GCE Mathematics A H240/03: Pure Mathematics and Mechanics Advanced GCE Mark Scheme for Autumn 2021

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GCE Mathematics A H240/03: Pure Mathematics and Mechanics Advanced GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Mathematics A H240/03: Pure Mathematics and Mecha... nics Advanced GCE Mark Scheme for Autumn 2021Oxford Cambridge and RSA Examinations OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. © OCR 2021H240/03 Mark Scheme October 2021 2 Text Instructions 1. Annotations and abbreviations Annotation in RM assessor Meaning ✓and  BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread BP Blank Page Seen Highlighting Other abbreviations in mark scheme Meaning dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working AG Answer given awrt Anything which rounds to BC By Calculator DR This question included the instruction: In this question you must show detailed reasoning.H240/03 Mark Scheme October 2021 3 2. Subject-specific Marking Instructions for A Level Mathematics A a Annotations must be used during your marking. For a response awarded zero (or full) marks a single appropriate annotation (cross, tick, M0 or ^) is sufficient, but not required. For responses that are not awarded either 0 or full marks, you must make it clear how you have arrived at the mark you have awarded and all responses must have enough annotation for a reviewer to decide if the mark awarded is correct without having to mark it independently. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. Award NR (No Response) - if there is nothing written at all in the answer space and no attempt elsewhere in the script - OR if there is a comment which does not in any way relate to the question (e.g. ‘can’t do’, ‘don’t know’) - OR if there is a mark (e.g. a dash, a question mark, a picture) which isn’t an attempt at the question. Note: Award 0 marks only for an attempt that earns no credit (including copying out the question). If a candidate uses the answer space for one question to answer another, for example using the space for 8(b) to answer 8(a), then give benefit of doubt unless it is ambiguous for which part it is intended. b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. If you are in any doubt whatsoever you should contact your Team Leader.H240/03 Mark Scheme October 2021 4 c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually 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. In some cases the nature of the errors allowed for the award of an M mark may be specified. A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words “Determine” or “Show that”, or some other indication that the method must be given explicitly. 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). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep*’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. e The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only – differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. If this is not the case please, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.H240/03 Mark Scheme October 2021 5 f We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so. • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value. • When a value is not given in the paper accept any answer that agrees with the correct value to 3 s.f. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range. NB for Specification B (MEI) the rubric is not specific about the level of accuracy required, so this statement reads “2 s.f”. Follow through should be used so that only one mark in any question is lost for each distinct accuracy error. Candidates using a value of 9.80, 9.81 or 10 for g should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric. g Rules for replaced work and multiple attempts: • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others. • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete. • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately. h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors. If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error. i If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold “In this question you must show detailed reasoning”, or the command words “Show” or “Determine”. Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader. j If in any case the scheme operates with considerable unfairness consult your Team Leader.H240/03 Mark Scheme October 2021 6 Question Answer Marks AO Guidance 1 B1 B1 B1 1.1 1.1 2.2a 2 y x = drawn correctly – must pass through the origin and (roughly) symmetrically about the y-axis x y + = 2 drawn correctly – must have positive y-intercept and negative gradient Correct identification of region (dependent on both previous B marks) – ignore shading if R correctly labelled Line need not intersect the x-axis and intercepts need not be labelled [3] Condone dashed line/curve for full marksH240/03 Mark Scheme October 2021 7 Question Answer Marks AO Guidance 2 (a) (p h h h h 2 = + + − − + − )(4 4 2 4 4 cos60 )2 2 ( ) ( )( ) M1 1.1 Correct application of cosine rule = + + + − + − − (16 8 16 8 16 h h h h h 2 2 2 ) ( ) ( ) 2 2 p h = + 16 3 A1 2.2a AG – at least one line of intermediate working (must have p2 ...) Any errors or missing brackets then A0 [2] 2 (b) (16 3 4 1 ... + = + h2)12 ( )12 B1 1.1 For reference: 1 2 3 2 4 1 16     + h   or for 1 1 16 (1 ...) 2 2 (1 1 ... + = + + kh kh 2 2 )12 12 M1 1.1 Correct first two terms for their k k 1 ( 2 )2 1 1 2 2 ... 2! kh       −    + A1ft 1.1 Correct third term following through their k ( p h h = + − + )4 ... 8 512 3 9 2 4 A1 1.1   = = − 8 512 3 9 , (oe) SC if candidates assume that p h h = + + 4   2 4 and then substitute into p h 2 2 = + 16 3 to find  and then B1 for correct  and B1 for correct (so 2/4 max.) [4]H240/03 Mark Scheme October 2021 8 Question Answer Marks AO Guidance 3 2 2 2 + = d r B1 1.1 Or for a d ar 2 2 2 12 2 + = d r B1 1.1 Or for a d ar 12 2 1 6 1 + = + d d ( )2 or 2 12 2 1 + = + d d ( )2 M1* 1.1 S dependent on one etting up an equation in B markd or r only – 2 12( 1) 2 r r2 2 d d d − =  = 4 0 ... M1dep* 1.1 Solving their two-term quadratic equation in d (or three-term quadratic in r) 2 r r 6 5 0 0 ... ( 5)( 1) r r r d = 4 and as the common difference is positive the progression is an increasing sequence A1 2.4 Correct value for d and link to increasing sequence – must either say that d is positive (oe) or state at least the correct first four terms and comment that they are increasing Condone no mention of d  0 [5] Alternative method 3 2 2 12 2 2 u d u d B1 or for 3 1 u u 2 1 2 2 2 u d u B1 2 12 2 2 2 2 2 d d d M1* Setting up an equation in d only – dependent on one B mark 2 d d d − =  = 4 0 ... M1dep* Solving their two-term quadratic equation in d d = 4 and as the common difference is positive the progression is an increasing sequence A1 As aboveH240/03 Mark Scheme October 2021 9 Question Answer Marks AO Guidance 4 (a) B1 B1 [2] 1.1 1.1 y x = −1 drawn correctly – must touch (but not intersect) the positive x-axis 1 y kx = − drawn correctly – must not intersect axes Intercepts with axes need not be labelled 4 (b) The graphs in (a) intersect at only point (for any negative values of k) and therefore x x x k 1 1 k x − =  − = has exactly one real root B1 2.4 Dependent on both marks in (a) – must mention that the solution of the equation x x k 1 corresponds to where the two graphs in (a) intersect (so just stating that the graphs in (a) intersect at only one point is B0) [1] 4 (c) x x x x − = −  − = − 1 6 1 6 ( ) 2 x x − − = 6 0 M1 3.1a Uses graph and sets up quadratic (oe) – allow if x x 2 6 0stated as well (but M0 if this is the only quadratic (oe) considered) Or setting up a fourterm quartic from 2 2 ( 1) x 36x x = −2 A1 2.2a BC x = −2 only (www) [2] SC If no marks awarded, then B1 for x = −2 only and then B1 for explicitly showing that 2 2 1 2(3) 6H240/03 Mark Scheme October 2021 10 Question Answer Marks AO Guidance 5 (a) R =13 B1 1.1 B0 for 169 or for 13 cos 12 5 tan sin 5 12 R R   =    = =  M1 1.1 M1 for tan = k where k =   12 5 5 12 , or for sin cos5 12 R ,  = 0.3948 A1 1.1 A0 if in degrees (must be stated to exactly 4 significant figures at some point) For reference: 0.3947911197… [3] 5 (b) 13cos 0.3948 3 0.3948 arccos ( ) 3 13 t t    − =   = +     M1 1.1 Sets R t cos( −) (with their R and ) equal to either 3 or −3, and attempt to solve (with correct order of operations) M1 only if in degrees 1.73 A1 1.1 awrt 1.73 1.7327192… 2.20 A1 1.1 awrt 2.20 (condone 2.2) – ignore other values that are greater than 2.20 2.1984556… [3]H240/03 Mark Scheme October 2021 11 Question Answer Marks AO Guidance 6 (a) Considers both f 0.5 ( ) and f 0.6 ( ) where f 6arcsin 2 1 (x x x ) =  − −  ( ) 2 M1 1.1 With at least one should be given to at least 2 sf (rot) correct value – values Allow degrees for M1 only: f(0.6) 68.8617... f 0.5 0.25 0, f 0.6 0.8481... 0 ( ) = −  =  ( ) change of sign indicates that the root lies between 0.5 and 0.6 A1 2.4 Correct values together with explanation in words (change of sign) and conclusion [2] 6 (b) 6arcsin 2 1 0 arcsin 2 1 ( ) 2 2 ( ) 1 6 x x x x − − =  − = So 2 1 sin 1 2 6 x x   − =     M1 1.1 Correct order of operations to get 2 1 sin x kx − = ( 2 ) k  0 1 1 1 2 sin 2 2 6 x x   = +     A1 2.2a p q = = 1 1 2 2 , and r = 16 (oe) [2] 6 (c) ( ) ( ) ( ) ( ) ( ) 0 1 2 3 4 0.5 0.5208273057... 0.5225973903... 0.5227511445.... 0.5227645245... x x x x x = = = = = M1 1.1 Uses their iterative formula with correct starting value to produce terms up to at least x2 to at least 4 significant figures Allow degrees for M1 only: For reference: 1 2 3 0.5003636... 0.5003641... 0.5003641... x x x = = = 0.5228 A1 1.1 Must be stated to exactly 4 significant figures [2]H240/03 Mark Scheme October 2021 12 Question Answer Marks AO Guidance 7 (a) d 2 3 d 3 y y x x   − =     − M1 A1 2.5 1.1 Correct left-hand side (including equals sign) and right-hand side must be of the form f( ) g( ) y x or g( ) f( ) x y cao (oe) [2] 7 (b) 1 1 d 3 d 2 3 y x y x = − −      M1* 1.1a Separation of variables the M mark in (a) – dependent on With an indication of integration ln 2 3ln 3 ( y x c − = − + ) ( )( ) A1ft 1.1 Follow from (a)through their differential eq. Condone no constant (4, 3 0, 2 3 )  = − = − c y x ( )3 M1dep* 1.1 Attempt to find c and eliminate logs y x = − + ( 3 2 )3 A1 2.2a oe (but must be of the form y = f(x)) [4] 7 (c) Translation B1 1.1 B0 if another type of transformation stated or if shift/move etc. used 3 2       B1ft 1.1 Follow through their y x p q = − + ( )3 - B0 if this transformation is given as a stretch/rotation/reflection/enlargement etc. (but condone no transformation stated or shift/move etc.) – need not be given as a vector p q , 0 [2]H240/03 Mark Scheme October 2021 13 Question Answer Marks AO Guidance 8 (a) M1* 2.1 Differentiates y with respect to x – answer of the form   e e − − 2 2 x x x   0 y x  = − e 1 2 −2x ( ) A1 1.1 y x  = − + e 4 4 −2x ( ) A1ft 1.1 Follow through their first derivative M1dep* 3.1a Solves y = 0 (or attempts to verify y = 0by substituting x = 1) or considers sign change either side of y y = 0 at x = 1 and y(0.5) 2e 0 = −  −1 , y(1.5 2e 0 ) =  −3 (so change of sign indicates a point of inflection at x = 1) A1 [5] 2.2a Conclusion not required for this mark 8 (b) 2 2 2 1 1 e d e e d 2 2 x x x x x x x − − −   = − + M1* 2.1 Integration by parts     x x e e d − − 2 2 x x  – of the form Where   , 2, = 12 2 2 2 1 1 e d e e 2 4 x x x x x x − − −  = − − A1 1.1 1 1 2 2 2 0 0 2 2 1 1 e d e e 2 4 1 1 1 e e 0 2 4 4 x x x x x x − − − − −   = − −         = − − − −          M1dep* 1.1 Use of correct limits in their fully integrated expression – need not be simplified (or equivalent) 1 3 2 e 4 4 − − A1 1.1 Allow unsimplified Area of triangle below OP = 1 e 2 2 − B1 1.1 Or by correctly evaluating 1 2 0  e d − x x Allow unsimplified = − 14 (1 5e−2) A1 2.2a a b = = − 1, 5 (must be in this form) [6]H240/03 Mark Scheme October 2021 14 Question Answer Marks AO Guidance 9 (a) 18 14 20 0.2 = +  = a a (ms−2) B1 1.1 [1] 9 (b) 14 2 0.2 330 2 2 = + u ( )( ) M1 3.3 Use of SUVAT equation to determine u with their a or other complete method to find u M0 if using 18 for v u = 8 m s ( −1) A1 1.1 [2]H240/03 Mark Scheme October 2021 15 Question Answer Marks AO Guidance 10 (a) B1 1.2 All three forces added correctly – need not be labelled Extra forces is B0 [1] 10 (b) F =15cos B1 M1 3.3 3.3 Not for just seeing the expression 15cos M1 for resolving vertically – allow sign errors and sin/cos confusion (three terms) Condone inclusion of g with the 50 for the M mark only R = + 15sin 50  A1 1.1 15cos 15sin 50 1 ( ) 5    + M1 3.4 Use of 1 5 F R with their F and R 50 Allow F R 15 for the M mark only 15cos 3sin 10   −  A1 2.2a AG [5]H240/03 Mark Scheme October 2021 16 Question Answer Marks AO Guidance 11 (a) M1 3.3 Use of s ut at = + 12 2 with a g =  and s = 4 Allow sin/cos confusion 4 25sin15 10 ( ) 1 ( ) 2 2 − = − t t A1 1.1 t =1.75(s) A1 2.2a BC (1.750981765…) 1.75 only For reference: 1.779296952… (if g = 9.8 used) [3] Penalise g = 9.8 only once in the question Alternative method 2 0 (25sin15) 2( 10)s1 and 0 25sin15 ( 10)t1 M1* Finding the maximum height s1( 2.093353...) above A and corresponding time t1( ) 0.647047... Using v = 0 and a = 10 2 1 2 1 4 10 s t 2 and t t t 1 2 M1dep* Complete correct method to find t Using u = 0 and where t2 ( ) 1.1039341... is the time from the maximum height to the ground t 0.6470476... 1.1039341... 1.75(s) A1 11 (b) (25cos15)t M1 3.4 Use of s ut = with their t from (a) Allow sin/cos confusion 42.3 (m) A1FT 1.1 42.2829627… - ft their positive value of t from (a) but must be using (25cos15)t For reference: 42.96672196… (if g = 9.8 used) [2]H240/03 Mark Scheme October 2021 17 Question Answer Marks AO Guidance 11 (c) vh = 25cos15 B1 1.2 Correct expression for horizontal velocity component (soi) 24.14814… vv = − 25sin15 10 1.5 ( ) B1 3.3 Correct expression for vertical velocity component at t = 1.5 (condone positive value) 8.529523... tan v h v v  = M1 3.1b Use of tan to find angle (allow reciprocal) – dependent on one B mark earned M0 if using expressions for displacements 19.5 below the horizontal A1 3.2a oe (e.g., 70.5to the downward vertical) For reference: 18.8(if g = 9.8 used) [4] 11 (d) e.g., a less accurate value of g was used B1 3.5a Any valid reason (do not accept mention of resistance e.g., air/wind resistance) e.g., no consideration of the wind e.g., no consideration of (back)spin on the ball (but not topspin) [1]H240/03 Mark Scheme October 2021 18 Question Answer Marks AO Guidance 12 (a) M1* 3.3 Moments (with correct number of terms) about A, C, D, B or com x is the centre of mass of the rod from A Allow g to be absent 0.5 4 0.7 20 T T gx C D + − = ( ) (moments about A) (x g T − = − 0.5 20 4 1.2 )( ) ( ) D (moments about C) (4 1.2 20 4 0.7 − = − − )T g x C ( ) (moments about D) 0.7 4 0.5 20 4 T T g x D C + − = − ( ) ( ) (moments about B) (x T x T − = − − 0.5 4 0.7 ) C D ( ) (moments about com) A1FT 1.1 Follow though their TC and TD (allow just T) and allow W for 20g 3 T T C D = or T T C = 3 and T T D = B1 3.3 Correct relationship(s) for the tensions at C and D (soi) T g C = = 147 15 ( ) and/or T g D = = 49 5 ( ) (from 20 T T g C D + = ) used in relevant moment equation(s) M1dep* 1.1 Equation in x only, e.g., • W = 4T and any one mom. eq. • 4T = 20g and any one mom. eq. • W = 20g and any two mom. eq. M0 if tension used are the same in both ropes x =1.2 (m) A1 [5] 2.2a Must be 1.2 as question asks for com from A Alternative: B1 as main scheme (soi), then M1 for splitting 2.8 in the ratio 1: 3 or 3:1 then A1 for correct 1:3 then B2 for 0.5 + 0.25(2.8) = 1.2(m) 12 (b) 0.7 20 4 0.7 mg g x = − − ( ) (moments about D) 20 0.5 3.5 4 0.5 0.7 20 g x mg g mg ( − + = − − + ) ( )( ) (moments about C) M1 3.1b Moments about D (oe) – correct number of terms – if taking moments about another point then must set TC 0 and 20 T g mg D Allow g to be absent m = 60 A1 2.2a [2]H240/03 Mark Scheme October 2021 19 Question Answer Marks AO Guidance 13 (a) F i j = − + − − (4 8 6 2 1 9 t t ) (( )2 ) M1 3.1b Combines given forces and considers either component equal to zero – allow M1 for either 4 8 0 t or for 2 6(2 1) 54 0 t When t = 2, the forces are in equilibrium A1 1.1 t = 2 only – need only consider i or j component but any contradictory working/answers scores A0 [2] 13 (b) m t t =  = − + − − 2 2 4 3 2 1 9 a i j ( ) (( )2 ) B1 3.3 Using F = 2a correctly Allow 2a = … M1* 3.1b Attempt to integrate a (or F) wrt t – two of their terms integrated correctly M0 if only considering one force or one component for a or F v i j c = − + − − + (t t t t 2 4 3 2 1 9 )       16 ( )3 ( ) A1 1.1 Condone no 2 3 2 +c for this mark v i j ( 4 ) (4 6 24 ) t t t t t (oe) Allow 2v =… 1 0, 2 t = =  = v 0 c j M1dep* 3.4 Uses correct initial conditions to find c (or c = 0 if expanded version used) Moving parallel to j = i 0 therefore t t ( − = 4 0 ) M1 3.1a Sets obtain i-component of a quadratic equation in v equal to 0 to t Dependent on first mark M 1 t =  = 4 64 (ms ) v − A1 2.2a Must have found +c for this mark [6]H240/03 Mark Scheme October 2021 20 Question Answer Marks AO Guidance 13 (c) M1* 1.1 Attempt to integrate v wrt t – two of their terms integrated correctly – dependent on first M mark in (b) no vector constant of integration required in (c) 1 1 9 1 3 2 2 2 3 2 1 ( )4 3 48 2 6 = − + − − +         t t t t t r i j     A1 1.1 r i j ( 2 ( 2 12 13 t t t t t 3 2 4 3 2 ) ) 1 0 16 t =  = r j , 3 9 1295 16 =  = − − r i j M1dep* 1.1 Attempt to find r at both t = 0 and t = 3 Dist. = ( ) 2 2 1295 1 9 16 16   − + − −     M1 1.1 Correct expression for the distance between given times (dependent on both previous M marks) 81.5 (m) A1 2.2a (For reference: 81.49846624…) 6642 9 82 = [5]H240/03 Mark Scheme October 2021 21 Question Answer Marks AO Guidance 14 (a) T g a AB − = 2 sin30 2 M1* 3.3 N2L parallel to plane for A – correct number of terms, allow cos/sin confusion Dimensionally consistent equations for M marks 4 sin60 4 g T a − = BC M1* 3.3 N2L parallel to plane for C – correct number of terms, allow cos/sin confusion M1M0M0 if T used in both equations 3 T T F a BC AB B − − = M1* 3.3 N2L parallel to plane for B 4 sin60 4 2 sin30 2 3 g a g a F a − − − − = B M1dep* 2.1 Eliminates both tensions Allow in terms of FB 9 4sin60 2sin30 3 a g = − − ( ) A1 3.3 Use of F g B =  (3 ) to get a correct equation in a and  9 (2 3 1 3 ) a g  ( )1 2 3 1 9 0 3 a g    = − −      M1 3.1b Explicitly uses   0 to get a strict inequality in a and g only. If 19 a g(2 3 1 3 )  19 a g(2 3 1) without justification is M0 Dependent on all previous M marks a g  − 19 (2 3 1) A1 [7] 2.2a AG – must follow from a correct equation involving ,aand g SC considering whole system with no friction B2 only for deriving 9 (2 3 1) a g 14 (b) a g =  = − 1 2 9 3  ( 3 1) B1 1.1 Correct (soi) value of  (oe) using given a  = 0.488033... F g B = − 2 3 1 ( ) B1 3.4 Correct value for FB FB =14.34819... (3 2 3 1 g g )2 + − ( ( ) )2 M1 3.1a (3g F )2 + B2 allow any value for FB or even just the expression B 32.7 (N) A1 2.2a (For reference: 32.71438099…) [4]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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