Cambridge A Level Mathematics (US) 9280 — 2015 May/June Paper 4 · Variant 1

*5975161927* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS (US) 9280/41 Paper 4 Mechanics 1 (M1) May/June 2015 1 hour 15 minutes Additional Materials: Answer Booklet/Paper Graph Paper List of Formulas (MF9) (US) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Center number, candidate number, and name on the work you hand in. 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. Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying larger numbers of marks later in the paper. This document consists of 3 printed pages and 1 blank page. JC15 06_9280_41/FP © UCLES 2015 [Turn over

2 1 A block B of mass 2.7 kg is pulled at constant speed along a straight line on a rough horizontal floor. The pulling force has magnitude 25 N and acts at an angle of 1 above the horizontal. The normal component of the contact force acting on B has magnitude 20 N. (i) Show that sin 1 = 0.28. [2] (ii) Find the work done by the pulling force in moving the block a distance of 5 m. [2] 2 x y O 25 N F N 63 N 1 tan−1 0.75 Three horizontal forces of magnitudes F N, 63 N, and 25 N act at O, the origin of the x-axis and y-axis. The forces are in equilibrium. The force of magnitude F N makes an angle 1 anticlockwise with the positive x-axis. The force of magnitude 63 N acts along the negative y-axis. The force of magnitude 25 N acts at tan−1 0.75 clockwise from the negative x-axis (see diagram). Find the value of F and the value of tan 1. [5] 3 A block of weight 6.1 N slides down a slope inclined at tan−111 60  to the horizontal. The coefficient of friction between the block and the slope is 1 4. The block passes through a point A with speed 2 m s−1. Find how far the block moves from A before it comes to rest. [5] 4 A lorry of mass 14 000 kg moves along a road starting from rest at a point O. It reaches a point A, and then continues to a point B which it reaches with a speed of 24 m s−1. The part OA of the road is straight and horizontal and has length 400 m. The part AB of the road is straight and is inclined downwards at an angle of 1Å to the horizontal and has length 300 m. (i) For the motion from O to B, find the gain in kinetic energy of the lorry and express its loss in potential energy in terms of 1. [3] The resistance to the motion of the lorry is 4800 N and the work done by the driving force of the lorry from O to B is 5000 kJ. (ii) Find the value of 1. [3] © UCLES 2015 9280/41/M/J/15

3 5 A cyclist and her bicycle have a total mass of 84 kg. She works at a constant rate of P W while moving on a straight road which is inclined to the horizontal at an angle 1, where sin 1 = 0.1. When moving uphill, the cyclist’s acceleration is 1.25 m s−2 at an instant when her speed is 3 m s−1. When moving downhill, the cyclist’s acceleration is 1.25 m s−2 at an instant when her speed is 10 m s−1. The resistance to the cyclist’s motion, whether the cyclist is moving uphill or downhill, is R N. Find the values of P and R. [8] 6 Two particles A and B start to move at the same instant from a point O. The particles move in the same direction along the same straight line. The acceleration of A at time t s after starting to move is a m s−2, where a = 0.05 −0.0002t. (i) Find A’s velocity when t = 200 and when t = 500. [4] B moves with constant acceleration for the first 200 s and has the same velocity as A when t = 200. B moves with constant retardation from t = 200 to t = 500 and has the same velocity as A when t = 500. (ii) Find the distance between A and B when t = 500. [6] 7 A B 0.5 m Particles A and B, of masses 0.3 kg and 0.7 kg respectively, are attached to the ends of a light inextensible string. Particle A is held at rest on a rough horizontal table with the string passing over a smooth pulley fixed at the edge of the table. The coefficient of friction between A and the table is 0.2. Particle B hangs vertically below the pulley at a height of 0.5 m above the floor (see diagram). The system is released from rest and 0.25 s later the string breaks. A does not reach the pulley in the subsequent motion. Find (i) the speed of B immediately before it hits the floor, [9] (ii) the total distance traveled by A. [3] © UCLES 2015 9280/41/M/J/15

4 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 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 2015 9280/41/M/J/15

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the May/June 2015 series 9280 MATHEMATICS (US) 9280/41 Paper 4, maximum raw mark 50 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 2015 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 41 © Cambridge International Examinations 2015 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 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.

Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 41 © Cambridge International Examinations 2015 The following abbreviations may be used in a mark scheme or used on the scripts: AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) 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 MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) 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.

Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 41 © Cambridge International Examinations 2015 1 (i) [20 + 25sinθ = 2.7g] M1 For resolving forces vertically sinθ = 0.28 A1 2 AG (ii) [25 × 5 × √(1 – 0.282)] M1 For using WD = Fdcosθ Work done is 120 J A1 2 2 M1 For resolving components of F in x and y directions Fx = F cosθ = 25 × 0.8 = 20, Fy = F sinθ = 63 – 25 × 0.6 = 48 A1 M1 For using F = √( Fx 2 + Fy 2) or for using tanθ = Fy ÷ Fx F = 52 N or tanθ = 2.4 A1 tanθ = 2.4 or F = 52 N B1 5 3 F = 0.25       × 61 60 1.6 [ = 1.5] B1 Allow F = 0.25(6.1cos10.4) [Wsinα – F = ma] M1 For using Newton’s 2nd law 6.1 ×       × −       61 60 1.6 25 .0 61 11 = 0.61a or 6.1 sin 10.4 – 0.25 × 6.1 cos 10.4 = 0.61a A1     − = − = 656 .0 61 40 a The value of a may be seen but is not a required answer. M1 For using 0 = vA 2 + 2as Distance is 4 ÷      × 61 40 2 = 3.05 m A1 5 Alternative method F = 0.25       × 61 60 1.6 [ = 1.5] B1 Allow F = 0.25(6.1 cos 10.4) KE loss = 1 2 × 0.61 × 22 B1 Finding loss of KE PE loss = 0.61 × 10 ×       61 11 x B1 Finding loss of PE [1.5x = 1.22 + 1.1x] M1 Using WD against F = KE loss + PE loss 0.4x = 1.22 → distance = 3.05 m A1 5

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 41 © Cambridge International Examinations 2015 4 (i) M1 For using KE gain = 1 2 mvB 2 or PE loss = mg × ABsinθ For KE gain = 4032 × 103 or PE loss = 42 × 106sinθ A1 PE loss = 42 × 106sinθ or KE gain = 4032 × 103 B1 3 (ii) M1 For using WD by DF = KE gain – PE loss + WD by resistance 5000 = 4032 – 42000sin θ + 3360 A1 θ = 3.3° A1 3 5 M1 For using DF = P v for DF up and down M1 For applying Newton’s 2nd law up and down 3 P – R – 84g × 0.1 = 84 × 1.25 A1 10 P – R + 84g × 0.1 = 84 × 1.25 A1     = −       − 0 168 10 1 3 1 P M1 For solving equations for P P = 720 A1     − − = 105 84 3 720 R M1 For substitution for P to obtain R R = 51 A1 8 6 (i) M1 For integrating a(t) to find v(t) v(t) = 0.05t – 0.0001t2 (+ 0) A1 v(200) = 10 – 4 = 6 ms–1 A1 v(500) = 25 – 25 = 0 A1 4 (ii) M1 For integrating v(t) between limits 0 to 500 to obtain the distance A travels

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 41 © Cambridge International Examinations 2015 ( )dt t t 0001 .0 05 .0 2 500 0 − ∫ 500 0 2 3 0.05 0.0001 2 3 t t −         A1 Distance = 0.025 × 5002 – 0.0001 × 5003 ÷ 3 = 2083 m A1 Accept 2080 M1 For using area property of graph or s = 1 2 (u + v)t or s = ut + 1 2 at2 to find distance travelled by B Distance = 1 2 × 6 × 500 = 1500 m or distance = 1 2 (0+6)×200+ 1 2 (6+0)×300 or distance =       × + 2 200 03 .0 2 1 0 ( )       − + × + 2 300 02 .0 2 1 300 6 A1 Distance between A and B is 2083 – 1500 = 583 m B1 6 Can only be scored if distance travelled by A has been found using integration 7 (i) M1 For using Newton’s 2nd law for both particles T – 0.2 × 3 = 0.3a and 7 – T = 0.7a A1 Acceleration = 6.4 ms-2 A1 [v = 0 + 6.4 × 0.25] M1 For using v = 0 + at to find speed when string breaks v = 1.6 ms-1 A1 2 1 Distance 0 6.4 0.25 2   = + ×     M1 For using s = ut + 1 2 at2 to find distance moved before break Distance = 0.2 m A1 [v2 = 1.62 + 2g × (0.5 – 0.2)] M1 For using v2 = u2 + 2gs to find speed when B hits floor Speed is 2.93 ms-1 A1 9

Page 7 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 41 © Cambridge International Examinations 2015 (ii) M1 For finding distance travelled by A after break from v2 = u2 + 2as Distance travelled after break = (0 – 1.62) ÷ (2 × –2) = 0.64 A1 For A, F = 0.2 × 3 and so – 0.2 × 3 = 0.3a so a = –2 Total distance travelled = 0.2 + 0.64 = 0.84 B1 3 Distance = 0.84 m Alternative method for 7(ii) (ii) T = 2.52 , F = 0.2 × 3 WD by T = 2.52 × 0.2 WD by F = 0.2 × 3 × d B1 For stating WD by T on A and WD by F [0.6d = 2.52 × 0.2] M1 Using WD by F = WD by T (No change in KE or PE for A) WD by T = WD by F → d = 0.84 A1 3 Distance = 0.84 m