Cambridge A Level Mathematics 9709 — 2017 May/June Paper 4 · Variant 3

*2044311977* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER MATHEMATICS 9709/43 Paper 4 Mechanics 1 (M1) May/June 2017 1 hour 15 minutes Candidates answer on the Question Paper. Additional Materials: List of Formulae (MF9) READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. Where a numerical value for the acceleration due to gravity is needed, use 10 m s−2. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 50. This document consists of 12 printed pages. JC17 06_9709_43/2R © UCLES 2017 [Turn over

2 1 A man pushes a wheelbarrow of mass 25 kg along a horizontal road with a constant force of magnitude 35 N at an angle of 20Å below the horizontal. There is a constant resistance to motion of 15 N. The wheelbarrow moves a distance of 12 m from rest. (i) Find the work done by the man. [2] … … … … … … … … … … (ii) Find the speed attained by the wheelbarrow after 12 m. [3] … … … … … … … … … … … … © UCLES 2017 9709/43/M/J/17

3 2 3P N P N P N 20 N 55Å 1Å 1Å The four coplanar forces shown in the diagram are in equilibrium. Find the values of P and 1. [5] … … … … … … … … … … … … … … … … … … … © UCLES 2017 9709/43/M/J/17 [Turn over

4 3 A train travels between two stations, A and B. The train starts from rest at A and accelerates at a constant rate for T s until it reaches a speed of 25 m s−1. It then travels at this constant speed before decelerating at a constant rate, coming to rest at B. The magnitude of the train’s deceleration is twice the magnitude of its acceleration. The total time taken for the journey is 180 s. (i) Sketch the velocity-time graph for the train’s journey from A to B. [1] t (s) v (m s−1) O (ii) Find an expression, in terms of T, for the length of time for which the train is travelling with constant speed. [2] … … … … … (iii) The distance from A to B is 3300 m. Find how far the train travels while it is decelerating. [3] … … … … … … … … … … © UCLES 2017 9709/43/M/J/17

5 4 A particle P moves in a straight line starting from a point O. At time t s after leaving O, the velocity, v m s−1, of P is given by v = 2t −53. (i) Find the values of t when the acceleration of P is 54 m s−2. [3] … … … … … … … … … … … (ii) Find an expression for the displacement of P from O at time t s. [3] … … … … … … … … … … … © UCLES 2017 9709/43/M/J/17 [Turn over

6 5 A particle is projected vertically upwards from a point O with a speed of 12 m s−1. Two seconds later a second particle is projected vertically upwards from O with a speed of 20 m s−1. At time t s after the second particle is projected, the two particles collide. (i) Find t. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9709/43/M/J/17

7 … … … … … … … … … … (ii) Hence find the height above O at which the particles collide. [1] … … … … … … … … … … … … … … © UCLES 2017 9709/43/M/J/17 [Turn over

8 6 A car of mass 1200 kg is travelling along a horizontal road. (i) It is given that there is a constant resistance to motion. (a) The engine of the car is working at 16 kW while the car is travelling at a constant speed of 40 m s−1. Find the resistance to motion. [2] … … … … … … … … … … (b) The power is now increased to 22.5 kW. Find the acceleration of the car at the instant it is travelling at a speed of 45 m s−1. [3] … … … … … … … … … … … © UCLES 2017 9709/43/M/J/17

9 (ii) It is given instead that the resistance to motion of the car is 590 + 2v N when the speed of the car is v m s−1. The car travels at a constant speed with the engine working at 16 kW. Find this speed. [3] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9709/43/M/J/17 [Turn over

10 7 B A 30Å 0.5 m 4 kg m kg Two particles A and B of masses m kg and 4 kg respectively are connected by a light inextensible string that passes over a fixed smooth pulley. Particle A is on a rough fixed slope which is at an angle of 30Å to the horizontal ground. Particle B hangs vertically below the pulley and is 0.5 m above the ground (see diagram). The coefficient of friction between the slope and particle A is 0.2. (i) In the case where the system is in equilibrium with particle A on the point of moving directly up the slope, show that m = 5.94, correct to 3 significant figures. [6] … … … … … … … … … … … … … … … … … … © UCLES 2017 9709/43/M/J/17

11 (ii) In the case where m = 3, the system is released from rest with the string taut. Find the total distance travelled by A before coming to instantaneous rest. You may assume that A does not reach the pulley. [8] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9709/43/M/J/17 [Turn over

12 … … … … … … … … … … … … … … … … … … … … … … … 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 International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. © UCLES 2017 9709/43/M/J/17

® IGCSE is a registered trademark. This document consists of 12 printed pages. © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS 9709/43 Paper 4 May/June 2017 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9709/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 2 of 12 Mark Scheme Notes Marks are of the following three types: 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. • 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 (or dep*) 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. • The symbol 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 and B marks are not given for fortuitously “correct” answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

9709/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 3 of 12 The following abbreviations may be used in a mark scheme or used on the scripts: 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 – often written by a ‘fortuitous’ answer ISW Ignore Subsequent Working SOI Seen or implied SR Special Ruling (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) Penalties MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become “follow through” marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR –2 penalty may be applied in particular cases if agreed at the coordination meeting. PA –1 This is deducted from A or B marks in the case of premature approximation. The PA –1 penalty is usually discussed at the meeting.

9709/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 4 of 12 Question Answer Marks Guidance 1(i) WD = 35 cos 20 × 12 M1 Uses WD = Fd cos θ 395 J A1 Total: 2 1(ii) EITHER: WD against resistance = 15 × 12 (B1 35cos20 × 12 = 15 × 12 + ½ (25v2) M1 Uses WDman = WDresistance + KE gain v = 4.14 ms–1 A1) OR: 35 cos 20 – 15 = 25 a [a = 0.716] (B1 Applies Newton’s Second Law v2 = 2 × 0.7155. × 12 M1 Uses v2 = u2 + 2as 4.14 v = ms–1 A1) Total: 3

9709/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 5 of 12 Question Answer Marks Guidance 2 EITHER: 3P sin 55 + P sin θ = 20 + P sin θ or 3P sin 55 = 20 (M1 Resolves forces vertically P = 8.14 A1 3P cos 55 = 2P cos θ M1 Resolves forces horizontally cos θ = 1.5 cos 55 → θ = … M1 Attempt to solve for θ θ = 30.6 A1) OR: 3 sin90 P = 20 sin125 (M1 Uses Lami’s Theorem (forces 3P and 20) P = 8.14 A1 3 sin90 P = 2 cos sin145 P θ M1 Uses Lami’s Theorem (forces 3P and 2P cos θ) cos θ = 1.5 sin 145→ θ = … M1 Attempt to solve for θ θ = 30.6 A1) Total: 5

9709/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 6 of 12 Question Answer Marks Guidance 3(i) Trapezium, right-hand steeper than left-hand slope B1 Total: 1 3(ii) Deceleration 0.5 T B1 May be implied Constant speed 180 – 1.5 T B1 Total: 2 3(iii) 0.5[180 + (180 – 1.5T)] × 25 = 3300 M1 Uses area property T = 64 A1 Distance decelerating = [0.5 × 32 × 25 =] 400 m B1 Total: 3 4(i) a = 3 × 2 × (2t – 5)2 [= 54] *M1 Uses a = dv/dt 6(2t – 5)2 = 54 → t = ... DM1 Solves for t t = 1, 4 A1 Total: 3

9709/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 7 of 12 Question Answer Marks Guidance 4(ii) ( ) 4 2 5 4 2 t s − = × (+ C) *M1 Uses s = ∫ v dt C = − 625 8 DM1 Uses s = 0 at t = 0 ( ) 4 2 5 625 8 8 t s − − = A1 Total: 3 Alternative method for Question 4 4(i) v = 8t3 – 60t2 + 150t – 125 → a = 24t2 – 120t + 150 *M1 Uses a = dv/dt 24t2 – 120t + 150 = 54 → t = ... DM1 Solves for t t = 1, 4 A1 Total: 3 4(ii) s = ∫ 8t 3 – 60t 2 + 150t – 125 dt → s = 8 4 t 4 – 60 3 t 3 + 150 2 t 2 – 125t (+ C) *M1 Uses s = ∫ v dt C = 0 DM1 Uses s = 0 at t = 0 (may be implied) s = 2t 4 – 20t 3 + 75t 2 – 125t A1 Total: 3

9709/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 8 of 12 Question Answer Marks Guidance 5(i) s2 = 20t – 0.5gt2 B1 Second particle M1 Uses s = ut + ½ at2 for first particle s1 = 12(t + 2) – 0.5g(t + 2)2 *A1 12(t + 2) – 0.5g(t + 2)2 = 20t – 0.5gt2 → t=… DM1 Solves s1 = s2 t = 1 7 = 0.143 A1 Total: 5 5(ii) [s = 20 × 1 7 – 5 × 2 1 ( ) 7 = 2.755…] Height is 2.76 m B1 Total: 1

9709/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 9 of 12 Question Answer Marks Guidance 6(i)(a) 16 000 = F × 40 M1 Using P = Fv Resistance is 400 N A1 Total: 2 6(i)(b) 22 500 = F × 45 F = 500 B1 500 – 400 = 1200a M1 Applying Newton’s Second Law a = 1 12 = 0.0833 (ms–2) A1 Total: 3 6(ii) 16 000 = (590 + 2v)v M1 Using P = Fv [2v2 + 590 v – 16 000 = 0] → v = … M1 Solving for v v = 25 (ms–1) A1 Total: 3

9709/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 10 of 12 Question Answer Marks Guidance 7(i) R = mg cos30 B1 Resolves normally F = 2m cos 30 [= m√3] M1 Uses F = µR T = 4g [= 40] B1 Particle B T = mgsin30 + F M1 Resolves parallel to plane for particle A 40 = 5m + m√3 A1 Equation in m m = 40 5 3 +√ = 5.94 A1 AG All correct and no errors seen Total: 6

9709/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 11 of 12 Question Answer Marks Guidance 7(ii) EITHER: [R = 3g cos30] F = 0.2 × 3g cos30 (3√3 = 5.196) (B1 4g – T = 4a or T – 3gsin30 – F = 3a or 4g – 3gsin30 –F =7a M1 Applies Newton’s Second Law to one of the particles or forms system equation in a (mBg – mAgsin30 – F = (mA + mB)a) T – 3gsin30 – 3√3 = 3a or 40 – T = 4a or 4g – 3gsin30 – 3√3 = 7a →a = … M1 Applies Newton’s Second Law to form second equation in T and a and solves for a or solves system equation for a 25 3 3 7 a −√ = = 2.83. A1 v2 = 2 × 2.83 × 0.5 v = 1.68… B1 FT v as T becomes zero FT on a –3gsin30 – 0.2(3gcos30) = 3a –15 – 3√3 = 3a →a = …(–5–√3 = –6.73) M1 Applies Newton’s Second Law and solves for a 0 = 1.682 – 2 × 6.73s s = …(0.210) M1 Uses v2 = u2 + 2as and solves for s Total distance = 0.710 m A1) OR: [R = 3g cos30] F = 0.2 × 3g cos30 (3√3 = 5.196) (B1

9709/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 12 of 12 Question Answer Marks Guidance M1 For 4kg mass, uses PE loss – WDT = KE gain M1 For 3kg mass, uses WDT = KE gain + PE gain + WDFr 4g(0.5) – 0.5T = ½ (4v2) and 0.5T = ½ (3v2) + 3g(0.5sin30) + 3√3(0.5) A1 v2 = (25 – 3√3)/7 or v = 1.68 B1 ½ (3)(1.68)2 = 3g(s sin30) + 3√3s M1 For 3kg mass, uses KE loss = PE gain + WDFr s = …(0.210) M1 Solves for s Total distance = 0.710 m A1) Total: 8