Cambridge A Level Mathematics - Further 9231 — 2015 May/June Paper 2 · Variant 2

*7039328109* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/22 Paper 2 May/June 2015 3 hours Additional Materials: Answer Booklet/Paper Graph Paper List of Formulae (MF10) 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 is necessary, take the acceleration due to gravity to be 10 m s−2. The use of a calculator is expected, where appropriate. Results obtained solely from a graphic calculator, without supporting working or reasoning, will not receive credit. 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. This document consists of 5 printed pages and 3 blank pages. JC15 06_9231_22/FP © UCLES 2015 [Turn over

2 1 A particle P is moving in a circle of radius 0.25 m. At time t seconds, its velocity is 2t2 −4t + 3 m s−1. At a particular time T seconds, where T > 0, the magnitude of the transverse component of the acceleration of P is 6 m s−2. Find the magnitude of the radial component of the acceleration of P at this instant. [4] 2 A particle P moves on a straight line AOB in simple harmonic motion. The centre of the motion is O, and P is instantaneously at rest at A and B. The point C is on the line AOB, between A and O, and CO = 10 m. When P is at C, the magnitude of its acceleration is 0.625 m s−2 and it is moving towards O with speed 6 m s−1. Find (i) the period of the motion, in terms of 0, [2] (ii) the amplitude of the motion. [2] The point M is the mid-point of OB. Find the time that P takes to travel directly from C to M. [4] 3 A particle P, of mass m, is placed at the highest point of a fixed solid smooth sphere with centre O and radius a. The particle P is given a horizontal speed u and it moves in part of a vertical circle, with centre O, on the surface of the sphere. When OP makes an angle 1 with the upward vertical, and P is still in contact with the surface of the sphere, the speed of P is v and the reaction of the sphere on P has magnitude R. Show that R = mg3 cos 1 −2 −mu2 a . [5] The particle loses contact with the sphere at the instant when v = 2u. Find u in terms of a and g. [4] 4 C A B 2a 30Å 30Å A uniform rod BC of length 2a and weight W is hinged to a fixed point at B. A particle of weight 3W is attached to the rod at C. The system is held in equilibrium by a light elastic string of natural length 3 5a in the same vertical plane as the rod. One end of the elastic string is attached to the rod at C and the other end is attached to a fixed point A which is at the same horizontal level as B. The rod and the string each make an angle of 30Å with the horizontal (see diagram). Find (i) the modulus of elasticity of the string, [5] (ii) the magnitude and direction of the force acting on the rod at B. [5] 5 Three uniform small smooth spheres A, B and C have equal radii and masses 3m, 2m and m respectively. The spheres are at rest in a straight line on a smooth horizontal surface, with B between A and C. The coefficient of restitution between A and B is e and the coefficient of restitution between B and C is e′. Sphere A is projected directly towards B with speed u. Show that, after the collision between B and C, the speed of C is 2 5u1 + e1 + e′ and find the corresponding speed of B. [7] After this collision between B and C it is found that each of the three spheres has the same momentum. Find the values of e and e′. [5] © UCLES 2015 9231/22/M/J/15

3 6 The independent random variables X and Y have distributions with the same variance 32. Random samples of N observations of X and 10 observations of Y are taken, and the results are summarised by Σx = 5, Σx2 = 11, Σy = 10, Σy2 = 160. These data give a pooled estimate of 12 for 32. Find N. [4] 7 A random sample of 8 sunflower plants is taken from the large number grown by a gardener, and the heights of the plants are measured. A 95% confidence interval for the population mean, - metres, is calculated from the sample data as 1.17 < - < 2.03. Given that the height of a sunflower plant is denoted by x metres, find the values of Σx and Σx2 for this sample of 8 plants. [7] 8 (a) For a random sample of ten pairs of values of x and y taken from a bivariate distribution, the equations of the regression lines of y on x and of x on y are, respectively, y = 0.38x + 1.41 and x = 0.96y + 7.47. (i) Find the value of the product moment correlation coefficient for this sample. [2] (ii) Using a 5% significance level, test whether there is positive correlation between the variables. [4] (b) For a random sample of n pairs of values of u and v taken from another bivariate distribution, the value of the product moment correlation coefficient is 0.507. Using a test at the 5% significance level, there is evidence of non-zero correlation between the variables. Find the least possible value of n. [2] 9 Cotton cloth is sold from long rolls of cloth. The number of flaws on a randomly chosen piece of cloth of length a metres has a Poisson distribution with mean 0.8a. The random variable X is the length of cloth, in metres, between two successive flaws. (i) Explain why, for x ≥0, PX > x = e−0.8x. [2] (ii) Find the probability that there is at least one flaw in a 4 metre length of cloth. [2] (iii) Find (a) the distribution function of X, [1] (b) the probability density function of X, [2] (c) the interquartile range of X. [4] © UCLES 2015 9231/22/M/J/15 [Turn over

4 10 Young children at a primary school are learning to throw a ball as far as they can. The distance thrown at the beginning of the school year and the distance thrown at the end of the same school year are recorded for each child. The distance thrown, in metres, at the beginning of the year is denoted by x; the distance thrown, in metres, at the end of the year is denoted by y. For a random sample of 10 children, the results are shown in the following table. Child A B C D E F G H I J x 5.2 4.1 3.7 5.4 7.6 6.1 3.2 4.0 3.5 8.0 y 6.2 4.8 5.0 5.6 7.7 7.0 4.0 4.5 3.6 8.5 [Σx = 50.8, Σx2 = 284.16, Σy = 56.9, Σy2 = 347.59, Σxy = 313.28.] A particular child threw the ball a distance of 7.0 metres at the beginning of the year, but he could not throw at the end of the year because he had broken his arm. By finding the equation of an appropriate regression line, estimate the distance this child would have thrown at the end of the year. [5] The teacher suspects that, on average, the distance thrown by a child increases between the two throws by more than 0.4 metres. Stating suitable hypotheses and assuming a normal distribution, test the teacher’s suspicion at the 5% significance level. [8] © UCLES 2015 9231/22/M/J/15

5 11 Answer only one of the following two alternatives. EITHER A B C D O a 2a A uniform disc, with centre O and radius a, is surrounded by a uniform concentric ring with radius 3a. The ring is rigidly attached to the rim of the disc by four symmetrically positioned uniform rods, each of mass 3 2m and length 2a. The disc and the ring each have mass 2m. The rods meet the ring at the points A, B, C and D. The disc, the ring and the rods are all in the same plane (see diagram). Show that the moment of inertia of this object about an axis through O perpendicular to the plane of the object is 45ma2. [4] Find the moment of inertia of the object about an axis l through A in the plane of the object and tangential to the ring. [3] A particle of mass 3m is now attached to the object at C. The object, including the additional particle, is suspended from the point A and hangs in equilibrium. It is free to rotate about the axis l. The centre of the disc is given a horizontal speed u. When, in the subsequent motion, the object comes to instantaneous rest, C is below the level of A and AC makes an angle sin−11 4 with the horizontal. Find u in terms of a and g. [7] OR Each of 200 identically biased dice is thrown repeatedly until an even number is obtained. The number of throws, x, needed is recorded and the results are summarised in the following table. x 1 2 3 4 5 6 ≥7 Frequency 126 43 22 3 5 1 0 State a type of distribution that could be used to fit the data given in the table above. [1] Fit a distribution of this type in which the probability of throwing an even number for each die is 0.6 and carry out a goodness of fit test at the 5% significance level. [8] For each of these dice, it is known that the probability of obtaining a 6 when it is thrown is 0.25. Ten of these dice are each thrown 5 times. Find the probability that at least one 6 is obtained on exactly 4 of the 10 dice. [5] © UCLES 2015 9231/22/M/J/15

6 BLANK PAGE © UCLES 2015 9231/22/M/J/15

7 BLANK PAGE © UCLES 2015 9231/22/M/J/15

8 BLANK PAGE 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 2015 9231/22/M/J/15

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the May/June 2015 series 9231 FURTHER MATHEMATICS 9231/22 Paper 2, maximum raw mark 100 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 2015 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 A Level – May/June 2015 9231 22 © Cambridge International Examinations 2015 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 A Level – May/June 2015 9231 22 © Cambridge International Examinations 2015 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 Cambridge International A Level – May/June 2015 9231 22 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 1 Find T by equating dv/dt at t = T to 6: 4T – 4 = 6, T = 2.5 M1 A1 Find radial component v2/r of acceln. at t = T: v2/r = (2T 2 – 4T + 3)2 / 0.25 (M0 if T not given a value) = (11/2)2 × 4 = 121 [m s–2] M1 A1 SR: Max M1 (1/4) if linear and angular confused 4 4 2 (i) Find ω2 from SHM eqn. d2x/dt2 = – ω2x at C: 0.625 = 10ω2, ω2 = 0.0625 or 1/16 B1 Find period T [s] from T = 2π/ω: (ft on ω2) T = 2π/¼ = 8π (not 25.1) B1 2 (ii) Find amplitude a [m] from vC 2 = ω2(a2 – 102): 62 = ω2(a2 – 102) a2 = 62 × 16 + 102, a = √676 = 26 M1 A1 Find time from C to M, e.g.: ω–1 sin–1 (10/a) + ω–1 sin–1 ½ or ω–1 cos–1 (–10/a) – ω–1 cos–1 ½ or ½T – ω–1 cos–1 (10/a) – ω–1 cos–1 ½ M1 (AEF throughout) = ω–1 {0.3948 + π/6 [= 0.5236]} or ω–1 {1.9656 – π/3 [= 1.0472]} or ω–1 {π – 1.760 – π/3 [= 1.0472]} A1 = 1.579 + 2.094 or 7.862 – 4.189 or 12.567 – 4.704 – 4.189 or 4 × 0.9184; = 3.67 [s] A1; A1 2 4 8 3 Find v2 from conservation of energy: ½mv2 = ½mu2 + mga(1 – cos θ ) M1 A1 Find R by using F = ma radially: R = mg cos θ – mv2/a B1 Eliminate v2 to find R: AG R = mg(3 cos θ – 2) – mu2/a M1 A1 Find u2 or v2 in terms of cos θ when R = 0: u2 = ag(3 cos θ – 2) or v2 = ag cos θ B1 EITHER: Replace cos θ in energy eqn with v = 2u: 4u2 = u2 + 2ag – ⅔(u2 + 2ag) or u2 + 2ag – 8u2 M1 A1 OR: Find cos θ and substitute in energy eqn: [v2/ag =] 4(3 cos θ – 2) = cos θ cos θ = 8/11 4u2 = u2 + 2ag (1 – 8/11) (M1 A1) Hence find u: u = √(2ag/11) or 0.426√(ag) A1 5 4 9 4 (i) Take moments for rod about B: W × a cos 30° + 3W × 2a cos 30° (or with cos 30° = √3/2) = T × 2a cos 30° M1 A1 Hence find tension T: T = 7W/2 A1 (Can earn M1 A0 A1 if e.g. sin 30° wrongly used) Find modulus λ using Hooke’s Law: T = λ (2a – 3a/5) / (3a/5) λ = (3/7) (7W/2) = 3W/2 M1 A1 5

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 22 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total (ii) EITHER: Find horizontal component of force F at B: X = T cos 30° = (7√3/4) W or 3.03 W B1 Find vertical component: (X, Y ft on T) Y = 4W – T sin 30° = 9W/4 B1 Find magnitude of F: F = √(X2 + Y2) = (√57/2) W or 3.77[5] W B1 Find direction of F (AEF): Upward force at angle to AB of tan–1 Y/X = tan–1 3√3/7 (A0 if direction unclear) = 36.6° or 0.639 radians M1 A1 OR: Find component along CB: F1 = (4W + T) sin 30° = 15W/4 (B1 ) Find normal component: (F1, F2 ft on T) F2 = (4W – T) cos 30° = (√3/4)W (B1 ) Find magnitude of F: F = (√57/2) W or 3⋅77[5] W (B1) Find direction of F (AEF): Upward force at angle to CB of tan–1 F2/F1 = tan–1 √3/15 (A0 if direction unclear) = 6.6° or 0.115 radians (M1 A1) OR: Find component parallel to string CA: ±F1 = T – 4W sin 30° = 3W/2 (B1 ) Find normal component: (F1, F2 ft on T) ±F2 = 4W cos 30° = 2√3 W (B1 ) Find magnitude of F: F = (√57/2) W or 3.77[5] W (B1) Find direction of F (AEF): Upward force at angle to AC of tan–1 F2/F1 = tan–1 4/√3 (A0 if direction unclear) = 66.6° or 1.16 radians (M1 A1) 5 10 5 For A & B use conservation of momentum, e.g.: 3mvA + 2mvB = 3mu M1 (m may be omitted here and below) Use Newton’s law of restitution (consistent signs): vB – vA = eu M1 Combine to find vB : vB = 3(1 + e) u/5 A1 For B & C use conservation of momentum, e.g.: 2mvB′ + mvC = 2mvB M1 Use Newton’s law of restitution (consistent signs): vC – vB′ = e′ vB M1 Combine to find vC and vB′: vC = 2(1 + e′) vB /3 = 2(1 + e) (1 + e′) u/5 AG A1 vB′ = (2 – e′) vB /3 = (1 + e) (2 – e′) u/5 A1 Find ratios or values of vA, vB′, vC from momentum: 3vA = 2vB′ = vC [= u] B1 Find e from first collision eqns, e.g.: vA = (3 – 2e) u/5 = u/3 (or find e′ and then use 3vA = 2vB′) or vB = ½ (3u – u) or (⅓ + e) u = 3(1 + e) u/5, e = ⅔ M1 A1 Find e′ from second collision eqns, e.g.: 2vB′ = vC so 2(2 – e′) = 2(1 + e′) or vC = 2(1 + ⅔) (1 + e′) u/5 = u or vB′ = (1 + ⅔) (2 – e′) u/5 = u/2 e′ = ½ M1 A1 7 5 12 6 Equate pooled estimate of σ2 to 12: (11 – 52/N + 160 – 102/10) / (N +10 – 2) = 12 M1 A1 Formulate and solve relevant quadratic eqn. for N: 12 N2 – 65 N + 25 = 0, N = 5 M1 A1 4 4

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 22 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 7 Find Σ x via sample mean x : Σ x = 8 x = 8 × ½ (1.17 + 2.03) = 8 × 1.6 = 12.8 M1 A1 Find estimate of population variance s2: t √(s2/8) = ½ (2.03 – 1.17) [= 0.43] M1 Use of correct tabular value (1⋅96 leads to 23⋅2): t7, 0.975 = 2.36[5] A1 (to 3 d.p.) s2 = 0.2645 or 32/121 or 0.51432 A1 Find Σ x2 from s2: s2 = {Σ x2 – (Σ x)2/8}/7 (M0 for s2 = {…}/8) Σ x2 = 7 × 0.2645 + 12.82/8 = 22.3 M1 A1 7 7 8 (a) (i) Find correlation coefficient r from r2 = b1b2: r = √(0.38 × 0.96) = 0.604 M1 *A1 2 (ii) State both hypotheses (B0 for r …): H0: ρ = 0, H1: ρ > 0 B1 State or use correct tabular one-tail r-value: r10, 5% = 0.549 *B1 State or imply valid method for reaching conclusion: Reject H0 if |r| > tab. value (AEF) M1 Correct conclusion (AEF, dep *A1, *B1): There is positive correlation A1 4 (b) State or use relevant tabular two-tail r-value: r16, 5% = 0.497 (or r15, 5% = 0.514) M1 Find least possible value of n: nmin = 16 A1 SR M1 A1 for stating 16 without explanation B1 for stating 15 without explanation B1 for finding or stating one-tail result 12 2 8 9 (i) Relate P(X > x) to number of flaws (AEF): P(X > x) = P(zero flaws in x m) B1 Relate this to Poisson distn. (AEF): = P0(0.8x) = e–0⋅8x A.G. B1 2 (ii) Find P(number of flaws ≥ 1): 1 – P0(0.8 × 4) = 1 – e–3⋅2 (M0 if “1 – “ omitted) = 1 – 0.0408 = 0.959 M1 A1 2 (iii) (a) Find or state distribution function F(x): F(x) = P(X ≤ x) = 1 – P(X > x) = 1 – e–0⋅8x B1 1 (b) Find or state probability density function f(x): f(x) = dF/dx = 0⋅8 e–0⋅8x M1 A1 S.R. Deduct A1 if (a), (b) interchanged 2 (c) Formulate equation for either quartile value Q : F(Q) = 1 – e–0⋅8Q = ¼ or ¾ M1 Find lower quartile Q1 : (AEF) Q1 = 1.2 ln 4/3 [= 0.360 ] A1 Find upper quartile Q3 : (AEF) Q3 = 1.2 ln 4 [= 1.733] A1 Find interquartile range (allow Q1 – Q3): Q3 – Q1 [= 1.2 ln 3] = 1.37 A1 4 11

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 22 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 10 Calculate gradient b1 in y – y = b1(x – x ) : Sxy = 313.28 – 50.8 × 56.9/10 = 24.228 Sxx = 284.16 – 50.82/10 = 26.096 (PA –1 so max 4/5 for 0.93, giving y = 7.48) b1 = Sxy / Sxx = 0.928 M1 A1 Find y when x = 7 from regression line of y on x: y = 56.9/10 + b1 (7 – 50.8/10) M1 A1 = 5.69 + 0.928 (7 – 5.08) [ = 0.928 ×7 + 0.976 ] y = 7.47 (allow 7.48 or 7.5) A1 SR If regression line of x on y used: Syy = 347.59 – 56.92/10 = 23.829 b2 = Sxy / Syy = 1.017 (M1) 7 = 50.8/10 + b2 (y – 56.9/10) (M1 A1) = 1.017 y – 0.707 (can earn at most 4/5) y = 7.58 (allow 7.6) (A1) Find differences (e.g. y – x) and sample mean: 1 0.7 1.3 0.2 0.1 0.9 0.8 0.5 0.1 0.5 d = 6⋅1 / 10 = 0.61 M1 A1 Estimate population variance (to 3 s.f.): s2 = (5.19 – 6.12/10) / 9 (allow biased here: 0.1469 or 0.38332) = 0.1632 or 0.4042 B1 State hypotheses (AEF; B0 for x ), e.g.: H0: µy – µx = 0.4, H1: µy – µx > 0.4 B1 Calculate value of t: t = (d – 0.4)/(s/√10) = 1.64 M1 A1 State or use correct tabular t-value: t9, 0.95 = 1.83[3] B1 (or can compared with 0.634) Consistent conclusion (AEF, ft on both t-values): [Accept H0:] No improvement of more than 0.4 B1 Wrong test can earn only B1 for hypotheses and B1 for conclusion 5 8 13

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 22 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 11A Find MI of disc about O: Idisc = ½ 2ma2 = ma2 B1 Find MI of ring about O: Iring = 2m (3a)2 = 18 ma2 B1 EITHER: Find MI of any rod about O: Irod = ⅓ (3m/2)a2 + (3m/2)(2a)2 = (13/2) ma2 B1 OR: Find MI of 2 collinear rods about O: ⅓ (9m/2)(3a)2 – ⅓ (3m/2)a2 = 13 ma2 (B1) Find MI of object about O: IO = ma2 + 18 ma2 + 4(13/2) ma2 = 45 ma2 AG B1 Find MI of object about axis at O // to tangent: IO′ = ½IO M1 Find MI of object about tangential axis at A: IA = IO′ + 10m (3a)2 = (225/2) ma2 M1 A1 Find new MI when particle is attached at C: IA′ = IA + 3m (6a)2 = (441/2) ma2 M1 Find and use initial angular speed: ω0 = u/3a B1 Find gain in P.E. at instantaneous rest: (10mg × 3a + 3mg × 6a)(1 – sin θ) M1 A1 = 45mga/2 + 27mga/2 or 48mga (1 – sin θ) = 36mga A1 Find u by equating to rotational KE: = ½ IA′ ω0 2 M1 u2 = (36/441) × 36ag [441=212] u = (12/7) √(ag) or 1.71 √(ag) A1 SR: Taking AC at sin–1(¼) to vertical: P.E. = 48mga (1 – cos sin–1(¼)) = 1.524 mga (A0) (max 6/7) u = 0.35[3] √(ag) (A1) SR: Overlooking added particle can earn M0 B1 M1 A0 A0 M1 A0 (max 3/7) 4 3 7 14 11B State suitable distribution: Geometric B1 State (at least) null hypothesis: (AEF) H0: Distn. fits data or p = 0.6 B1 (B0 for “It is a good fit”) Find exp. values using 200pqx–1 with p = 0.6, q = 0.4: 120 48 19.2 7.68 (ignore incorrect final value here, 3.072 1.2288 0.8192 M1 e.g. 0.4915 which can earn max 5/8 Combine last 3 cells since exp. value < 5: O: . . . 6 E: . . . 5.12 B1 Calculate χ2 (result correct to 3 s.f.): χ2 = 0.3 + 0.5208 + 0.4083 + 2.8519 + 0.15125 = 4.23[2] M1 A1 State or use consistent tabular value (to 3 s.f.): 5 cells: χ4 ,0.95 2 = 9.488 [or if 2 or no cells combined: 6 cells: χ5, 0.95 2 = 11.07 7 cells: χ6, 0.95 2 = 12.59 or if 4 cells combined, as with 0⋅4915: 4 cells: χ3, 0.95 2 = 7.815 ] B1 Valid method for reaching conclusion: Accept H0 if χ2 < tabular value M1 Conclusion (AEF, requires both values correct): 4.23 < 9.49 so distn fits or p = 0.6 A1 (Allow A1 for “It is a good fit”) Find prob. p of at least one 6 on 5 throws of one die: p = 1 – 0.755 = 0.7627 (4 s.f.) M1 A1 Find prob. of at least one 6 on exactly 4 of 10 dice: 10C4 × ; p4 (1 – p) 6 M1; M1 = 210 × 6.043×10–5 = 0.0126 or 0.0127 A1 1 8 5 14