Cambridge A Level Mathematics - Further 9231 — 2024 May/June Paper 3 · Variant 3

This document has 16 pages. Any blank pages are indicated. [Turn over DC (JP/FC) 331456/2 © UCLES 2024 * 4 4 0 9 7 7 2 2 2 8 * FURTHER MATHEMATICS 9231/33 Paper 3 Further Mechanics May/June 2024 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 [ ]. Cambridge International AS & A Level ,

 

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2 9231/33/M/J/24 © UCLES 2024 1 A m 2m u B i u 2 1 Two smooth uniform spheres A and B of equal radii have masses m and 2m respectively. The two spheres are moving on a smooth horizontal surface when they collide with speeds u and u 2 1 respectively. Immediately before the collision, A’s direction of motion is along the line of centres, and B’s direction of motion makes an angle i with the line of centres (see diagram). As a result of the collision, the direction of motion of A is reversed and its speed is reduced to u 1 4 . The direction of motion of B again makes an angle i with the line of centres, but on the opposite side of the line of centres. The speed of B is unchanged. Find the value of the coefficient of restitution between the spheres. [4] … … … … … … … … … … … … … … … … … … * 0019655479002 * ,

 

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3 9231/33/M/J/24 © UCLES 2024 [Turn over 2 A particle P of mass m is attached to one end of a light elastic string of natural length a and modulus of elasticity 2mg. A particle Q of mass km is attached to the other end of the string. Particle P lies on a smooth horizontal table. The string has part of its length in contact with the table and then passes through a small smooth hole H in the table. Particle P moves in a horizontal circle on the surface of the table with constant speed ga 2 1 . Particle Q hangs in equilibrium vertically below the hole with HQ a 4 1 = . (a) Find, in terms of a, the extension in the string. [4] … … … … … … … … … … … … … … … … … … (b) Find the value of k. [2] … … … … * 0019655479003 * ,

 

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4 9231/33/M/J/24 © UCLES 2024 3 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. When the particle is hanging vertically below O, it is projected horizontally with speed u so that it begins to move along a circular path. When P is at the lowest point of its motion, the tension in the string is T. When OP makes an angle i with the upward vertical, the tension in the string is S. (a) Show that ( ) cos S T mg 3 1 i = - + . [5] … … … … … … … … … … … … … … … … … … … … … … … … * 0019655479004 * ,

 

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5 9231/33/M/J/24 © UCLES 2024 [Turn over (b) Given that u ag 4 = , find the value of cosi when the string goes slack. [2] … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655479005 * ,

 

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6 9231/33/M/J/24 © UCLES 2024 4 A B C O 2a i a 2 3 a 4 3 A light spring of natural length a and modulus of elasticity kmg is attached to a fixed point O on a smooth plane inclined to the horizontal at an angle i, where sin 4 3 i = . A particle of mass m is attached to the lower end of the spring and is held at the point A on the plane, where OA a 2 = and OA is along a line of greatest slope of the plane (see diagram). The particle is released from rest and is moving with speed V when it passes through the point B on the plane, where OB a 2 3 = . The speed of the particle is V 2 1 when it passes through the point C on the plane, where O a C 3 4 = . Find the value of k. [7] … … … … … … … … … … … … … … … … … … * 0019655479006 * ,

 

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7 9231/33/M/J/24 © UCLES 2024 [Turn over … … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655479007 * ,

 

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8 9231/33/M/J/24 © UCLES 2024 5 A x 18a B C O 24a A uniform lamina is in the form of a triangle OBC, with OC a 18 = , O a B 24 = and angle ° COB 90 = . The point A on OB is such that OA x = (see diagram). The triangle OAC is removed from the lamina. (a) Find, in terms of a and x, the distance of the centre of mass of the remaining object ABC from OC. [3] … … … … … … … … … … … … … … … … … … * 0019655479008 * ,

 

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9 9231/33/M/J/24 © UCLES 2024 [Turn over The object ABC is suspended from C. In its equilibrium position, the side AB makes an angle i with the vertical, where tan 5 6 i = . (b) Find x in terms of a. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655479109 * ,

  
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10 9231/33/M/J/24 © UCLES 2024 6 A particle P is projected with speed u m s–1 at an angle i above the horizontal from a point O and moves freely under gravity. After 5 seconds the speed of P is u 4 3 . (a) Show that sin u u 100 2500 0 16 7 2 i - + = . [3] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655479110 * ,

  
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11 9231/33/M/J/24 © UCLES 2024 [Turn over (b) It is given that the velocity of P after 5 seconds is perpendicular to the initial velocity. Find, in either order, the value of u and the value of sini. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655479111 * ,

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12 9231/33/M/J/24 © UCLES 2024 7 A parachutist of mass m kg opens his parachute when he is moving vertically downwards with a speed of 50 m s–1. At time t s after opening his parachute, he has fallen a distance x m from the point where he opened his parachute, and his speed is v m s–1. The forces acting on him are his weight and a resistive force of magnitude mv N. (a) Find an expression for v in terms of t. [6] … … … … … … … … … … … … … … … … … … … … … … … … * 0019655479112 * ,

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13 9231/33/M/J/24 © UCLES 2024 (b) Find an expression for x in terms of t. [3] … … … … … … … … … … … … … … … (c) Find the distance that the parachutist has fallen, since opening his parachute, when his speed is 15 m s–1. [2] … … … … … … … … … … … * 0019655479113 * ,

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14 9231/33/M/J/24 © UCLES 2024 Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655479114 * ,

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15 9231/33/M/J/24 © UCLES 2024 BLANK PAGE * 0019655479115 * ,

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16 9231/33/M/J/24 © UCLES 2024 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 * 0019655479116 * ,

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This document consists of 15 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/33 Paper 3 Further Mechanics May/June 2024 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 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 2 of 15 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.

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 3 of 15 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.

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 15 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.

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 5 of 15 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

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 6 of 15 Question Answer Marks Guidance 1 Along line of centres, PCLM: 1 1 1 2 cos 2 cos 2 2 4          m u mu m u m u M1 Masses must be included. Allow sign errors. 5 cos 8  A1 NEL: 1 1 1 cos cos 2 4 2            u u e u u M1 Allow sign errors, e must be on correct side. 3 7  e A1 AEF 4

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 15 Question Answer Marks Guidance 2(a) Hooke’s law: 2mgx T a  B1 2mgx a seen anywhere. 2 4     mga T a a x B1 RHS seen anywhere. May be in terms of radius or extended length, for example 2 1 4 mga l a  , 2 mga r . Equate: 2 2 2 2 , 4 3 0 3 4 mga mgx x ax a a a x      M1 Equate two expressions for T and obtain a simplified homogeneous quadratic equation 2 4 5 0, l al   2 2 4 3 0, r ar a    2 2 3 2 0 k k    4 a x  A1 Single correct answer only. 4 2(b) T kmg   B1 kmg seen anywhere in an equation. This may be seen in part (a). Note that no response in part (b) can earn B1 if kmg seen in part (a). 2 1 , 2 mgx T k a   B1 CWO. Part (a) needs to be correct. 2

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 8 of 15 Question Answer Marks Guidance 3(a) At lowest point, 2 mu T mg a   B1 Condone r used consistently instead of a throughout this question. When string makes angle  with upward vertical, 2 cos mv S mg a    B1 Energy:   2 2 1 1 1 cos 2 2     mu mv mga M1 Must include m. Allow sin instead of cos for this mark, allow sign errors. Eliminate 2 u and 2 v M1 Need to see at least one line of working.   3 1 cos S T mg     A1 AG 5 3(b) When string goes slack, S = 0 so   3 1 cos T mg    M1 May use 2 cos v ag   substituted into energy equation. But 2 4 5 ,      mu T mg mg mg mg a so 2 cos 3   A1 2

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 9 of 15 Question Answer Marks Guidance 4 Consider one situation: A to B: Loss in EPE = 2 2 1 2 2 kmg a a a                 = 3 8 kmga B1 Accept unsimplified. Energy: 2 1 3 sin 2 2 8 mga mV kmga    (   2 3 1 4 V ga k         M1A1 KE, GPE, EPE terms present. Must be dimensionally correct. Must include sin or cos in GPE. Consider a second situation: A to C: Loss in EPE = 2 2 2 4 kmg a a a                 = 15 32 kmga B1 Accept unsimplified. Energy: 2 1 1 5 15 sin 2 2 4 32 mg a m V kmga           2 15 2 4 V ag k         M1 KE, GPE, EPE terms present. Must be dimensionally correct. Must include sin or cos in GPE. Third possible situation: B to C : Loss in EPE = 2 2 2 2 4 kmg a a a                       = 3 32 kmga (B1) This may be used in combination with either of the first two situations. Mark to the candidate’s benefit, Energy: 2 2 1 1 1 1 3 3 sin 2 2 2 4 4 32 mg a m V m V kmga                   2 1 6 4 V ag k         (M1) KE, GPE, EPE terms present. Must be dimensionally correct . Must include sin or cos in GPE. Eliminate 2 V from two energy equations to obtain expression involving only k, a and possibly sin M1 At least one of the energy equations must have scored M1. 9 4 k  A1 7

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 10 of 15 Question Answer Marks Guidance 5(a) OBC OAC ABC Area 1 24 18 2 a a   9ax 2 216 9 a ax  Centre of mass from OC 8a 1 3 x x Moments about OC   2 2 1 216 9 216 8 9 3      a ax x a a ax x M1 A1 Note that moments about OB is M0 ( 6 y a  ). Moments equation about OC with all terms present, allow sign error, dimensionally correct. All correct for A1. 2 2 576 72 3 a x x a x    or 24 3 x a  A1 Accept any equivalent form. Alternative solution to question 5(a) Consider system as equivalent to particles at (0, 18a), (x, 0) and (24a, 0) B1 Then the x-coordinate of the centre of mass is at   1 3 24 x a  M1 A1 3

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 11 of 15 Question Answer Marks Guidance 5(b) 18 6 tan a a x    or 18 a y x  or 1 5a y x  or 1 2 18 1 2 a a x  M1 A1 Either way up (their value for x may be substituted in). 2 2 30 144 0    x ax a Or, with simplified form, 30 24   a x a M1 Obtain homogeneous (quadratic) equation Note that if simplified form of x is used, equation will be linear. 6 x a  A1 Single correct answer only. 4 Question Answer Marks Guidance 6(a) Components of velocity are cos , sin u u gt    B1     2 2 2 3 cos sin 4 u u gt u           M1 Square and add components of velocity and equate to 2 3 4u       . 2 7 100 sin 2500 0 16 u u     A1 AG At least one correct line of working seen. 3

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 12 of 15 Question Answer Marks Guidance 6(b) Let  be angle of direction of motion with horizontal at 5, t  then   sin 5 tan cos u g u      B1 Either way up. tan tan 1    , so sin 5 tan 1 cos u g u           M1 Must be – 1 not +1. FT their expression for tan. 50sin u   A1 Use in result from part (a) to form equation in u or sin M1 2 1600, 40 u u   and 4 sin 5  A1 Both. Alternative method for question 6(b) 3 cos sin 4     u u M1 4 tan 3  or 4 sin 5  A1 3 4 cos 50 4 5     u u M1 A1 Allow sign error. 40 u  and 4 sin 5  A1 Both seen.

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 13 of 15 Question Answer Marks Guidance 6(b) Alternative method for question 6(b) 3 cos sin 4     u u M1 4 tan 3  or 4 sin 5  A1 Use in result from part (a) to form equation in u M1 1000 40 and 7        u A1   40 only u  and 4 sin 5  A1 Both seen. 5

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 14 of 15 Question Answer Marks Guidance 7(a)   d 10 d v m m v t   B1 No marks in this part if suvat used. Must have sight of m (for example in F = ma). ln 10 v t A    or   ln 10 v t A    *M1 A1 Separate variables and integrate to obtain a ln term. Constant may be omitted. Constant needed for A1 Use 0, 50: ln 40 t v A     DM1 Find constant, dependent on previous M1. May use limits instead. 0.1 e 4 e t t v   M1 Remove all logs . 10 40e t v    A1 Correct work only . 6 7(b) 10 40e t x t B     M1 No marks in this part if suvat used in part (a) or part (b). Integrate their answer to part (a). Constant may be omitted. Use 0, 0 : 40 t x B    M1 Use initial condition in their expression for x in terms of t . 10 40e 40 t x t     A1 3

9231/33 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 15 of 15 Question Answer Marks Guidance 7(c) When 1 15, e , 2.08 8 t v t     or ln 8 M1 No marks in this part if suvat used in part (a) part (b) or part (c). Find value of t from their answer to part (a). x 55.8 (metres) A1 Note 35 + 10 ln 8 scores A0. 2