Y411/01 Mark Scheme June 2023 6 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. Y411/01 Mark Scheme June 2023 7 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. Y411/01 Mark Scheme June 2023 8 Question Answer Marks AOs Guidance 1 (a) Let the driving force on the car be D N 5000 20 D = ( = 250) B1 3.4 soi 𝐷= 𝑘(20)2 M1 3.3 Using driving force = resistance e.g. 5000 = 𝑘(20)2 × 20 implies B1 and M1 3 5000 5 8 20 k  = = A1 1.1 AG Must be correctly obtained [3] (b) Power is 5 8 (28)3 = 13720 (W) B1 1.1 Accept 13 700 www Accept 14 000 from correct method [1] (c) Let the mass of the car be m kg 13000 20 = 5 8 (20)2 + 𝑚𝑔sin 2° 650 = 250 + 𝑚𝑔sin 2° B1 M1 1.1 1.1 Correct weight component 𝑚𝑔sin 2° (B0 for 𝑊sin 2° unless 𝑊= 𝑚𝑔 used later) Resolving parallel to the plane – attempt at driving force (using 13000), resistance, and attempt to resolve weight (condone missing g) Driving force 13000 is M0 Mass is 1170 (kg) A1 1.1 1169.53911… [3] Y411/01 Mark Scheme June 2023 9 Question Answer Marks AOs Guidance 2 (a) Let the speed of the ball just before and just after hitting the ground be u and v m s-1. Let the coefficient of restitution be e. 𝑢2 = 02 + 2𝑔× 12.8 or 1 2 2 12.8 =  mu mg 𝑢2 = 128 5 𝑔= 250.88, 𝑢= 15.84 M1 3.3 Attempt at equation for u 02 = 𝑣2 −2𝑔× 5 or 1 2 2 5 =  mv mg 𝑣2 = 10𝑔= 98, 𝑣= 9.90 M1 3.3 Attempt at equation for v OR 5 = 12.8𝑒2 M2 𝑒= 𝑣 𝑢= 5 8 (= 0.625) A1 1.1 Accept 0.62 to 0.63 from correct method (e.g. 9.9/15.8 = 0.627) A0 for −5/8 (as final answer) [3] (b) ‘Perfectly elastic’ means no energy loss B1 1.2 Or Speed is unchanged by the impact Just 𝑒= 1 is not sufficient ( ) 12.8 12.8 5 5  − + =  m mg g E M1 3.3 WEP Allow sign errors M1 normally implies B1 OR 1 2 𝑚𝑣1 2 = 𝑚𝑔× 12.8 −12.8𝐸 , 1 2 𝑚𝑣2 2 = 𝑚𝑔× 5 + 5𝐸 𝑣2 = 𝑣1 M1 B1 WEP in two stages ⇒17.8𝐸= 7.8𝑚𝑔⇒𝐸= 39 89 𝑚𝑔 A1 2.2a AG Must be convincingly shown. Note that (39/89)g = 1911/445 [3] (c) Model A. Speed just after second bounce is 𝑤= 𝑒𝑣= 5 8 × √98 = 6.187, 𝑤2 = 125 32 𝑔= 38.28125 M1 3.1b Attempt to find w or w2 OR Maximum height is 5𝑒2 = 5 × ( 5 8) 2 or 5 × 5 12.8 M1 Maximum height is 125 64 (= 1.953125) m A1 1.1 Accept 1.9 to 2.0 from correct method Model B. Let maximum height be h m after second bounce. ( ) 39 89 5 5 − + = mg h mgh mg M1 1.1 WEP ( ) 128 250 125 89 89 64 1.953125  =  = = h h (which is the same) A1 2.2a Both A1’s can only be awarded if values agree to at least 3 sf [4] Y411/01 Mark Scheme June 2023 10 Question Answer Marks AOs Guidance 3 (a)  2 3 M T L   = G M1 1.1 Allow units (kg, m, s)  1 3 2 M L T − −  = G A1 1.1 Accept L3/(MT2) isw Do not allow units [2] (b) [v] = 1 LT− B1 1.1 Correct dimensions for v Allow units 1 LT− ( ) ( ) ( )( )( ) 1 3 2 1 2 M L T LT L M M    − − − = M1 1.1 Setting up an equation in M, L and T using given equation and their [G] Condone (M + M), 2M etc Allow units M0 for [𝑚1𝑚2(𝑚1 + 𝑚2)] = M or M2 [ −𝛼+ 3 = 0 ⟹ ] 3 = B1 1.1 cao 3 1    + + = 1 2  − − = − M1 1.1a Setting up equations using L and T FT their dimensions equation. Allow one error 5 = − and 3 = − A1 1.1 cao [5] (c) ( ) 3 8.64 6.13 ( 2.8  ) or ( ) 3 6.13 8.64 ( 0.357  ) M1 3.4 For ( 6.13 8.64) ± their 𝛾 M0 if their 𝛾= 0 So the stars approach 2.8 times faster when closer together. A1 2.2b cao M1A0 for 2.8 obtained from 𝛾= 3 [2] Y411/01 Mark Scheme June 2023 11 Question Answer Marks AOs Guidance 4 (a) Let the velocity of B after collision be u m s-1 in the direction AB . 8 0.5 1.6 0. 0. 5 3−  = u M1 3.3 COLM – all terms present but allow sign errors 3.2 = u A1 1.1 So coefficient of restitution 3.2 1 8 1.6 3 + = = A1 1.1 AG Must be correctly obtained [3] (b) Let B reach the lower section with speed v m s-1. 2 1 2 2 2 1 0.5 3.2 9.8 0.45 0 5 0.5 .     +  = v M1 3.3 Attempt at WEP: require two KE terms and a GPE term. M0 if 𝑣2 = 𝑢2 + 2𝑎𝑠 used ⇒𝑣2 = 19.06 , 𝑣= 4.36577 … A1 1.1 AG Must be correctly obtained [2] (c) Let the speed of C after collision be w m s-1. 0.7 0.5 = w v ( ⇒𝑤= 5 7 𝑣= 3.1184 ) M1 3.4 COLM KE of C is 1 2 × 0.7 × (3.1184)2 (= 3.40357 … ) J A1 1.1 C needs to gain 9.8 0.45 3.087 0.7  = J of GPE so C will collide with A next. A1 2.1 Argument must be clear. OR Assume C reaches the top section with speed x ms−1 1 2 × 0.7 × 3.11842 −0.7 × 9.8 × 0.45 = 1 2 × 0.7 × 𝑥2 A1 For KE of C 𝑥2 (= 0.9044) > 0 ( x = 0.951 ) So C will collide with A next A1 Or C can reach a height of 0.496 m ( > 0.45 ) [3] (d) OR OR [ If 𝑣B = 0, ] by COLM, 𝑣C > 𝑢B giving 𝑒> 1 (or increase of KE) which is impossible [contradicting 𝑒≤1 ] [ If 𝑣B = 0, since 𝑒≤1 ] 𝑣C ≤𝑢B and so 𝑚C𝑣C < 𝑚B𝑢B contradicting COLM 𝑚B𝑣B + 𝑚C𝑣C = 𝑚B𝑢B and 𝑣C −𝑣B ≤𝑢B ⟹(𝑚B + 𝑚C)𝑣B ≥(𝑚B −𝑚C)𝑢B ⟹ 𝑣B > 0 B2 3.5a Correct explanation; must mention momentum Give B1 for an explanation which includes at least one of the following (ignore incorrect statements) • This requires 𝑣C > 𝑢B • This requires 𝑒> 1 • If 𝑣C = 𝑢B then 𝑚C𝑣C < 𝑚B𝑢B [2] Y411/01 Mark Scheme June 2023 12 Question Answer Marks AOs Guidance 5 (a) Area of S ( ) 2 2 2 π 50 15 0 0 3 10 0  = = − − B1 1.1 Or ratio of masses (e.g. 25:9:1:15) used throughout 1 2 0 50 0 0 20 1500 0 900 0 40 0 0   −       = + +             x y M1 1.1 For (𝜋× 302)(20) or (𝜋× 102)(40) seen ⟹𝑥̅ = 12 A1 1.1 AG Needs convincing working 𝑦̅ = − 8 3 A1 1.1 Allow −2.67 [4] (b) arctan (12 ÷ 8 3) Angle is 77.47119…° M1 A1 FT 3.1b 1.1 tan 𝛼= 12 ÷ their |𝑦̅| (allow reciprocal) Just arctan(𝑥̅/𝑦̅) is not sufficient Accept 102.5°, −77.5° FT their |𝑦̅| [2] (c) 50 12  =  W F M1 1.1 Taking moments about O – both moments present 0.24  = F W A1 1.1 λ = 6/25 (or 0.24) [2] (d) Let the angle between A R and the horizontal be . 𝑅A cos 𝜃+ 𝐹sin 𝜃= 𝑅B cos 𝜃 0.6𝑅A + 0.8(0.24𝑊) = 0.6𝑅B Resolving horizontally. Must attempt to resolve all three forces. F may be their 0.24W 𝑅A sin 𝜃+ 𝑅B sin 𝜃= 𝑊+ 𝐹cos 𝜃 0.8𝑅A + 0.8𝑅B = 𝑊+ 0.6(0.24𝑊) Resolving vertically. Must include W and attempt to resolve the other three forces 60𝑅B sin 𝜃= 42𝑊 0.8 × 60𝑅B = 42𝑊 Moments about A. Both moments present. Must attempt to ‘resolve’ RB (or its distance) M1 M1 3.3 Any two of the above. See first lines and notes OR 18𝑊+ 60𝐹cos 𝜃= 60𝑅A sin 𝜃 18𝑊+ 0.6 × 60(0.24𝑊) = 0.8 × 60𝑅A M2 Moments about B. Must attempt to ‘resolve’ RA and F (or their distances) M1 1.1 Correct values for cos and sin used in at least one equation for which M1 has been earned. See second lines above. Allow cos(53.1°) etc Note Only sin needed in the third equation ⇒𝑅A = 0.555𝑊 ( 𝑅B = 0.875𝑊 ) A1 1.1 cao ( ) 0.24 16 0.555 37 0.432432   = = = W W A1 FT 2.2a FT their F and RA given as multiples of W, dependent on M3 [5] Y411/01 Mark Scheme June 2023 13 Question Answer Marks AOs Guidance 6 (a) Each support must exert ( ) 1 2 4 10 7 + = g g g N on the beam. B1 1.1 soi Taking moments about the left-most point: 𝑅A𝑥+ 𝑅B(𝑥+ 3.8) = 4𝑔× 1.95 + 10𝑔× 3 7𝑔𝑥+ 7𝑔(𝑥+ 3.8) = 4𝑔× 1.95 + 10𝑔× 3 M1 A1 3.3 1.1 Attempt to take moments about some point. All moments present. Allow one error Correct equation for x FT their 7g Correct equation www implies B1 ⇒𝑥= 0.8 A1 1.1 cao [4] (b) Moments about B, 3 5.3 sin60 10 2. L g T =  M1 3.1b Attempt at moments. No force acting at A; allow sin/cos switch. If moments taken about A, 𝑅A = 0 must be soi (e.g. by 𝑇𝐿sin 60° + 𝑅B = 10𝑔 ) Allow inequalities in (b) and (c) 49.10745 L T  = A1 1.1 AG Must be convincingly shown. [2] (c) Let the force at the supports be A R and B R N, and the friction at B be F N. B 0.4 = F R B1 3.4 Modelling friction. Must clearly be reaction at B 𝑇𝑆cos 60° = 𝐹  (⇒𝑇𝑆= 0.8𝑅B) M1 3.3 Resolving horizontally 𝑇𝑆sin 60° + 𝑅A + 𝑅B = 10𝑔 M1 1.1 Resolving vertically Moments about LH end, A B 1.5 5.3 3 10 + =  R R g M1 3.1b Attempt at moments. All moments present. Allow one error OR 3.8𝑅B = 1.5 × 10𝑔+ 1.5 × 𝑇𝑆sin 60° M2 Moments about A (⇒𝑅B = 53.2460 … 𝑅A = 7.864 …) 42.5968  = S T so beam will slide first (since 42.6 < 49.1) A1 2.2a cao [5] 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. 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