Cambridge A Level Mathematics 9709 — 2004 Oct/Nov Paper 5 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level HIGHER MATHEMATICS MATHEMATICS 8719/05 9709/05 Paper 5 Mechanics 2 (M2) October/November 2004 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. 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. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. This document consists of 3 printed pages and 1 blank page. © UCLES 2004 [Turn over

2 1 A light elastic string has natural length 1.5 m and modulus of elasticity 60 N. The string is stretched between two fixed points A and B, which are at the same horizontal level and 2 m apart. (i) Find the tension in the string. [2] A particle of weight W N is now attached to the mid-point of the string and the particle is in equilibrium at a point 0.75 m vertically below the mid-point of AB. (ii) Find the value of W. [4] 2 A uniform rod AB of length 1.2 m and weight 30 N is in equilibrium with the end A in contact with a vertical wall. AB is held at right angles to the wall by a light inextensible string. The string has one end attached to the rod at B and the other end attached to a point C of the wall. The point C is 0.5 m vertically above A (see diagram). Find (i) the tension in the string, [3] (ii) the horizontal and vertical components of the force exerted on the rod by the wall at A. [3] 3 A car of mass 1000 kg is moving on a straight horizontal road. The driving force of the car is 28 000 v N and the resistance to motion is 4v N, where v m s−1 is the speed of the car t seconds after it passes a fixed point on the road. (i) Show that dv dt = 7000 −v2 250v . [2] The car passes points A and B with speeds 10 m s−1 and 40 m s−1 respectively. (ii) Find the time taken for the car to travel from A to B. [4] 4 A particle is projected from a point O on horizontal ground with speed 50 m s−1 at an angle θ to the horizontal. Given that the speed of the particle when it is at its highest point is 40 m s−1, (i) show that cos θ = 0.8, [2] (ii) find, in either order, (a) the greatest height reached by the particle, (b) the distance from O at which the particle hits the ground. [5] 9709/05/O/N/04

3 5 One end of a light elastic string of natural length 0.4 m and modulus of elasticity 16 N is attached to a fixed point O of a horizontal table. A particle P of mass 0.8 kg is attached to the other end of the string. The particle P is released from rest on the table, at a point which is 0.5 m from O. The coefficient of friction between the particle and the table is 0.2. By considering work and energy, find the speed of P at the instant the string becomes slack. [7] 6 A horizontal turntable rotates with constant angular speed ω rad s−1 about its centre O. A particle P of mass 0.08 kg is placed on the turntable. The particle moves with the turntable and no sliding takes place. (i) It is given that ω = 3 and that the particle is about to slide on the turntable when OP = 0.5 m. Find the coefficient of friction between the particle and the turntable. [3] (ii) Given instead that the particle is about to slide when its speed is 1.2 m s−1, find ω. [5] 7 A light container has a vertical cross-section in the form of a trapezium. The container rests on a horizontal surface. Grain is poured into the container to a depth of y m. As shown in the diagram, the cross-section ABCD of the grain is such that AB = 0.4 m and DC = (0.4 + 2y) m. (i) When y = 0.3, find the vertical height of the centre of mass of the grain above the base of the container. [5] (ii) Find the value of y for which the container is about to topple. [5] 9709/05/O/N/04

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary and Advanced Level MARK SCHEME for the November 2004 question papers 9709 MATHEMATICS 8719 HIGHER MATHEMATICS 9709/05, 8719/05 Paper 5 (Mechanics 2), 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 initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. 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. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the November 2004 question papers for most IGCSE and GCE Advanced Level syllabuses.

Grade thresholds taken for Syllabus 9709/8719 (Mathematics and Higher Mathematics) in the November 2004 examination. minimum mark required for grade: maximum mark available A B E Component 5 50 41 37 22 The thresholds (minimum marks) for Grades C and D are normally set by dividing the mark range between the B and the E thresholds into three. For example, if the difference between the B and the E threshold is 24 marks, the C threshold is set 8 marks below the B threshold and the D threshold is set another 8 marks down. If dividing the interval by three results in a fraction of a mark, then the threshold is normally rounded down.

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. PA -1 This is deducted from A or B marks in the case of premature approximation.

November 2004 GCE A AND AS LEVEL MARK SCHEME MAXIMUM MARK: 50 SYLLABUS/COMPONENT: 9709/05, 8719/05 MATHEMATICS AND HIGHER MATHEMATICS Paper 5 (Mechanics 2)

Page 1 Mark Scheme Syllabus Paper A AND AS LEVEL – NOVEMBER 2004 9709/8719 5 © University of Cambridge International Examinations 2005 1 (i) For using T = L x λ (T = 60(0.5)/1.5) M1 Tension is 20 N A1 2 (ii) For using extension 0.5 m (or 1 m) in Hooke’s Law M1 T = 60(0.5)/0.75 or T = 60(1)/1.5 (T = 40) A1 For resolving forces vertically (W = 2T x       5 3 ) M1 (2 components of T necessary) W = 48 N A1 4 2 (i) For taking moments about A (N.B. A moment = a force is M0) M1 30 x 0.6 = T x 1.2 x 3 . 1 5 . 0 or 30 x 0.6 = T x 1.2 sin 22.62o A1 Tension is 39 N A1 3 (ii) Horizontal component is 36 N B1 ft For taking moments about B (1.2Y = 30 x 0.6) or for resolving forces vertically (30 – T sin 22.62o) M1 Vertical component is 15 N A1 ft 3 3 (i) For using Newton’s second law and a = dt dv M1 Correct working to obtain the given answer A1 2 (ii) For separating the variables and attempting to integrate M1 t = −125ln(7000 – v2) (+C) (aef) A1 For attempting to find t(40) – t(10) (or equivalent) M1 Time taken is 30.6 s A1 4 4 (i) For ‘speed at greatest height = x’ used (40 = 50 cosθ ) M1 cosθ = 0.8 A1 2 (ii) For using y& = 0 at maximum height or y = 0 at impact [At max height t = 50 x 0.6/10 or at impact t = 50 x 0.6 (½ 10)] M1 [if (b) is answered before (a) using the range formula, this mark my be scored using R/2 = (50t cosθ ) for t at max ht or R = (50t cosθ ) for t at impact]

Page 2 Mark Scheme Syllabus Paper A AND AS LEVEL – NOVEMBER 2004 9709/8719 5 © University of Cambridge International Examinations 2005 (a) For substituting for t at max height (= 3) into M1 H = 50t sinθ − 2 1 gt2 or H = 2 0 sin 50 + θ t Greatest height is 45m A1 (b) For substituting t at max ht (= 3) into R = 2(50t cosθ ) or t at impact (= 6) into R = (50t cosθ ) or for using R = V2sin2θ /g (directly or by substituting y = 0 in the trajectory equation) M1 Distance is 240m A1 5 Alternative for methods not involving t at max ht. or t at impact: For using y& = 0 in 2 y& = 2 0 y& – 2gH or using the principle of M1 conservation of energy (2 non-zero KE terms and a GPE term needed) or using the trajectory equation with x = R/2 if (b) is answered before (a). 0 = (50 x 0.6)2 – 2 x 10H or ½m502 = ½m402 + mgH or A1 H = 120(0.75) – 10(120)2/[2(50)2(0.8)2] Greatest height is 45m A1 For using R – V2sin2θ /g (directly or by substituting y = 0 in the trajectory equation) M1 Distance is 240m A1 5 Loss in EPE = 0.4 x 2 )1 . 0 ( 16 2 (= 0.2) B1 For using F = µ R and R = mg M1 F = 0.2 x 0.8 x 10 (=1.6) A1 WD against friction = 1.6 x 0.1 (= 0.16) B1 ft For using KE gained = EPE lost – WD against friction M1 2 1 0.8v2 = 0.2 – 0.16 A1 ft Speed is 0.316 ms-1 A1 7 6 (i) For using F = mω2r (0.36 N) M1 For using F = µ R (0.36 = 0.8 µ ) M1 µ = 0.45 A1 3

Page 3 Mark Scheme Syllabus Paper A AND AS LEVEL – NOVEMBER 2004 9709/8719 5 © University of Cambridge International Examinations 2005 (ii) ( µ , R fixed →) F = 0.36 B1 ft For using F = mv2/r or F = mr(1.2/r)2 (r ≠0.5) M1 (0.36 = 0.08 x 1.22/r or 0.36 = 0.08 r(1.2/r)2 ) [for F = m(v/ω) x ω2 treat as for alternative case below.] For using v = ωr (r ≠0.5) M1 r = 0.32 or unsimplified equation in ω only A1 ft Alternative for the above last three marks: For using F = mvω M2 0.08 x 1.2ω = 0.36 A1 ω = 3.75 A1 5 7 (i) Centre of mass of triangle is at height 0.2m B1 For taking moments about the base M1 (0.4 × 0.3 + ½ 0.6 × 0.3) y = 0.4 × 0.3 × 0.15 + ½ 0.6 × 0.3 × 0.2 For correct total area and correct (unsimplified) moment of the rectangle A1 For the correct (unsimplified) moment of triangle A1 ft Distance of centre of mass is 0.171 or 6/35 m A1 5 (ii) Centre of mass of triangle is 3 2y from interface B1 For using ‘moment about the interface = 0’ or equivalent or ‘moment about AD = 0.4A’ AND with areas in terms of y M1 [0.4y × 0.2 = ½ 2 y × y × 3 2y or 0.4(0.4y + ½ 2y × y) = 0.4y × 0.2 + ½2y × y (0.4 + 3 2y )] For LHS of the above A1 For RHS of the above A1 ft Value of y is or 5 3 0.346 A1 5