Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 4 · Variant 3

*2275268691* Cambridge International AS & A Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER MATHEMATICS 9709/43 Paper 4 Mechanics October/November 2021 1 hour 15 minutes You must answer on the question paper. You will need: List of formulae (MF19) INSTRUCTIONS ³ Answer all questions. ³ Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ³ Write your name, centre number and candidate number in the boxes at the top of the page. ³ Write your answer to each question in the space provided. ³ Do not use an erasable pen or correction fluid. ³ Do not write on any bar codes. ³ 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. ³ You should use a calculator where appropriate. ³ You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator. ³ 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. ³ Where a numerical value for the acceleration due to gravity (g) is needed, use 10 m s–2. INFORMATION ³ The total mark for this paper is 50. ³ The number of marks for each question or part question is shown in brackets [ ]. This document has 12 pages. Any blank pages are indicated. JC21 11_9709_43/2R © UCLES 2021 [Turn over

2 BLANK PAGE © UCLES 2021 9709/43/O/N/21

3 1 A metal post is driven vertically into the ground by dropping a heavy object onto it from above. The mass of the object is 120 kg and the mass of the post is 40 kg (see diagram). The object hits the post with speed 8 m s−1 and remains in contact with it after the impact. (a) Calculate the speed with which the combined post and object moves immediately after the impact. [2] … … … … … … (b) There is a constant force resisting the motion of magnitude 4800 N. Calculate the distance the post is driven into the ground. [3] … … … … … … … … … © UCLES 2021 9709/43/O/N/21 [Turn over

4 2 A particle of mass 8 kg is suspended in equilibrium by two light inextensible strings which make angles of 60Å and 45Å above the horizontal. (a) Draw a diagram showing the forces acting on the particle. [1] (b) Find the tensions in the strings. [6] … … … … … … … … … … … … … … … … … … © UCLES 2021 9709/43/O/N/21

5 3 A ball of mass 1.6 kg is released from rest at a point 5 m above horizontal ground. When the ball hits the ground it instantaneously loses 8 J of kinetic energy and starts to move upwards. (a) Use an energy method to find the greatest height that the ball reaches after hitting the ground. [3] … … … … … … … … … … (b) Find the total time taken, from the initial release of the ball until it reaches this greatest height. [3] … … … … … … … … … … … © UCLES 2021 9709/43/O/N/21 [Turn over

6 4 A car of mass 1400 kg is moving on a straight road against a constant force of 1250 N resisting the motion. (a) The car moves along a horizontal section of the road at a constant speed of 36 m s−1. (i) Calculate the work done against the resisting force during the first 8 seconds. [2] … … … … … … … … … … (ii) Calculate, in kW, the power developed by the engine of the car. [2] … … … … … … … … … … … … © UCLES 2021 9709/43/O/N/21

7 (iii) Given that this power is suddenly increased by 12 kW, find the instantaneous acceleration of the car. [3] … … … … … … … … … … … … (b) The car now travels at a constant speed of 32 m s−1 up a section of the road inclined at 1Å to the horizontal, with the engine working at 64 kW. Find the value of 1. [2] … … … … … … … … … … © UCLES 2021 9709/43/O/N/21 [Turn over

8 5 A particle P moves in a straight line, starting from rest at a point O on the line. At time t s after leaving O the acceleration of P is k16 −t2 m s−2, where k is a positive constant, and the displacement from O is s m. The velocity of P is 8 m s−1 when t = 4. (a) Show that s = 1 64t296 −t2. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2021 9709/43/O/N/21

9 (b) Find the speed of P at the instant that it returns to O. [3] … … … … … … … … … … … … (c) Find the maximum displacement of the particle from O. [3] … … … … … … … … … … … … © UCLES 2021 9709/43/O/N/21 [Turn over

10 6 The diagram shows a particle of mass 5 kg on a rough horizontal table, and two light inextensible strings attached to it passing over smooth pulleys fixed at the edges of the table. Particles of masses 4 kg and 6 kg hang freely at the ends of the strings. The particle of mass 6 kg is 0.5 m above the ground. The system is in limiting equilibrium. (a) Show that the coefficient of friction between the 5 kg particle and the table is 0.4. [2] … … … … … … … The 6 kg particle is now replaced by a particle of mass 8 kg and the system is released from rest. (b) Find the acceleration of the 4 kg particle and the tensions in the strings. [5] … … … … … … … … © UCLES 2021 9709/43/O/N/21

11 … … … … … … … … … (c) In the subsequent motion the 8 kg particle hits the ground and does not rebound. Find the time that elapses after the 8 kg particle hits the ground before the other two particles come to instantaneous rest. (You may assume this occurs before either particle reaches a pulley.) [5] … … … … … … … … … … … … … © UCLES 2021 9709/43/O/N/21

12 Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … 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 the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge. © UCLES 2021 9709/43/O/N/21

This document consists of 11 printed pages. © UCLES 2021 [Turn over Cambridge International AS & A Level MATHEMATICS 9709/43 Paper 4 Mechanics October/November 2021 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 2021 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 2021 © UCLES 2021 Page 2 of 11 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 descriptors 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 2021 © UCLES 2021 Page 3 of 11 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, 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 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 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 2021 © UCLES 2021 Page 4 of 11 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 2021 © UCLES 2021 Page 5 of 11 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 2021 © UCLES 2021 Page 6 of 11 Question Answer Marks Guidance 1(a) 120 8 120 40 × = + v v M1 Applying conservation of momentum. 1 6 ms− = v A1 2 1(b) 1600 4800 160 − = a leading to a = –20 M1 Applying Newton’s 2nd law to the system. 0 = 62 + 2 × (–20) × s M1 Use of constant acceleration equations such as v2 = u2 + 2as. Distance travelled by post = 0.9 m A1 Alternative method for question 1(b) Initial KE = 1 2 × 160 × 62 M1 Use of KE = 1 2 mv2 for combined mass. 1 2 × 160 × 62 + 160 × 10 × s = 4800s M1 Forms work/energy equation. Distance travelled by post = 0.9 m A1 3

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 7 of 11 Question Answer Marks Guidance 2(a) Correct 3 force diagram, including angles shown B1 1 2(b) 1 2 cos60 cos45 = T T M1 Resolving forces horizontally. 1 2 sin60 sin 45 8 + = T T g M1 Resolving forces vertically. 1 2 cos60 cos45 = T T and 1 2 sin60 sin 45 8 + = T T g A1 Attempting to solve for either T1 or T2 M1 T1 = 58.6 N A1 T2 = 41.4 N A1 Alternative method for question 2(b) 1 2 T T 80 sin135 sin150 sin75 = = M1 Applies Lami’s Theorem – at least two terms correct. A1 T1 = 80sin135 sin75 M1 Solves for T1. T1 = 58.6 N A1 T2 = 80sin150 sin75 M1 Solves for T2. T2 = 41.4 N A1 6

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 8 of 11 Question Answer Marks Guidance 3(a) PE = 1.6 × 10 × 5 [= 80J] or v↓= ( ) 2 10 5 ´ ´ [= 10] KE = 1 2 × 1.6 × 102 [= 80 J] B1 Either finds PE loss or uses v2 = u2 + 2as to find the velocity and hence the kinetic energy on reaching the ground 1.6 × 10 × 5 = 1.6 × 10 × h + 8 or 1 2 × 1.6 × v2 = 80 – 8, v↑ = 90 , 0 = 90 + 2× (–10) × h leading to h = … or 1 2 × 1.6 × v2 = 80 – 8, v↑ = 90 , 1 2 m × 90 = m × 10 × h leading to h = … M1 Using Initial PE = Final PE + Loss in KE or using KE = 1 2 mv2 to find initial velocity upwards and either v2 = u2 + 2as or KE loss = PE gain to form equation in h. 4.5 = h m A1 3 3(b) 5 = 0 + 1 2 × 10 × t2 leading to t = 1 or 5 = 1 2 × (0 + 10) × t leading to t = 1 or 10 = 10t leading to t = 1 M1 Use of s = ut + 1 2 gt2 for downward motion or use of s = 1 2 (u + v)t for downward motion or use of v = u + gt for downward motion. 4.5 = 0 – 1 2 × (–10) × t2 leading to t = 0.9 or 4.5 = 1 2 × ( 90 + 0) × t leading to t = 0.9 or 0 = 90 – 10t leading to t = 0.9 M1 Use of s = vt – 1 2 (– g)t2 for upward motion or use of s = 1 2 (u + v)t for upward motion or use of v = u – gt for upward motion. t = 1.95 s A1 3

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 9 of 11 Question Answer Marks Guidance 4(a)(i) [WD = 1250 × 36 × 8] M1 For using Work Done = Force × Distance. WD = 360000 J A1 or 360 kJ 2 4(a)(ii) Power = 1250 × 36 or P = 360000 8 [= 45000 J] B1 FT FT Work Done from ( ) 8 a i . = 45 kW B1 2 4(a)(iii) DF = 57000 36 [= 1583.3..] M1 Use changed Power in P = DF × v. 57000 1250 1400 36 − = a M1 For using Newton’s 2nd law applied to the car. 2 0.238 ms− = a A1 3 4(b) 64000 32 = 1250 + 1400g sin θ M1 For using DF = resistance + component of the weight of the car. θ = 3.1 [3.0708…] A1 2

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 10 of 11 Question Answer Marks Guidance 5(a) 2 3 1 16 , 16 3 = − = − a k kt v kt kt M1 Uses v = ∫a dt. 8 = 16k × 4 – 1 3 k × 43 leading to k = … M1 Substitutes t = 4, v = 8. v = 3 16 3 kt kt - and 3 16 = k A1 OE 2 4 1 8 12 = − s kt kt leading to 2 4 24 3 16 192 = − s t t M1 Uses s = ∫vdt and attempts to find s in terms of t only. May be using v = 3t – 1 16 t 3. 2 2 1 (96 ) 64 = − s t t A1 AG, no errors seen. 5 5(b) s = 0, t2 = 96, t = 4 6 M1 Attempt to find t when s = 0. 3 3 3 1 16 96 96 16 16 3 = × × − × × v M1 Attempt to find v at this t value Speed is 29.4 ms-1 A1 Do not condone 29.4 = − v . 3 5(c) v = 0, t2 = 48, t = 4 3 M1 Determine the time, t (or t2) at which v = 0 1 48 (96 48) 64 = × × − s M1 Use substitution of the t or t2 value to find s. 36 m = s A1 3

9709/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 11 of 11 Question Answer Marks Guidance 6(a) R = 5g, F = 6g – 4g M1 For resolving forces to find F and R. 2 0.4 5 μ = = g g A1 AG 2 6(b) T1 – 4g = 4a or 8g – T2 = 8a M1 For applying Newton’s 2nd law to the 4 kg particle or the 8 kg particle. T1 – 4g = 4a and 8g – T2 = 8a A1 Both equations correct. T2 – T1 – F = 5a and F = 0.4 × 5g B1 Adding gives 8g – 4g – 2g = 17a leading to a = …. M1 Attempt to solve for a, T1 or T2. a = 1.18 ms–2, T1 = 44.7 N, T2 = 70.6 N A1 5 6(c) T – 4g = 4a, -T – F = 5a, F = 2g or –4g – 2g = 9a M1 Applying Newton’s 2nd law to both active particles. a = 60 9 - A1 2 20 20 2 0.5 17 17 v = ´ ´ = leading to v = … [v = 1.0846…] M1 Use of v2 = u2 + 2as or equivalent to find v or v2 when the 8 kg particle reaches the ground. 20 60 0 17 9 t = − M1 Use of v = u + at or equivalent to find t. 0.163 = t s A1 From t = 0.1626978… 5