Cambridge A Level Mathematics 9709 — 2016 Oct/Nov Paper 4 · Variant 1

*0685093323* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS 9709/41 Paper 4 Mechanics 1 (M1) October/November 2016 1 hour 15 minutes Additional Materials: List of Formulae (MF9) READ THESE INSTRUCTIONS FIRST An answer booklet is provided inside this question paper. You should follow the instructions on the front cover of the answer booklet. If you need additional answer paper ask the invigilator for a continuation booklet. 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. This document consists of 4 printed pages and 1 insert. JC16 11_9709_41/2R © UCLES 2016 [Turn over

2 1 P Q Two particles P and Q, of masses 0.6 kg and 0.4 kg respectively, are connected by a light inextensible string. The string passes over a small smooth light pulley fixed at the edge of a smooth horizontal table. Initially P is held at rest on the table and Q hangs vertically (see diagram). P is then released. Find the tension in the string and the acceleration of Q. [4] 2 A particle of mass 0.1 kg is released from rest on a rough plane inclined at 20Å to the horizontal. It is given that, 5 seconds after release, the particle has a speed of 2 m s−1. (i) Find the acceleration of the particle and hence show that the magnitude of the frictional force acting on the particle is 0.302 N, correct to 3 significant figures. [3] (ii) Find the coefficient of friction between the particle and the plane. [2] 3 A particle P is projected vertically upwards from a point O. When the particle is at a height of 0.5 m, its speed is 6 m s−1. Find (i) the greatest height reached by the particle above O, [3] (ii) the time after projection at which the particle returns to O. [3] 4 2F N F N 15 N P 60Å !Å Three coplanar forces of magnitudes F N, 2F N and 15 N act at a point P, as shown in the diagram. Given that the forces are in equilibrium, find the values of F and !. [6] © UCLES 2016 9709/41/O/N/16

3 5 t (s) v (m s−1) 90 70 40 30 10 0 5 V The diagram shows a velocity-time graph which models the motion of a cyclist. The graph consists of five straight line segments. The cyclist accelerates from rest to a speed of 5 m s−1 over a period of 10 s, and then travels at this speed for a further 20 s. The cyclist then descends a hill, accelerating to speed V m s−1 over a period of 10 s. This speed is maintained for a further 30 s. The cyclist then decelerates to rest over a period of 20 s. (i) Find the acceleration of the cyclist during the first 10 seconds. [1] (ii) Show that the total distance travelled by the cyclist in the 90 seconds of motion may be expressed as 45V + 150 m. Hence find V, given that the total distance travelled by the cyclist is 465 m. [3] (iii) The combined mass of the cyclist and the bicycle is 80 kg. The cyclist experiences a constant resistance to motion of 20 N. Use an energy method to find the vertical distance which the cyclist descends during the downhill section from t = 30 to t = 40, assuming that the cyclist does no work during this time. [4] 6 A block of mass 25 kg is pulled along horizontal ground by a force of magnitude 50 N inclined at 10Å above the horizontal. The block starts from rest and travels a distance of 20 m. There is a constant resistance force of magnitude 30 N opposing motion. (i) Find the work done by the pulling force. [2] (ii) Use an energy method to find the speed of the block when it has moved a distance of 20 m. [2] (iii) Find the greatest power exerted by the 50 N force. [2] 5Å 50 N 10Å After the block has travelled the 20 m, it comes to a plane inclined at 5Å to the horizontal. The force of 50 N is now inclined at an angle of 10Å to the plane and pulls the block directly up the plane (see diagram). The resistance force remains 30 N. (iv) Find the time it takes for the block to come to rest from the instant when it reaches the foot of the inclined plane. [4] © UCLES 2016 9709/41/O/N/16 [Turn over

4 7 A racing car is moving in a straight line. The acceleration a m s−2 at time t s after the car starts from rest is given by a = 15t −3t2 for 0 ≤t ≤5, a = −625 t2 for 5 < t ≤k, where k is a constant. (i) Find the maximum acceleration of the car in the first five seconds of its motion. [3] (ii) Find the distance of the car from its starting point when t = 5. [3] (iii) The car comes to rest when t = k. Find the value of k. [5] 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. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. 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 2016 9709/41/O/N/16

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 7 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS 9709/41 Paper 4 October/November 2016 MARK SCHEME Maximum Mark: 50 Published 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 October/November 2016 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2016 9709 41 © UCLES 2016 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 Cambridge International AS/A Level – October/November 2016 9709 41 © UCLES 2016 The following abbreviations may be used in a mark scheme or used on the scripts: AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) 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 SOI Seen or implied 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 Cambridge International AS/A Level – October/November 2016 9709 41 © UCLES 2016 1 [0.4g – T = 0.4a T = 0.6a System equation 0.4g = (0.4 + 0.6)a] M1 For applying Newton’s 2nd law to either particle or to the system M1 For applying Newton’s 2nd law to the other particle and attempt to solve for a and T a = 4 m s–2 A1 T = 2.4 N A1 [4] 2 (i) 2 = 5a → a = 0.4 ms-2 B1 [0.1g sin20 – F = 0.1 × 0.4] M1 For applying Newton’s 2nd law to the particle F = 0.302 N AG A1 [3] (ii) [R = 0.1g cos20 ( = 0.9397)] M1 For attempting to find R and using µ = F/R µ = 0.3020/0.9397 = 0.321 A1 [2] 3 (i) [0 = 62 – 2g × s] M1 For using v2 = u2 + 2as s = 1.8 A1 Total height = 2.3 m B1 [3] Alternative for 3(i) [62 = u2 – 2g × 0.5] M1 For using v2 = u2 + 2as to find the initial velocity u2 = 46 A1 02 = 46 – 2gs → s = total height = 2.3 m B1 [3]

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2016 9709 41 © UCLES 2016 (ii) [2.3 = 0 + 0.5gt2] M1 For using s = ut + 0.5gt2 to find time to reach the ground t = 0.678 A1 Total time = 2 × 0.678 = 1.36 s B1 [3] Alternative for 3(ii) [ 0 46 gt = − ] M1 Using v = u – gt to find time taken to the highest point 46 0.678 10 t = = A1 Total time = 2 × 0.678 = 1.36 s B1 [3] 4 M1 For resolving forces horizontally 2F + Fcos60 = 15cosα A1 M1 For resolving forces vertically Fsin60 = 15sinα A1 M1 For using Pythagoras or for using tan α to find F and α F = 5.67 and α = 19.1 A1 [6] Allow 15 7 / 7 F = 5 (i) a = 0.5 m s–2 B1 [1] (ii) [Distance = 25 + 100 + 5(5 + V) + 30V + 10V] M1 For attempting to find the distance travelled 150 + 45V AG A1 150 + 45V = 465 → V = 7 m s–1 B1 [3] (iii) ½ × 80 × 72 – ½ × 80 × 52 [= 960] M1 For change in KE 20 × (5 + 7)/2 × 10 [=1200] M1 For work done against friction using F × d [80gh = 960 + 1200] M1 For using PE loss = KE gain + WD against Res. h = 2.7 m A1 [4]

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2016 9709 41 © UCLES 2016 6 (i) [Work done = 50 cos10 × 20] M1 Using WD = Fd cos θ = 984.8 J A1 [2] (ii) [984.8 = ½ × 25v2 + 30 × 20] M1 Using WD by DF = KE gain + WD against Res. v = 5.55 m s–1 A1 [2] (iii) M1 For using Power = Fv Max power = 50cos10 × 5.55 = 273 W A1 [2] Greatest power is at vmax (iv) [50 cos10 – 30 – 25g sin5 = 25a] M1 For using Newton’s 2nd law up the plane a = – 0.102 m s–2 A1 [0 = 5.55 – 0.102t] M1 For using v = u + at Time t = 54.4 s A1 [4] Alternative for 6(iv) M1 For using WD by DF + KE loss = PE gain + WD against Res to find distance s up plane 50 cos 10 × s + ½ × 25 × 5.552 = 25g × s sin 5 + 30 × s A1 s = 151 m M1 For using s = ½(u + v)t t = 302/5.55 = 54.4 s A1 [4] 7 (i) [15 – 6t = 0] M1 For differentiation Max acceleration when t = 2.5 s A1 May be stated from an a-t diagram Max acceleration = 18.75 m s–2 A1 [3] (ii) [Speed = 7.5t2 – t3 (+ c)] M1 For using integration to obtain speed [Distance = 2.5t3 – 0.25t4 (+ ct + d)] M1 For using integration to obtain distance = 2.5×125 – 0.25×625 = 156.25 m A1 [3] Allow distance = 625/4

Page 7 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2016 9709 41 © UCLES 2016 (iii) v(5) = 7.5×25 – 125 = 62.5 m s–1 B1 Allow v(5) = 125/2 2 5 5 625 625 d k k t t t   − =     ∫ M1 Integral with correct limits 625 625 625 125 5 k k = − = − A1 ( ) ( ) 625 125 5 62.5 v k v k − = − = − M1 Use of v(5) = 62.5 and v(k) = 0 k = 10 A1 [5] Alternative for 7(iii) v(5) = 7.5×25 – 125 = 62.5 m s–1 B1 ( ) 2 625 625 d v t t c t t = ∫− = + M1 Using indefinite integration [c = –62.5] ( ) 625 62.5 v t t = − A1 For using v(5) = 62.5 to find c and setting v(k) = 0 ( ) 625 62.5 0 v k k = − = M1 k = 10 A1 [5]