Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 4 · Variant 3

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level MATHEMATICS 9709/43 Paper 4 Mechanics October/November 2025 1 hour 15 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 [ ]. * 3 0 5 1 5 4 3 9 4 3 * DC (PQ/CGW) 344080/4 © UCLES 2025 ,










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3 9709/43/O/N/25 © UCLES 2025 [Turn over 1 v (m s−1) t (s) 8 5 10 15 20 25 30 – 8 0 The diagram shows the velocity-time graph for the motion of an athlete. The athlete runs in a straight line from point A to point B, runs back to point A and finishes at point B. The graph consists of four straight line segments. (a) Find the deceleration of the athlete 7 seconds after leaving A. [1] … … … … … (b) Find the total distance travelled by the athlete. [2] … … … … … … … … … … … … … * 0000800000003 * ,










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4 9709/43/O/N/25 © UCLES 2025 2 The engine of a motorcycle can generate a maximum power of 105 kW. The mass of the motorcycle and its rider is 400 kg. The total resistance to the motion of the motorcycle and its rider is cv N, where m v s 1 - is the motorcyclist’s speed and c is a constant. The motorcyclist travels along a straight horizontal road under maximum engine power. When the motorcyclist’s speed is , m 35 s 1 - his acceleration is . . m 0 5 s 2 - (a) Show that c 80 = . [3] … … … … … … … … … … The motorcyclist now travels up a straight hill under maximum engine power. The hill makes an angle of sin 1 20 3 - b l with the horizontal. (b) Find the steady speed at which the motorcyclist travels up the hill. [3] … … … … … … … … … … * 0000800000004 * ,










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5 9709/43/O/N/25 © UCLES 2025 [Turn over 3 A block of mass 4 kg is pulled along a rough horizontal road by a constant force of magnitude 25 N acting at an angle of 36° above the horizontal. The block moves in a straight line passing through two points A and B on the road, where m AB 120 = . The coefficient of friction between the block and the road is 0.4. (a) Find the work done against friction in moving the block from A to B. [3] … … … … … … … … The speed of the block at A is . m 7 s 1 - (b) Use an energy method to find the speed of the block at B. [4] … … … … … … … … … … … … … … … * 0000800000005 * ,










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6 9709/43/O/N/25 © UCLES 2025 4 A particle P moves in a straight line. At time t s after passing through a point O on the line, the displacement of P from O is s m, where . . . s t t t 0 01 0 3 2 07 3 2 = - - . (a) Find the value of s when P has its minimum velocity, and find also the speed of P at this instant. [6] … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000006 * ,










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7 9709/43/O/N/25 © UCLES 2025 [Turn over (b) Find the acceleration of P at the instant when the direction of motion of P changes. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000007 * ,










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8 9709/43/O/N/25 © UCLES 2025 5 3.2 m 2 kg 6 kg A B Particles A and B, of masses 2 kg and 6 kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley and the particles hang vertically below the pulley. Both particles are initially held at rest at a height of 3.2 m above horizontal ground (see diagram). Particle A is projected vertically downwards with a speed of . . m 1 2 s 1 - (a) Find the tension in the string and the magnitude of the acceleration of the particles. [4] … … … … … … … … … … … … … … … … … … * 0000800000008 * ,










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9 9709/43/O/N/25 © UCLES 2025 [Turn over In the subsequent motion, A does not hit the ground and neither particle reaches the pulley. When B hits the ground, it does not rebound. (b) Find the time that B is in motion. [2] … … … … … … … (c) Find the distance between the lowest and highest points that A reaches. [4] … … … … … … … … … … … … … … … … … … * 0000800000009 * ,










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10 9709/43/O/N/25 © UCLES 2025 6 Particle P of mass m kg and particle Q of mass 2 kg are free to move on a smooth horizontal plane. P and Q are moving directly towards each other with speeds u m s 1 - and 3u m s 1 - respectively. P and Q collide and the direction of motion of each particle is reversed by the collision. Immediately after the collision the speed of P is . s u 3 m 1 - (a) Find, in terms of m and u, an expression for the velocity of Q after the collision and hence show that . m 4 5 2 . [3] … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000010 * ,










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11 9709/43/O/N/25 © UCLES 2025 [Turn over Q subsequently hits a vertical wall which is perpendicular to the direction of motion of Q. The speed of Q after the impact with the wall is a quarter of its speed before the impact with the wall. There are no further collisions between P and Q. (b) Given that m is an integer, determine the largest possible value of m. [4] … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000011 * ,









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12 9709/43/O/N/25 © UCLES 2025 7 12 N i P A particle P of mass 6 kg lies on a rough plane inclined at an acute angle i to the horizontal. A horizontal force of magnitude 12 N acts on P as shown in the diagram. The line of action of this horizontal force lies in a vertical plane which contains the line of greatest slope of the plane that passes through P. The coefficient of friction between P and the plane is n. P is in equilibrium and on the point of sliding down the plane. (a) Show that tan tan 5 5 1 n i i = + - . [5] … … … … … … … … … … … … (b) Hence find the set of possible values of i. [2] … … … … … … * 0000800000012 * ,









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13 9709/43/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. … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000013 * ,









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16 9709/43/O/N/25 © UCLES 2025 BLANK PAGE 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. * 0000800000016 * ,









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This document consists of 21 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level MATHEMATICS 9709/43 Paper 4 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.

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 2 of 21 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.

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 3 of 21 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 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.

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 4 of 21 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

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 5 of 21 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

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 of 21 Annotation Meaning Error in number of significant figures Correct Transcription error Correct answer from incorrect working

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 21 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.

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 21 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

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 21 Question Answer Marks Guidance 1(a) 1.6 2 ms− B1 oe e.g. 8 5 etc. Condone 1.6 − . 1 1(b)  1 3 10 8 2   or  1 6 5 8 2   M1 Correct expression for the area of either one of the right- angled triangles with base length 5 or one of the isosceles triangles of base length 10. Sight of 40 or 20 as part of a total distance calculation can score this mark. 120 m A1 2 Question Answer Marks Guidance 2(a) Driving force 105000 35 = F [=3000] B1 OE e.g. 35 105000  = F . 35 400 0.5 − =  F c M1 Use of N2L; allow sign errors but must be 3 terms. Allow F or any non-zero value for the driving force (but not 105000). If assuming a = 0 then e.g. 2 105000 35 =  c ( 85.714...)  = c scores B1 M0. 105000 35 400 0.5 80 35 − =   = c c ( ) 2800 35 80 =  = c c A1 AG – a correct equation for c e.g. 3000 35 400 0.5 − =  c followed by c = 80 scores A1. 3

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 21 Question Answer Marks Guidance 2(b) 2 105000 3 80 400 [ 0] 20 80 600 105000[ 0] − −  =  + − = v g v v v B1 B1 for correct weight component: 3 400 20   g or ( ) 1 3 20 400 sin sin−         g .   600 =  . OR 2 80 600 105000 + = v v M1 OE. M1 for applying N2L with the correct number of dimensionally correct terms (with a = 0) leading to a three term quadratic equation/expression in v. Allow sign errors and sin/cos mix on the weight component. Must be using c = 80 for this mark. Speed = 32.7 m 1 s− ONLY A1 CWO. Allow exact 15 5 849 4 − + or AWRT 32.7. (For reference: 32.6720057…). If negative value ( 40.172...) − stated, then must be rejected for this mark. 3

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 21 Question Answer Marks Guidance 3(a) 25sin36 4 + = R g   25.30536...  = R *M1 Attempt to resolve vertically to form an equation for the normal contact force for the block; correct number of dimensionally correct terms, condone sign errors, and sin/cos mix on the 25 N force term. Use of R = 4g scores no marks in this part. Work done against friction ( ) 2 120 40 25sin36 5 =   − DM1 Use of Work done = 0.4 120  R ONLY so M0 if subtracting/adding the Work done by the pulling force to this value e.g. a final answer of 1212 (or 1210) coming from ( ) 120 25cos36 0.4 (40 25sin36)  −  − scores M1 M0 A0. 1210 or 1215 J A1 CWO 1210 or final AWRT 1215 or 1214 (usually coming from using R = 25.3). For reference: 1214.6576… 3

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 21 Question Answer Marks Guidance 3(b) KE change = 2 2 1 1 4 4 7 2 2     −      B v B1 SOI, by KE terms on opposite sides of an equation with the same sign so must imply a change in KE (e.g. 2 2v on one side and 98 on the other). Work Done by pulling force = 25cos36 120     2427.050... =  B1 2 2 1 1 4 4 7 25cos36 120 '1214.65...' 2 2  −  =  − B v 2 2 98 2427.050... '1214.65...'   − = −   B v M1 Attempt at work-energy equation using their work done against friction from 3(a) with four dimensionally correct terms but allow sign errors and sin/cos mix on the pulling force term. Dependent on both M1 marks in 3(a) unless the total work done is recovered in this part e.g. a candidate might, in 3(a), find the total work done by the block (which if correct is 1212.39…) rather than the required work done against friction. Therefore, an answer of 1212.39… 3(a) would score M1 M0 A0 but then seeing 2 2 98 1212.39... − = v (or even 1210 but only if coming from 1212.39…) followed by 25.6 would score full marks in this part. M0 if using 25 120  for the pulling force. 25.6 = B v m 1 s− A1 AWRT 25.6. (For reference: 25.59680…). Special case for using N2L Maximum 2 marks ( ) 25cos36 0.4 40 25sin36 4 2.53 − − =  = a a B1 This is for getting 2.53 = a . (For reference: a = 2.525819…). 2 2 7 2 2.5258... 120 25.6 = +    = v v m 1 s− B1 AWRT 25.6. (For reference: 25.59680…). 4

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 21 Question Answer Marks Guidance 4(a) Attempt to differentiate 3 2 0.01 0.3 2.07 = − − s t t t *M1 Decrease power by 1 and change of coefficient on the same term; = v t a is M0. 2 [ ] 0.03 0.6 2.07 = − − v t t A1 Allow un-simplified. [ ]0.06 0.6 = − a t OR 2 [ ] 0.03( 10) 5.07 = − − v t A1 Allow un-simplified. Must follow from the correct 3 term quadratic in t for v so A0 if coming from 2 0.03 0.6 = − v t t . Or the correct completing the square form for the correct expression for v. 0.06 0.6 0 − = t ( ) 10 = t DM1 Setting their a = 0 where a is a two-term linear expression (of the form 1 2 + k t k where 1 2 , 0  k k ) in t. Or setting their t – 10 (from completing the square on a three term quadratic in t) equal to 0. 40.7 = − s A1 CWO (so first four marks must have been awarded). Must be negative so do not allow 40.7 40.7 = − = s s so no ISW. Saying that 40.7 = − s or 40.7 is A0. 5.07 = −  v speed is 5.07 m 1 s− A1 CWO (so first four marks must have been awarded or first three marks if completing the square). Must be positive – allow A1 if negative sign is dropped without justification but saying speed is either 5.07 − or 5.07 is A0. 6

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 14 of 21 Question Answer Marks Guidance 4(b) 2 0.03 0.6 2.07 0 − − = t t *M1 Set their 3 term quadratic expression in t for v equal to zero – dependent on first M1 mark in 4(a). This mark can be implied by a correct value of t = 23 appearing in this part. 23 = t A1 CWO Ignore mention of t = – 3. Working might appear in 4(a), but this value must be stated in 4(b) to earn this mark. 0.06 23 0.6 =  − a DM1 Their positive t substituted into their two-term linear expression for a – this mark can be implied by a correct answer. 0.78 = a m 2 s− ONLY A1 OE e.g. 39 50 , etc. if 0.78 − found then must be explicitly rejected. 4

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 15 of 21 Question Answer Marks Guidance 5(a) Attempt at N2L for either A or B or the system *M1 Correct number of dimensionally correct terms, allow sign errors. Condone using 10 =  a (or another incorrect value of a) in either the equation for A or B for this mark only. A: 2 2 − = T g a B: 6 6 − = g T a System: 6 2 (2 6) − = + g g a A1 For any two correct equations in a and T (or implied by correct answers) – note that T must be positive in one equation and negative in the other. Note that the three equations: 2 2 , 6 6 − = − = g T a T g a and 2 6 8 − = g g a score M1 A1 and lead to 3 = T g and 0.5 = − a g . Solving for T DM1 From two N2L equations (both containing a) with the correct number of dimensionally correct terms OE. 0.5 = a g or 5 m 2 s− and 3 = T g or 30 N A1 a must be positive – allow those who say 0.5 0.5 = −  = a g a g (or 5 5 = − = a a ) without justification (but must be from correct working). 4

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 16 of 21 Question Answer Marks Guidance 5(b) 2 1 3.2 1.2 ( 5) 2 − = + −  t t OR 2 1 3.2 1.2 5 2 = − +  t t M1 Using 2 1 2 = + s ut at with 3.2, =  s u = 1.2 and their a from 5(a) (not g but allow 10 if come from wrong working in 5(a)) – dependent on either a system equation with the correct number of dimensionally correct terms from 5(a) or from equations for A and B that both contain the correct number of dimensionally correct terms – or other complete method to find t. 1.4(0) s A1 AWRT 1.4 or 6 2 209 25 + . (For reference: 1.39654…). 2 5(c) Distance A travels from initial position to first time at rest: 2 1 0 1.2 2 ( 5) = + − s ( ) 1 0.144  = s OR 2 1 0 ( 1.2) 2 (5) = − +  s ( ) 1 0.144  = − s OR 2 1 1 1.2 0.24 ( 5) 0.24 2 =  + −  s ( ) 1 0.144  = s M1 Attempt at finding the distance of A below its starting position when it first comes to instantaneous rest. Using 1.2 and their 5 from 5(a) – same dependency as in 5(b) for the value of a from 5(a). If using 2 1 1 1.2 0.5 =  + s t at then their 1t must have come from 1 0 1.2 = +  a t with their value of a from 5(a). Energy for A: 2 1 1 1 0 2 1.2 2 10 30 2 −  =   −  s s . Energy for B: 2 1 1 1 0 6 1.2 6 10 30 2 −  = −  +  s s . Energy for system: . 2 2 1 1 1 1 0 2 1.2 6 1.2 2 10 6 10 2 2   −  +  =   −      s s . If using energy – must have the correct number of dimensionally correct terms but allow sign errors. If considering either A or B then their value of T (= 30) is dependent on both M1 marks in 5(a).

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 17 of 21 Question Answer Marks Guidance 2 2 1.2 2 5 3.2 = +  w or 2 2 0 2 5 (3.2 0.144) = +  + w ( ) 2 33.44, 5.7827...  = = w w OR 1.2 5(1.1565... 0.24) = + − w *M1 Correct method to find the speed or speed2 of A/B after travelling 3.2 metres (oe) – using 1.2 and their a from 5(a) – same dependency as in part (b) for the value of a from 5(a). If using = + v u at then t must come from considering 2 1 − t t (see reference on previous page). Distance A travels from a point 6.4 metres above the ground to rest: 2 0 33.44 2 ( 10) = + − s ( ) 2 1.672  = s DM1 Attempt to find the distance travelled by A after B hits the ground, using their w, =− a g and v = 0. Difference in distances is 1 2 3.2 5.016 + + = s s m A1 AWRT 5.02 or 5.01 CWO. 4

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 18 of 21 Question Answer Marks Guidance 6(a) 1 2 ( 3 ) 2 3    + − = − +      m u u m u v M1 Attempt at conservation of momentum with correct number of terms. Allow sign errors ONLY so no misreads in this part. However, if using g with either (or both) masses, then allow this mark but do not allow subsequent A marks. 2 3 3 = − v mu u A1 A correct (un-simplified) expression for the velocity of Q after the collision e.g. (2 9) 4 , 3 , 3 6 − − m u mu u 1 6 3 2 − + mu u mu , etc. 2 0 3 0 4.5 3   −   v mu u m A1 AG CWO – accept a correct expression for v put > 0 followed by 4.5  m . Note that 2 3 0 4.5 4.5 3 − =  =   mu u m m is A0. Using trial and improvement to show that v is negative with value(s) of m that are less than 4.5 is A0 unless they also show that v = 0 when m = 4.5. Similarly, if attempting to show that v is positive with value(s) of m that are greater than 4.5 they must also show that v = 0 when m = 4.5 for A1. This A1 mark is dependent on the previous A1 mark. 3

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 19 of 21 Question Answer Marks Guidance 6(b) [Q’s speed is reduced by 75% by its impact with the wall ] speed after impact is 1 2 1 3 3 4 3 6 4    − = −       mu u mu u *B1FT Follow through their v of the form 1 2 + k mu k u (if simplified) for 1 2 , 0  k k from 6(a) OE e.g. ( ) 2 9 12 − m u or possibly implied by later working e.g. ( ) 9 2 1 3 12 − − m u u . Allow e.g. 1 2 3 4 3   − −     mu u . Look out for this expression embedded in later (possibly incorrect) working. No further collisions therefore 1 1 3 3 6 4 − u mu u  DM1 For 1 1 3 4 u v  with v of the form 1 2 + k mu k u for 1 2 , 0  k k from 6(a) – allow > or = for this mark Note that 4 2 3 3 3 = − u mu u (which leads to m = 6.5) scores B1 M1 ONLY. Solve inequality to obtain 6.5 m  DM1 Solve to obtain an inequality of the form ‘m <’ or ‘m ’. Dependent on both previous marks - M0 if an equation for m found rather than an inequality for a range of values of m. ( ) 4.5 6.5 6   = m m  A1 CWO – condone m = 6 from m < 6.5. 4

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 20 of 21 Question Answer Marks Guidance 7(a) For resolving in any direction to form an equation *M1 Correct number of relevant dimensionally correct terms; allow sin/cos mix; allow sign errors. Forces that need resolving should be resolved, forces that do not need resolving should not be resolved. Perpendicular to plane: 6 cos 12sin   = + R g A1 Or 12 cos sin   + = F R (resolving horizontally). Parallel to plane: 12cos 6 sin   + = F g or 6 sin 12cos (...)    − = g A1 Or cos sin 6   + = R F g (resolving vertically). For use of  = F R to get an equation in  and  only DM1 From dimensionally correct equations with the correct number of relevant terms (forces that need resolving should be resolved, forces that do not need resolving should not). ( ) 6 sin 12cos 6 cos 12sin      − = + g g leading to 5tan 1 5 tan    − = + With some evidence of intermediate step e.g. 5sin 12cos 1 cos sin 12cos 5 cos         −       +     OR ( ) 12cos 5tan 1 12cos (5 tan )     − + OR ‘divide by cosine’ OR use/mention of sin tan cos    = OR sin tan cos ,    = etc. A1 AG – CWO must give some indication of how the equation in sin/cos leads to the given answer e.g. 6 sin 12cos 6 cos 12sin      − = + g g then ‘divide through by cos’ gives 5tan 1 5 tan    − = + scores A1 (so therefore mentioning ‘dividing by cos’). Seeing 6 sin 12cos 5tan 1 6 cos 12sin 5 tan        − − = = + + g g only is A0. 5

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 21 of 21 Question Answer Marks Guidance 7(b) 1 0 tan 5      M1 Set the numerator of  either =, > or  to 0 and obtain tan k where  is =, > or  and 0  k - ignore any consideration of an upper bound (e.g. saying that 1 ). ( ) 11.3 90    A1 AWRT 11.3 (11.30993…). Must be strict inequality. Allow 0.197(39...)  - if anything other than < 90 then A0. 2