Y431/01 Mark Scheme June 2023 4 2. Subject-specific Marking Instructions for A Level Mathematics B (MEI) a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded. 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 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. 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 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. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. 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. Y431/01 Mark Scheme June 2023 5 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. f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.) 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. This rule should be applied to each case. When a value is not given in the paper accept any answer that agrees with the correct value to 2 s.f. Follow through should be used so that only one mark is lost for each distinct accuracy error, except for errors due to premature approximation which should be penalised only once in the examination. There is no penalty for using a wrong value for g. E marks will be lost except when results agree to the accuracy required in the question. g Rules for replaced work: if a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests; if there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook. 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 mark in the question. Marks designated as cao may be awarded as long as there are no other errors. E marks are lost unless, by chance, the given results are established by equivalent working. ‘Fresh starts’ will not affect an earlier decision about a misread. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error. i If a graphical calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). 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. Y431/01 Mark Scheme June 2023 6 Question Answer Marks AOs Guidance 1 (a)   2 Force MLT− = B1 1.1   1 Velocity LT− = B1 1.1   3 Density ML− = B1 1.2 [3] (b)  ( ) ( ) 1 3 2 1 3 2 RHS LT ML M − − =  M1 1.1 Use expressions from (a) to express 1 3 2 2 ( ) u m  dimensionally. ( ) ( )   1 3 2 1 3 3 2 2 1 2 LT M L L T ML MLT LHS − − − − − = =  = = A1 1.1 Convincingly reached. [2] (c) Model predicts that when air density is doubled, drag force should increase by factor of 3 2 . oe B1 3.5b Or by contradiction [1] (d) ( ) ( ) ( ) 2 3 1 2 MLT ML LT L    − − − = : 1 : 3 2 1 : 2 M L T      = − + + = − = − M1 3.3 Setting up equations in 𝛼, 𝛽 𝑎𝑛𝑑 𝛾 using given equation and their dimensions for (a) 1 = and 𝛽= 2 A1ft 3.4 Follow through from (a) −3 + 2 + 2𝛾= 1. 2𝛾= 2 𝛾= 1 A1ft 1.1 Follow through from (a) [3] Y431/01 Mark Scheme June 2023 7 Question Answer Marks AOs Guidance 2 (a) Driving force = resistance = 12 P B1 3.3 soi 12 8 1 0 400 sin7 14 0 0. P g − =  M1 B1 3.4 1.1 Attempt at N2L, correct number of terms and weight resolved (but condone sign errors and sin/cos confusion) Correct weight component if seen in (a) or (b) 6624.568 P  = A1 1.1 [4] (b) 6624.568 12 16000 0 1400 sin7 v g − − = M1 3.3 Attempt at N2L with zero acceleration – correct number of terms (condone sign errors) FT their value of P or resistive force from (a) 7.1939 v = A1 1.1 [2] Y431/01 Mark Scheme June 2023 8 Question Answer Marks AOs Guidance 3 (a) Let the speed of P when it reaches B be u m s-1. Frictional force 2 3 0.5g =  B1 3.3 Either 2 1 2 2 3 0.5 0.5 0.6 0 u g  −   = Or 2 3 a g = − , so 2 2 2 3 6 0 2 0. g u =  − M1 3.4 If using WEP, both KE and WD against friction terms need to be present. Let the speed of Q be v m s-1 after collision with P. COLM: 5 0 6 0. 2 8 2 .5 . v = − + M1 3.3 Three terms. Condone sign error. 2.8 u = and 2.2 v = A1 1.1 or if both seen in (b) and (c) Coefficient of restitution between P and Q 2.2 2.8 6 + = M1 3.3 Attempt at sep app v v  . Allow sign errors but not num/dem switch. 2.2 2.8 5 6 6 + = = ( 0.8333 = ) A1 1.1 [6] (b) ( ) ( ) 0.5 2.8 6 4.4 −− = N s B1 3.3 Use “their” value for 𝑢 towards A. B1 1.1 oe eg ‘towards B’, ‘in the direction CB’ [2] (c) Let the coefficient of friction between Q and AB be . Speed of Q when it reaches B is 2.2 m s-1 B1 3.4 Use “their” value for 𝑣 Either 1 2 2 2 2.2 0.3 0 2g   = − Or a g = − , so 2 2 3 0 2.2 . 2 0 g = − M1 3.1b 121 147 = ( 0.823129 = ) A1 1.1 cao [3] Y431/01 Mark Scheme June 2023 9 Question Answer Marks AOs Guidance 4 (a) Let the tension in the string be T N. 5 T g = B1 1.1 10 sin T g  = M1 1.1 Attempt to resolve tangentially. Condone cos for sin 1 2 sin 30   =  = A1 2.2a [3] (b) 1 2 2 2 1 2 10 10 (2cos35 2cos45) 5 10 5 2 2 360 g v v g    = − −      +    M1 B1 B1 3.4 3.1a 1.1 Attempt at conservation of energy; two PE terms present and at least one KE term. Correct PE term for Q Correct PE term for P 0.805 v = (m s-1) A1 2.2b 0.8047065… [4] (c) e.g. we have assumed that in the subsequent motion that Q does not reach the pulley before P arrives at the point where 45  or the length of string from AP is equal to the arc length AP as P is small and or the pulley size is negligible B1 2.4 [1] Y431/01 Mark Scheme June 2023 10 Question Answer Marks AOs Guidance 5 (a) 5 co 0 c 1 o 4 1 s s g g H + = 5 sin40 sin H g =  M1 A1 3.3 1.1 Attempt to resolve both horizontally and vertically Correct number of terms 5 sin 40 11 5 cos 40 t 4 an 2 .14487 g g g    −  =  =  A1 1.1 and 77.0 H  A1 1.1 77.000260 H = so accept 77. Or triangle of forces method: Force triangle with • sides H, 5g and 11g, • 40° between 5g and 11g, •  between H and 11g. M1 ( ) ( ) ( )( ) 2 2 2 5 11 2 5 11 cos40 H g g g g = + −  A1 77.0 H   A1 ( ) ( ) 2 2 2 5 11 2 11 24.14487 T g g T g  +  − = =  A1 [4] (b) Friction 5 sin30 g  = Normal contact 7 5 cos30 g g  = − M1 A1 3.3 1.1 Resolving vertically and horizontally for the beam 5 sin30 7 5 cos30 min 0.936374    − = = g g g A1 1.1 [3] (c) Taking moments about the point of contact between beam and floor: 5g cos 30° ∙3𝑐𝑜𝑠𝜙= 5g𝑠𝑖𝑛 30° ∙3𝑠𝑖𝑛𝜙+ 7g ∙3 2𝑐𝑜𝑠 𝜙 or 5g𝑐𝑜𝑠(𝜙+ 30 °) ∙3 = 7g ∙3 2𝑐𝑜𝑠 𝜙 M1* A2 3.1b 1.1 1.1 Taking moments: dimensionally consistent; correct number of terms for their method. All correct. A1 for one error. 5 3 5tan 7   = + M1dep* 1.1 Obtaining tan (oe) 5 3 7 5 tan 18.36878   −  =  =  A1 2.2a [5] Y431/01 Mark Scheme June 2023 11 Question Answer Marks AOs Guidance 6 (a) Let the centre of mass lie x cm above the base. Total area ( ) 2 1 2 4 4 8 160     = +  = B1 1.1 Expression for correct total area of container 9 1 16 0 14 0 4 6 x     = +  M1 1.1 Table of values idea with correct number of terms 8.1 x  = A1 1.1 AG must be convincingly shown. [3] (b) DR Total mass ( ) 6 400 16 1 400 1 h h    = + = + B1 1.1 Expression for correct total mass of container and water ( ) 1 2 1 4 0 00 1 4 0 8.1 6 6 h y h h   = +  +  M1 1.1 Table of values idea with correct number of terms ( ) 2 4 3 00 4 16 2 0 8 h y h   + + = 2 2 405 400 16 3240 0 2 8 5 h h y h h     +  + + = + = A1 2.2a ( ) ( ) ( ) 2 2 50 2 405 50 2 2 d 2 d h h h y h h      + + +  −  = M1* 3.1a Attempt to differentiate using quotient rule. Least value of y occurs when ( ) ( ) 2 0 50 2 4 2 05 2 h h h     + +  −  = M1dep* 2.1 Setting the numerator of their expression to zero. 2 0 50 405 h h  + −  = 5.90736 h  = A1 2.2a [6] (c) Let the centre of mass of the water lie d cm above the base. ( ) 3 3 4 4 3 3 13.5 6.75 16 13.5 3 3 3 16 d       =  −  +   M1 3.1b 180 1350 7.5 d d    =  = A1 2.2a AG must be convincingly shown. [2] (d) Total mass ( ) ( ) 3 3 4 4 3 3 13.5 3 3 4 0 40 3 40 1 2 6 0 4     =  −  +   = + + B1 1.1 ( ) 3 4 3 3 4 3 00 324 4 1 00 8.1 50 3 4 z    + +  +   = M1 3.1b Table of values idea with correct number of terms Y431/01 Mark Scheme June 2023 12 Question Answer Marks AOs Guidance 6.233491 z = A1 2.2a [3] Need to get in touch? If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in touch with our customer support centre. Call us on 01223 553998 Alternatively, you can email us on support@ocr.org.uk For more information visit ocr.org.uk/qualifications/resource-finder ocr.org.uk Twitter/ocrexams /ocrexams /company/ocr /ocrexams OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge. 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