Cambridge A Level Mathematics 9709 — 2014 May/June Paper 4 · Variant 3

*9189996608* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS 9709/43 Paper 4 Mechanics 1 (M1) 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_43/FP © UCLES 2014 [Turn over

2 1 B X N 15Å A block B of mass 7 kg is at rest on rough horizontal ground. A force of magnitude X N acts on B at an angle of 15Å to the upward vertical (see diagram). (i) Given that B is in equilibrium find, in terms of X, the normal component of the force exerted on B by the ground. [2] (ii) The coefficient of friction between B and the ground is 0.4. Find the value of X for which B is in limiting equilibrium. [3] 2 A car of mass 1250 kg travels up a straight hill inclined at an angle ! to the horizontal, where sin ! = 0.02. The power provided by the car’s engine is 23 kW. The resistance to motion is constant and equal to 600 N. Find the speed of the car at an instant when its acceleration is 0.5 m s−2. [5] 3 P 2.24 N O 0.9 m 0.5 m 1.3 m 1.5 m 1.2 m S1 S2 A particle P of weight 1.4 N is attached to one end of a light inextensible string S1 of length 1.5 m, and to one end of another light inextensible string S2 of length 1.3 m. The other end of S1 is attached to a wall at the point 0.9 m vertically above a point O of the wall. The other end of S2 is attached to the wall at the point 0.5 m vertically below O. The particle is held in equilibrium, at the same horizontal level as O, by a horizontal force of magnitude 2.24 N acting away from the wall and perpendicular to it (see diagram). Find the tensions in the strings. [6] © UCLES 2014 9709/43/M/J/14

3 4 A small ball of mass 0.4 kg is released from rest at a point 5 m above horizontal ground. At the instant the ball hits the ground it loses 12.8 J of kinetic energy and starts to move upwards. (i) Show that the greatest height above the ground that the ball reaches after hitting the ground is 1.8 m. [4] (ii) Find the time taken for the ball’s motion from its release until reaching this greatest height. [3] 5 A lorry of mass 16 000 kg travels at constant speed from the bottom, O, to the top, A, of a straight hill. The distance OA is 1200 m and A is 18 m above the level of O. The driving force of the lorry is constant and equal to 4500 N. (i) Find the work done against the resistance to the motion of the lorry. [3] On reaching A the lorry continues along a straight horizontal road against a constant resistance of 2000 N. The driving force of the lorry is not now constant, and the speed of the lorry increases from 9 m s−1 at A to 21 m s−1 at the point B on the road. The distance AB is 2400 m. (ii) Use an energy method to find F, where F N is the average value of the driving force of the lorry while moving from A to B. [3] (iii) Given that the driving force at A is 1280 N greater than F N and that the driving force at B is 1280 N less than F N, show that the power developed by the lorry’s engine is the same at B as it is at A. [2] 6 A particle starts from rest at a point O and moves in a horizontal straight line. The velocity of the particle is v m s−1 at time t s after leaving O. For 0 ≤t < 60, the velocity is given by v = 0.05t −0.0005t2. The particle hits a wall at the instant when t = 60, and reverses the direction of its motion. The particle subsequently comes to rest at the point A when t = 100, and for 60 < t ≤100 the velocity is given by v = 0.025t −2.5. (i) Find the velocity of the particle immediately before it hits the wall, and its velocity immediately after its hits the wall. [2] (ii) Find the total distance travelled by the particle. [4] (iii) Find the maximum speed of the particle and sketch the particle’s velocity-time graph for 0 ≤t ≤100, showing the value of t for which the speed is greatest. [4] [Question 7 is printed on the next page.] © UCLES 2014 9709/43/M/J/14 [Turn over

4 7 P Q 30 cm 40 cm 160 cm A smooth inclined plane of length 160 cm is fixed with one end at a height of 40 cm above the other end, which is on horizontal ground. Particles P and Q, of masses 0.76 kg and 0.49 kg respectively, are attached to the ends of a light inextensible string which passes over a small smooth pulley fixed at the top of the plane. Particle P is held at rest on the same line of greatest slope as the pulley and Q hangs vertically below the pulley at a height of 30 cm above the ground (see diagram). P is released from rest. It starts to move up the plane and does not reach the pulley. Find (i) the acceleration of the particles and the tension in the string before Q reaches the ground, [4] (ii) the speed with which Q reaches the ground, [2] (iii) the total distance travelled by P before it comes to instantaneous rest. [3] 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 9709/43/M/J/14

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2014 series 9709 MATHEMATICS 9709/43 Paper 4 (Mechanics 1), 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 AS/A LEVEL – May/June 2014 9709 43 © 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 AS/A LEVEL – May/June 2014 9709 43 © 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 AS/A LEVEL – May/June 2014 9709 43 © Cambridge International Examinations 2014 1 (i) [N + component of X = Weight of B] M1 For resolving forces acting on the block vertically (3 terms required) Normal component is (70 – Xcos15o) N A1 [2] (ii) F = Xsin15o B1 [Xsin15o = 0.4(70 – Xcos15o)] M1 For using F = µR Value of X is 43.4 A1 [3] 2 M1 For using Newton’s 2nd law DF – 600 – 1250 × 0.02g = 1250 × 0.5 A1 M1 For using DF = 23000/v v = 23000 ÷ (625 + 600 + 250) A1ft ft error in one term for DF above (1st A mark) Speed of car is 15.6 ms–1 A1 [5] Alternative Method M1 For using WD by driving force = KE gain + PE gain + WD against resistance WD = 1250 × 0.5s + 1250g × 0.02s + 600s A1 M1 For using WD by driving force = DF × s and DF=23000/v v = 23000 ÷ (625 + 600 + 250) A1ft ft error in one term for WD above (1st A mark) Speed of car is 15.6 ms–1 A1 [5] 3 M1 For resolving forces acting on P horizontally. 0.8T1 +12T2/13 = 2.24 A1 M1 For resolving forces acting on P vertically. 0.6T1 – 5T2/13 = 1.4 A1 M1 For solving for T1 and T2 T1 = 2.5 and T2 = 0.26 A1 [6] SR for using Lami’s Rule for T1,T2 and 2.24 N (weight missing) (max 3/6) T1/sin157.38 = 2.24/sin59.49 B1 T2/sin143.13 = 2.24/sin59.49 B1 T1 = 1(.00) and T2 = 1.56 B1

Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9709 43 © Cambridge International Examinations 2014 4 (i) PE loss = 0.4g × 5 J = 20 J B1 Initial KEup = 0.4g × 5 – 12.8 = 7.2 J B1 [0.4gh = 2g – 12.8] M1 Uses PE gain = KE loss to form equation in h Height reached is 1.8 m A1 [4] AG (ii) 5 = 0 + 2 1 gtdown 2 (tdown = 1) B1 0 = 6 – gtup or 1.8 = 2 1 gtup 2 (tup = 0.6) B1 Total time is 1.6 s B1 [3] First Alternative for part (i) v2 = 2 × 10 × 5 → (v = 10) B1 KE loss = 2 1 0.4(102 – vup 2) = 12.8 B1 [vup = 60, 0 = 62– 2gh] M1 Uses v2 = u2 – 2gs to form equation in h Height reached is 1.8 m A1 [4] AG Second Alternative for part (i) 0.4gh = 12.8 M1 Uses PE gain = KE loss h = 3.2 m A1 [Height reached = 5 – 12.8/0.4g] M1 Uses height reached = 5 – ‘height not reached’ Height reached is 1.8 m A1 [4] AG Third Alternative for part (i) 2 1 × 0.4v2 = 12.8 (v=8) and [82 = 02 + 2gh] M1 Uses KE loss = 12.8 and v2=u2+2gs h = 3.2 m A1 [Height reached = 5 – 3.2] M1 Uses height reached = 5 – ‘height not reached’ Height reached is 1.8 m A1 [4] AG

Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9709 43 © Cambridge International Examinations 2014 5 (i) M1 For using WD by driving force = Gain in PE + WD against resistance WD against resistance = 4500 × 1200 – 16000g × 18 A1 WD against resistance = 2.52 × 106 J A1 [3] Alternative Method for part (i) [R + 16000g × 18/1200 = 4500] M1 For resolving along the plane [WD=(4500 – 16000g × 18/1200) × 1200] M1 For using WD against resistance = Rs WD against resistance = 2.52 × 106 J A1 [3] (ii) KE gain = 2 1 16000(212 – 92) J B1 M1 For using F = (KE gain + 2000 × 2400) ÷ 2400 F = 7680000 ÷ 2400 = 3200 A1 [3] SR (max 1/3) for using v2=u2+2as and Newton’s 2nd law 212 – 92 = 2a × 2400, a = 0.075 F –2000 = 16000 × 0.075 F = 3200 B1 (iii) [PA = (3200 + 1280) × 9 and PB = (3200 – 1280) × 21] M1 For using P = Fv to find PA and PB PA = PB = 40320 W A1 [2] 6 (i) Velocity immediately before is 1.2 ms–1 B1 Velocity immediately after is – 1 ms–1 B1 [2] (ii) M1 For using distance OW = ʃvdt with limits 0 to 60 (W is wall) or For using distance WA = – ʃvdt with limits 60 to 100 Distance OW = 0.025 × 602 – 0.0005 × 603 ÷ 3 A1 Distance WA = – [(0.0125 × 1002 – 2.5 × 100) – (0.0125 × 602 – 2.5 × 60] A1 Distance is 54 + 20 = 74 m A1 [4]

Page 7 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9709 43 © Cambridge International Examinations 2014 (iii) [dv/dt = 0.05 – 0.001t = 0 or 0.0005t(100 – t) = 0 → t = 0 or 100] M1 For using vmax occurs when dv/dt = 0 or when t = the midpoint of the roots of the quadratic equation v = 0. Maximum speed (= 0.05 × 50 – 0.0005 × 502) is 1.25 ms–1 A1 Plausible quadratic curve starting at (0,0), with max. at (50, 1.25) and terminating at (60, 1.2) B1 Straight line segment from (60,–1) to (100,0) B1 [4] 7 (i) M1 For applying Newton’s 2nd law to P or to Q For T – (40÷160) × 0.76g = 0.76a or 0.49g – T = 0.49a A1 For 0.49g – T = 0.49a or T – (40 ÷ 160) × 0.76g = 0.76a or 0.49g – (400 ÷ 160) × 0.76g = (0.49 + 0.76)a B1 Acceleration is 2.4 ms–2 and tension is 3.72 N (3.724 exact) A1 [4] (ii) [v2 = 2 × 2.4 × 0.3] M1 For using v2 = 0 + 2as Speed is 1.20 ms–1 A1ft [2] ft a from (i) (a ≠ ±g) (iii) M1 For using v2 = u2 + 2as with v = 0 and a = – (40 ÷ 160)g Distance while Q is on the ground = (2 × 2.4 × 0.3) ÷ 2(40g ÷ 160) (= 0.288 m) A1ft ft a from (i) and/or s = 30 Distance travelled is 0.588 m A1 [3]