Cambridge A Level Mathematics (US) 9280 — 2015 May/June Paper 5 · Variant 1

*0208552611* Cambridge International Examinations Cambridge International Advanced Level MATHEMATICS (US) 9280/51 Paper 5 Mechanics 2 (M2) 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 4 printed pages. JC15 06_9280_51/FP © UCLES 2015 [Turn over

2 1 One end of a light elastic string of natural length 0.7 m is attached to a fixed point A on a smooth horizontal surface. The other end of the string is attached to a particle P of mass 0.3 kg which is held at a point B on the horizontal surface, where AB = 1.2 m. It is given that P is released from rest at B and that when AP = 0.9 m, the particle has speed 4 m s−1. Calculate the modulus of elasticity of the string. [3] 2 A stone is projected from a point O on horizontal ground. The equation of the trajectory of the stone is y = 1.2x −0.15x2, where x m and y m are respectively the horizontal and vertically upwards displacements of the stone from O. Find (i) the greatest height of the stone, [2] (ii) the distance from O of the point where the stone strikes the ground. [2] 3 P A O 5 rad s−1 1 One end of a light inextensible string is attached to a fixed point A and the other end of the string is attached to a particle P. The particle P moves with constant angular speed 5 rad s−1 in a horizontal circle which has its center O vertically below A. The string makes an angle 1 with the vertical (see diagram). The tension in the string is three times the weight of P. (i) Show that the length of the string is 1.2 m. [3] (ii) Find the speed of P. [4] © UCLES 2015 9280/51/M/J/15

3 4 B 15 m s−1 O 30Å A small ball B is projected from a point O above horizontal ground, with initial speed 15 m s−1 at an angle of projection of 30Å above the horizontal (see diagram). The ball strikes the ground 3 s after projection. (i) Calculate the speed and direction of motion of the ball immediately before it strikes the ground. [5] (ii) Find the height of O above the ground. [2] 5 A particle P of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.9 m and modulus of elasticity 18 N. The other end of the string is attached to a fixed point O which is 3 m above the ground. (i) Find the extension of the string when P is in the equilibrium position. [2] P is projected vertically downwards from the equilibrium position with initial speed 6 m s−1. At the instant when the tension in the string is 12 N the string breaks. P continues to descend vertically. (ii) (a) Calculate the height of P above the ground at the instant when the string breaks. [2] (b) Find the speed of P immediately before it strikes the ground. [4] 6 A particle P of mass 0.1 kg moves with decreasing speed in a straight line on a smooth horizontal surface. A horizontal resisting force of magnitude 0.2e−x N acts on P, where x m is the displacement of P from a fixed point O on the line. The velocity of P is v m s−1 when its displacement from O is x m. (i) Show that vdv dx = ke−x, where k is a constant to be found. [2] P passes through O with velocity 2.2 m s−1. (ii) Calculate the value of x at the instant when the velocity of P is 2 m s−1. [4] (iii) Show that the speed of P does not fall below 0.917 m s−1, correct to 3 significant figures. [2] [Question 7 is printed on the next page.] © UCLES 2015 9280/51/M/J/15 [Turn over

4 7 45Å B A O E D C 0.8 m 0.6 m The diagram shows the cross-section OABCDE through the center of mass of a uniform prism on a rough inclined plane. The portion ADEO is a rectangle in which AD = OE = 0.6 m and DE = AO = 0.8 m; the portion BCD is an isosceles triangle in which angle BCD is a right angle, and A is the mid-point of BD. The plane is inclined at 45Å to the horizontal, BC lies along a line of greatest slope of the plane, and DE is horizontal. (i) Calculate the distance of the center of mass of the prism from BD. [3] The weight of the prism is 21 N, and it is held in equilibrium by a horizontal force of magnitude P N acting along ED. (ii) (a) Find the smallest value of P for which the prism does not topple. [2] (b) It is given that the prism is about to slip for this smallest value of P. Calculate the coefficient of friction between the prism and the plane. [3] The value of P is gradually increased until the prism ceases to be in equilibrium. (iii) Show that the prism topples before it begins to slide, stating the value of P at which equilibrium is broken. [5] 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 9280/51/M/J/15

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the May/June 2015 series 9280 MATHEMATICS (US) 9280/51 Paper 5 (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 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 A Level – May/June 2015 9280 51 © 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 A Level – May/June 2015 9280 51 © 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 A Level – May/June 2015 9709 51 © Cambridge International Examinations 2015 1 EE(B) = 2 0.5 2 0.7 λ × ×    = 28 5λ OR EE = 2 0.2 2 0.7 λ × ×    = 35 λ B1 Correct EE when AP = 1.2 m Correct EE when AP = 0.9 m 2 0.5 2 0.7 Λ× × – 2 0.2 2 0.7 λ × × = 2 0.3 2 4 × M1 Using EE loss = KE gain λ = 16 N A1 [3] 2 (i) d d y x = 1.2 – 2(0.15x) = 0 M1 Solving trajectory derivative = 0 y = 2.4 A1 [2] From x = 4 OR Greatest height at half range (ii) M1 Uses y = 1.2x – 0.15 x2 with x = distance in (ii) ÷ 2 y = 2.4 A1 (ii) Twice 'x' for greatest height M1 Using 'x' from (i) Distance = 8 m A1 [2] No ft OR 1.2x – 0.15 x2 = 0 M1 Solves quadratic with y = 0 Distance = 8 m A1 3 (i) Tsinθ = m 2 ω r M1 Newton’s 2nd law, acceleration = 52r and component of T 3ωsinθ 2 5       = g ω (Lsinθ) A1 3mgsinθ = m 52 (Lsinθ) L = 1.2 m AG A1 [3] (ii) 3ω cosθ = ω M1 Resolves vertically for P θ = 70.53° A1 OR θ = cos –1       3 1 , θ = sin–1 8 9 etc. v = 5×1.2sinθ M1 v = ωr v = 5.66 ms –1 A1 [4]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9709 51 © Cambridge International Examinations 2015 4 (i) x' = 15cos30 ( = 12.990..) B1 y' = 15sin30 – 3g ( = – 22.5 ) B1 Or with signs reversed v2 = (15cos30)2 + (15sin30 – 3g)2 or tanθ = 3 – 15sin30 15cos30 g M1 v = 26(.0) ms –1 A1 θ = 60° to the horizontal A1 [5] Or 30° to the vertical (ii) y = 3(15sin30) – 23 2 g M1 Uses s = ut + 2 1at2 Height = 22.5 m A1 [2] N.B. this is also y' OR ( – 22.5)2 = (15sin30)2 – 2 × 10y M1 Uses v2 = u2 + 2as Height = 22.5 m A1 5 (i) 0.3g = 18 0.9 e M1 Uses T = x l λ e = 0.15 m A1 [2] (ii) (a) 12 = 18ext 0.9 and ht = 3 – 0.9 – ext M1 Both ideas needed, ext = 0.6 ht = 1.5 m A1 [2] (ii) (b) 2 0.3 6 2 × – 2 0.3 2 u + 0.3g(0.6 – 0.15) = 2 18 0.6 2 0.9 × × – 2 18 0.15 2 0.9 × × M1 A1 KE/PE/EE balance up to string breaking       = 375 .3 2 3.0 2 u u2 = 22.5 0.3v2 = 0.3u2 + 0.3g(3 – 0.6 – 0.9) OR v2 = u2 + 2g(3 – 0.6 – 0.9) M1 KE/PE balance after string breaks or v2 = u2 + 2g(ht) using ht from (ii)(a) v = 7.25 ms–1 A1 4 7.2456 6 (i) 0.1v d d v x = – 0.2 e–x v d d v x = – 2e–x M1 Newton’s 2nd law, 1 force Must have negative coefficient k = –2 A1* [2]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9709 51 © Cambridge International Examinations 2015 (ii) d v v ∫ = x e− − ∫ 2 dx M1 Integrates 2 2 v = 2e–x ( + c ) D*A1 Needs first A1 in (i) 42 .0 2 2 2.2 0 2 =       − = e c M1 Or uses limits of 2 and 2.2 for v and x and 0 for x 2 2 2 = 2 x e − + 0.42 x = 0.236 A1 [4] (iii) 2 2 v = 2 ∞ − e + 0.42 M1 OR finds x when v = 0.9165 v = 0.917 ms–1 AG A1 [2] No solution when v = 0.917 7 (i) d(0.6 × 0.8 + 0.62) = 0.4(0.6 × 0.8) – (0.6/3) × 0.62 M1 A1 Moments about BAD d = 0.143 m A1 [3] Exact 1/7 (ii) (a) 21 × 0.143 = 1.2 P M1 Moments about B P = 2.5(0) A1 [2] (ii) (b) r F = 21sin45 – 2.5cos45 and R = 21cos45 + 2.5sin45 B1 M1 For using r F = µR µ = 0.787 A1 [3] (iii) P × 0.6 = 21 × (0.143 + 0.6) M1 Moments about C P = 26(.0) A1 Required r F = 26sin45 – 21sin45 M1 3.5355.. Max r F = 26(.155) A1 0.787 × (26cos45 + 21cos45) As actual r F < max r F , no sliding A1 [5]