Cambridge A Level Mathematics 9709 — 2014 May/June Paper 5 · Variant 3

*0717637935* Cambridge International Examinations Cambridge International Advanced Level MATHEMATICS 9709/53 Paper 5 Mechanics 2 (M2) May/June 2014 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. JC14 06_9709_53/FP © UCLES 2014 [Turn over

2 1 A light elastic string has modulus of elasticity 5 N and natural length 1.5 m. One end of the string is attached to a fixed point O and a particle P of mass 0.1 kg is attached to the other end of the string. P is released from rest at the point 2.4 m vertically below O. Calculate the speed of P at the instant the string first becomes slack. [3] 2 50Å 50Å 10Å A B C F N 12 cm A uniform lamina ABC in the shape of an isosceles triangle has weight 24 N. The perpendicular distance from A to BC is 12 cm. The lamina rests in a vertical plane in equilibrium, with the vertex A in contact with a horizontal surface. Angle BAC = 100Å and AB makes an angle of 10Å with the horizontal. Equilibrium is maintained by a force of magnitude F N acting along BC (see diagram). Show that F = 8. [3] 3 A small block B of mass 0.2 kg is placed at a fixed point O on a smooth horizontal surface. A horizontal force of magnitude 0.42 N is applied to B. At time t s after the force is first applied, the velocity of B away from O is v m s−1. (i) Find the value of v when t = 1. [2] For t > 1 an additional force, of magnitude 0.32t N and directed towards O, is applied to B. The force of magnitude 0.42 N continues to act as before. (ii) Find the value of v when t = 2. [3] For t > 2 a third force, of magnitude 0.06t2 N and directed away from O, is applied to B. The other two forces continue to act as before. (iii) Show that the velocity of B is the same when t = 2 and when t = 3. [3] 4 One end of a light inextensible string of length 2.4 m is attached to a fixed point A. The other end of the string is attached to a particle P of mass 0.2 kg. P moves with constant speed in a horizontal circle which has its centre vertically below A, with the string taut and making an angle of 60Å with the vertical. (i) Find the speed of P. [4] The string of length 2.4 m is removed, and P is now connected to A by a light inextensible string of length 1.2 m. The particle P moves with angular speed 4 rad s−1 in a horizontal circle with its centre vertically below A. (ii) Calculate the angle between the string and the vertical. [4] © UCLES 2014 9709/53/M/J/14

3 5 A small ball is thrown horizontally with speed 5 m s−1 from a point O on the roof of a building. At time t s after projection, the horizontal and vertically downwards displacements of the ball from O are x m and y m respectively. (i) Express x and y in terms of t, and hence show that the equation of the trajectory of the ball is y = 0.2x2. [3] The ball strikes the horizontal ground which surrounds the building at a point A. (ii) Given that OA = 18 m, calculate the value of x at A, and the speed of the ball immediately before it strikes the ground at A. [6] 6 O 1.5 m R 0.4 m B P 1Å A A particle P of mass 0.6 kg is attached to one end of a light elastic string of natural length 1.5 m and modulus of elasticity 9 N. The string passes through a small smooth ring R fixed at a height of 0.4 m above a rough horizontal surface. The other end of the string is attached to a fixed point O which is 1.5 m vertically above R. The points A and B are on the horizontal surface, and B is vertically below R. When P is on the surface between A and B, RP makes an acute angle 1Å with the horizontal (see diagram). (i) Show that the normal force exerted on P by the surface has magnitude 3.6 N, for all values of 1. [3] P is projected with speed 2.5 m s−1 towards B from its initial position at A where 1 = 30. The speed of P when it passes through B is 3 m s−1. (ii) Find the work done against friction as P moves from A to B. [4] (iii) Calculate the value of the coefficient of friction between P and the surface. [2] [Question 7 is printed on the next page.] © UCLES 2014 9709/53/M/J/14 [Turn over

4 7 0.3 m 0.3 m O A B The diagram shows a container which consists of a bowl of weight 14 N and a handle of weight 8 N. The bowl of the container is in the form of a uniform hemispherical shell with centre O and radius 0.3 m. The handle is in the form of a uniform semicircular arc of radius 0.3 m and is freely hinged to the bowl at A and B, where AB is a diameter of the bowl. (i) Calculate the distance of the centre of mass of the container from O for the position indicated in the diagram, where the handle is perpendicular to the rim of the bowl. [3] (ii) Show that the distance of the centre of mass of the container from O when the handle lies on the rim of the bowl is 0.118 m, correct to 3 significant figures. [5] In the case when the handle lies on the rim of the bowl, the container rests in equilibrium with the curved surface of the bowl on a horizontal table. (iii) Find the angle which the plane containing the rim of the bowl makes with the horizontal. [2] 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. 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 2014 9709/53/M/J/14

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2014 series 9709 MATHEMATICS 9709/53 Paper 5, 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 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9709 53 © Cambridge International Examinations 2014 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 GCE A LEVEL – May/June 2014 9709 53 © Cambridge International Examinations 2014 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 GCE A LEVEL – May/June 2014 9709 53 © Cambridge International Examinations 2014 1 ( ) 1.5 2.4 5 EE − × = 2 / ( ) 15 2 × ( ) 5.1 4.2 5 − × 2 / (2 × 1.5) – 0.1g × (2.4 – 1.5) R0.1v 2 / 2 v R 3 ms –1 B1 M1 A1 [3] EE/KE/PE conservation 2 ( ) ( ) 2 / 100 10 cos 3 / 12 2 24 F 12 + × × = FR8 AG M1 A1 A1 [3] Moments about vertex attempt 3 (i) 0.2a = 0.42 ( ) 2.1 1 2.1 = × = v M1 A1 [2] Newton's Second Law and at u v + = (ii) 0.2dv / dt = 0.42 – 0.32t [ ] [ ] 2 1 2 2.1 0.8t 2.1t − = v v v = 1.8 M1 M1 A1 [3] Newton's Second Law with t v d / d a = (iii) ( ) ∫ + − = 2.0 / d t 06 .0 32 .0 0.42 2 t t v [ ] 2.0 / 02 .0 16 .0 42 .0 3 0 3 2 t t t v + − = or [ ] [ ] 2.0 / 02 .0 16 .1 42 .0 3 2 3 2 8.1 t t t v v + − = 8.1 = v , so no change M1 M1 A1 [3] For attempt to integrate and correct limits seen 4 (i) g 2.0 60 cos T = 4N T = ( ) 60 sin 4.2 / 2.0 60 sin T 2 v = 1 ms 6 − = v M1 A1 M1 A1 [4] Resolve vertically for P Newton's Second Law with r v a 2 = (ii) θ θ sin 2.1 4 2.0 sin T 2 × × = T = 3.84 g 2.0 cos 84 .3 = θ
= 58.6° M1 A1 M1 A1 [4] Newton's Second Law with r ω a 2 = Resolve vertically with new tension

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9709 53 © Cambridge International Examinations 2014 5 (i) 2 / 2tg y = t x 5 = [ ] 2 2 x 2.0 2 /) ) 5 / ( 10 = = x y B1 M1 A1 [3] Must be positive (ii) 2 2 2 18 = + y x 2 2 2 2 18 ) 2.0 ( = + x x ( ) 0 324 04 .0 2 2 2 = − + x x 85 .8 = x 2 ... 8523 .8 2.0 × = y ... 67 . 15 2 52 2 × + = g v 1 ms 4. 18 − = v OR 2 2 ... 8523 .8 2.0 2 × × = g V 2 2 2 ... 7. 17 5 + = v 1 ms 4. 18 − = v OR 7. 17 77 .1 10 , 77 .1 5 / 85 .8 = × = = = V t 2 2 2 5 7. 17 + = v 1 ms 4. 18 − = V M1* D*M1 A1 B1 M1 A1 B1 M1 A1 B1 M1 A1 [6] Sets up and tries to solve a 3 term quadratic equation ... 3639 . 78 from ... 8523 .8 2 = x y = 15.67… V = 17.7046 6 (i) ( ) 5.1 / sin / 4.0 9 T θ × = θ sin T 6.0 R − = g 3.6N R = AG B1 M1 A1 [3] Must use the general angle θ (ii) ( ) ( ) 4.0 B ext , 30 sin / 4.0 A Ext = = ( ) ( ) 5.1 2 / 4.0 8.0 9 change EE 2 2 × − × = ( ) 2 / 5.2 3 6.0 44 .1 WD 2 2 − × − = J 615 .0 WD= B1 B1 M1 A1 [4] Uses EE in two positions. EE=1.44 (iii) ( ) 6.3 / 30 tan / 4.0 / 615 .0 = µ 247 .0 = µ M1 A1 [2] Accept 0.246 without wrong working

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9709 53 © Cambridge International Examinations 2014 7 (i) ) 2 / π/2)/(π sin( 3.0 (handle) OG = π / 6.0 8 15 .0 14 14 8 × − × = + d ) ( ( )m 0 026 .0 = d B1 M1 A1 [3] ... 19098 .0 ,π / 6.0 (ii) 191 .0 8 ) 14 8 ( × = = x ... 0694545 .0 = x 15 .0 14 14) (8 × = + y .. 0.0954545. = y Distance = 0.118 AG M1 A1 M1 A1 A1 [5] Horizontal distance from bowl axis Vertical distance from bowl rim 2 2 2 095454 .0 06945 .0 Distance + = (iii) 0954545 0 0694545 0 tan . / . = θ or 118 .0 / 0694545 .0 sin = θ or 118 .0 / 094545 .0 cos = θ
= 36(.0)° M1 A1 [2] Any trig ratio correctly using the two answers in (ii) Accept 36.1, 35.9 Alternative for 7(ii), 7(iii) (diagram below) 7 (ii) 98 0.15/0.190 tanH = ( ) 2 2 19098 .0 15 .0 22 14 HG + × = 2 2 2 15454 .0 19098 .0 + = OG – 147 . 38 cos 15454 .0 19098 .0 2 × × Distance = 0.118 AG M1 A1 M1 A1 A1 [5] H = 38.147° 0.15454 m Cosine rule on triangle OHG (iii) sinHOG = 0.15454sin38.147/0.118 ( ) 9. 35 HOG 90 Angle = − = M1 A1 [2] 54.06° Accept 36(.0), 36.1 O 0.19098 m 0.15 m H Handle (C of M) B Bowl (C of M) 8 G 14