Y543/01 Mark Scheme June 2022 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. Y543/01 Mark Scheme June 2022 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. 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. Y543/01 Mark Scheme June 2022 5 • 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. Y543/01 Mark Scheme June 2022 6 Question Answer Marks AO Guidance 1 (a) 10000 = D  5 => D = 2000 M1 3.4 Using “P = Fv” to find the ‘driving force’ 2000 – 250 = 1200a M1 1.1 Using NII with two force terms in opposite directions to derive an equation in a a = 1.458... so acceleration is awrt 1.46 ms–2 A1 1.1 [3] (b) At max speed, 250 = 20000/v M1 3.4 Using NII with a = 0 oe and “P = Fv” with maximum power => v = 80 so max speed is 80 ms–1 A1 1.1 [2] (c) 20000/v = 250 + 1200gsin M1 3.4 Using NII with a = 0 oe and three force terms and “P = Fv” with maximum power 20000 = 838v Allow sign slip v = 23.86... so max speed is awrt 23.9 ms–1 A1 1.1 [2] Question Answer Marks AO Guidance 2 (a) Displacement = (5 – –1)i + (12 – 6)j m M1 1.1 Subtracting the position vectors of the points to find the displacement 6i + 6j m (7i – 2j).(6i + 6j) = 42 + –12 = 30 so 30 J A1 1.1 [2] (b) P = F.v = (7i – 2j).(–i – 5j) M1 1.1 Use of P = F.v = –7 + 10 = 3 so 3 W A1 1.1 [2] (c) OB = (52 + 122) = 13 so extension is 5m M1 3.1b Calculating OB and hence finding the extension T = 24  (“OB”-8) / 8 (= 15) or EPE = 24  (“OB”-8) 2 / (28) (= 37.5) M1 1.1 T = 24  5 / 8 = 15 so 15 N and EPE = 24  52 / (28) = 37.5 so 37.5 J A1 1.1 If M0M1A0 SC1 for T = 1.46N and EPE = 0.353J [3] Y543/01 Mark Scheme June 2022 7 Question Answer Marks AO Guidance 3 (a) 2 2 0 1 d 5 4e t I t − = −  M1 1.1 Correct form of integral and correct limits. = 0.81365... so impulse is awrt 0.814 Ns A1 1.1 BC Accept awrt 0.81 or 0.82 if correct integral shown [2] (b) Impulse = change in momentum so “0.81365...” = ±(6v – 61.9) M1 1.1 Use of impulse-momentum principle v = 1.9 + 0.81365.../6 = 2.0356... so P’s speed is awrt 2.04 ms–1 A1 1.1 Accept awrt 2.03 or 2.04 [2] (c) WD = change in KE = ½6(“2.0356...”2 – 1.92) J M1 1.1 Use of work-energy principle with values for v, m and u substituted in = awrt 1.60 J A1 1.1 Accept anything awrt 1.60 – 1.62 [2] Y543/01 Mark Scheme June 2022 8 Question Answer Marks AO Guidance 4 (a) m1 and m2 have the same dimensions and NIII says that the magnitude of the force that each applies to the other is the same. So m1 and m2 must be interchangeable which means that the dependency on each must be the same and hence  = . B1 2.4 Argument must be based on the idea that the dimensions/units of the quantities are equal and that they are interchangeable Must reference at least 2 of the 3 bold statements [1] (b) M–1L3T–2 B1 1.1 cao but condone change in order of M, L and/or T Don’t accept square brackets [1] (c) [F] = MLT–2, [mi] = M and [r] = L => M: 1 = –1 + 2 M1 1.1 Setting up equation in  (and/or ) by considering mass dimension Ignore attempt to set up equation involving time dimension or 1 = –1 +  +  L: 1 = 3 +  M1 1.1 Setting up equation in  by considering length dimension Ignore attempt to set up equation involving time dimension =>  =  = 1 and  = –2 A1 1.1 or F = Gm1m2r–2 [3] (d) Gm1m2r–2 = m2r2 M1* 3.3 Using NII with a correct form for the centripetal acceleration  = (6.6710–11  5.971024/(3.84108)3) =>  = 2.65 10–6 = 2 / T M1dep* 3.4 Using 𝑣= 2𝜋𝑟 𝑇or 𝜔= 2𝜋 𝑇 with values substituted in 𝑇2 = 4𝜋2(3.84 × 108)3 (6.67 × 10−11)(5.97 × 1024) => T = awrt 2370000 so period is 2370000/(606024) = 27 days to the nearest day A1 1.1 [3] Y543/01 Mark Scheme June 2022 9 Question Answer Marks AO Guidance 5 (a) DR ( ) 4 4 2 2 0 0 15 d 15 ln 9 9 15(ln9 ln3) 15ln3 x x x x   = + +     + = − =  B1 1.1 If this is not seen in (a) the mark can be awarded here if integral correctly evaluated in (b). 15 ∫1 2 × 2𝑥(𝑥2 + 9)−1 2 4 0 𝑑𝑥 = 15 [(𝑥2 + 9) 1 2] 0 4 M1 1.1 Integrating. Ignore limits here. Must be in the form 𝑘[(𝑥2 + 9) 1 2]   1 1 2 2 2 2 15 (4 9) (0 9) 15 5 3 30 30 2 15ln3 ln3 x   = + − +       = − =  = = A1 1.1 AG. Some intermediate working must be seen. Award for fully complete proof only [3] (b) DR 1 2 ∫( 15 √𝑥2 + 9 ) 4 0 2 𝑑𝑥= 225 2 ∫ 1 𝑥2 + 9 4 0 𝑑𝑥 = 225 2 × 1 3 [𝑡𝑎𝑛−1 𝑥 3] 0 4 M1 M1 1.1 1.1 Correct integral and correct limits Integrating into the form 𝑘[𝑡𝑎𝑛−1 𝑥 3] May have used substitution 1 1 75 4 tan 2 3 75 4 tan 2 3 awrt2.11 15ln3 y − − =  = = A1 1.1 [3] Y543/01 Mark Scheme June 2022 10 (c) DR x = 3 => 15 18 y = oe B1 3.4 Finding the y coord of P. 5 5 2 2 2 = = or awrt 3.54 𝑡𝑎𝑛𝜃= 3 −2 𝑙𝑛3 " 15 √18 " −"2.1101. . . " = 1.17952. . . 1.42538. . . = 0.8275. . . 𝜃= awrt39.6° M1 A1 2.1 2.2a tan 𝜃= ∆𝑥 ∆𝑦 or ∆𝑦 ∆𝑥 0.691rads [3] Y543/01 Mark Scheme June 2022 11 Question Answer Marks AO Guidance 6 (a) I = mv – mu = 2.5(18 – –5) = 57.5 B1 1.1 [1] (b) 2 5 d 60 2.5 d F ma v v v v x x =  − = M1 3.3 Using F = ma to express a differential form of the acceleration in terms of the given forces with the signs correct d 2 24 d v v x x  = − so a = 2 and b = –24 A1 2.2a [2] (c) 𝑑𝑣 𝑑𝑥= 2𝑣 𝑥−24 ⇒𝑑𝑣 𝑑𝑥−2 𝑥𝑣= −24 ⇒IF = 𝑒∫−2 𝑥𝑑𝑥= 𝑒−2 𝑙𝑛𝑥= 𝑥−2 B1FT 1.1 Correctly determining the integrating factor 𝑥−𝑎 𝑎≠0 FT their value of a provided 𝑎≠0 ( ) 2 3 2 2 d d 2 24 d d v x x v vx x x x − − − − − = = − M1* 1.1 Multiplying by the IF and writing LHS as an exact derivative. IF must be in the form 𝑥𝑘 2 2 1 24 d 24 vx x x x c − − − = − = +  M1dep* 1.1 Integrating BS with a constant of integration or using a definite integral with suitable limits x = 1, v = 18 => 181–2 = 241–1 + c => c = –6 => v = 24x – 6x2 A1 3.4 Substituting initial conditions to find c or using the limits correctly in a definite integral with suitable limits [4] (d) • As t → , x → 4 OR P moves from its initial position to the point on the x-axis where x = 4 • P initially accelerates and then decelerates (to 0) • P reaches its max speed of 24ms-1 or reaches max speed when 𝑡= 1 24 ln 3 (0.0458) or reaches max speed when x = 2 B1 B1 B1 2.4 2.4 2.2a [3] Y543/01 Mark Scheme June 2022 12 Question Answer Marks AO Guidance 7 (a) ½m72 - ½m5.52 M1 3.3 Change in KE considered in conservation of energy equation Assuming the zero PE level is set at ground level. Any level can be used but if not ground level then it must be clearly defined. PE at general angle  = mg(6.5 – 4.5cos) M1 3.1b Expressing height difference in terms of 𝜃 which may be embedded Which must be correct Here  is angle with downward vertical. Allow sin/cos 3.5mg + ½m72 = mg(6.5 – 4.5cos) + ½mv2 M1 3.3 Using their h in a conservation of energy equation involving KE and PE (could see 5.5 for v) May see h = 4.45633 Allow sin/cos –69.8 + 72 – 5.52 = –99.8cos => cos = 89/196 = 0.45408... A1 1.1 Correct value for cos (or )  = 63.0 or 1.099rads may see =27.0 or 0.472 NII => 2 cos mv T mg r  − = M1 3.3 NII with a correct form for centrepetal acceleration used Allow sin/cos. Must see their values substituted in and a weight component => T = 905.52/4.5+909.889/196 = 1005.5 so magnitude of tension is 1005.5 N A1 1.1 AG. Intermediate step must be seen [6] (b) vx = 5.5cos = 5.50.45408 = 2.4974... and vy = 5.5sin = 5.50.89096 = 4.90028 B1FT 3.4 FT their cos/sin –(6.5 – 4.5cos) = –4.45663 = 4.90028t –½9.8t2 M1 1.1 Using suvat to set up three term quadratic equation in t Must use a component of 5.5 for u Height must be -ve if taken upwards as positive => t = 1.57685... or –0.57679... but t > 0 => t = 1.57685... A1 2.3 Could be BC. Must eliminate unfeasible solution OT = (4.52 – 32) + 4.50.89096 + 1.576852.4974 = 3.354 + 4.009 + 3.938 M1 3.4 Complete method for finding OT using their values. = awrt 11.3 m A1 2.2a [5] Y543/01 Mark Scheme June 2022 13 (c) Vy = 4.90028 – 9.81.57685 = –10.5528... => Init KE = ½90(2.49742 + (–10.5528)2) B1FT 3.1b FT their vx, vy or their initial energy from (a) 5291.98 Or use initial energy from (a) tan = 10.5528/2.4974 = 4.2255... =>  = 76.685... so 5291.98 + 909.80.3sin76.685 = 0.32/(20.5) M1 3.1b Conservation of energy with KE, GPE and EPE terms (two of these correct (can FT their KE) Give M mark if only two components considered 58800 implies B1M1 909.80.3sin76.685 B1 3.1b GPE term correct =>  = 61661 so MoE is 61661 N A1cao 1.1 61740 comes from not considering angle [4] Question Answer Marks AO Guidance 8 (a) Because if the collision where inelastic then the velocity of A after the collision would be 0 (ie A would ‘stick to the wall’) and so it would not collide with B again (which we know it does). B1 2.4 [1] (b) Recollide at same point => vAx = 0 B1 3.1b Seen in solution Com: usin = mvB M1 3.3 or vAx + mvB on RHS NEL: sin B v e u  = M1 3.3 or sin B Ax v v e u  − = If vAx is used must be consistent in both equations => sin 1 sin u m e u m         = = A1 1.1 AG (vB = eusin) [4] Y543/01 Mark Scheme June 2022 14 (c) Velocity of A in y direction unchanged B1 3.4 vAy = ucos FT if sin and cos confusion in part b Either A’s velocity multiplied by (–)e or B’s velocity multiplied by (–)(5/9)e after collision with wall. B1 3.4 Ignore inconsistency between velocity and speed here. Must be the correct component cos cos d d u eu   + or 2 5 sin sin 9 d d eu e u   + M1 3.1b Calculating time of travel for either A or B from point of collision to wall and back Could see eg d – r 2 5 cos cos sin sin 9 d d d d u eu eu e u     + = + M1 2.1 Equating expressions that are in terms of theta, e or m and a single velocity 2 2 2 5tan 5 9 5tan 25 25 5 9 0 2 2 25 15 18 0 e e e e e e e e  + = + + − − = + − = M1 1.1 Reducing equation to 3 term quadratic in e or m (5e + 6)(5e – 3) = 0 => 6 3 or 5 5 e = − A1 1.1 Could be BC. –1.2 or 0.6 15 2025 50 −  0  e  1 3 1 5 5 3 e m e  =  = = A1 3.2a AG. Correctly rejecting negative root [7] (d) In practice the plane cannot be smooth or the discs may not be the same size/radius or air resistance may be significant or NEL is only an approximation B1 3.5b Any sensible limitation [1] Need to get in touch? 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