Cambridge A Level Mathematics 9709 — 2014 May/June Paper 4 · Variant 1

*2506497633* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS 9709/41 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_41/RP © UCLES 2014 [Turn over

2 1 A train is moving at constant speed V m s−1 along a horizontal straight track. Given that the power of the train’s engine is 1330 kW and the total resistance to the train’s motion is 28 kN, find the value of V. [3] 2 A rough plane is inclined at an angle of
 to the horizontal. A particle of mass 0.25 kg is in equilibrium on the plane. The normal reaction force acting on the particle has magnitude 2.4 N. Find (i) the value of
, [2] (ii) the least possible value of the coefficient of friction. [2] 3 5 N 7 N 3 N 4 N 60
30
Four coplanar forces act at a point. The magnitudes of the forces are 5 N, 4 N, 3 N and 7 N, and the directions in which the forces act are shown in the diagram. Find the magnitude and direction of the resultant of the four forces. [6] 4 A particle is projected vertically upwards with speed 9 m s−1 from a point 3.15 m above horizontal ground. The particle moves freely under gravity until it hits the ground. For the particle’s motion from the instant of projection until the particle hits the ground, find the total distance travelled and the total time taken. [6] 5 O A B 1760 m 160 m A car of mass 1100 kg starts from rest at O and travels along a road OAB. The section OA is straight, of length 1760 m, and inclined to the horizontal with A at a height of 160 m above the level of O. The section AB is straight and horizontal (see diagram). While the car is moving the driving force of the car is 1800 N and the resistance to the car’s motion is 700 N. The speed of the car is v m s−1 when the car has travelled a distance of x m from O. (i) For the car’s motion from O to A, write down its increase in kinetic energy in terms of v and its increase in potential energy in terms of x. Hence find the value of k for which kv2 = x for 0 ≤x ≤1760. [4] (ii) Show that v2 = 2x −3200 for x ≥1760. [4] © UCLES 2014 9709/41/M/J/14

3 6 A B h m Fig. 1 Particles A of mass 0.25 kg and B of mass 0.75 kg are attached to opposite ends of a light inextensible string which passes over a fixed smooth pulley. The system is held at rest with the string taut and its straight parts vertical. Both particles are at a height of h m above the floor (see Fig. 1). The system is released from rest, and 0.6 s later, when both particles are in motion, the string breaks. The particle A does not reach the pulley in the subsequent motion. (i) Find the acceleration of A and the distance travelled by A before the string breaks. [4] t (s) v
m s−1 O V 0.6 T 1.6 Fig. 2 The velocity-time graph shown in Fig. 2 is for the motion of particle A until it hits the floor. The velocity of A when the string breaks is V m s−1 and T s is the time taken for A to reach its greatest height. (ii) Find the value of V and the value of T. [3] (iii) Find the distance travelled by A upwards and the distance travelled by A downwards and hence find h. [3] [Question 7 is printed on the next page.] © UCLES 2014 9709/41/M/J/14 [Turn over

4 7 Two cyclists P and Q travel along a straight road ABC, starting simultaneously at A and arriving simultaneously at C. Both cyclists pass through B 400 s after leaving A. Cyclist P starts with speed 3 m s−1 and increases this speed with constant acceleration 0.005 m s−2 until he reaches B. (i) Show that the distance AB is 1600 m and find P’s speed at B. [3] Cyclist Q travels from A to B with speed v m s−1 at time t seconds after leaving A, where v = 0.04t −0.0001t2 + k, and k is a constant. (ii) Find the value of k and the maximum speed of Q before he has reached B. [6] Cyclist P travels from B to C, a distance of 1400 m, at the speed he had reached at B. Cyclist Q travels from B to C with constant acceleration a m s−2. (iii) Find the time taken for the cyclists to travel from B to C and find the value of a. [4] 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 9709/41/M/J/14

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2014 series 9709 MATHEMATICS 9709/41 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 A LEVEL – May/June 2014 9709 41 © 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 A LEVEL – May/June 2014 9709 41 © 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 A LEVEL – May/June 2014 9709 41 © Cambridge International Examinations 2014 1 DF = 28000 B1 [1330 000 = 28000V] M1 For using P = (DF)V V = 47.5 A1 [3] 2 (i) 2.4 = 0.25g cosα B1 α = 16.3 B1 [2] (ii) [µ = 0.28÷0.96] M1 For using µ = F/R or µ = tanα Least possible value of µ is 7/24 or 0.292 A1 [2] 3 M1 For finding the component of the forces in the x direction X = 5 – 7cos 60o – 3cos 30o (= – 1.098) A1 M1 For finding the component of the forces in the y direction Y = 7sin 60o – 3sin 30o – 4 (= 0.5622) A1 M1 For using R2 = X2 + Y2 and tan θ = Y/X Resultant is 1.23 N and Direction is 152.9o anticlockwise from +ve x-axis oe A1 [6] 4 M1 For using 0 = u2 – 2gs for the upwards motion For s = 4.05 A1 Total distance = 4.05 + (3.15 + 4.05) = 11.25 m B1 tupwards = 0.9 B1 For downwards motion (3.15 + 4.05) = 2 1 gt2 → t = 1.2 B1 Time taken is 2.1 s B1 [6]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9709 41 © Cambridge International Examinations 2014 Alternative Mark Scheme for final 3 marks [–3.15 = 9T + 2 1 (–g) T2] M1 For using s = ut + 2 1 at2 for the total displacement and time [100t2 – 180t – 63 = 0] M1 For solving a quadratic equation for the total time T (10T – 21)(10T + 3) = 0 A1 T = 2.1 only 5 (i) KE gain = 550v2 B1 PE gain = 1000x B1 [1800x = 550v2 + 1000x + 700x] M1 k = 5.5 A1 [4] ft for incorrect coeff(s) of v2 and/or of x (ii) At A 5.5v2 = 1760 → v2 = 320 B1 M1 For using from A, KEgain= WD by DF –WD against R 550(v2 – 320) = 1800(x – 1760) – 700(x – 1760) A1 v2 = 2x – 3200 (cwo) A1 [4] AG Alternative for part (ii) [1800 – 700 = 1100a and 5.5v2 = 1760] M1 For applying Newton’s 2nd Law to find acceleration along AB and for using kv2 = x to find v2 at A a = 1 and v2 = 320 A1 [v2 = 320 + 2 × 1 × (x – 1760)] M1 For using v2 = u2 + 2as for motion from A to B v2 = 2x – 3200 A1 [4] 6 (i) M1 For using Newton’s second law for both particles and eliminating T, or using (M + m)a = (M – m)g Acceleration is 5 ms–2 A1 M1 For using s = 0 + 2 1 at2 Distance is 0.9 m A1 [4] (ii) 2 1 0.6 × V = 0.9 → V = 3 B1 ft distance in (i) M1 For using 0 = V – g(T – 0.6) T = 0.9 A1 [3]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9709 41 © Cambridge International Examinations 2014 (iii) [sup = 2 1 0.9 × 3 and sdown = 0 + 2 1 g(1.6 – 0.9)2] M1 For using area property in graph or equivalent Distance upwards is 1.35 m and distance downwards is 2.45 m A1 h = 1.1 B1 [3] ft sdown – sup 7 (i) M1 For using s = ut + 2 1 at2 to find the distance AB, or for using v = u + at to find P’s speed at B AB = 3 × 400 + 2 1 0.005 × 4002 = 1600 m (AG) or vB = 3 + 0.005 × 400 = 5 ms–1 A1 vB = 3 + 0.005 × 400 = 5 ms–1 or AB = 3 × 400 + 2 1 0.005 × 4002 = 1600 m (AG) B1 [3] (ii) M1 For using ʃ0 400v dt = 1600 [0.02t2 – 0.0001t3/3 + kt]0 400 = 1600 A1 400k = 1600 – 0.02 × 4002 + 0.0001 × 4003 ÷ 3 → k = 4 – 8 + 16/3 = 4/3 A1 [dv/dt = 0.04 – 0.0002t (= 0 when t = 200) M1 For differentiating and solving dv/dt = 0 vmax = 0.04 × 200 – 0.0001 × 2002 + 4/3 A1 ft incorrect k or incorrect value of t from dv/dt = 0 Maximum speed is 5.33 ms–1 A1 [6] (iii) M1 For using constant speed 5 ms–1 = 1400/T Time taken is 280 s A1 [1400 = 4/3 × 280 + 2 1 2802a] M1 For using s = ut + 2 1 at2 to find a a = 0.0262 A1 [4]