Cambridge A Level Mathematics - Further 9231 — 2025 Oct/Nov Paper 3 · Variant 4

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/34 Paper 3 Further Mechanics October/November 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 [ ]. * 1 9 9 8 4 0 1 3 6 4 * DC (PQ/TC) 360447/2 © UCLES 2025 ,










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2 9231/34/O/N/25 © UCLES 2025 1 i R A B A light inextensible string of length 12a is threaded through a fixed smooth ring R. One end of the string is attached to a particle A of mass m. The other end of the string is attached to a particle B of mass 0.5m. Particle A hangs in equilibrium vertically below the ring. Particle B moves with constant angular speed ~ in a horizontal circle with particle A at its centre. The angle between AR and BR is i (see diagram). Express ~ in terms of g and a. [5] … … … … … … … … … … … … … … … … … … … … * 0000800000002 * ,










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3 9231/34/O/N/25 © UCLES 2025 [Turn over 2 A particle P is projected with speed ms u 1 - at an angle i, where tan 2 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 time t s are denoted by x m and y m respectively. (a) Use the equation of the trajectory given in the list of formulae (MF 19) to show that y x u x 2 25 2 2 = - . [1] … … … … … (b) In the subsequent motion, P passes through the point with coordinates (8,12). The particle then hits a fixed vertical barrier 7 m high that is at a horizontal distance of D m from the point of projection. Find the set of possible values of D. [5] … … … … … … … … … … … … … … … … * 0000800000003 * ,










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4 9231/34/O/N/25 © UCLES 2025 3 2a D A C B F E h a The lamina BFDE is obtained by removing triangles AED and BCF from a uniform square lamina ABCD of side 2a. The length of side AE is a and the length of side FC is h (see diagram). The centre of mass of BFDE is at a distance x from AD, and at a distance y from AB. (a) Show that ( ) x a h h ah a 3 3 6 11 2 2 = - - + and find a corresponding expression for y . [5] … … … … … … … … … … … … … … … … … … … * 0000800000004 * ,










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5 9231/34/O/N/25 © UCLES 2025 [Turn over (b) The lamina BFDE is placed vertically on its edge EB on a smooth horizontal surface. Find, in terms of a, the set of possible values of h for which the lamina remains in equilibrium. [2] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000005 * ,










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6 9231/34/O/N/25 © UCLES 2025 4 i O a A v A fixed smooth spherical shell has centre O and radius a. A particle of mass m moves in complete vertical circles on the smooth inner surface of the shell, where the plane of the circular motion is vertical and passes through O. The particle has speed v when it is at point A, where OA makes an angle i with the upward vertical through O, and cos 18 1 i = (see diagram). (a) Show that v ag 26 3 1 H . [5] … … … … … … … … … … … … … … … … … … … * 0000800000006 * ,










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7 9231/34/O/N/25 © UCLES 2025 [Turn over It is given that v ag 26 3 1 = . (b) Find, in terms of m and g, an expression for the greatest possible value of the normal reaction between the shell and the particle. [3] … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000007 * ,










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8 9231/34/O/N/25 © UCLES 2025 5 A particle P of mass m kg is projected vertically upwards from a point O with an initial speed of ms 20 1 - and moves under gravity. There is a resistive force of magnitude . N mv 0 025 2 , where ms v 1 - is the speed of P at time t s after projection. The displacement of P from O is x m at time t s after projection. (a) Find an expression for v in terms of x, while P is moving upwards. [5] … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000008 * ,










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9 9231/34/O/N/25 © UCLES 2025 [Turn over (b) Find an expression for v in terms of t, while P is moving upwards. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000009 * ,










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10 9231/34/O/N/25 © UCLES 2025 6 A and B are two fixed points at a distance 22a apart, with B vertically below A. A light elastic string of natural length 4a and modulus of elasticity 5mg has one end attached to A and the other end attached to a particle P of mass km. Another light elastic string of natural length 8a and modulus of elasticity 6mg has one end attached to B and the other end attached to P. Particle P is vertically above B. (a) Show that, when the system is in equilibrium, BP a ak 4 57 2 = - . [5] … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000010 * ,










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11 9231/34/O/N/25 © UCLES 2025 [Turn over The particle P is pulled vertically upwards so that BP a 18 = , and is then released from rest. In its subsequent motion, P first comes to instantaneous rest at the point where BP a 8 = . (b) Find the value of k. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000011 * ,









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12 9231/34/O/N/25 © UCLES 2025 7 i a u d Wall Y Wall X P3 P1 P2 X and Y are two fixed smooth vertical walls on a smooth horizontal surface. The walls are parallel and at a distance d apart. The points P1, P2 and P3 all lie on the surface. A particle Q is projected horizontally from the point P1 on Wall X with speed u, and moves along the surface. The particle Q strikes Wall Y at the point P2. Immediately before the collision, the direction of motion of Q makes an angle a with Wall Y, where sin 5 4 a = . Immediately after the collision, the direction of motion of Q makes an angle i with Wall Y. The particle Q then strikes Wall X at the point P3 (see diagram). The time that it takes Q to travel the distance P P 1 2 is T. The time that it takes Q to travel the distance P P 2 3 is kT. Find, in terms of k, the coefficient of restitution between Q and Wall Y. [6] … … … … … … … … … … … … … … … * 0000800000012 * ,









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14 9231/34/O/N/25 © UCLES 2025 Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000014 * ,









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16 9231/34/O/N/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. BLANK PAGE * 0000800000016 * ,









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This document consists of 18 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/34 Paper 3 Further Mechanics October/November 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 October/November 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 2 of 18 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.

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 3 of 18 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.

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 4 of 18 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

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 5 of 18 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

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 of 18 Annotation Meaning Indicates that work/page has been seen Error in number of significant figures Correct Transcription error Correct answer from incorrect working

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 18 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.

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 18 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

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 18 Question Answer Marks Guidance 1 For A: T mg = For B vertically: cos 0.5 T mg = M1 Both equations. Allow sin/cos mix. 60 =  A1 Equate and solve for , SOI. For B horizontally: ( ) 2 sin 0.5 sin T m BR   = M1 Allow sin/cos mix. Allow their stated BR. cos BR AR =   8 BR a = B1 SOI 1 2 g a = A1 AEF in terms of g and a. 5

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 18 Question Answer Marks Guidance 2(a) Use trajectory equation from MF19: 2 2 2 co tan s 2 gx y x u   = − . 2 2 2 2 1 5 10 25 2 2 2 x x y x x u u  = − = − B1 Allow equation derived from first principles. Substitute tan 2 = , 2 1 5 cos = , 10 g = and simplify. AG, shown convincingly. 1 2(b) 2 2 25 8 12 2 8 u  = − M1 Substitute correctly and solve equation from part 2(a). 2 400 u =   20 u = A1 2 25 0 2 400 D D = − leading to   0, 32 D D = = . M1 Solve the equation from part 2(a) with 0 y = . 2 25 7 2 400 D D = − leading to   4, 28 D D = = . M1 Solve the equation from part 2(a) with 7 y = . 28 32 D ≤ ≤ only. A1 Accept strict inequalities. 5

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 18 Question Answer Marks Guidance 3(a) Area From AD From AB ABCD 2 4a a a AED 2 a 1 3 a 2 3 a BCF ah 1 3 2 a h − 4 3 a Shape BFDE 2 3a ah − x y B1 Areas and distances from AD correct. Does not need to be presented in a table. Taking moments about AD: ( ) ( ) 2 2 2 1 1 3 3 3 4 2 a ah x a a a a ah a h − =  −  −  − M1 Dimensionally correct moments equation with at least three correct terms. ( ) 3 2 2 2 2 2 2 11 1 1 3 3 3 2 2 ( 6 11 ) 6 11 (3 ) 3 3 3 a a h ah a h ah a h ah a x a a h a h a ah − + − + − + = = = − − − A1 AG, shown convincingly. Taking moments about AB: ( ) 2 2 2 2 4 3 3 3 4 a ah y a a a a ah a − =  −  −  M1 Dimensionally correct moments equation with at least three correct terms. ( ) ( ) 2 5 2 3 3 a a h y a h − = − A1

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 18 Question Answer Marks Guidance 3(a) Alternative method for question 3(a) Area From AD From AB DEF (2 ) a a h − 1 3 (3 ) a h − 4 3 a EBF 2 a 1 3 (5 ) a h − 2 3 a Shape BFDE 2 3a ah − x y B1 Areas and distances from AD correct. Does not need to be presented in a table. Taking moments about AD: ( ) ( ) 2 2 1 1 3 3 3 2 (3 ) (5 ) a ah x a a h a h a a h − = −  − +  − M1 Dimensionally correct moments equation with at least two correct terms. ( ) 2 2 2 2 2 2 2 1 1 3 3 2 (6 5 5 ) ( 6 11 ) 6 11 (3 ) 3 3 3 a a ah h a ah a h ah a h ah a x a a h a h a ah − + + − − + − + = = = − − − A1 AG, shown convincingly. Taking moments about AB: ( ) ( ) 2 2 4 2 3 3 3 2 a ah y a a h a a a − = −  +  M1 Dimensionally correct moments equation with at least two correct terms. ( ) ( ) 2 5 2 3 3 a a h y a h − = − A1 5 3(b) x a ≥ meaning 2 2 3 2 0 h ah a − + ≥ M1 Use condition x a ≥ and simplify to three-term homogeneous quadratic inequality. [0 ] h a  ≤, 2 h a = . A1 Accept 0 h = . 2

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 18 Question Answer Marks Guidance 4(a) At top 0 R≥, so 2 mw mg a ≥ . B1 Accept equality. ( ) 2 2 1 1 2 2 1 cos mv mw mga  − − ≥ M1 Energy inequality with correct number of terms and correct dimensions, accept equality. A1 Fully correct energy inequality, accept equality. 2 13 1 2 9 mv mga ≥ M1 Substitute for cos and 2 w . 1 3 26 v ag ≥ A1 AG, shown convincingly. If equality used throughout, introduction of inequality must be convincing. 5 4(b) Maximum reaction is at bottom: ( ) 2 2 1 1 2 2 1 cos mv mga mV  + + = 2 5 V ag   =   M1 Energy equation with correct number of terms and correct dimensions and correct signs. Allow sin/cos mix. 2 max mV R mg a − = B1 max 6 R mg = A1 3

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 14 of 18 Question Answer Marks Guidance 5(a) 2 d 0.025 10 d v mv v m m x = − − B1 m must be seen. 2 d d 10 0.025 v v x v = − +     ( ) 2 1 2 0.025 ln 400 x c v − + = + *M1 Separate variables and integrate to a term involving ( ) 2 ln av b + or ln av b pv q   +   +   . A1 Ignore modulus signs. Condone missing + c . OE, for example ( ) 2 20ln 400 x c v −+ = + . 1 2 ln800 c = DM1 Use initial condition to find constant of integration. 1 20 20 2e 1 x v − = − A1 AEF 5 5(b) 2 d 0.025 10 d v m v m m t = − − 2 1 d d 10 0.025 t v v − = +     1 2tan 20 v t k − −+ = M1 Equation and integration to a term involving 1 tan−. Accept missing m . A1 Condone missing + k . 1 20 1 20 2 2tan π k − = = M1 Use initial condition with a term involving 1 tan v a −      . ( ) 1 1 4 2 20tan π v t = − A1 4

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 15 of 18 Question Answer Marks Guidance 6(a) N2L: 1 2 T T kmg = + B1 ( ) 1 5 22 4 4 mg a BP a T a − − = B1 Apply Hooke’s law to string AP. ( ) 2 6 8 8 mg BP a T a − = B1 Apply Hooke’s law to string BP. ( ) ( ) 5 22 4 6 8 4 8 57 2 4 8 114 4 mg a BP a mg BP a k k mg P BP a a ak a a a B = − − − = + − − = M1 Substitute 1T and 2 T into N2L and obtain linear expression for BP in terms of a and k . A1 AG, shown convincingly. Alternative method for question 6(a) N2L: 1 2 T T kmg = + B1 1 5 4 mgx T a = B1 Apply Hooke’s law to string AP, where x denotes the extension in string AP. ( ) ( ) 2 6 18 8 6 10 8 8 mg a x a mg a x T a a − −  −  = =     B1 Apply Hooke’s law to string BP. ( ) 6 10 5 4 8 15 2 4 15 2 57 2 18 18 4 4 mg a x mgx kmg a a a ak x a ak a ak BP a x a − = + + = + − = − = − = M1 Substitute 1T and 2 T into N2L and obtain linear expression for x in terms of a and k AND substitute their x into 18 BP a x = − . A1 AG, shown convincingly.

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 16 of 18 Question Answer Marks Guidance 6(a) Alternative method for question 6(a) N2L: 1 2 T T kmg = + B1 ( ) 1 5 4 4 mg x a T a − = B1 Apply Hooke’s law to string AP, where x denotes the length of string AP. ( ) ( ) 2 6 22 8 6 14 8 8 mg a x a mg a x T a a − −  −  = =     B1 Apply Hooke’s law to string BP. ( ) ( ) 5 4 6 14 4 8 31 2 31 2 57 2 , 22 22 4 4 4 mg x a mg a x kmg a a a ak a ak a ak x BP a x a − − = + + + − = = − = − = M1 Substitute 1T and 2 T into N2L and obtain linear expression for x in terms of a and k AND substitute their x into 22 BP a x = −. A1 AG, shown convincingly. Alternative method for question 6(a) N2L: 1 2 T T kmg = + B1 ( ) 1 5 10 4 mg a x T a − = B1 Apply Hooke’s law to string AP, where x denotes the extension in string BP. 2 6 8 mgx T a = B1 Apply Hooke’s law to string BP. ( ) 5 10 6 4 8 25 2 25 2 57 2 , 8 8 4 4 4 mg a x mgx kmg a a a ak a ak a ak x BP a x a − = + − − − = = + = + = M1 Substitute 1T and 2 T into N2L and obtain linear expression for x in terms of a and k AND substitute their x into 8 BP a x = + . A1 AG, shown convincingly.

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 17 of 18 Question Answer Marks Guidance 6(a) Alternative method for question 6(a) N2L: 1 2 T T kmg = + B1 1 5 4 A mgx T a = or 2 6 8 B mgx T a = B1 Apply Hooke’s law to string AP or BP, where A x denotes the extension in string AP and B x denotes the extension in string BP. 1 5 4 A mgx T a = , 2 6 8 B mgx T a = , 10 A B x x a + = B1 Apply Hooke’s law to string AP and BP and recognise that the sum of the extensions in both strings is 10a. 5 6 , 5 3 4 4 8 A B A B mgx mgx kmg x x k a a = + − = 15 2 4 A a ak x + = OR 25 2 4 B a ak x − = Either 15 2 57 2 18 18 4 4 A a ak a ak BP a x a + − = − = − = OR 25 2 57 2 8 8 4 4 A a ak a ak BP a x a − − = + = + = M1 Substitute 1T and 2 T into N2L and obtain linear expression in A x and B x in terms of a and k AND use the expression and 10 A B x x a + = to obtain an expression for A x or B x only AND substitute their A x into 18 A BP a x = − or their B x into 8 B BP a x = + . A1 AG, shown convincingly. 5 6(b) ( ) ( ) 2 2 6 18 8 5 14 4 18 8 2 8 2 4 mg a a mg a a kmga kmga a a − − + = +   B1 At least one correct EPE term. M1 Dimensionally correct energy equation with two EPE terms and at least one GPE term (no KE terms). A1 Fully correct energy equation. 5 2 k = A1 4

9231/34 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 18 of 18 Question Answer Marks Guidance 7 Perpendicular to the wall: sin sin eu v   = B1 1 2 PP uT = 2 3 P P vkT = B1 1 2 sin d PP = 2 3 sin d P P = M1 sin d uT = sin d vkT = M1 Eliminate 1 2 PP and 2 3 P P . eud vd uT vkT = M1 1 e k = A1 6