Cambridge A Level Mathematics 9709 — 2006 Oct/Nov Paper 5 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level MATHEMATICS 9709/05 Paper 5 Mechanics 2 (M2) October/November 2006 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 on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. 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 ms−2. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. 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. At the end of the examination, fasten all your work securely together. This document consists of 4 printed pages. © UCLES 2006 [Turn over

2 1 A stone is projected horizontally with speed 8 m s−1 from a point O at the top of a vertical cliff. The horizontal and vertically upward displacements of the stone from O are x m and y m respectively. (i) Find the equation of the stone’s trajectory. [2] The stone enters the sea at a horizontal distance of 24 m from the base of the cliff. (ii) Find the height above sea level of the top of the cliff. [2] 2 A horizontal turntable rotates with constant angular speed 3 rad s−1. A particle of mass 0.06 kg is placed on the turntable at a point 0.25 m from its centre. The coefficient of friction between the particle and the turntable is µ. As the turntable rotates, the particle moves with the turntable and no sliding takes place. (i) Find the vertical and horizontal components of the contact force exerted on the particle by the turntable. [3] (ii) Show that µ ≥0.225. [1] 3 A hollow cylinder of radius 0.35 m has a smooth inner surface. The cylinder is fixed with its axis vertical. One end of a light inextensible string of length 1.25 m is attached to a fixed point O on the axis of the cylinder. A particle P of mass 0.24 kg is attached to the other end of the string. P moves with constant speed in a horizontal circle, in contact with the inner surface of the cylinder, and with the string taut (see diagram). (i) Find the tension in the string. [2] (ii) Given that the magnitude of the acceleration of P is 8 m s−2, find the force exerted on P by the cylinder. [3] © UCLES 2006 9709/05/O/N/06

3 4 A stone is projected from a point on horizontal ground with speed 25 m s−1 at an angle θ above the horizontal, where sin θ = 4 5. At time 1.2 s after projection the stone passes through the point A. Subsequently the stone passes through the point B, which is at the same height above the ground as A. Find the horizontal distance AB. [5] 5 A non-uniform rod AB of length 2.5 m and mass 3 kg has its centre of mass at the point G of the rod, where AG = 1.5 m. The rod hangs horizontally, in equilibrium, from strings attached at A and B. The strings at A and B make angles with the vertical of α◦and 15◦respectively. The tension in the string at B is T N (see diagram). Find (i) the value of T, [3] (ii) the value of α. [3] 6 A large uniform lamina is in the shape of a right-angled triangle ABC, with hypotenuse AC, joined to a semicircle ADC with diameter AC. The sides AB and BC have lengths 3 m and 4 m respectively, as shown in the diagram. (i) Show that the distance from AB of the centre of mass of the semicircular part ADC of the lamina is
2 + 2 π  m. [3] (ii) Show that the distance from AB of the centre of mass of the complete lamina is 2.14 m, correct to 3 significant figures. [5] © UCLES 2006 9709/05/O/N/06 [Turn over

4 7 A cyclist starts from rest at a point O and travels along a straight path. At time t s after starting, the displacement of the cyclist from O is x m, and the acceleration of the cyclist is a m s−2, where a = 0.6x0.2. (i) Find an expression for the velocity v m s−1 of the cyclist in terms of x. [4] (ii) Show that t = 2.5x0.4. [3] (iii) Find the distance travelled by the cyclist in the first 10 s of the journey. [2] 8 The diagram shows a light elastic string of natural length 0.6 m and modulus of elasticity 5 N with one end attached to a fixed point O. A particle P of mass 0.2 kg is attached to the other end of the string. P is held at the point A, which is 0.9 m vertically above O. The particle is released from rest and travels vertically downwards through O to the point C, where it starts to move upwards. B is the point of the line AC where the string first becomes slack. (i) Find the speed of P at B. [4] (ii) The extension of the string when P is at C is x m. (a) Show that x2 −0.48x −0.81 = 0. [3] (b) Hence find the distance AC. [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. University of Cambridge International Examinations is part of the University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. © UCLES 2006 9709/05/O/N/06

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2006 question paper 9709 MATHEMATICS 9709/05 Paper 5, maximum raw mark 50 This mark scheme is published as an aid to teachers and students, 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. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the report on the examination. The grade thresholds for various grades are published in the report on the examination for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2006 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

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.

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/AS LEVEL - OCT/NOV 2006 9709 5 © UCLES 2006 1 (i) y = ± 2 2 2 1 8 2 10 × × x or x = 8t, y = ± 2 1 gt 2 M1 For correct substitution into the trajectory formula ( in booklet) For an attempt to eliminate t y = - 64 5 x² A1 2 (ii) M1 For using h = - y(24) or substituting t=3 in y = 2 1 gt 2 Height = 45m A1 2 4 2 (i) R = 0.6N B1 F = 0.06 x 3² x 0.25 M1 For using Newton’s second law and a = rω ² F = 0.135 A1 3 (ii) µ ≥ 0.225 B1 1 From using µ ≥ F/R 4 3 (i) Tcosα = 0.24g where sinα = 0.35/1.25 M1 For resolving forces vertically (α = 16.26) Tension is 2.5N A1 2 (ii) M1 For using Newton’s second law (3 terms required) R + Tsinα = 0.24 x 8 A1ft ft for their T only Force exerted is 1.22N A1 3 5 4 Height at A = 16.8 B1 25 x 0.8t - ½10t² = 25 x 0.8 x 1.2 - ½10 x 1.2² M1 t = (1.2), 2.8 A1 AB = 25(2.8 – 1.2)0.6 M1 For using Distance = V(t B - t A )cosθ Distance = 24m A1 5 5 OR 4 R = 2 x 25² x 0.8 x 0.6 ÷ 10 ( = 60) B1 Using R = 2V 2 sinθcosθ/g (θ = 53.13) M1 For using Distance=R-2x(1.2) DM1 For using x = Vtcosθ AB = 2 x 25² x 0.8 x 0.6 ÷ 10 - 2 x 25 x 1.2 x 0.6 A1ft Distance is 24m A1 5 5 OR 4 Height at A = 16.8 B1 16.8 = 4x/3 - x 2 /45 or equivalent M1 For using trajectory equation from formula booklet x = 18 or x = 42 A1 Distance = 42 - 18 M1 Distance is 24m A1 5 5

Page 5 Mark Scheme Syllabus Paper GCE A/AS LEVEL - OCT/NOV 2006 9709 5 © UCLES 2006 5 (i) M1 For taking moments about A 3g x 1.5 = 2.5(Tcos15º) A1 T = 18.6 A1 3 (ii) T A sinα =Tsin15º T A cosα + Tcos15º = 3g or T A cosα x 2.5 = 3g x 1 M1 For resolving forces horizontally and vertically or taking moments about B. tanα = Tsin15º ÷ (3g-Tcos15º) M1 For eliminating T A α = 21.8 or 21.9 A1 3 6 6 (i) Distance of centre of mass of semicircle from centre = 10/3π B1 2.5x0.8 + (10/3π )x0.6 M1 rcosθ + dsinθ Distance is (2 + π 2 )m B1 3 From correct working (AG) (ii) M1 For using A x = A 1 x 1 + A 2 x 2 (3 terms required) (½4x3 +½π x2.5²) x = ½4x3x(4/3)+ (½π x2.5²)(2+ π 2 ) A1 B1 A1 Distance is 2.14m A1 5 From correct working (AG) 8 7 (i) v dx dv = 0.6x 2.0 B1 M1 For separating and integrating 0.5v² = 0.5x 2.1 (+C 1) A1 v = x 6.0 A1 4 (ii) ∫ −6.0 x dx = ∫dt M1 For using v = dx/dt, separating and integrating x 4.0 /0.4 = t (+C 2 ) A1 t = 2.5x 4.0 A1 3 (iii) x = (10/2.5) 5.2 M1 For substituting for t and solving for x Distance is 32m A1 2 9

Page 6 Mark Scheme Syllabus Paper GCE A/AS LEVEL - OCT/NOV 2006 9709 5 © UCLES 2006 8 (i) EE at A = (5x0.3²)÷ (2x0.6) M1 For using EE = L x 2 2 λ PE loss at B = 0.2x10x0.3 B1 0.375 = ½0.2v² - 0.6 M1 For using EE at A = KE gain – PE loss (Allow sign errors) Velocity is 3.12ms 1 − A1 4 (ii) (a) M1 For using EE at A = EE at C – PE loss 0.375 = 5x² ÷ (2x0.6) – 0.2x10(1.5+x) A1 x² - 0.48x – 0.81 = 0 A1 3 From correct working (AG) OR (ii) (a) (M1) For using KE at B + PE at B = EE at C 0.975 + 0.2x10(1.2 + x) = 5x 2 /1.2 (A1) x 2 - 0.48x – 0.81 = 0 (A1) (3) (b) M1 For solving and adding 1.5 to +ve root Distance is 2.67m A1 2 From 1.5 + 1.17 9