Cambridge A Level Mathematics - Further 9231 — 2025 May/June Paper 3 · Variant 2

This document has 16 pages. Any blank pages are indicated. [Turn over * 7 7 1 7 3 4 0 8 7 6 * Cambridge International AS & A Level FURTHER MATHEMATICS 9231/32 Paper 3 Further Mechanics May/June 2025 1 hour 30 minutes You must answer on the question paper. You will need: List of formulae (MF19) INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● If additional space is needed, you should use the lined page at the end of this booklet; the question number or numbers must be clearly shown. ● You should use a calculator where appropriate. ● You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator. ● Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. ● Where a numerical value for the acceleration due to gravity (g) is needed, use 10 m s–2. INFORMATION ● The total mark for this paper is 50. ● The number of marks for each question or part question is shown in brackets [ ]. DC (DE/CGW) 347112/3 © UCLES 2025 ,










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3 9231/32/M/J/25 © UCLES 2025 [Turn over 1 A particle P of mass 8 kg is moving in a straight horizontal line. At time t s, P has displacement x m from a fixed point O on the line and velocity m s v 1 - . The only horizontal force acting on P has magnitude ( ) x x 4 3 + N and acts in the direction OP. When t 0 = , x 0 = and v 1 = . (a) Find an expression for v in terms of x, giving your answer in the form v ax b 2 = + , where a and b are constants to be determined. [3] … … … … … … … … … … … … … … … (b) Find an expression for x in terms of t. [2] … … … … … … … … … * 0000800000003 * ,










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4 9231/32/M/J/25 © UCLES 2025 2 A particle P is projected with speed m s 24 1 - at an angle i° above the horizontal from a point O on a horizontal plane, and moves freely under gravity. At a horizontal distance 35 m from O, there is a vertical wall of height 10 m which is perpendicular to the vertical plane of motion of P. (a) Determine the two values of i for which P just clears the wall. [4] … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000004 * ,










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5 9231/32/M/J/25 © UCLES 2025 [Turn over (b) Given that P clears the wall, find the minimum distance from point O where P can land. [2] … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000005 * ,










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6 9231/32/M/J/25 © UCLES 2025 3 a A P 1.5 m O A rough horizontal disc, centre O, rotates with constant angular speed s rad 1 ~ - . A particle P of mass 1.6 kg lies on the disc at a distance 1.5 m from O, and is attached to a point A vertically above O by a light elastic string. The string has natural length 2 m, modulus of elasticity 32 N and makes an angle a with the vertical OA (see diagram). Particle P moves in a horizontal circle also at a constant angular speed s rad 1 ~ - . Particle P is on the point of slipping in the direction OP. The coefficient of friction between the particle and the disc is 0.5. (a) Given that the tension in the string is 8 N, show that . sin 0 6 a = . [2] … … … … … … … … … … … … … … … … … … * 0000800000006 * ,










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7 9231/32/M/J/25 © UCLES 2025 [Turn over (b) Find the number of revolutions per minute made by the disc and the particle P. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000007 * ,










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8 9231/32/M/J/25 © UCLES 2025 4 θ 3r 2r An object is formed by removing a solid hemisphere, radius 2r, from a uniform solid cone, radius 3r and semi-vertical angle i, where tan 2 1 i = . The axes of symmetry of the cone and the hemisphere coincide. The base of the cone and the base of the hemisphere are in the same plane as each other (see diagram). (a) Find, in terms of r, the distance of the centre of mass of the object from its base. [4] … … … … … … … … … … … … … … … … * 0000800000008 * ,










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9 9231/32/M/J/25 © UCLES 2025 [Turn over The object is placed such that its circular base makes contact with a rough plane which is inclined to the horizontal at an angle a. The object is on the point of toppling. The plane is sufficiently rough to prevent sliding. (b) Find the value of a. [3] … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000009 * ,










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10 9231/32/M/J/25 © UCLES 2025 5 One end of a light elastic string of natural length 0.5 m and modulus of elasticity 14 N is attached to a fixed point A on a smooth plane. The plane makes an angle a to the horizontal, where tan 24 7 a = . A particle P of mass 2 kg is attached to the other end of the string. The string lies along a line of greatest slope of the plane. The particle P is initially held on the plane above the level of A, where AP = 0.8 m. The particle P is then released from rest. Find the maximum velocity of P during the subsequent motion. [6] … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000010 * ,










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12 9231/32/M/J/25 © UCLES 2025 6 Two uniform smooth spheres A and B of equal radii have masses 2m and m respectively. Sphere A is moving in a straight horizontal line with speed u, and sphere B is stationary. Sphere A collides directly with B, and they both then move in the same direction with speeds Av and B v respectively. After the collision, the kinetic energy of B is 2 9 times the kinetic energy of A. (a) Show that B v u 5 6 = . [3] … … … … … … … … … Sphere B then collides with a fixed vertical barrier. Immediately before the collision, the direction of motion of B makes an angle a with the barrier. Immediately after the collision, the direction of motion of B makes an angle b with the barrier. The coefficient of restitution between B and the barrier is 5 4. As a result of the collision, the velocity of B is reduced to u 5 25 12 . (b) Find the value of ( ) sin a b + . [6] … … … … … … … … … … … … * 0000800000012 * ,









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14 9231/32/M/J/25 © UCLES 2025 7 θ A B C a u ka O A fixed hollow sphere has radius a and centre O. The points A, B and C lie on the inner surface of the sphere with OA and BC horizontal. A portion of the sphere has been removed by a horizontal cut through points B and C at a vertical distance ka above the centre of the sphere, where k is a positive constant and k 1 1 . The points O, A, B and C all lie in the same vertical plane. OB makes an angle i with the upward vertical through O (see diagram). A particle P of mass m is free to move on the smooth inner surface of the sphere. The particle P is projected vertically downwards from A with speed u and begins to move in a vertical circle. (a) In the case where u ga 5 6 = , the reaction on P at B is half the reaction on P at A. Find the value of k. [5] … … … … … … … … … … … … … … * 0000800000014 * ,









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15 9231/32/M/J/25 © UCLES 2025 … … … … … (b) Find an expression for u, in terms of a and g, in the case that the particle just reaches B. [1] … … … (c) Find an expression for u, in terms of a and g, in the case that the particle passes through B and in its subsequent motion reaches C. [4] … … … … … … … … … … … … … … … … … * 0000800000015 * ,









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16 9231/32/M/J/25 © UCLES 2025 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … * 0000800000016 * ,









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This document consists of 17 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/32 Paper 3 Further Mechanics May/June 2025 MARK SCHEME Maximum Mark: 50 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the May/June 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 17 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors. GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 3 of 17 Mathematics Specific Marking Principles 1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required then no marks will be awarded for a scale drawing. 2 Unless specified in the question, non-integer answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the degree of accuracy is not affected. 3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points. 4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (ISW). 5 Where a candidate has misread a number or sign in the question and used that value consistently throughout, provided that number or sign does not alter the difficulty or the method required, award all marks earned and deduct just 1 A or B mark for the misread. 6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 17 Annotations guidance for centres Examiners use a system of annotations as a shorthand for communicating their marking decisions to one another. Examiners are trained during the standardisation process on how and when to use annotations. The purpose of annotations is to inform the standardisation and monitoring processes and guide the supervising examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and is understood by all examiners who mark the component. We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the response. The annotations listed below were available to examiners marking this component in this series. Annotations Annotation Meaning More information required Accuracy mark awarded zero Accuracy mark awarded one Independent accuracy mark awarded zero Independent accuracy mark awarded one Independent accuracy mark awarded two Benefit of the doubt Blank Page Incorrect Dep Used to indicate DM0 or DM1

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 17 Annotation Meaning DM1 Dependent on the previous M1 mark(s) Follow through Indicate working that is right or wrong Highlighter Highlight a key point in the working Ignore subsequent work Judgement Judgement Method mark awarded zero Method mark awarded one Method mark awarded two Misread Omission or Other solution Off-page comment Allows comments to be entered at the bottom of the RM marking window and then displayed when the associated question item is navigated to. On-page comment Allows comments to be entered in speech bubbles on the candidate response. Judgment made by the PE Premature approximation Special case Indicates that work/page has been seen

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 17 Annotation Meaning Error in number of significant figures Correct Transcription error Correct answer from incorrect working

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 17 Mark Scheme Notes The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Mark for a correct result or statement independent of method marks. DM or DB When a part of a question has two or more ‘method’ steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly, when there are several B marks allocated. The notation DM or DB is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. FT Implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. • A or B marks are given for correct work only (not for results obtained from incorrect working) unless follow through is allowed (see abbreviation FT above). • For a numerical answer, allow the A or B mark if the answer is correct to 3 significant figures or would be correct to 3 significant figures if rounded (1 decimal place for angles in degrees). • The total number of marks available for each question is shown at the bottom of the Marks column. • Wrong or missing units in an answer should not result in loss of marks unless the guidance indicates otherwise. • Square brackets [ ] around text or numbers show extra information not needed for the mark to be awarded.

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 17 Abbreviations AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed) CWO Correct Working Only ISW Ignore Subsequent Working SOI Seen Or Implied SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) WWW Without Wrong Working AWRT Answer Which Rounds To

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 17 Question Answer Marks Guidance 1(a) 3 d 4 8 d v x x v x + = B1 Correct use of N2L. 3 ( 4 ) d 8 d x x x v v + =   2 4 2 1 4 4 2 v x x c = + + M1 Attempt to integrate by separating variables. Must have d d v v x . Must include c + . ( ) 2 2 4 2 2 1 1 1 16 2 4 1 1 v x x x   = + + = +     2 1 4 1 v x = + A1 Use initial conditions to find c and rearrange to obtain an expression for v in terms of x . Condone 2 4 4 x v + = . Condone omitted explanation for choosing positive v . 3 1(b) 2 1 4 d 1 d x x t = + leading to 2 1 4 1 d d 1 x t x = +   . ( )  1 1 2 2 1 4 d 2tan 4 t x x C x − = = + +  M1 Separate variables and integrate to 1 tan− form. Only FT their v of the form 2 ax b + . 0 C = so ( ) 1 2 2tan x t = A1 Must have C + and statement that 0 C = . 2

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 17 Question Answer Marks Guidance 2(a) ( ) 2 2 2 10 35 35 1 2 24 g t t = − +  B1 Use trajectory equation to set up quadratic in tan t  = using ( ) 35,10 . 2 6125 20160 11885 0 t t − + = M1 Form three-term quadratic in t and equate to 0. OE, for example 2 1225 4032 2377 0 t t − + = . tan 0.769 = leading to 37.6 = . A1 If B0M0 earned, award B1 for correct answer. tan 2.52 = leading to 68.4 = . A1 If B0M0 earned, award B1 for correct answer. 4 2(b) Use trajectory equation with y = 0: 2 2 2 0 tan (1 tan ) 2 24 g x x   = − +  M1 Complete method to find the range, x , using the larger of their two values found in part 2(a). 39.5 A1 Accept 39.4 39.5 x ≤ ≤ . Alternative method for question 2(b) Time of flight, 4.8sin T  = [= 4.4629…] 24 cos x T  = M1 Complete method to find the range, x , using the larger of their two values found in part 2(a). 39.5 A1 Accept 39.4 39.5 x ≤ ≤ . Alternative method for question 2(b) Use range formula: 2 sin2 u x g  = M1 Complete method to find the range, x , using the larger of their two values found in part 2(a). 39.5 A1 Accept 39.4 39.5 x ≤ ≤ . 2

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 17 Question Answer Marks Guidance 3(a) 32 8 2 x = leading to 0.5 x = . M1 Use λx T l = to find x . May be implied by correctly derived value for sin 0.6 = . Hence, 2.5 PA = and 1.5 2.5 sin 0.6 = = . A1 AG, shown convincingly. 2 3(b) ( ) 1.6 8cos R g   = − ( ) 2 8sin mr R     = + *M1 For either equation (resolving horizontally or vertically). Allow sin/cos wrong way round. Correct number of terms and correct signs. Dimensionally correct. Numbers or algebraic. A1 Both equations correct. ( ) ( ) ( ) 2 3 4 5 5 8 0.5 1.6 8 1.6 1.5 g  + − =  DM1 Eliminate R to obtain equation involving 2  with no other unknown quantities. 2 4 = leading to 2 = A1 Number of revolutions per minute = π 60 19.1 = . A1 19.0985… Accept final answer of 19. Allow final answer of π 60 . 5

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 17 Question Answer Marks Guidance 4(a) Volume CoM above base Cone 3 π 18 r 3 2 r Hemisphere 3 16 3 πr 3 4 r Total object 3 3 16 3 π π 18 r r − y M1 Allow one error for cone or hemisphere. SOI, may be seen in moments equation. ( ) ( ) ( ) 3 3 3 3 16 3 16 3 3 2 3 4 π π π π 18 18 r r y r r r r − = − M1 A moments equation with 3 terms, allow sign errors, must be dimensionally correct. A1 Fully correct equation, does not need to be simplified. 69 38 y r = A1 4 4(b) When on point of toppling: 38 23 3 tan r their y    = =   M1 For attempt at tan using 3 r their y or 3 their y r . A1 FT FT their y . = 58.8o A1 CAO 3

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 13 of 17 Question Answer Marks Guidance 5 At max v , 14 25 2 sin T g g    = =   M1 Allow sin T mg  = . 14 14 0.5 25 x g = leading to 1 5 x = . A1 Find extension at maximum velocity. EPE at top:   2 14 0.3 1.26 2 0.5  =  EPE at bottom: ( )   2 14 " " 0.56 2 0.5 their x  =  B1 FT One correct EPE term seen anywhere. FT only on their x found using a correct method (dependent on first M mark). ( ) ( ) 2 2 2 1 2 14 " " 14 0.3 2 sin 0.8 0.5 " " 2 2 0.5 2 0.5 their x g their x v    + + + = +    M1 Dimensionally correct, including KE, GPE, and at least one EPE. A1 FT Correct equation. FT only on their x found using a correct method (dependent on first M mark). 2 0.91 9.1 v g = = leading to 3.02 v = m s–1. A1 3.016620625 v =  6

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 14 of 17 Question Answer Marks Guidance 6(a) 2 2 9 1 1 2 2 2 2 A B mv mv  =   3 A B v v = B1 KE equation linking A v and B v . Needs to be KE equation with 1 2 and m . 2 2 A B mu mv mv = + B1 Conservation of momentum. Must see m present. 6 2 5 5 A B v u v u   = =   B1 AG, shown convincingly. If a quadratic equation is formed, must show method for solving and obtain a single answer. 3 6(b) Let w denote the speed after the second collision. Perpendicular to wall: cos cos B v w  = B1 OE, for example 5cos 2cos   = . Parallel to wall: sin sin B ev w   = B1 OE, for example 2sin 5sin   = . 2 2 2 2 2 2 2 576 144 625 125 sin sin sin sin B e v w       = =   2 2 2 2 2 2 36 144 25 125 cos cos cos B v w cos       = =   *M1 Obtain an equation involving only one of , . Eliminate  or  by squaring both sides and adding. ( ) 2 2 576 36 144 625 25 125 sin 1 sin   + − = DM1 Use 2 2 sin cos 1   + = to reduce to equation with just sine or cosine of 1 variable. 2 5 1 9 3 sin sin 5     = =   A1 Award for any of the following: 2 3 cos= or 1 3 cos 5 = or 2 3 sin = . ( ) sin 1   + = A1 Exact answer. 6

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 15 of 17 Question Answer Marks Guidance 7(a) 2 A mu R a = 2 cos B mv R mg a  = − B1 Both equations correct. Use of N2L towards centre at A and B . May see 2 6 5 u ag = already substituted. May see k or cos. 2 2 cos 2 mu mv mg a a  = − leading to 2 2 1 2 v u agk = + . *M1 Use of 1 2 B A R R = . Condone use of 1 2 A B R R = for this mark. Two terms required in their expression for B R . May see 2 6 5 u ag = already substituted. May see k or cos. 2 2 1 1 2 2 mu mv magk = + leading to 2 2 2 v u agk = − . *M1 Energy equation, dimensionally correct. 6 3 5 5 2 ag agk ag agk − = + DM1 Dependent on both previous M marks. Combine N2L and energy equations to obtain an equation that simplifies to involve k only. 1 5 k = A1 5 7(b) 2 2 2 5 0 leading to v u ag   = =   2 5 u ag = B1 FT FT their 2agk . 1

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 16 of 17 Question Answer Marks Guidance 7(c) Vertically B to C: 2 sin v t g  = Horizontally B to C: 2 sin 2 sin cos cos v a v t v g     = =  M1 Complete method to find 2 v . Must use 2 sin a  and not 2a. May quote range formula. FT their k from part 7(a). 2 cos , sin 1 their k their k   = = − 2 cos ag v  = A1 May see 2 ag v k = or 2 5 v ag = . 2 s 2 co ag u agk  = + M1 Use their energy equation from part 7(a) ( ) 2 2 2 v u agk = − . Must substitute for 2 v . 2 27 27 5 5 , u ag u ag = = A1 CAO Alternative method for question 7(c) Equation of trajectory: ( ) ( ) ( ) 2 2 2 sin 0 2 sin tan 2 cos g a a v     = − ( ) ( ) 2 4 5 4 5 2 2 25 6 0 6 2 6 g a v     = −       (M1) Complete method to find 2 v . Must use 2 sin a  and not 2a. FT their k from part 7(a). 2 cos , sin 1 their k their k   = = − 2 cos ag v  = A1 May see 2 ag v k = or 2 5 v ag = . 2 s 2 co ag u agk  = + M1 Use their energy equation from part 7(a) ( ) 2 2 2 v u agk = − . Must substitute for 2 v . 2 27 27 5 5 , u ag u ag = = A1 CAO

9231/32 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 17 of 17 Question Answer Marks Guidance 7(c) Alternative method for question 7(c) 2 2 2 tan 2 cos gx y x v   = − leading to 2 2 d tan d cos y gx x v   = − Reaches maximum when sin x a  = : 2 2 sin 0 tan cos ga v    = − M1 Complete method to find 2 v . Must use sin a  and not a. FT their k from part 7(a). 2 cos , sin 1 their k their k   = = − 2 cos ag v  = A1 May see 2 ag v k = or 2 5 v ag = . 2 s 2 co ag u agk  = + M1 Use their energy equation from part 7(a) ( ) 2 2 2 v u agk = − . Must substitute for 2 v . 2 27 27 5 5 , u ag u ag = = A1 CAO 4