a 2 − 6ah 4a − 13 h 4a 18a 2 x = 36 a 2 − 6 ah 18 a 2 − 6 ah 4 a − 13 h ) EFH ( )( 3a − 12 h ) − ( )( AFED 18a 2 x y A1 Fully correct equation, does not need to be simplified. h 2 − 6 ah + 36 a 2 A1 AG, shown convincingly. x = 18a Moments about AB: M1 All terms present. All terms must be dimensionally correct. 18a 2 y = 36 a 2 − 6 ah 18a 2 − 6 ah 4 a ) ( )( 3a ) − ( )( y = 2 a + 13 h A1 Must expand brackets and collect like terms. 5 5(b) x B1 Correct expression for tan. tan = 6 a − y h 2 − 6 ah + 36 a 2 M1 FT Equate their expression for tan (must involve x and y ) to 157 , 7 18a = leading to 5h2 − 16ah + 12a 2 = 0 . substitute the correct x and their y and simplify to a 3-term 1 6 a − ( 2 a + 3 h ) 15 2 2  h − 6 ah + 36 a  1  = 7 ( 6 a − 2 a − 3 h ) .  18 a ( h − 2a )( 5h − 6a ) = 0  quadratic. May see 15  

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6 u A B i 5m 4m Two uniform smooth spheres A and B of equal radii have masses 5m and 4m respectively. Sphere A is moving with speed u on a horizontal surface when it collides with sphere B which is at rest. Immediately before the collision, A’s direction of motion makes an angle i with the line of centres (see diagram). The coefficient of restitution between the spheres is e. (a) Show that the speed of B after the collision is 5 u ( 1 + e)cos i . 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After the collision the kinetic energy of A is equal to the kinetic energy of B. (b) Given that tan i = 2 , find the value of e. 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Mark scheme: 6(a) Speeds for A and B along line of centres after the collision are M1 Momentum equation: m must be present. 1v and 2v , respectively. Allow minus sign between terms on LHS. 5mv1 + 4mv2 = 5mu cos v2 − v1 = eu cos M1 Restitution equation: directions must be consistent with those used in momentum equation (signs must be consistent).  A1 AG, shown convincingly. v2 = 95 u (1 + e ) cos 3  B1 May be found in part 6(a) but credit it in part 6(b).6(b) v1 = 19 u ( 5 − 4e ) cos Does not need to be simplified. Speed of V perpendicular to line of centres = u sin. *B1 Seen anywhere. 1 2 2 1 2 M1 Equate KEs after collision. = 2 (4 m) v2 1 2 (5m) ( v1 + ( u sin) ) Allow 2 m to be missing on both sides but must have the 5 and 4. KE of A must have two terms. Allow sin/cos mix. 5 1 2 2 4 5 2 DM1 Substitute expressions for speeds AND numerical values for = 2 ( 9 u (1 + e ) cos) ( 9 u ( 5 − 4e ) cos) + ( u sin ) ) 2 ( trigonometric functions. Accept un-simplified.  2 2  2 This mark is dependent on the second B1.       3 2 3       5 1 100   + u  5 u (1 + e )     2 9 u ( 5 − 4e ) = 24         OE may be seen. 5 u 2 ( 5 − 4e ) 2 + 20u 2 = 9 u 2 (1 + e ) 2     9  13 13 13            9    4e2 + 80e − 41 = 0 DM1 AEF, obtain 3-term quadratic equation in e. This mark is dependent on the second B1.

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7 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 moves in complete vertical circles with centre O, with the string taut. When the string makes an angle i with the downward vertical through O the speed of P is 4ag . The ratio of the greatest and least tensions in the string during the motion is 11 : 1. Find the value of cosi. 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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 Energy, highest to lowest: M1 Note that assuming the same speed at highest and lowest 1 2 mu 2 − 12 mv 2 = 2 amg  v 2 = u 2 − 4 ag  points could give the correct final answer but will not score full marks. Energy, start to lowest: One energy equation, all terms present, dimensionally correct, 1 allow sign error. u 2 = ag ( 6 − 2cos)  2 mu 2 − 12 m (4 ag ) = mg ( a − a cos)   M1 A second energy equation, all terms present, dimensionally Energy, start to highest, correct, allow sign error 1 v 2 = 2 ag (1 − cos)  2 m (4 ag ) − 12 mv 2 = mg ( a + a cos)   A1 One fully correct energy equation. mu 2 B1 Use of N2L at the lowest point, may see TL = mg ( 7 − 2cos) . At lowest point, TL − mg = a mv 2 B1 Use of N2L at the highest point, may see TH = mg (1 − 2cos) . At highest point, TH + mg = a Must use a different speed. TL = 11TH B1 mu 2 + mg a SOI, must be correct way around. For example, = 11 . mv 2 − mg a Combine to find equation involving cos only. M1 Must see an energy equation involving cos only. cos= 15 A1 8

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