Y421/01 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. Y421/01 Mark Scheme June 2023 7 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 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. Y421/01 Mark Scheme June 2023 8 Question Answer Marks AO Guidance 1 15 27000 D M1 1.1 Use of P = Dv with P = 27000 or 27 1800 D = (N) or 27000 15 A1 1.1 Or implied by later working 600 800 D a − = M1 3.3 N2L – correct number of terms with their D (or just D) – condone sign errors but must be correct mass of 800 (so dimensionally consistent) 1.5 a = (m s-2) A1 1.1 [4] 2 0.5(6) 2(2) 0.5(0.2) 2v + = + M1 3.3 Use of CLM – correct number of terms – allow sign errors and a slip in one value only – allow mgv for M marks only 3.45 v = (m s-1) A1 1.1 0.5(6) 2(2) 0.5( 0.2) 2v + = − + M1 3.1b Use of CLM (again) – correct number of terms. When compared to first application of CLM must be the same total momentum before collision but different sign of 0.2 in expression for total momentum after the collision 3.55 v = (m s-1) A1 1.1 [4] Y421/01 Mark Scheme June 2023 9 Question Answer Marks AO Guidance 3 (a) cos 0.2 T g = M1 1.1 Resolving vertically – correct number of terms and T resolved. M0 if 0.2 used as the weight. Allow with either T or 15 49 cos 82 375   =  = A1 1.1 For reference: 82.491881… (no indication of why this is the greatest value is required) allow awrt 82 and awrt 1.44 (radians) - condone 82  but not < 82 Ignore incorrect use of inequality symbols if 82 given as answer [2] (b) 0.75sin r  = B1 1.1 Allow a value of 0.74 (or better) to imply this mark 2 sin 0.2(0.75sin ) T   = M1 3.3 Applying N2L horizontally – correct number of terms and T resolved (allow with either T or 15). Allow with r for radius and 2 a r = or 2 v a r = but not just a 10 = A1 1.1 Condone awrt 10 from correct working (might see 10.018… from using r = 0.74 and 82 = ) condone 10  but not 10  Ignore incorrect use of inequality symbols if 10 given as answer [3] Y421/01 Mark Scheme June 2023 10 Question Answer Marks AO Guidance 4 (a) Horizontal component of F is 0 B1 1.1 This mark can be implied if direction is stated as vertical (oe) e.g. ( ) 2cos 4cos 3cos 5cos 2cos      + − − = − e.g. 2 2 1 1 2(1)(1)cos(2(90 ))  + − − M1 3.3 Resolving vertically (oe) (at least four terms required) where is the angle between AB and the upward vertical – condone if trig. terms are not consistent (but all four must have been resolved) 3 cos 5 = vertical component of F is 1.2 A1 1.1 Sight of 1.2  Magnitude of F is 1.2 Direction of F is in the same direction as AC A1 2.2a Magnitude must be positive – allow ‘vertically upwards’ or ‘upwards’ only (oe) for direction (‘vertically’ only though is A0, but may be clarified with a diagram) [4] (b) e.g. moments about A: G = ( ) ( ) 6 3sin 6 4sin   + or 3(4sin ) 4(4cos ) 3(6sin )    + + or 3(3sin ) 4(3cos ) 3(4sin ) 4(4cos )     + + + or e.g. moments about O: 3(3sin ) 3(2sin ) 4(4cos ) 4(5cos )     + + + or 4(4 2)cos 4(3 5)cos   + + + B1 3.1b Taking moments (resolved force distance (oe)) with correct terms for their chosen point (note that if their answer to (a) is correct then taking moments about D or B, their F must be included too). A correct answer implies this mark 4 42sin 42 33.6 5 G    = = =     B1 1.1 No direction required but must be positive (seen anywhere in their solution so ISW if 33.6 is then stated as –33.6) [2] Y421/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 5 (a)  2 MLT F − = or  1 LT v − = B1 1.2 For either the dimensions of force or velocity correct (possibly seen in an expression for the dimensions of k)  2 1 MLT LT k − − = M1 1.1 Use of  F k v =  with at least one of [F] or [v] correct 1 [ ] MT k − = A1 2.2a [3] (b) e.g. 2 [ ] T k m u F      =   M1* 2.1 For correctly equating the dimensions of T (or either of the other two terms on the rhs) to the dimensions of 2 1 3 k m u F   - this mark can be implied by setting up an equation in M, L and T – if k is missing or assumed to be dimensionless then no marks in this part. Must be powers of k, m, u and F and no other (incorrect) terms 1 2 1 2 T (MT ) M (LT ) (MLT )    − − − = M1dep* 1.1 Setting up an equation in M, L and T using their [k] and at least one of [F] or [v] correct 2 3 2 0       + = − + = − + = M1 1.1 Setting up three equations in , and from their MLT (dependent on both previous M marks) – allow one slip from their MLT equation 1, 3, 3    = = = − A1 2.2a [4] Y421/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 6 (a) 2 (4 12) (3 6 8) t t t = − + + − v i j B1 1.1 Condone lack of brackets and accept column vector form [1] (b) 2 1(3 6 8) 4(4 12) t t t + − = − − M1* 3.1b Setting up a quadratic equation in t only – allow sign errors (including on the 1 and the 4) and the 1 and the –4 on the wrong side (or multiples of 1 and –4) 2 3 22 56 0 t t + − = ... t = M1dep* 1.1 Re-arrange and solves their three term quadratic equation in t to obtain a positive value of t 2 t = A1 1.1 Ignore 28 3 t = − unless used 2 2 2 (4 12) (3 6 8) t t t = − + + − v with their positive value of t M1 3.4 Dependent on both previous M marks – if working not shown then numerical components of v must be consistent with their positive value of t 16.5 = v (m s-1) A1 1.1 awrt 16.5 (allow exact 4 17 ) – for reference: 16.4924225… [5] (c) 4 (6 6) t = + + a i j B1ft 1.1 Follow their v from (a) 2 2 2 4 (6 6) 20.2 t + + = M1 1.1 Setting up an equation in t using 20.2 and their a 2.3 t = A1 1.1 Positive value of t only [3] Y421/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 7 (a) 2 cos30 R g = M1* 3.3 Resolving perpendicular to the plane for A – correct number of terms – allow sin/cos confusion (M0 if 2 used for the weight) 1 2 1 (2 cos30) ( ) 2 3 F g g = = M1dep* 3.4 Use of F R  = with correct and their R Work done against friction is 1 2 gd A1 1.1 Must be in terms of g and d [3] 7 (b) PE (lost by) B: 4gd B1 1.1 Allow 4gd  or 39.2d  PE (gained by) A: 2 ( sin30) (= ) g d gd B1 1.1 Allow gd  or 9.8d  KE (gained by) A and B: 2 2 1 1 2 2 (2)(1.75) (4)(1.75) + B1 1.1 KE (gained) for either A or B (so B1 for either 3.0625 or 6.125 or 9.1875) 2 2 1 1 1 2 2 2 (2)(1.75) (4)(1.75) (4 ) gd gd gd + = − − M1 3.4 Work energy principle – condone sign errors and slips but must be the correct number of terms (so must be considering KE for A and B and three terms in d) and dimensionally correct d = 0.375 A1 2.2a [5] 7 (c) • Consider the dimensions of the pulley or block(s) • Consider the weight/mass of the rope • More accurate value of g • Friction at the pulley • Elastic rope B1 3.5c Must be suggesting an improvement so B0 for ‘do not modelled blocks as particles’ (oe) B0 for ‘include air resistance/resistance/wind’ or ‘friction/resistance should be proportional to speed or speed squared’ [1] Y421/01 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 8 (a) Area below the curve 4 112 9 0 3 4 d x x = + =  B1 1.1 BC or 128 16 9 9 − (possibly embedded in other calculations) 4 1 2 0 3 4d 20 Ay x x = + =  B1 1.1 BC - y-coordinate of centre of mass of the lamina below the curve is 45 28 which implies B1B1 Might be seen in the evaluation of 4 4 2 1 1 2 2 0 (3 4)d 16 d 4 k x k Ay x x x k −   = + −   −          Possibly embedded in other calculations Centre of mass of triangle is at a distance of 4 3 from the x-axis B1 1.2 Used in a moment calculation (so not just stated) - possibly implied in later working If done by integration then this mark is awarded for correctly evaluating 4 2 1 2 16 d 4 k x k x k −     −     as 8 (4 ) 3 k − (or unsimplified) M1 2.1 Table of values idea – correct number of terms (dimensionally consistent) – so of the form 1 2 f( ) ( g( )) k k k k y   =   where f(k) and g(k) are equivalent to linear functions of k 1 4 112 1 20 (4)(4 ) (4)(4 ) 2 3 9 2 k k y     − − = − −         B1 1.1 For either the correct lhs or rhs seen (e.g. 40 2 9 k + or 28 8 3 3 k + or unsimplified equivalent expressions seen anywhere in their solution so do not need to be seen in a complete equation for y ) 42 12 20 9 k y k + = + A1 2.2a a = 42, b = 12, c = 20, d = 9 (or correct integer multiplies e.g. a = 84, b = 24, c = 40 and d = 18 etc.) [6] Y421/01 Mark Scheme June 2023 15 Question Answer Marks AO Guidance (b) 42 12 3 ... 20 9 2 k k k + =  = + M1 3.1a Sets their expression for y equal to 1.5 (which when fractions are cleared must be a linear equation in k) and attempts to solve for k 8 k = but 0 < k < 4 so y cannot be 1.5 A1 2.3 k = 8 plus reason (so some mention of k < 4) [2] Y421/01 Mark Scheme June 2023 16 Question Answer Marks AO Guidance 9 (a) sin , cos x g y g   = − = − B1 1.1 Possibly implied by later working 2 (20sin ) 0.5( ) y t y t  = + M1* 3.3 Applying 2 0.5 s ut at = + perpendicular to the plane – allow sin/cos confusion with the component of 20 (but not in terms of ) ( ) ( ) 2 3 1 12 5 2 13 20 0 ... t g t t − = = M1dep* 3.4 Sets y = 0 and solves for t (acceleration must have been a component of either sin or cos) 2.6 t = A1 1.1 Allow 26 g (provided g not later used as 9.8). Could be implied by correct values for the velocity components (provided g = 10) or correct value of OA ( ) ( ) 5 4 5 13 20cos ( sin ) 20 10 x g t t   = + − = − M1 3.3 Applying v u at = + (oe) parallel to the plane – allow sin/cos confusion (but correct use of theta/alpha for those components) – allow with their t or just t but must have substituted trigonometric values ( ) ( ) 3 12 5 13 20sin ( cos ) 20 10 y g t t   = + − = − M1 3.3 Applying v u at = + (oe) perpendicular to the plane – allow sin/cos confusion (but correct use of theta/alpha for those components) – allow with their t or just t but must have substituted trigonometric values 6, 12 x y = = − A1 2.2a Allow 12 ( ) ( ) 2 2 5 1 4 2 5 13 OA (20cos ) ( sin ) 20 5 t g t t t   = + − = − M1 3.4 Applying 2 0.5 s ut at = + parallel to the plane – allow with their t or just t (or equivalent e.g. 2 2 (20cos ) 2( sin )(OA) x g   = + − ) but must have substituted trigonometric values but allow sign errors OA = 28.6 (m) A1 2.2a If g missing from acceleration terms can score first two M marks only If using g = 9.8 then first and last A marks cannot be awarded (so max. 7/9 – look out for t = 2.653… and OA = 29.183…) [9] Y421/01 Mark Scheme June 2023 17 Question Answer Marks AO Guidance (b) Maximum value of 1 e  = B1 3.1b Indication that the greatest angle occurs when the collision is elastic (award if using the values for ,x y found in part (a)) 1 1 5 tan tan 12 y x  − −     = +         M1 3.4 Correct method for finding - all values substituted 86 = A1 3.2a For reference: 86.05481… - allow awrt 86 (but not from incorrect working) or awrt 1.50 (radians) [3] Y421/01 Mark Scheme June 2023 18 Question Answer Marks AO Guidance 10 (a) Initially, PE = ( sin30) ( 1.5 ) mg r r mgr + = B1 1.1 Or if reference level is through O then for sin30 mgr At angle , PE = ( cos ) mg r r  − B1 1.1 Or if reference level is through O then for cos mgr  − Or B2 for (sin30 cos ) mgr  + 2 2 1 1 2 2 ( sin30) ( cos ) mu mg r r mg r r mv  + + = − + M1* 3.3 Use of conservation of energy between initial position and when P is at B (at least 2 KE and 2 PE terms) 2 2 2 cos v u gr gr  = + + A1 1.1 Correct expression for the speed or speed-squared when P is at B (so must have been re-arranged or implied by later working) – correct expression for v or 2 v implies the first four marks 2 cos mv R mg r  − = M1* 3.3 N2L radially with correct number of terms and weight resolved (but allow sin/cos confusion and sign errors) – acceleration must be either 2 v r or 2 r   2 cos 2 cos m R mg u gr gr r   − = + + M1dep* 3.4 Substitute expression for 2 v (with correct number of terms) 2 3 cos mu R mg mg r  = + + A1 2.2a Must be simplified to three terms [7] (b) At A, 2 cos60 mu R mg r + = B1 3.3 2 2 3 cos cos60 4 mu mu mg mg mg mg r r      + + − − =             M1* 1.1 The difference of two expressions for R (their expression from (a) with three terms, and their two term expression for R at A with the equivalent of a single term for the resolved weight) equated to 4mg  3 5 2 6 3cos 4 cos   + =  = M1dep* 1.1 Obtain cos k = where 0 1 k  (or corresponding value of ) Vertical distance = 5 1 6 6 r r r − = A1 2.2a [4] Y421/01 Mark Scheme June 2023 19 Question Answer Marks AO Guidance (c) Diameter of the rim of the bowl is 2 cos30 ( 3) r r = B1 1.1 2 3 3 ( cos60) r r u t t u   = =       M1* 3.1b Realise that P leaves the inner surface with speed u and applies s ut = horizontally with s = their diameter ( ) ,2 r r  and component of u 2 2 2 2 3 1 2 3 6 ( sin60) 3 2 r r gr y u g r u u u       = − = −                   M1dep* 3.4 Applies 2 0.5 s ut at = + vertically with their value of t 2 2 6 0 3 0 gr y r u   −  M1 3.1b Sets their expression for y (in terms of r, g and u) > 0 – dependent on both previous M marks (condone ‘equals’ or ‘greater than or equal to’ zero for this mark) ( ) 2 2 3 2 0 2 r u gr u gr −    A1 2.2a AG (if using non-strict inequality or equals then argument for strict inequality must be convincing) [5] Diameter of the rim of the bowl is 2 cos30 ( 3) r r = B1 2 1 2 cos30 0 ... u t gt t − = = M1* Finding the time (from 2 0.5 s ut at = + with s = 0, a component of u and a g =  ) when P is at the same horizontal level as the rim of the bowl – if correct then 3 u t g = 1 2 3 ( sin30) u x u t u g   = =       M1dep* Using their t and s = ut horizontally (with a component of u) to find the horizontal distance travelled by P when it is at the same horizontal level as the rim 1 2 3 3 u u r g         M1 Sets their x > their 3 r (not just r or 2r) – condone ‘equals’ or ‘greater than or equal to’ – dependent on both previous M marks 2 2 u gr  A1 AG Y421/01 Mark Scheme June 2023 20 Question Answer Marks AO Guidance 11 (a) For reference: M1* 3.3 Attempt to resolve horizontally or vertically (with correct number of terms) allow if using 15g for the M mark (condone in words ‘P = Fr at A + Reaction at C’ for M1 only – for the A mark must be using symbols) A C A C 15, R F P F R = + = + A1 1.1 Both correct (Least value of P implies) A A C C 0.5 , 0.5 F R F R = = B1* 3.1b For both (no justification required) Taking moments to form an equation containing all relevant forces M1* 3.3 Taking moments about e.g. A, B, C etc.– correct number of terms (at least 4 terms if A or C and 5 terms if B) – one for each force), dimensionally consistent – allow sin/cos confusion. Must have numerical values for all distances and each term must be resolved with a term in sin / cos   C C (3cos ) 4(15cos ) 1(15sin ) 12( sin ) 12( cos ) P R F      + = + − A A 8(15cos ) 1(15sin ) (12sin 3cos ) (12cos ) (12sin ) P R F       + + − = + A2 1.1 1.1 Award A1 for any two terms correct A2 for a fully correct equation C C (3cos ) 15 17 sin( arctan4) 12( sin ) 12( cos ) P R F     = − + − 3 cos 60cos 15sin 4 30 4 30 12sin 6cos 5 5 P P P      + − − −     = −         M1dep* 1.1 Deriving an equation containing P and only (Reference: C C C 0.5 15 2( ) 0.5 15 A R R P R R = +  − = + C 4 30 5 P R −  = ) 120cos 285sin 40cos 95sin 48sin 39cos 16sin 13cos P         + +  = = − − A1 2.2a AG Possible intermediate step is 15 cos 300cos 75sin 12(4 30)sin 6(4 30)cos P P P      + − = − − − [8] Y421/01 Mark Scheme June 2023 21 Question Answer Marks AO Guidance 11 (b) 0 16sin 13cos 0 P     −  M1 3.4 Considers denominator of given answer in (a) either > 0 or = 0 or 0 13 tan 39.09... 16      A1 1.1 awrt 39 For the prism to be in equilibrium tan 4  M1 3.1b Considers tan y x = (or reciprocal) for triangular lamina 75.96...  A1 1.1 awrt 76 (For full marks must imply an interval between these two values) [4] Y421/01 Mark Scheme June 2023 22 Question Answer Marks AO Guidance 12 (a) (i) Spheres are smooth and so the impulse acts in a direction parallel to the line of centres (which is in the i direction) B1 2.4 [1] (a) (ii) M1* 3.3 Use of conservation of linear momentum (parallel to the line of centres) – correct number of terms (condone no masses present for this mark) but condone sign errors M1* 3.3 Use of Newton’s experimental law (parallel to the line of centres) – correct number of terms 1 A B mu mv mv  = + A B 1 v v eu − = − A1 1.1 Use of NEL must be consistent with CLM e.g. 1 A B mu mv mv  = − + and A B 1 v v eu + = ( A v and B v are the horizontal components of the velocities of A and B after impact parallel to the line of centres) Solving simultaneously to find either A v or B v in terms 1, u e and  M1dep* 1.1 For reference: 1 1 A 1 (1 ) (1 ) 1 1 u e u e v eu    + − = − = + + 1 B (1 ) 1 u e v  + = + M1 3.3 Using either mw  =  I i or ( ) A 1 m v u =  − I i (condone lack of i) to get I in terms of , , m e  and 1u only. Dependent on all previous M marks. Must be using the correct mass Impulse = 1 1 1 e m u   +     +   i A1 2.5 Allow unsimplified e.g. 1 1 (1 ) 1 u e m u   −   − −   +   i - condone lack of i [6] Y421/01 Mark Scheme June 2023 23 Question Answer Marks AO Guidance (b) 2 2 1 1 1 2 2 2 ( ) mu mu + B1 1.1 KE before collision - or just considering energy in the i direction (so B1 for 2 1 1 2 mu ) ( ) ( ) ( ) 2 2 2 1 2 1 1 1 2 2 2 1 1 1 1 u e e m m u       − +    +     +   +   2 1 2 2 ( ) mu + B1 B1 1.1 KE after collision – B1 for each correct term (need only consider the energy in the i direction) For reference: 1 1 A 1 (1 ) (1 ) 1 1 u e u e v eu    + − = − = + + 1 B (1 ) 1 u e v  + = + 2 2 1 1 1 1 1 1 2 2 1 2 1 8 e e     − +     − − =     + +     M1* 1.1 Using KE before – KE after = 2 1 1 8 mu  to obtain an equation in and e only For reference only : ( ) 2 2 3 2 1 4 1 e    + − = + and 2 2 2 (4 3) (4 2) 1 0 e e   − + − + = 2 1 1 1 4 e     − =     +    2 3 1 4 e   − = 3 1 0 4 1   −  M1dep* 3.1b Re-arranging to make 2 e (or e) the subject from an equation containing terms in  (possibly 2 ) and 2 e (not e) only and considering either 2 0 e or 2 1 e (condone ‘equals’ or ‘strict’ comparison). Or obtaining an equation in (possibly 2 ) and 2 e (not e) only and setting e = 1 or e = 0 1 3  A1 2.2a A0 if strict inequality or an upper limit for the value of  stated [6] Y421/01 Mark Scheme June 2023 24 Question Answer Marks AO Guidance 13 (a) 0 T mg = M1 1.1 Resolving vertically for P 2 0 0 man e T a = M1 3.3 Apply Hooke’s law – where 0e is the extension of the spring from its natural length 2 2 0 0 man e mg e gn a − =  = A1 1.1 [3] (b) 0 y a e a x z + + = + + M1 3.1b Attempt an equation containing z where z is the upward displacement of P from its initial position (at time t) – must include terms in y, x and 0e (allow sign errors only) 2 ( ) z y gn x − = + − A1 1.1 [2] (c) ( ) z kt x = − B1 1.1 [1] (d) T mg mz − = M1* 2.1 Apply N2L vertically – correct number of terms but allow sign errors (allow for T mg ma − = ) – implied by a correct equation 2 ( ) man x mg m kt x a − = − or 2 mx mkt mg mn x = + − M1dep* 3.4 Use Hooke’s law with extension = x and acceleration equal to their expression from part (c) – if (c) blank then must be correct 2 2 n x g kt x x n x kt g − = −  + = + A1 2.2a AG – so sufficient working must be shown – any errors then A0 [3] Y421/01 Mark Scheme June 2023 25 Question Answer Marks AO Guidance (e) 3 1 ( sin( )) x knt gn k nt n For reference: 2 x n x kt g + = + and 3 1 ( sin( )) x knt gn k nt n = + − 3 1 ( cos( )) x kn kn nt n = − B1 1.1 Correct derivative 2 3 1 ( sin( )) x kn nt n = B1ft 1.1 Follow through their first derivative 2 2 3 3 1 1 ( sin( )) ( sin( ) sin( ) sin( ) kn nt n knt gn k nt n n k nt ktn gn k nt kt g n n   + + −     + − = + = + B1 1.1 Correctly confirm given differential equation - at least one stage of working before obtaining kt + g When t = 0, 2 3 1 ( ) x gn gn n and 3 1 ( cos(0)) 0 x kn kn n = − = B1 3.4 Confirm initial conditions [4] CF: cos sin A nt B nt + B1* GS: 2 2 cos sin kt g x A nt B nt n n = + + + B1* From a trial solution of the form 2 2 , 0 , g k x t x x n n      = +  = =  = = Initial conditions: 2 0, g t x n = = and 0 x = B1dep* 3 3 2 2 0, sin k k kt g A B x nt n n n n = = −  = − + + B1 Y421/01 Mark Scheme June 2023 26 Question Answer Marks AO Guidance (f) 2 (1 cos( )) k x nt n = − but cos( ) 1 nt for all values of t so 0 x B1* 3.1b Using correct x (possibly seen in part (c) and allow if correctly stated there) to conclude that x is always greater than or equal to 0 or between 0 and 2 2k n (must imply that it could be zero) We can therefore infer that P does not move closer to the ceiling of the lift in the subsequent motion B1dep* 2.2b or equivalent comment e.g. in the subsequent motion P is never nearer the ceiling of the lift as it was when t = 0, or ‘is always moving away from the ceiling’ [2] Need to get in touch? 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