Cambridge A Level Mathematics 9709 — 2016 May/June Paper 5 · Variant 2

*3261705408* Cambridge International Examinations Cambridge International Advanced Level MATHEMATICS 9709/52 Paper 5 Mechanics 2 (M2) May/June 2016 1 hour 15 minutes Additional Materials: Answer Booklet/Paper Graph Paper List of Formulae (MF9) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all 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 4 printed pages. JC16 06_9709_52/2R © UCLES 2016 [Turn over

2 1 A small ball B is projected with speed 12 m s−1 at an angle of 30Å above the horizontal from a point O on horizontal ground. At the instant 0.8 s after projection, B is 0.5 m vertically above the top of a vertical post. (i) Calculate the height of the top of the post above the ground. [3] (ii) Show that B is at its greatest height 0.2 s before passing over the post. [2] 2 One end of a light elastic string of natural length 0.4 m is attached to a fixed point O. The other end of the string is attached to a particle of weight 5 N which hangs in equilibrium 0.6 m vertically below O. (i) Find the modulus of elasticity of the string. [2] The particle is projected vertically upwards from the equilibrium position and comes to instantaneous rest after travelling 0.3 m upwards. (ii) Calculate the speed of projection of the particle. [3] (iii) Calculate the greatest extension of the string in the subsequent motion. [3] 3 The point O is 8 m above a horizontal plane. A particle P is projected from O. After projection, the horizontal and vertically upwards displacements of P from O are x m and y m respectively. The equation of the trajectory of P is y = 2x −x2. (i) Find the value of x for the point where P strikes the plane. [2] (ii) Find the angle and speed of projection of P. [3] (iii) Calculate the speed of P immediately before it strikes the plane. [2] © UCLES 2016 9709/52/M/J/16

3 4 0.1 m 0.2 m A B C D 0.4 m 0.7 m A uniform object is made by drilling a cylindrical hole through a rectangular block. The axis of the cylindrical hole is perpendicular to the cross-section ABCD through the centre of mass of the object. AB = CD = 0.7 m, BC = AD = 0.4 m, and the centre of the hole is 0.1 m from AB and 0.2 m from AD (see diagram). The hole has a cross-section of area 0.03 m2. (i) Show that the distance of the centre of mass of the object from AB is 0.212 m, and calculate the distance of the centre of mass from AD. [4] The object has weight 70 N and is placed on a rough horizontal surface, with AD in contact with the surface. A vertically upwards force of magnitude F N acts on the object at C. The object is on the point of toppling. (ii) Find the value of F. [2] The force acting at C is removed, and the object is placed on a rough plane inclined at an angle 1Å to the horizontal. AD lies along a line of greatest slope, with A higher than D. The plane is sufficiently rough to prevent sliding, and the object does not topple. (iii) Find the greatest possible value of 1. [2] 5 A particle P of mass 0.4 kg is placed at rest at a point A on a rough horizontal surface. A horizontal force, directed away from A and with magnitude 0.6t N, acts on P, where t s is the time after P is placed at A. The coefficient of friction between P and the surface is 0.3, and P has displacement from A of x m at time t s. (i) Show that P starts to move when t = 2. Show also that when P is in motion it has acceleration 1.5t −3 m s−2. [3] (ii) Express the velocity of P in terms of t, for t ≥2. [4] (iii) Express x in terms of t, for t ≥2. [3] [Question 6 is printed on the next page.] © UCLES 2016 9709/52/M/J/16 [Turn over

4 6 P 30Å 30Å O A B 0.4 m OA is a rod which rotates in a horizontal circle about a vertical axis through O. A particle P of mass 0.2 kg is attached to the mid-point of a light inextensible string. One end of the string is attached to the rod at A and the other end of the string is attached to a point B on the axis. It is given that OA = OB, angle OAP = angle OBP = 30Å, and P is 0.4 m from the axis. The rod and the particle rotate together about the axis with P in the plane OAB (see diagram). (i) Calculate the tensions in the two parts of the string when the speed of P is 1.2 m s−1. [6] The angular speed of the rod is increased to 5 rad s−1, and it is given that the system now rotates with angle OAP = angle OBP = 60Å. (ii) Show that the tension in the part AP of the string is zero. [6] 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 2016 9709/52/M/J/16

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 5 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International Advanced Level MATHEMATICS 9709/52 Paper 5 May/June 2016 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 2016 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 A Level – May/June 2016 9709 52 © Cambridge International Examinations 2016 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 A Level – May/June 2016 9709 52 © Cambridge International Examinations 2016 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 A Level – May/June 2016 9709 52 © Cambridge International Examinations 2016 Qu Answer Part Marks Marks Notes 1 (i) M1 For attempting to find the height above the ground at the position of the post s = (12sin30) × 0.8 – g × 2 0.8 /2 A1 1.6 m H ( = 1.6 − 0.5) = 1.1 m A1 3 (ii) 0 = 12sin30 − g t M1 t = 0.6 Time ( = 0.8 − 0.6) = 0.2 AG A1 2 2 (i) 5 = 0.2λ/0.4 M1 Tension = λext/l λ = 10 N A1 2 (ii) B1 Correct EE term 10( 2 0.2 /(2 × 0.4) + (5/g) 2v )/2 = 0.3 × 5 M1 PE/KE/EE 3 terms v = 2 m 1 s− A1 3 (iii) B1 Correct EE term 10 2e /(2 × 0.4) = 5(e + 0.1) M1 Energy equation e = 0.483 A1 3 3 (i) −8 = 2x − 2 x M1 Sub y = −8 in the given equation with an attempt to solve x = 4 A1 2 (ii) θ = 63.4° B1 From ta 1 n−2 −g 2 x /(2 2 2 cos v θ) = − 2 x M1 v = 5 m 1 s− A1 3 Accept 4.99 (iii) 2 V = (vcosθ 2) + (vsinθ 2) + 2 × 8g M1 V = 13.6 m 1 s− A1 2 4 (i) (0.7 × 0.4) × 0.2= (0.28 − 0.03) x + 0.03 × 0.1 M1 x = 0.212 AG A1 (0.7 × 0.4) × 0.35 = (0.28 − 0.03)y + 0.03 × 0.2 M1 y = 0.368 A1 4 (ii) 0.4F = 0.212 × 70 M1 Topples about A F = 37.1 A1 2 (iii) θ = -1 tan [(0.4-0.212)/0.368] M1

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9709 52 © Cambridge International Examinations 2016 Qu Answer Part Marks Marks Notes θ = 27.1 A1 2 5 (i) t = 2 AG B1 From 0.6t = 0.3 × 4 or 1.5t − 3=0 0.4a = 0.6t − 0.3 × 0.4 g M1 a = 1.5t − 3AG A1 3 (ii) v = ( ) 1.5 3 d t t ∫ − M1 v = 0.75 2t − 3t ( + c) A1 0 = 0.75 × 22 −3 × 2 + c (so c=3) M1 Or uses limits with 2 and t v = 0.75 2t − 3t + 3 A1 4 (iii) x = 2 (0.75t ∫ −3t + 3)dt M1 x = 0.25 3t −1.5 2t + 3t (+ k) A1 ft c from (ii) x = 0.25 3t − 1.5 2t + 3t − 2 A1 3 6 (i) Bcos30 − Asin30 = 0.2g B1 Resolving vertically for P Bcos60 − Acos30 = 0.2× 2 1.2 /0.4 M1 A1 2 components of tension, N2L with accn = 2v /r M1 Attempts to eliminate one unknown A = 0.753 N A1 B = 2.74 N A1 6 (ii) r = 0.8sin60 B1 M1 Resolves vertically or uses N2L horizontally Bcos60 − Acos30 = 0.2g A1 Bcos30 − Acos60 = 0.2 × 25 × 0.8sin60 A1 M1 For solving to find A A = 0 A1 6