Cambridge A Level Mathematics 9709 — 2009 May/June Paper 4 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level MATHEMATICS 9709/04 Paper 4 Mechanics 1 (M1) May/June 2009 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 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 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. © UCLES 2009 [Turn over *4159902989*

2 1 P B A block B of mass 5 kg is attached to one end of a light inextensible string. A particle P of mass 4 kg is attached to other end of the string. The string passes over a smooth pulley. The system is in equilibrium with the string taut and its straight parts vertical. B is at rest on the ground (see diagram). State the tension in the string and find the force exerted on B by the ground. [3] 2 15° C F N P Q 100 m A crate C is pulled at constant speed up a straight inclined path by a constant force of magnitude F N, acting upwards at an angle of 15◦to the path. C passes through points P and Q which are 100 m apart (see diagram). As C travels from P to Q the work done against the resistance to C’s motion is 900 J, and the gain in C’s potential energy is 2100 J. Write down the work done by the pulling force as C travels from P to Q, and hence find the value of F. [3] 3 x y 15 N 10 N 7 N 50° 80° Forces of magnitudes 7 N, 10 N and 15 N act on a particle in the directions shown in the diagram. (i) Find the component of the resultant of the three forces (a) in the x-direction, (b) in the y-direction. [3] (ii) Hence find the direction of the resultant. [2] © UCLES 2009 9709/04/M/J/09

3 4 20° A block of mass 8 kg is at rest on a plane inclined at 20◦to the horizontal. The block is connected to a vertical wall at the top of the plane by a string. The string is taut and parallel to a line of greatest slope of the plane (see diagram). (i) Given that the tension in the string is 13 N, find the frictional and normal components of the force exerted on the block by the plane. [4] The string is cut; the block remains at rest, but is on the point of slipping down the plane. (ii) Find the coefficient of friction between the block and the plane. [2] 5 2.6° A B C D 5 m s–1 9 m s–1 250 m d m A cyclist and his machine have a total mass of 80 kg. The cyclist starts from rest at the top A of a straight path AB, and freewheels (moves without pedalling or braking) down the path to B. The path AB is inclined at 2.6◦to the horizontal and is of length 250 m (see diagram). (i) Given that the cyclist passes through B with speed 9 m s−1, find the gain in kinetic energy and the loss in potential energy of the cyclist and his machine. Hence find the work done against the resistance to motion of the cyclist and his machine. [3] The cyclist continues to freewheel along a horizontal straight path BD until he reaches the point C, where the distance BC is d m. His speed at C is 5 m s−1. The resistance to motion is constant, and is the same on BD as on AB. (ii) Find the value of d. [3] The cyclist starts to pedal at C, generating 425 W of power. (iii) Find the acceleration of the cyclist immediately after passing through C. [3] © UCLES 2009 9709/04/M/J/09 [Turn over

4 6 0.36 m A B Particles A and B are attached to the ends of a light inextensible string which passes over a smooth pulley. The system is held at rest with the string taut and its straight parts vertical. Both particles are at a height of 0.36 m above the floor (see diagram). The system is released and A begins to fall, reaching the floor after 0.6 s. (i) Find the acceleration of A as it falls. [2] The mass of A is 0.45 kg. Find (ii) the tension in the string while A is falling, [2] (iii) the mass of B, [3] (iv) the maximum height above the floor reached by B. [3] 7 A particle P travels in a straight line from A to D, passing through the points B and C. For the section AB the velocity of the particle is (0.5t −0.01t2) m s−1, where t s is the time after leaving A. (i) Given that the acceleration of P at B is 0.1 m s−2, find the time taken for P to travel from A to B. [3] The acceleration of P from B to C is constant and equal to 0.1 m s−2. (ii) Given that P reaches C with speed 14 m s−1, find the time taken for P to travel from B to C. [3] P travels with constant deceleration 0.3 m s−2 from C to D. Given that the distance CD is 300 m, find (iii) the speed with which P reaches D, [2] (iv) the distance AD. [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. University of 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 2009 9709/04/M/J/09

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2009 question paper for the guidance of teachers 9709 MATHEMATICS 9709/04 Paper 4, 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2009 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9709 04 © UCLES 2009 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: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9709 04 © UCLES 2009 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: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9709 04 © UCLES 2009 1 Tension is 40 N B1 [R + T = W] M1 For resolving forces on B vertically Force exerted is 10 N A1 [3] 2 Work done is 3000 J B1 [3000 = F×100cos15°] M1 For using WD = Fdcosα F = 31.1 N A1ft [3] ft only from WD = 1200 (F = 12.4) 3 (i) [X = 7 + 10cos50° – 15cos80°, Y = 10sin50° + 15sin80°] M1 For obtaining an expression for X or Y (a) x-component is 10.8 N A1 (b) y-component is 22.4 N A1 [3] (ii) [θ = tan–1(22.4/10.8)] M1 For using θ = tan–1(Y/X) Direction 64.2° anticlockwise from x-axis A1 [2] Accept 64.3° 4 (i) [F + T = 8×10sin20°] M1 For resolving forces parallel to the plane Frictional component is 14.4 N A1 [R = 80cos20°] M1 For resolving forces normal to the plane Normal component is 75.2 N A1 [4] SR (max 3 out of 4) for consistent sin/cos exchange – method marks as above and A1 (only) for F = 62.2 and R = 27.4 Alternative scheme for part (i) [Tcos20° + Fcos20° = Rsin20° and Tsin20° + Fsin20° + Rcos20° = 8 g] (M1) For resolving forces horizontally and vertically [tan20° = (13cos20° + Fcos20°) ÷ (80 – 13sin20° – Fsin20°) → F = 80sin20° – 13 or tan20° = (80 – Rcos20° – 13sin20°) ÷ (Rsin20° – 13cos20°) → R = 80cos20°] (M1) For attempting to solve for F or R Frictional component is 14.4 N (A1) Normal component is 75.2 N (A1) (ii) F = 8×10sin20° or µ = tan20° B1ft ft following consistent sin/cos mix in (i) for F = 8×10cos20° or µ = tan70° Coefficient is 0.364 (accept 0.36) B1 [2] ft following consistent sin/cos mix in (i) for µ = 2.75 5 (i) Gain in KE is 3240 J B1 Loss in PE is 9070 J B1 SR (max 1 out of 2) for answers –3240 and –9070 B1 Work done is 5830 J B1ft [3] ft WD = loss of PE – gain in KE (subject to loss of PE ≠gain in KE) (ii) R = 5830/250 (= 23.3) B1ft [23.3d = ½80(92 – 52) or –23.3 = 80a and 52 = 92 + 2(–23.3/80)d] M1 For using WD = Loss of KE or for using –R = 80a and v2 = u2 + 2as d = 96.0 A1ft [3] Accept 96 or 96.1; ft 560000/WD(i) or 2240/R (iii) Driving force = 425/5 B1 [DF – 23.3 = 80a] M1 For using Newton’s second law Acceleration is 0.771 ms–2 A1 [3]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9709 04 © UCLES 2009 6 (i) [0.36 = ½a(0.6)2] M1 For using s = (ut) + ½ at2 Acceleration is 2 ms–2 A1 [2] (ii) [0.45 g – T = 0.45×2] M1 For applying Newton’s second law to A Tension is 3.6 N A1ft [2] ft T = 0.45(10 – a) (iii) [T – mg = 2m or 0.9 + 2m = 4.5 – 10m] M1 For applying Newton’s second law to B or for using (M + m)a = (M – m)g (2 + g)m = 3.6 (must have m terms combined) A1ft ft a and/or a non-zero value of T Mass is 0.3 kg A1 [3] (iv) u = 1.2 B1ft ft u = 0.6a [0 = 1.44 – 20s → 0.072] M1 For using 0 = u2 – 2gs Maximum height is 0.792 A1ft [3] ft 0.72 + 0.05u2 7 (i) a = 0.5 – 0.02t B1 [0.5 – 0.02t = 0.1] M1 For solving dt dv = 0.1 Time taken is 20 s A1 [3] (ii) u = 0.5×20 – 0.01×202 (= 6) B1ft ft 0.5t1 – 0.01t1 2 [14 = 6 + 0.1t] M1 For using v = u + at Time taken is 80 s A1ft [3] ft t = 10(14 – 0.5t1 + 0.01t1 2) (iii) [v2 = 142 – 2×0.3×300] M1 For using v2 = u2 + 2as Speed is 4 ms–1 A1 [2] (iv) M1 For using ∫ = vdt s s = 0.25t2 – 0.01t3/3 (+ C) A1 DM1 For using limits 0 to 20 or equivalent AB = 0.25×202 – 0.01×203/3 (= 73.3) A1ft ft 0.25t1 2 – 0.01t1 3/3 BC = ½ (6 + 14) ×80 or 6×80 + ½ 0.1×802 or (142 – 62)/(2×0.1) (= 800) B1 Distance AD is 1170 m A1 [6]