Cambridge A Level Mathematics - Further 9231 — 2014 Oct/Nov Paper 2 · Variant 1

*7107187544* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/21 Paper 2 October/November 2014 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. JC14 11_9231_21/3R © UCLES 2014 [Turn over

2 1 Two smooth spheres A and B, of equal radii and masses 2m and m respectively, lie at rest on a smooth horizontal table. The spheres A and B are projected directly towards each other with speeds 4u and 3u respectively. The coefficient of restitution between the spheres is e. Find the set of values of e for which the direction of motion of A is reversed in the collision. [5] 2 4 m s−1 P ! A small smooth ball P is moving on a smooth horizontal plane with speed 4 m s−1. It strikes a smooth vertical barrier at an angle ! (see diagram). The coefficient of restitution between P and the barrier is 0.4. Given that the speed of P is halved as a result of the collision, find the value of !. [5] 3 ! A B O a A smooth cylinder of radius a is fixed with its axis horizontal. The point O is the centre of a circular cross-section of the cylinder. The line AOB is a diameter of this circular cross-section and the radius OA makes an angle ! with the upward vertical (see diagram). It is given that cos ! = 3 5. A particle P of mass m moves on the inner surface of the cylinder in the plane of the cross-section. The particle passes through A with speed u along the surface in the downwards direction. The magnitude of the reaction between P and the inner surface of the sphere is RA when P is at A, and is RB when P is at B. It is given that RB = 10RA. Show that u2 = ag. [6] The particle loses contact with the surface of the cylinder when OP makes an angle 1 with the upward vertical. Find the value of cos 1. [4] © UCLES 2014 9231/21/O/N/14

3 4 A B C D l 1 4l ! A uniform rod AB, of length l and mass m, rests in equilibrium with its lower end A on a rough horizontal floor and the end B against a smooth vertical wall. The rod is inclined to the horizontal at an angle !, where tan ! = 3 4, and is in a vertical plane perpendicular to the wall. The rod is supported by a light spring CD which is in compression in a vertical line with its lower end D fixed on the floor. The upper end C is attached to the rod at a distance 1 4l from B (see diagram). The coefficient of friction at A between the rod and the floor is 1 3 and the system is in limiting equilibrium. (i) Show that the normal reaction of the floor at A has magnitude 1 2mg and find the force in the spring. [7] (ii) Given that the modulus of elasticity of the spring is 2mg, find the natural length of the spring. [4] 5 The points A and B are on a smooth horizontal table at a distance 8a apart. A particle P of mass m lies on the table on the line AB, between A and B. The particle is attached to A by a light elastic string of natural length 3a and modulus of elasticity 6mg, and to B by a light elastic string of natural length 2a and modulus of elasticity mg. In equilibrium, P is at the point O on AB. (i) Show that AO = 3.6a. [4] The particle is released from rest at the point C on AB, between A and B, where AC = 3.4a. (ii) Show that P moves in simple harmonic motion and state the period. [6] (iii) Find the greatest speed of P. [2] 6 A random sample of 50 observations of a random variable X and a random sample of 60 observations of a random variable Y are taken. The results for the sample means, x and y, and the unbiased estimates for the population variances, s2 x and s2 y, respectively, are as follows. x = 25.4 y = 23.6 s2 x = 23.2 s2 y = 27.8 A test, at the !% significance level, of the null hypothesis that the population means of X and Y are equal against the alternative hypothesis that they are not equal is carried out. Given that the null hypothesis is not rejected, find the set of possible values of !. [5] © UCLES 2014 9231/21/O/N/14 [Turn over

4 7 The time, T seconds, between successive cars passing a particular checkpoint on a wide road has probability density function f given by ft = T 1 100e−0.01t t ≥0, 0 otherwise. (i) State the expected value of T. [1] (ii) Find the median value of T. [3] Sally wishes to cross the road at this checkpoint and she needs 20 seconds to complete the crossing. She decides to start out immediately after a car passes. Find the probability that she will complete the crossing before the next car passes. [2] 8 The numbers of a particular type of laptop computer sold by a store on each of 100 consecutive Saturdays are summarised in the following table. Number sold 0 1 2 3 4 5 6 7 ≥8 Number of Saturdays 7 20 39 16 14 2 1 1 0 Fit a Poisson distribution to the data and carry out a goodness of fit test at the 2.5% significance level. [9] 9 A random sample of 10 pairs of values of x and y is given in the following table. x 4 6 6 8 2 7 12 14 9 5 y 2 4 6 8 6 10 9 8 6 5 (i) Find the equation of the regression line of y on x. [4] (ii) Find the product moment correlation coefficient for the sample. [2] (iii) Find the estimated value of y when x = 10, and comment on the reliability of this estimate. [2] (iv) Another sample of N pairs of data from the same population has the same product moment correlation coefficientas the first sample given. A test, at the 1% significance level, on this second sample indicates that there is sufficient evidence to conclude that there is positive correlation. Find the set of possible values of N. [3] 10 The continuous random variable X has probability density function f given by fx = T 1 2 1 ≤x ≤3, 0 otherwise. The random variable Y is defined by Y = X3. Find the distribution function of Y. [5] Sketch the graph of the probability density function of Y. [3] Find the probability that Y lies between its median value and its mean value. [4] © UCLES 2014 9231/21/O/N/14

5 11 Answer only one of the following two alternatives. EITHER A B W X Y Z r r r A uniform plane object consists of three identical circular rings, X, Y and Z, enclosed in a larger circular ring W. Each of the inner rings has mass m and radius r. The outer ring has mass 3m and radius R. The centres of the inner rings lie at the vertices of an equilateral triangle of side 2r. The outer ring touches each of the inner rings and the rings are rigidly joined together. The fixed axis AB is the diameter of W that passes through the centre of X and the point of contact of Y and Z (see diagram). It is given that R = 1 + 2 3ï3r. (i) Show that the moment of inertia of the object about AB is 7 + 2ï3mr2. [8] The line CD is the diameter of W that is perpendicular to AB. A particle of mass 9m is attached to D. The object is now held with its plane horizontal. It is released from rest and rotates freely about the fixed horizontal axis AB. (ii) Find, in terms of g and r, the angular speed of the object when it has rotated through 60Å. [6] OR Fish of a certain species live in two separate lakes, A and B. A zoologist claims that the mean length of fish in A is greater than the mean length of fish in B. To test his claim, he catches a random sample of 8 fish from A and a random sample of 6 fish from B. The lengths of the 8 fish from A, in appropriate units, are as follows. 15.3 12.0 15.1 11.2 14.4 13.8 12.4 11.8 Assuming a normal distribution, find a 95% confidence interval for the mean length of fish in A. [5] The lengths of the 6 fish from B, in the same units, are as follows. 15.0 10.7 13.6 12.4 11.6 12.6 Stating any assumptions that you make, test at the 5% significance level whether the mean length of fish in A is greater than the mean length of fish in B. [7] Calculate a 95% confidence interval for the difference in the mean lengths of fish from A and from B. [2] © UCLES 2014 9231/21/O/N/14

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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. 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 9231/21/O/N/14

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the October/November 2014 series 9231 FURTHER MATHEMATICS 9231/21 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 October/November 2014 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 – October/November 2014 9231 21 © 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 Cambridge International A Level – October/November 2014 9231 21 © 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 Cambridge International A Level – October/November 2014 9231 21 © Cambridge International Examinations 2014 Question Number Mark Scheme Details Part Mark Total 1 Use conservation of momentum, e.g.: 2mvA + mvB = 8mu – 3mu B1 Use restitution (must be consistent with prev. eqn.): vA – vB = –e (4u + 3u) B1 Solve for vA (or 3vA): vA = 1 (5 7 ) 3 e u − M1 [vB = 1 (5 14 ) 3 e u + ] Find lower limit on e for which vA < 0: 5 – 7e < 0, e > 5 7 or 0⋅714 M1 A1 5 [5] 2 Find speed component along barrier: V cos β = 4 cos α B1 Find speed component normal to barrier: V sin β = 0⋅4 × 4 sin α B1 Find β by eliminating α with V = 2: V2 = 22 = 1⋅62 sin2 α + 16 cos2 α M1 1 – sin2 α + 0⋅16 sin2 α = 0⋅25 sin2 α = 0.75 0.84 = 25 28 = 0⋅8929 or cos2 α = 3 28 = 0⋅1071 α = 1⋅24 rad or 70⋅9° M1 A1 5 [5] 3 Use conservation of energy: 1 2 mvB 2 = 1 2 mu2 + 2mga cosα B1 [vB 2 = u2 + 12 5 ag ] Use F = ma radially at A and B (B1 for either): 2 cos A mu R mg a α = − B1 2 cos B B mv R mg a α = + Equate RB to 10 RA: 2 2 cos 10 cos B mv mu mg mg a a α α −   + =     M1 A1 Eliminate vB 2: u2 + 4ag cos α = 10u2 – 11ag cosα [ 2 17 5 B ag v = ] u2 = ( 5 3 ag ) cosα = ag A.G. M1 A1 6

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 21 © Cambridge International Examinations 2014 Use conservation of energy for loss of contact: 1 2 mv2 = 1 2 mu2 + mga (cosα – cosθ) or 1 2 mvB 2 – mga (cosα + cosθ) B1 Use F = ma radially with R = 0: 2 mv a = mg cosθ B1 Eliminate v2 with u2 = ag to find cosθ: ga + 2ga (cosα – cosθ) = ga cosθ cosθ = 1 3 (2cosα + 1) = 11 15 M1 A1 4 [10] 4 (i) Relate FA and RA using F = µR: FA = 1 3 RA B1 Resolve horizontally: RB = FA [= 1 3 RA] B1 Resolve vertically (may not be needed here): S = mg – RA B1 EITHER: Take moments about C: RB 1 4 l sin α + FA 3 4 l sin α + mg 1 4 l cos α = RA 3 4 l cos α M1 A1 Combine, using tanα = 3 4 , to find RA: RA + 3RA + 4mg = 12RA RA = 1 2 mg A.G. M1 A1 OR: Take moments about A: RB l sin α + S 3 4 l cos α = mg 1 2 l cos α (M1 A1) Combine, using tanα = 3 4 , to find RA: RA + 3(mg – RA) = 2mg RA = 1 2 mg A.G. (M1 A1) OR: Take moments about B: FA l sin α + mg 1 2 l cos α = RA l cos α + S 1 4 l cos α (M1 A1)

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 21 © Cambridge International Examinations 2014 Combine, using tanα = 3 4 , to find RA: RA + 2mg = 4RA + mg – RA RA = 1 2 mg A.G. (M1 A1) OR: Take moments about D: RA 3 4 l cos α = RB l sin α + mg 1 4 l cos α (M1 A1) Combine, using tanα = 3 4 , to find RA: 3RA = RA + mg RA = 1 2 mg A.G. (M1 A1) 7 (ii) Use Hooke’s Law to relate extn. e and nat. length L: S = 1 2 mg = 2mge L , e = 1 4 L B1 Find length of CD: CD = 3 4 l sin α = 9 20 l B1 Combine to find L: L – 1 4 L = 9 20 l , L = 3 5 l M1 A1 4 [11]

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 21 © Cambridge International Examinations 2014 5 (i) Find extn. of either string by equating equil. tensions: a e a mg a mge A A 2 ) 3 ( 3 6 − = or a mge a e a mg B B 2 3 ) 3 ( 6 = − M1 A1 eA = 5 3a or eB = 5 12a A1 Find AO: A.G. AO = 3a + eA or 6a – eB = 3⋅6a B1 4 (ii) Apply Newton’s law at general point, e.g.: m 2 2 d d t x = a x e a mg A 2 ) 3 ( − − (lose A1 for each incorrect term) – a x e mg A 3 ) ( 6 + or m 2 2 d d y t = (3 ) 2 A mg a e y a − + − + a y e mg A 3 ) ( 6 − M1 A2 Simplify to give standard SHM eqn, e.g.: a gx t x 2 5 d d 2 2 − = A1 S.R.: B1 if no derivation (max 3/6) Find period T using SHM with ω = 5 2 g a       : T = 2 2 5 a g π       (A.E.F.) M1 A1 6 (iii) Find max speed using ωA with A = 0⋅2 a: vmax = 5 2 g a       × 0⋅2a = 10 ag       or √a (A.E.F.) M1 A1 2 [12] 6 Estimate population variance for combined sample: s2 = 2 2 50 60 X Y s s + = 1500 1391 or 0⋅9273 or 0⋅96302 M1 Calculate value of z (to 2 d.p., either sign): z = s 8.1 = 1⋅869 M1 A1 Find Φ(z) and set of possible values of α (to 1 d.p.): Φ(z) = 0⋅9692 [or 96⋅92%] (M1 A0 for α < 3⋅1 or α > 93⋅8) α < (or <) 6⋅2 (allow 6⋅1) M1 A1 5 [5]

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 21 © Cambridge International Examinations 2014 S.R. Assuming equal population variances: Explicit assumption (B1) Find pooled estimate of common variance s2 2 2 (49 59 ) 108 X Y s s + = 2777 108 or 25⋅71 or 5⋅0712 and calculate value of z (to 2 d.p.): z = 1 1 1.8 50 60 s − − + = 1⋅854 (M1 A1) Find Φ(z) and values of α (to 1 d.p.): Φ(z) = 0⋅9681 [or 96⋅81%] (M1 A0 for α < 3⋅2 or α > 93⋅6) α < (or <) 6⋅4 (M1 A1) 7 (i) State or find E(T): E(T) = 01 .0 1 = 100 B1 1 (ii) State or use eqn. for median m of T: [ ] m t e 0 01 .0 − − = 1 2 (A.E.F.) M1 Find value of m: e-0.01m = 1 2 , m = 100ln 2 = 69⋅3 M1 A1 Find P(T > 20): P(T > 20) = 1 – (1 – e–0.2) S.R. B1 for 0⋅181 = e–0.2 or 0⋅819 M1 A1 3 2 [6] 8 Find mean of sample data for use in Poisson distn.: λ = 100 225 = 2⋅25 B1 State (at least) null hypothesis (A.E.F.): H0: Poisson distn. fits data B1 Find expected values ! 100 r e r λ λ − (to 1 d.p.): 10⋅540 23⋅715 26⋅679 20⋅009 11⋅255 (ignore incorrect final value here for M1) 5⋅065 1⋅899 0⋅6105 0⋅2275 M1 A1 Combine last four cells so that exp. value > 5: Oi : . . . 16 14 4 Ei : . . . 20⋅009 11⋅255 7⋅802 *M1 Calculate value of χ2 (to 1 d.p.; A1 dep *M1): χ2 = 1⋅189 + 0⋅582 + 5⋅690 + 0⋅803 + 0⋅6695 + 1⋅853 = 10⋅8 (allow 10⋅7) M1 A1 State or use consistent tabular value (to 1 d.p.): χ4, 0.975 2 = 11⋅14 (if cells combined) B1 [χ7, 0.975 2 = 16⋅01, χ5, 0.975 2 = 12⋅83] Consistent conclusion (A.E.F., on two χ2 values): χ2 < 11⋅1 so Poisson distn. fits B1 9 [9]

Page 9 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 21 © Cambridge International Examinations 2014 9 (i) Calculate gradient b in ( ) y y b x x − = − : Sxy = 513 – 73 × 64 10 = 45⋅8 Sxx = 651 – 2 73 10 = 118⋅1 b = xx xy S S = 0⋅388 M1 A1 Find regression line of y on x: y = 64 10 + 0⋅388 (x – 73 10 ) M1 = 0⋅388 x + 3⋅57 or (458 4215) 1181 x + A1 4 (ii) Find correlation coefficient r: Syy = 2 64 462 10 − = 52⋅4 r = ( ) yy xx xy S S S = 0⋅582 M1 A1 2 (iii) Find y when x = 10: y = 7⋅45 B1 State valid comment on reliability, e.g.: Not reliable as value of r is small or reliable since x = 10 is in range or is near mean B1 2 (iv) Formulate condition for N: Require one-tail rN,1% < r [0⋅582] M1 Identify critical value near r using table: 15 or 16 ( on r) A1 State set of possible values of N: N > 16 A1 3 [11]

Page 10 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 21 © Cambridge International Examinations 2014 10 Find F(x) for 1 < x < 3: F(x) = 2 1 (x – 1) B1 Find G(y) from Y = X3 for 1 < x < 3: G(y) = P(Y < y) = P(X3 < y) = P(X < 3 1y ) = F( 3 1y ) (result may be stated) = 1 2 ( 3 1 y – 1); (1 < y < 27) M1 A1; B1 State G(y) for other values of y: 0 (y < 1) and 1 (y > 27) B1 Find g(y) for 1 < y < 27 ( on G(y)): g(y) = 6 3 2 − y or 3 2 6 1 y B1 Sketch g(y) for 1 < y < 27 B1 with g(y) = 0 on either side of this interval B1 Find mean of Y: E(Y) = ∫ y g(y) dy = ∫ ( 6 3 1y ) dy (no need to find median = 8) = 27 1 8 3 4        y = 10 M1 A1 Find probability Y lies between median and mean: G(10) – G(8) or G(10) – 1 2  = ( ) 1 1 3 3 1 2 10 8 − (2 s.f. sufficient) or ( ) 1 3 1 1 10 1 2 2 − − = 0⋅077 [2] M1 A1 5 3 4 [12]

Page 11 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 21 © Cambridge International Examinations 2014 11a (i) EITHER: State or find (by ⊥ axes) MI of X about AB: IX = 2 1 mr2 M1 A1 State or find MI of Y (or Z) about AB: IY = 2 1 mr2 + mr2 = 2 3 2 mr M1 A1 State or find (by ⊥ axes) MI of W about AB: IW = 2 1 3mR2 = 2 1 3mr2 (1 + 2 3 √3)2 = 1 2 (7 + 4√3) mr2 M1 A1 Find MI of object about AB: I = ( 1 2 + 2× 2 3 + 2 7 + 2√3) mr2 = (7 + 2√3) mr2 A.G. M1 A1 OR: State or find MI of X, Y or Z about centre O: IX = mr2 + m 2 2 3 r       = 2 7 3 mr (M1 A1) State or find MI of W about O: IW = 3mR2 = 3mr2(1 + 2 3 √3)2 = (7 + 4√3) mr2 (M1 A1) Find MI of object about O: IO = 3IX + IW = (14 + 4√3) mr2 (M1 A1) Find (by ⊥ axes) MI of object about AB: I = 2 1 IO = (7 + 2√3) mr2 A.G. (M1 A1) 8 (ii) Find new MI of object plus particle about AB: I′ = I + 9mR2 = I + 3 (7 + 4√3) mr2 = 14 (2 + √3) mr2 M1 A1 Find eqn for angular speed ω using energy: 2 1 I′ ω2 = 9mg × R sin 60° M1 A1 Substitute and simplify to find ω: ω2 = ' 3 9 I mgR (AEF) ω = 9 14 g r or 0⋅802 g r M1 A1 6 [14]

Page 12 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 21 © Cambridge International Examinations 2014 11b Estimate population variance using A’s sample: sA 2 = 2 106 1422.34 8 7 −       (allow use of biased: σ A,8 2 = 2⋅23 or 1⋅4932) = 446 175 or 2⋅549 or 1⋅5962 M1 A1 Find confidence interval: 106 8 ± t 2 8 Xs       M1 State or use correct tabular value of t: t7,0.975 = 2⋅36 [5] A1 Evaluate C.I. correct to 1 d.p.: 13⋅25 ± 1⋅335 or [11⋅9, 14⋅6] A1 5 State suitable assumptions (A.E.F.): Distribution of B is Normal with same population variance B1 State hypotheses (B0 for a …) e.g.: H0: µA = µB , H1: µA > µB B1 Estimate (or imply) B’s population variance: sB 2 = ( ) 5 53 . 971 6 9. 75 2 − (allow use of biased: σ B,6 2 = 1⋅899 or 1⋅3782) = 2⋅279 or 1⋅5102 and find pooled estimate of common variance s2: s2 = ( ) 2 2 12 7 5 A B s s + = (17.84 11.395) 12 + = 2⋅436 or 1⋅5612 M1 Calculate value of t (to 2 d.p., either sign): t = ( ) 1 1 13.25 12.65 ( ) 8 6 s − − − + = 0.6 0.8430 = 0⋅712 M1 A1 State or use correct tabular t-value (to 2 d.p.): t12,0.95 = 1⋅782 B1 (or compare 0⋅6 with 1⋅782 s√(8–1 + 6–1) = 1⋅50) Consistent conclusion (AEF, on two t values): [Accept H0]; mean lengths are the same B1 7 Find confidence interval for the difference: 13⋅25 – 12⋅65 ± t s√(8–1 + 6–1) M1 Evaluate C.I. with t12,0.975 = 2⋅179, to 2 d.p.: 0⋅6 ± 1⋅84 or [– 1⋅24, 2⋅44] A1 2 [14]