Cambridge A Level Mathematics - Further 9231 — 2016 Oct/Nov Paper 2 · Variant 2

*2088783440* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/22 Paper 2 October/November 2016 3 hours Additional Materials: List of Formulae (MF10) 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 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 6 printed pages, 2 blank pages and 1 insert. JC16 11_9231_22/FP © UCLES 2016 [Turn over

2 1 A B M C 4 m 2 m The point C is on the fixed line l. Points A and B on l are such that AC = 4 m and CB = 2 m, with C between A and B. The point M is the mid-point of AB (see diagram). A particle P of mass m oscillates between A and B in simple harmonic motion. When P is at C, its speed is 4 m s−1. Find (i) the magnitude of the maximum acceleration of P, [3] (ii) the number of complete oscillations made by P in one minute, [2] (iii) the time that P takes to travel directly from A to C. [3] 2 P !Å 75Å 60Å D E u v 1 4u Two smooth vertical walls each with their base on a smooth horizontal surface intersect at an angle of 60Å. A small smooth sphere P is moving on the horizontal surface with speed u when it collides with the first vertical wall at the point D. The angle between the direction of motion of P and the wall is !Å before the collision and 75Å after the collision. The speed of P after this collision is v and the coefficient of restitution between P and the first wall is e. Sphere P then collides with the second vertical wall at the point E. The speed of P after this second collision is 1 4u (see diagram). The coefficient of restitution between P and the second wall is 3 4. (i) By considering the collision at E, show that v = ï2 5 u. [5] (ii) Find the value of ! and the value of e. [5] © UCLES 2016 9231/22/O/N/16

3 3 1 a a 8a 7a A B C D P Q The end P of a uniform rod PQ, of weight kW and length 8a, is rigidly attached to a point on the surface of a uniform sphere with centre C, weight W and radius a. The end Q is rigidly attached to a point on the surface of an identical sphere with centre D. The points C, P, Q and D are in a straight line. The object consisting of the rod and two spheres rests with one sphere in contact with a rough horizontal surface, at the point A, and the other sphere in contact with a smooth vertical wall, at the point B. The angle between CD and the horizontal is 1. The point B is at a height of 7a above the base of the wall (see diagram). The points A, B, C, D, P and Q are all in the same vertical plane. (i) Show that sin 1 = 3 5. [1] The object is in limiting equilibrium and the coefficient of friction at A is -. (ii) Find the numerical value of -. [7] (iii) Given that the resultant force on the object at A is W65, show that k = 5. [3] 4 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle is held vertically above O with the string taut and then projected horizontally with speed / 13 3 ag ! . It begins to move in a vertical circle with centre O. When P is at its lowest point, it collides with a stationary particle of mass ,m. The two particles coalesce. (i) Show that the speed of the combined particle immediately after the impact is 5 , + 1 / 1 3ag ! . [4] In the subsequent motion, the string becomes slack when the combined particle is at a height of 1 3a above the level of O. (ii) Find the value of ,. [6] (iii) Find, in terms of m and g, the instantaneous change in the tension in the string as a result of the collision. [4] © UCLES 2016 9231/22/O/N/16 [Turn over

4 5 The distance, X km, completed by a new car before any mechanical fault occurs has distribution function F given by Fx = D 1 −e−ax x ≥0, 0 otherwise, where a is a positive constant. The mean value of X is 10 000. Find (i) the value of a, [1] (ii) the probability that a new car completes less than 15 000 km before any mechanical fault occurs. [2] The probability that a new car completes at least d km before any mechanical fault occurs is 0.75. (iii) Find the value of d. [3] 6 A random sample of 8 observations of a normal random variable X has mean x, where x = 6.246 and Σx −x2 = 0.784. Test, at the 5% significance level, whether the population mean of X is less than 6.44. [7] 7 The random variable X has probability density function f given by fx = T 1 6x 2 ≤x ≤4, 0 otherwise. (i) Find the distribution function of X. [3] The random variable Y is defined by Y = X3. Find (ii) the probability density function of Y, [2] (iii) the value of k for which PY ≥k = 7 12. [3] 8 The amounts spent on the weekly food shopping by families in the big city P and the small town Q are to be compared. The amounts spent, in dollars, in P and Q are denoted by x and y respectively. For a random sample of 60 families in P and a random sample of 50 families in Q, the amounts are summarised as follows. Σx = 9600 Σx2 = 1 560 000 Σy = 7200 Σy2 = 1 052 500 Assuming a common population variance, find (i) a pooled estimate for the population variance, [4] (ii) a 95% confidence interval for the difference in the population means in P and Q. [5] © UCLES 2016 9231/22/O/N/16

5 9 The number of visitors arriving at an art exhibition is recorded for each 10-minute period of time during the ten hours that it is open on a particular day. The results are as follows. Number of visitors in a 10-minute period 0 1 2 3 4 5 6 7 8 ≥9 Number of 10-minute periods 2 2 12 8 11 13 4 7 1 0 (i) Calculate the mean and variance for this sample and explain whether your answers support a suggestion that a Poisson distribution might be a suitable model for the number of visitors in a 10-minute period. [3] (ii) Use an appropriate Poisson distribution to find the two expected frequencies missing from the following table. [2] Number of visitors in a 10-minute period 0 1 2 3 4 5 6 7 8 ≥9 Expected number of 10-minute periods 1.10 8.79 11.72 9.38 6.25 3.57 1.79 1.28 (iii) Test, at the 10% significance level, the goodness of fit of this Poisson distribution to the data. [8] [Question 10 is printed on the next page.] © UCLES 2016 9231/22/O/N/16 [Turn over

6 10 Answer only one of the following two alternatives. EITHER O A B C a a 2a 3a A thin uniform rod AB has mass 2m and length 3a. Two identical uniform discs each have mass 1 2m and radius a. The centre of one of the discs is rigidly attached to the end A of the rod and the centre of the other disc is rigidly attached to the end B of the rod. The plane of each disc is perpendicular to the rod AB. A second thin uniform rod OC has mass m and length 2a. The end C of this rod is rigidly attached to the mid-point of AB, with OC perpendicular to AB (see diagram). The object consisting of the two discs and two rods is free to rotate about a horizontal axis l, through O, which is perpendicular to both rods. (i) Show that the moment of inertia of one of the discs about l is 13 4 ma2. [3] (ii) Show that the moment of inertia of the object about l is 52 3 ma2. [4] When the object is suspended from O and is hanging in equilibrium, the point C is given a speed of 2ag in the direction parallel to AB. In the subsequent motion, the angle through which OC has turned before the object comes to instantaneous rest is 1. (iii) Show that cos 1 = 8 21. [7] OR For a random sample, A, of 5 pairs of values of x and y, the equations of the regression lines of y on x and x on y are respectively y = 4.5 + 0.3x and x = 3y −13. Four of the five pairs of data are given in the following table. x 1 5 7 9 y 5 6 7 7 Find (i) the fifth pair of values of x and y, [5] (ii) the value of the product moment correlation coefficient. [2] A second random sample, B, of 5 pairs of values of x and y is summarised as follows. Σx = 20 Σx2 = 100 Σy = 17 Σy2 = 69 Σxy = 75 The two samples, A and B, are combined to form a single random sample of size 10. (iii) Use this combined sample to test, at the 5% significance level, whether the population product moment correlation coefficient is different from zero. [7] © UCLES 2016 9231/22/O/N/16

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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. 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 9231/22/O/N/16

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 11 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/22 Paper 2 October/November 2016 MARK SCHEME Maximum Mark: 100 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 A Level – October/November 2016 9231 22 © 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 A Level – October/November 2016 9231 22 © 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 A Level – October/November 2016 9231 22 © UCLES 2016 1 (i) Find ω2 from v2 = ω2 (a2 – x2), or ω : Find max. acceln. from d2x/dt2 = – ω2x, √ on ω2: 42 = ω2 (32 – 12), ω2 = 2 2 × 3 = 6 [m s–2] (allow –6) M1 A1 A1 [3] (ii) Find no. of oscillations in 60 s from T = 2π/ω : and hence no. of complete oscillations (allow M1 A0 for 60 / (π/ω) [= 27]) 60 / (2π/ω) [= 60 / 4⋅443 = 13⋅5] 13 M1 A1 [2] (iii) Find time from A to C, e.g.: ω –1 sin–1 (1) + ω –1 sin–1 ⅓ or ¼T + ω –1 sin–1 ⅓ [= 1⋅111 + 0⋅240] or ω –1 cos–1 (–⅓) or ½T – ω –1 cos–1 ⅓ [= 2⋅221 – 0⋅870] = 1⋅91 /ω ; = 1⋅35 [s] M1 A1; A1 [3] 2 (i) EITHER: Find comps. of speed after colln. at E: Relate v to u, or v2 to u2: OR: Relate angle β after colln. to u, v: Find tan β, or β: Eliminate β from either eqn. above, e.g.: v cos 45° // to wall and ¾ v sin 45° ⊥ to wall √{(v / √2)2 + (¾ v / √2)2} = ¼ u (5/4√2) v = ¼ u v = (√2/5) u A.G ¼ u cos β = v cos 45° and ¼ u sin β = ¾ v sin 45° tan β = ¾ or β = 36⋅9° ¼ u × (4/5) = v / √2 v = (√2/5) u A.G. M1 A1 M1 A1 A1 (M1 A1) (A1) (M1) (A1) [5] (ii) Relate comps. of speed // to wall after colln. at D: Find cos α: Find α: Relate comps. of speed ⊥ to wall after colln. at D: Find e: v cos 75° = u cos α cos α = (√2/5) cos 75° [= 0⋅0732] α = 85⋅8° or 1⋅50 rads v sin 75° = eu sin α e = (√2/5) sin 75° / sin α or = tan 75° / tan α = 0⋅274 M1 A1 A1 M1 A1 [5]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 22 © UCLES 2016 3 (i) Verify sinθ from triangle CDE where E is level with C and vertically below D: sin θ = 6a / 10a = 3/5 A.G. B1 [1] (ii) Resolve forces on object vertically: (may be needed in part (iii) only) Take moments about B : or D: or A : or C : or centre of rod: Find FA using RB = FA , sin θ = 3/5, cos θ = 4/5 as necessary: B: D: A or C: Centre of rod: Find µ : RA = (k + 2) W FA 7a + Wa + kWa(1 + 5 cos θ ) + (W – RA) a(1 + 10 cos θ ) = 0 FA 7a + kW 5a cos θ + (W – RA) 10a cos θ = 0 RB 7a = kW 5a cos θ + W 10a cos θ FA a + RB 6a = W 10a cos θ + kW 5a cos θ FA (5a sin θ + a) + RB 5a sin θ + W 5a cos θ = RA 5a cos θ + W 5a cos θ 7FA = 9RA – 5kW – 10W = 4kW + 8W 7FA = 8RA – 4kW – 8W = 4kW + 8W 7FA = 4kW + 8W 7FA = 4RA [so RA is not reqd. here] µ = FA / RA = 4/7 or 0⋅ M1 A1 M1 A1 A1 M1 A1 [7] (iii) Equate resultant force at A to W√(65): Solve to verify value of k: (k + 2)2 + (4/7)2 (k + 2)2 = 65 k2 + 4k – 45 = (k – 5)(k + 9) = 0 or (k + 2)2 = 49 so k = 5 A.G. M1 A1 A1 [3]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 22 © UCLES 2016 4 (i) Find v1 2 at lowest point from consvn. of energy: Verify new v2 from consvn. of momentum: ½mv1 2 = ½mu2 + 2mga ½mv1 2 = ½mu2 + 2mga Mv2 = mv1 with M = (λ + 1) m v2 = v1 / (λ + 1) = {5/(λ + 1)}√(⅓ ag) A.G. M1 A1 M1 A1 [4] (ii) Use F = ma radially at slack point with T = 0, e.g.: (θ is angle of string with upward vertical) Find v3 2 at slack point from consvn of energy: SR: Lose max 1 mark if mass is m in either eqn. Eliminate v3 2 [= ag/3] using cos θ = ⅓: Substitute for v2 to find λ: Mv3 2/a = Mg cos θ ½ Mv3 2 = ½ Mv2 2 – Mg (4a/3 ) ag/3 = v2 2 – 8ag/3 3ag = {25/(λ + 1)2} ag/3 (λ + 1)2 = 25/9, λ = ⅔ [rejecting –8/3] B1 M1 A1 M1 M1 A1 [6] (iii) Use F = ma radially just before collision: Use F = ma radially just after collision: Find change in tension (either sign, AEF): T1 = mv1 2/a + mg [= (25/3 + 1)mg = 28mg/3] T2 = Mv2 2/a + Mg [= (3 + 1)(5m/3)g = 20mg/3] (25mg/3){λ /(λ + 1)} – λmg = 8mg/3 or 2⋅67mg B1 B1 A1 A1 [4] 5 (i) Find a from mean: a = 1/10 000 or 10–4 B1 [1] (ii) Find P(X < 15 000): 1 – e–15 000a = 1 – e–1.5 = 0⋅777 M1 A1 [2] (iii) Formulate condition for d: (M0 for 1 – e–ad = 0⋅75, giving d = 13 900)) Rearrange and take logs to give d: 1 – (1 – e–ad ) = 0⋅75 d = – (ln 0⋅75) / a = 2877 or 2880 M1 A1 A1 [3]

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 22 © UCLES 2016 6 Estimate population variance: (allow biased here: 0⋅098 or 0⋅31302) State hypotheses (AEF; B0 for x ): Calculate value of t (either sign; to 3 s.f.): State or use correct tabular t-value (to 3 s.f.): (or can compare x with 6⋅4 – 0⋅224 = 6⋅176 ) State or imply valid method for conclusion e.g.: Conclusion (AEF, requires both values correct): s2 = 0⋅784 / 7 = 0⋅112 or 14/125 or 0⋅33472 H0: µ = 6⋅44, H1: µ < 6⋅44 t = (6⋅44 – x )/(s/√8) = 1⋅64 t7, 0.95 = 1⋅89[5] Accept H0 if t < tabular value 1⋅64 < 1⋅89 so popln. mean not less than 6⋅44 M1 B1 M1 A1 B1 M1 A1 [7] 7 (i) Find or state distribution function F(x) for 2 ⩽ x ⩽ 4: Use F(2) = 0 or F(4) = 1 to find F(x): F(x) = ∫ f(x) dx = x2/12 + c F(x) = x2/12 – ⅓ [(2 ⩽ x ⩽ 4)], 0 (x < 2), 1 (x > 4) M1 A1 A1 [3] (ii) Find or state G(y) from Y = X 3 for 2 ⩽ x ⩽ 4: (allow < or ⩽ throughout) (A0 if G(y) incorrect) Find g(y) by differentiation: G(y) = P(Y < y) = P(X 3 < y) = P(X < y1/3) = F(y1/3) = y2/3/12 – ⅓ g(y) = (1 / 18) y –1/3 [for 8 ⩽ y ⩽ 64, 0 otherwise] M1 A1 [2] (iii) Formulate condition for k: (M0 for 7/12 = G(k)) Find k: (AEF) 7/12 = 1 – G(k) = 1 – k2/3/12 + ⅓ k2/3 = 9, k = 27 M1 A1 A1 [3]

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 22 © UCLES 2016 8 (i) EITHER: Estimate P’s popln. variance (to 3 d.p.): (allow biased here: 400) Estimate Q’s popln. variance (to 3 d.p.): (allow biased here: 314) Find pooled estimate of common variance: OR: Find pooled estimate of common variance: sP 2 = (1 560 000 – 96002/60) / 59 [= 406⋅78] sQ 2 = (1 052 500 – 72002/50) / 49 [= 320⋅41] s2 = (59 sP 2 + 49 sQ 2) / 108 = 367⋅6 or 368 or 9925/27s2 = (1 560 000 – 96002/60 s2 = (1 560 000 – 96002/60 + 1 052 500 – 72002/50) / 108 = 367⋅6 or 368 or 9925/27 M1 M1 M1 A1 (M3 A1) [4] (ii) Find confidence interval for the difference: Use appropriate tabular value (to 2 d.p.): Evaluate confidence interval (AEF, to 1 d.p.): SR Using combined variance sP 2/60 + sQ 2/50 = 13⋅19: (i) M1, M1 as above; then M0 A0 (max 2/4) (ii) M1 A0 for 9600/60 – 7200/50 ± z s A1 for tabular value as above M1 A0 for evaluating interval 16 ± 7⋅1 (or 7⋅2) (max 3/5) 9600/60 – 7200/50 [= 160 – 144] ± z s√(60–1 + 50–1) z0.975 = 1⋅96 or t120, 0.975 = 1⋅98 16 ± 7⋅2 or [8⋅8, 23⋅2] or 16 ± 7⋅3 or [8⋅7, 23⋅3] M1 A1 A1 M1 A1 [5]

Page 9 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 22 © UCLES 2016 9 (i) Find mean and variance of sample data.: State valid reason why Poisson distn. suitable (AEF): (allow unsuitable since 4 ≠ 3⋅57) x = 240/60 = 4 σ2 = 1174/60 – 42 = 3⋅57 4 ≈ 3⋅57 (no √ onx, σ2) B1 B1 B1 [3] (ii) Find expected values 60λre-λ / r! with λ = 4: 4⋅40; 11⋅7[2] (to 1 d.p.) B1; B1 [2] (iii) State (at least) null hypothesis (AEF): Combine cells so that all exp. value ⩾ 5: Calculate value of χ2 (to 2 d.p.; A1 dep *M1): State or use consistent tabular value (to 2 d.p.): [or fewer or no cells combined: State or imply valid method for conclusion e.g.: Conclusion (AEF, requires both values correct): H0: [Poisson] distribution fits data Oi : 4 12 … 6 8 Ei : 5⋅50 8⋅79 … 6⋅25 6⋅64 χ2 = 0⋅409 + 1⋅172 + 1⋅181 + 0⋅044 + 1⋅397 + 0⋅81 + 0⋅279 = 5⋅29 7 cells: χ5, 0.9 2 = 9⋅236 8 cells: χ6, 0.9 2 = 10⋅64 9 cells: χ7, 0.9 2 = 12⋅02 10 cells: χ8, 0.9 2 = 13⋅36 ] Accept H0 if χ2 < tabular value 5⋅29 < 9⋅24 so distn. fits B1 *M1 A1 M1 DA1 B1√ M1 A1 [8]

Page 10 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 22 © UCLES 2016 10 (a) (i) Verify MI of either disc about axis l at O: Idisc = ½ (¼ma2) + ½m {(3a/2)2 + (2a)2} = ma2/8 + 25ma2/8 = 13ma2 / 4 A.G. M1 A1 A1 [3] (ii) Find or state MI of rod OC about l: Find MI of rod AB about l: Verify MI of object about l: IOC = (4/3) ma2 IAB = ⅓ 2m (3a/2)2 + 2m (2a)2 = (19/2) ma2 I = (2 × 13/4 + 4/3 + 19/2) ma2 = (52/3) ma2 A.G. B1 M1 A1 A1 [4] (iii) Find and use initial angular speed: Find initial rotational KE: Find gain in P.E. at instantaneous rest: Verify cos θ by equating KE and PE: ω0 = √(2ag) / 2a or √(g / 2a) ½ I ω0 2 = ½ (52/3) ma2 × (g/2a) = (13/3) mga (3mg × 2a + mga) (1 – cos θ ) or (4mg × 7a/4) (1 – cos θ ) = 7mga (1 – cos θ ) 1 – cos θ = (13/3) mga / 7mga = 13/21, cos θ = 8/21 A.G. B1 M1 A1 M1 A1 M1 A1 [7]

Page 11 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 22 © UCLES 2016 (b) (i) EITHER: Find x , y by solving simultaneous eqns: Hence find missing values x5, y5: OR: Formulate simultaneous eqns for x5, y5: (AEF, M1 for either) Hence find missing values x5, y5: x = 5; y = 6 22 + x5 = 5 × 5, x5 = 3 25 + y5 = 5 × 6, y5 = 5 25 + y5 = 5 × 4⋅5 + 0⋅3 (22 + x5) 22 + x5 = 3 (25 + y5) – 5 × 13 y5 = 0⋅3 x5 + 4⋅1, x5 = 3 y5 – 12 x5 = 3; y5 = 5 M1, A1; A1 A1 A1 (M1 A1) (A1) (A1 A1) [5] (ii) Find correlation coefficient r: (A0 for – 0⋅949 or ± 0⋅949) r = √(0⋅3 × 3) or 12/√(40 × 4) = 0⋅949 M1 A1 [2] (iii) Find corresponding summations for combined data: (B1 needs all 5 correct) Find correlation coefficient r′ : (M0 if based on B only) State both hypotheses (B0 for r …): State or use correct tabular two-tail r-value: State or imply valid method for conclusion e.g.: Correct conclusion (AEF, dep *A1, *B1): ∑ x = 25 + 20 = 45 ∑ x2 = 165 + 100 = 265 ∑ y = 30 + 17 = 47 ∑ y2 = 184 + 69 = 253 ∑ xy = 162 + 75 = 237 Sxy = 237 – 45 × 47/10 = 25⋅5 Sxy = 265 – 452/10 = 62⋅5 Syy = 253 – 472/10 = 32⋅1 r′ = Sxy / √(Sxx Syy) or √{( Sxy / Sxy)( Sxy / Syy)} = 25⋅5 / 44⋅79 or √ (0⋅408 × 0⋅7944) = 0⋅569 H0: ρ = 0, H1: ρ ≠ 0 r10, 5% = 0⋅632 Accept H0 if |r′ | < tab. value (AEF) Popln. pmcc not different from 0 B1 M1 *A1 B1 *B1 M1 DA1 [7]