H240/03 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. H240/03 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. H240/03 Mark Scheme June 2022 5 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. • 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 June 2022 6 Question Answer Marks AO Guidance 1 6 x = B1 1.1 B0 for 6 x = unless replaced with x = 6 2 3 9 2 3 9 x x − = ⇒ − = − M1 1.1a oe e.g.( ) 2 2 3 81 x − = or 2 3 9 x − = ± M0 for( ) 2 2 3 9 x − = 3 x = − A1 1.1 A0 for 3 x = - unless replaced with 3 x = - The correct answer implies the previous M mark Two marks maximum if more than two answers given [3] H240/03 Mark Scheme June 2022 7 Question Answer Marks AO Guidance 2 (a) Translation B1 2.5 Do not accept shift, move, transformation, etc. for first B1 8 - units parallel to the y-axis or ቀ0 −8ቁ B1 1.1 Correct description e.g. correct vector (not as a coordinate), ‘8 units down’. Do not allow second B1 after incorrect type of transformation e.g. stretch/rotation etc. but allow after shift/move etc. Condone lack of ‘units’ but do not accept ‘factor 8 - ’, ‘8 places/spaces/steps down’ etc. For ‘parallel to the y axis’ allow ‘vertically’, ‘in the y direction’. Do not accept across/up/along/to/in/ towards the y axis’ [2] If more than a single transformation, then no marks (unless two translations equivalent to the correct answer) Mark vector before description/words 2 (b) 3 3 8 8 y x y x = −⇒ + = ... x = M1 1.1 Attempt to make x the subject (allow sign errors only) May use either f(x) or y ( ) ( ) 1 3 1 f 8 x x − = + A1 1.1 Must be in terms of x. Allow the expression only e.g. ( ) 1 3 8 x + , 3 8 x + , oe Ignore what this expression is equated to [2] 2 (c) One graph is the reflection of the other graph in the line y x = B1 1.2 Must include both ‘reflection’ or ‘mirror image’ or ‘mirrored’ and ‘y = x’ B0 for ‘symmetrical’ unless clearly describing reflective symmetry [1] H240/03 Mark Scheme June 2022 8 Question Answer Marks AO Guidance 3 2 3 5 1 , 2 2 + −+       M1* 2.1 Attempt at the midpoint (3 out of 4 values used correctly or one correct coordinate) If correct 5 , 2 2   −     ( ) ( ) 2 2 2 2 3 5 1 d = − + −− M1* 1.1 Attempt to calculate diameter d or radius r (or the square of either of these two) e.g. ( ) ( ) ( ) 2 2 2 5 2 2 5 2 r = − + −−− M0 if implying that ( ) ( ) 2 2 2 2 3 5 1 r = − + −− . If the value is not labelled (as either r or d) then consider how this value is used in their circle equation 3 out of 4 values used correctly – if correct 2 37 d = or 2 37 4 r = ( ) 2 2 5 37 2 2 4 x y   − + + =     M1dep* 1.1 Correct form for the equation of a circle with their values for the centre and radius Dependent on both previous M marks 2 2 25 37 5 4 4 4 4 x x y y − + + + + = 2 2 5 4 1 0 x y x y + − + + = A1 2.2a Must have = 0 Order of terms on the lhs irrelevant a, b and c need not be explicitly stated [4] H240/03 Mark Scheme June 2022 9 Question Answer Marks AO Guidance 4 (a) GP: 𝑥𝑥, 15 𝑦𝑦,  𝑧𝑧⇒ 15 𝑦𝑦 𝑥𝑥= 𝑧𝑧 15 𝑦𝑦 M1* 3.1a M1 for un-simplified correct use of 1 n n u r u − = so allow this mark for stating that 15 r xy = or 15 yz r = or 2r x z = oe Or for the terms of the GP as 15 4, , 4 y y y + − in y only AP: , , 4 x y z y x ⇒ − = − or 4 z y − = − M1* 1.1 M1 for 4 y x − = ± or 4 z y − = ± or 8 z x − = ± oe Or for the terms of the AP as 4, , 4 y y y + − in y only For reference: 2 225 xy z = , 4 x y = + and 4 z y = − ( ) ( ) 2 4 4 225 y y y + − = M1dep* 1.1 Eliminate x and z to form an equation in y only (must be equivalent to a quartic in y) Or for 𝑦𝑦−4 15 𝑦𝑦 = 15 𝑦𝑦 𝑦𝑦+ 4 ൬⇒𝑦𝑦2 −16 = 225 𝑦𝑦2 ൰ 2 2 4 2 ( 4 4 16) 225 16 225 0 y y y y y y − + − = ⇒ − − = A1 2.2a AG so sufficient working must be shown www – note that 4 y x = + and 4 y z = - (from d = +4) can lead to the correct equation (which can score the M marks only) [4] H240/03 Mark Scheme June 2022 10 Question Answer Marks AO Guidance 4 (b) 4 2 16 225 0 5 y y y − − = ⇒ = ± but 0 5 y y > ⇒ = B1 1.1 BC Allow implied e.g. 5 y = ± then using y = 5 only (B0 if the sum to infinity for other values of y are not rejected) 9, 1 x z = = 2 1 1 9 3 z r r x ⇒ = = ⇒ = M1 1.1 Calculate r (soi) corresponding to y = 5 allow unsimplified e.g. (5)(1) 15 r = and allow if more than one value of y stated Possibly done implicitly in formula for S∞ 2 3 9 1 x S r ∞= = − M1 1.1 Using the correct formula for the sum to infinity of a GP with their value of x ( their 4 y = ± ) and a value of r where 1 1 r −< < 13.5 S∞= A1 1.1 cao oe eg 27 2 - do not award this mark if more than one value for S¥ stated A0 for a triple-decker fraction e.g. 2 3 9 [4] H240/03 Mark Scheme June 2022 11 Question Answer Marks AO Guidance 5 (a) DR ( ) ( ) ( )( )( ) ( ) 1 2 1 2 2 2 (2 3) 4 1 d 2 4 1 2 3 1 4 1 8 d y x x y x x x x x − − − = − + ⇒ = + + − − + M1* 1.1 Attempt use of quotient rule or equivalent (e.g. product rule). Condone one incorrect term only (of the five terms) but must be subtraction in the numerator (but allow subtraction the wrong way round); condone absence of brackets; no denominator (if using quotient rule) is M0 By the five terms we mean the four in the numerator and the fifth is the term in the denominator ( )( ) ( )( ) ( ) 2 2 2 2 4 1 2 2 3 8 2 3 d d 4 1 4 1 x x x x y y x x x + − − − = ⇒ = + + A1 1.1 cao must include brackets as necessary Any correct equivalent form ( ) ( ) 2 2 2 2 1 12 4 2 4 1 x x x + − = + M1dep* 3.1a Sets their derivative (in any form) equal to 2 (M0 if equating to normal gradient) May equate at any stage (even after incorrect manipulation of their derivative) ( ) 2 2 2 4 2 1 12 4 4 1 16 12 12 0 x x x x x x + − = + ⇒ + − = M1 1.1 Multiply both sides by ( ) 2 2 4 1 x + and simplify (so combining like terms) to obtain a quartic equation (must be expanded with at least three terms – condone lack of = 0 if all terms on the same side) – allow sign errors/minor slips but the expansion of ( ) 2 2 4 1 x + must be three terms of the form 4 2 16 1 x ax + + where 4, 8 a = ± ± Dependent on both previous M marks ( ) 3 3 4 3 3 0 4 3 3 0 x x x x x + − = ⇒ + − = as 0 x ≠ A1 2.3 AG with explicit rejection of x = 0 – as a minimum must indicate that x cannot equal 0 Just cancelling x is A0 [5] H240/03 Mark Scheme June 2022 12 Question Answer Marks AO Guidance 5 (b) DR Consider both ( ) f 0.5 and ( ) f 1 Where ( ) ( ) 3 f 4 3 3 x x x = ± + − M1 1.1 Working or correct answer for one value is sufficient evidence of correct method but both 0.5 and 1 must be seen Just stating that ( ) f 0.5 0 < and ( ) f 1 0 > is M0 ( ) f 0.5 1 0 = −< and ( ) f 1 4 0 = > (or ( ) f 0.5 1 0 = > and ( ) f 1 4 0 = −< ) Change of sign indicates that the x-coordinate lies between 0.5 and 1 A1 2.4 Correct values together with explanation (change of sign) and correct conclusion (as a minimum ‘root’ oe) [2] Alternative Considers both ( ) g 0.5 and ( ) g 1 where ( )( ) ( )( ) ( ) 2 2 2 4 1 2 2 3 8 g( ) 4 1 x x x x x + − − = + M1 Must be using the correct derivative. Working or correct answer for one value is sufficient evidence of correct method but both 0.5 and 1 must be seen Just stating that ( ) g 0.5 2 > and ( ) g 1 2 < is M0 ( ) g 0.5 3 2 = > and ( ) g 1 0.72 2 = < Values either side of 2 indicates that the x-coordinate lies between 0.5 and 1 A1 Correct values together with explanation (values either side of 2) and correct conclusion (as a minimum ‘root’ oe) [2] H240/03 Mark Scheme June 2022 13 Question Answer Marks AO Guidance 5 (c) DR Let ( ) 3 2 3 4 h h ( ) 4 3 x x x x − ′ = ⇒ = − B1* 2.1 Calculates correct derivative of rhs of given iterative formula As the root α lies in the interval (0.5, 1) ( ) h 1 α ′ ⇒ < − so iterative formula cannot converge to the x-coordinate of P B1dep* 2.2a Correct explanation that any value in the given interval gives a gradient which is less than –1 No marks for just showing that the iteration doesn’t converge using different starting values [2] 5 (d) DR ( ) ( ) 3 2 f 4 3 3 f 12 3 n n n n n x x x x x ′ = + −⇒ = + B1 1.1 Correct derivative (possibly seen in N-R formula) Condone x for nx oe 3 1 2 4 3 3 12 3 n n n n n x x x x x + + −   = −  +   M1 2.1 Correct N-R formula seen with correct ( ) f nx and their ( ) f nx ′ substituted Condone x for nx oe 29 2 0 1 2 3 45 3 0.5, or 0.666666..., or 0.644444..., ( 0.64395510...) x x x x = = = = A1 1.1 First two iterations correctly stated to at least 5 decimal places (or exact) (truncated or rounded) The correct first two iterations can imply B1 M1 x coordinate of P is 0.64395 A1 2.2a Independent of previous A mark (but must have scored B1 M1) – must be stated to exactly 5 decimal places This A mark does not imply the previous A mark y coordinate of P is –0.64395 B1 1.1 Independent of all previous marks – must be stated to exactly 5 decimal places The correct answers with no evidence of N-R (e.g. no iterations stated and no N-R formula) then B0M0A0A0B1 max. [5] H240/03 Mark Scheme June 2022 14 6 DR M1 2.1 M1 for ( ) 3 2 2 9 k x + with non-zero k 1 k ≠ ( ) ( ) 3 1 2 2 1 2 9 d 2 9 3 x x x + = + ∫ A1 1.1 cao (allow unsimplified) ( ) 0 3 2 9 2 2 9 9 3 x −   +   =       A1 1.1 Uses correct limits (or implies correct limits) to get 9. Condone limits the wrong way round leading to 9 − but must be changed to +9 2 2 1 4e 1 0 e 4 x x − − −= ⇒ = 1 2 ln 4 x   − =     M1* 3.1a Attempt to solve 2 4e 1 0 x − −= by correctly taking logs of both sides leading to 2 ln x α ± = ± where 0 α > Allow sign errors and other minor slips only 1 1 ln 2 4 x   = −     A1 1.1 Or equivalent exact value (soi possibly by correct exact value used later) e.g. 1 ln 4 2 or ln 2 ( ) 2 2 4e 1 d 2e x x x x − − − = − − ∫ M1* 1.1 Integrate 2 4e 1 x − − to obtain 2 e x c x − ± Where c is non-zero and 4 c ≠ ( ) ( ) 1 2 ln4 2 ln4 0 1 4e 1 d 2e ln 4 2 2 x x − −   − = − − −−     ∫ M1dep* 1.1 Uses limits correctly 1 2( ln 4) (0) F F - (with their 1 2 ln 4 ) – dependent on the previous two M marks (allow non-exact top limit). Condone limits the wrong way round only if the sign of their answer is subsequently changed If zero limit is assumed to give 0 (with no working) then M0 Area = 3 1 21 1 21 9 ln 4 ln 4 ln 2 2 2 2 2 2 + − = − = − A1 2.2a If the values of the integral(s) are changed from negative to positive (e.g. from limits the wrong way round) with no justification given then A0 p and q need not be explicitly stated. 21 2 p = (oe) and 1 q = − [8] H240/03 Mark Scheme June 2022 15 7 (a) DR sec 3cos 4sin m θ θ θ + = 3 sec 4sin sec m θ θ θ   ⇒ + =     2 sec 3 4sin sec m θ θ θ + = M1 2.1 Or for sec 3cos 4sin m θ θ θ + = sec sin 3 4 cos cos m θ θ θ θ   ⇒ + =     2 sec 3 4tan m θ θ + = The first M mark is for a valid method arriving at a three term equation containing 2 sec θ Squaring each individual term of the original equation scores no marks ( ) 2 1 tan 3 4tan m θ θ + + = M1 1.1 Correctly uses the identity 2 2 1 tan sec θ θ + ≡ to obtain an equation in tanθ only 2 tan 3 4tan m m θ θ + + = ( ) 2 tan 4tan 3 0 m m θ θ ⇒ − + + = A1 2.2a AG so sufficient working must be shown A0 if angle missing from any trigonometric terms [3] H240/03 Mark Scheme June 2022 16 7 (b) DR ( ) ( ) 2 4 4 3 m m ∆= − − + M1* 3.1a Considers discriminant of given quadratic equation in tan (c must be two terms) to get an expression in m only. M0 for embedded discriminant in quadratic formula unless explicitly considered Allow 2 4 4 ( 3) m m - + As the quadratic equation in tan has only one solution for θ in the given interval (and as the range of tan in the given interval is all non-zero real values) this implies that the given equation must only have one real root and therefore ( ) 2 2 4 4 ( 3) 0 3 4 0 m m m m − − + = ⇒ + − = M1dep* 1.1 Sets their discriminant equal to zero and obtains an expanded three-term quadratic in m Reasoning for setting the discriminant equal to zero is not required for this mark ( 4)( 1) 0 4 m m m + − = ⇒ = − only as m is a negative integer A1 1.1 State or imply 4 m = −only ( ) 2 4 2tan 1 0 m θ = −⇒ + = so tan 0.5 θ = − M1 1.1 Uses their negative integer value of m, and solves the equivalent of their three term quadratic equation in tan, to obtain (at least) tan k θ = - dependent on both previous M marks Allow 2 4tan 4tan 1 0 θ θ - - - = tan 0.5 θ Þ = - for this mark where k is non-zero. If no method shown for solving their quadratic, then award this mark if the solution is correct for their quadratic 2.68 θ = (3 sf) A1 2.4 For full marks must explain why the discriminant should be set equal to zero – must say that as there is only one value of θ or tan θ 0 ∆ Þ = (as a minimum must see explicit mention of ‘one’ together with ' ' or 'tan ' θ θ for this mark). Allow awrt 2.68 2.677945045... [5] H240/03 Mark Scheme June 2022 17 8 cos15 60 T = M1 1.1 Resolve horizontally, correct number of terms, allow sin/cos confusion (e.g. sin15 60 T = scores M1 but 60cos15 T = scores M0) Where T is the tension in the rope 62.1 T = (N) A1 1.1 awrt 62.1 – allow the exact answer of ( ) 60 6 2 − 62.11657082… [2] H240/03 Mark Scheme June 2022 18 9 (a) 8 20 30a = + M1 3.4 Use of v = u + at with given values (allow v = 20 and u = 8) 0.4 a = − so deceleration is 0.4 ( 2 ms−) A1 1.1 Allow 0.4 or –0.4 [2] 9 (b) Distance travelled by B: 12t B1 1.1 Or first 30 seconds: B travels 12(30) (= 360) Distance travelled by A: ( )( ) ( ) 1 8 20 30 8 30 2 t + + − or ( )( ) 1 30 12 8 2 t + B1 1.1 Or first 30 seconds: A travels ( ) 0.5 8 20 (30) ( 420) + = 12 '420' 8 '240' t t = + − M1 3.1b M1 for a (possibly unsimplified) correct equation/inequality in t where ‘360’ and ‘420’ must have been correct unsimplified e.g. ( )( ) '420' '360' 30 12 8 , t − = − − 12( 30) '360' 8( 30) '420', t t - + = - + '420' 8( 30) 12 t t + - = etc. Or for a correct equation in another time variable e.g. 12 '360' 8 '420' T T + = + and 30 T t = - seen or implied If an inequality used, then allow incorrect direction or strict inequality symbol for this mark 45 t = A1 1.1 Allow t > 45 Þ t = 45 [4] H240/03 Mark Scheme June 2022 19 10 (a) 3 16.8 3 g a − = M1 3.3 N2L for Q – correct number of terms with the correct mass. Condone sign errors M0 if using 3g for the mass but allow g missing from the net force 3 16.8 4.2 3 g a − = = ( 2 ms−) A1 [2] 1.1 10 (b) 16.8 2.5(4.2) P F − = M1* 3.3 N2L for P horizontally using T = 16.8 and their a (but not 9.8 ± ) from (a) – allow sign errors but must have correct number of terms and correct mass ( 3 ¹ ) – if correct 6.3 P F = P F is the friction between P and B ( ) 16.8 2.5(4.2) 2.5g µ − = M1dep* 3.4 Use of F R µ = for P with R = 2.5g 0.257 µ = A1 1.1 awrt 0.257 – condone working with F R µ „ provided that the value of µ is explicitly stated (and not left in an inequality) 0.257142857… allow 9 35 [3] 10 (c) 2.5 B R g Mg = + M1* 3.1b Resolving vertically for B – correct number of terms, allow sign errors and condone g’s missing Where M is the mass of B ( ) 5 6.3 2.5 49 g Mg + „ M1dep* 3.4 Use of F R µ „ or F R µ = with correct R and 5 49 µ = with F being their P F from (b) where 16.8 P F ¹ No g’s missing for this mark 3.8 M … so least possible value for the mass of B is 3.8 (kg) A1 2.2a 3.8 - allow use of ‘=’ throughout this part No justification required [3] H240/03 Mark Scheme June 2022 20 11 (a) M1 3.1b Taking moments about A – correct number of terms, allow sin/cos confusion and sign errors – may take moments about another point (and resolve) but must end up after elimination with an equation in T and x only Dimensionally correct. Must be xT and not cos(...) or sin(...) xT xT ( ) ( ) 3 2 4 cos60 3cos60 g g xT + = A1 1.1 Correct equation in g, x and T only – condone g replaced by 9.8 T is the tension in the string 9 2 g T x = (N) A1 2.2a An answer of 44.1 x (or with trigonometric terms) is A0 unless correct answer in terms of g seen oe exact answers in terms of g and x (condone correct triple decker fractions) [3] For reference for parts (a) and (b): Moments about C: ( cos60) (3 )cos60 4 ( 1.5)cos60 ( sin 60) A A R x g x g x F x + - = - + Moments about B: (3 ) (3cos60) 4 (1.5cos60) (3sin 60) A A T x R g F - + = + Moments about midpoint of AB: (1.5cos60) (1.5cos60) ( 1.5) (1.5sin 60) A A R g T x F + = - + Resolving perpendicular to AB: cos60 4 cos60 cos60 sin 60 A A T R g g F + = + + Resolving parallel to AB: sin 60 cos60 4 sin 60 sin 60 A A R F g g + = + H240/03 Mark Scheme June 2022 21 11 (b) M1 3.3 Resolve vertically or horizontally – correct number of terms with the tension at C in terms of cos/sin, condone sign errors, allow sin/cos confusion but forces that require resolving must be (and correspondingly those that don’t require resolving e.g. the weights if resolving vertically, should not be resolved) Or obtain an equation in A F and/or A R in terms of T (or their T) only (see list of equations below) cos60 4 A T R g g + = + 9 5 4 A g R g x   ⇒ = −     sin60 A F T = 9 3 4 A g F x   ⇒ =     A1 1.1 Both correct (unsimplified) – allow with T or their (possibly incorrect) T (oe eg two valid equations in A R and A F ) A R is the normal contact force at A A F is the frictional contact force at A 9 3 9 3 9 5 4 35 4 g g g x x   = −     M1dep* 3.4 Use of F R µ = with correct µ to form an equation in x only – no forces missing from their A R and A F and all required forces resolved accordingly or not e.g. if resolving vertically the two weights should not contain sin/cos 2.2 x = A1 2.2a awrt 2.2 www [4] For reference for parts (a) and (b): Moments about C: ( cos60) (3 )cos60 4 ( 1.5)cos60 ( sin 60) A A R x g x g x F x + - = - + Moments about B: (3 ) (3cos60) 4 (1.5cos60) (3sin 60) A A T x R g F - + = + Moments about midpoint of AB: (1.5cos60) (1.5cos60) ( 1.5) (1.5sin 60) A A R g T x F + = - + Resolving perpendicular to AB: cos60 4 cos60 cos60 sin 60 A A T R g g F + = + + Resolving parallel to AB: sin 60 cos60 4 sin 60 sin 60 A A R F g g + = + H240/03 Mark Scheme June 2022 22 12 (a) ( ) ( ) 2 1 2 2 13 t t t = − + + − v i j If P is stationary, then 1 2 0 t − = and 2 2 13 0 t t + − = M1 3.1b Considers either the i or j component equal to zero or forms a five-term quartic equation for 2 0 v = (oe) 4 3 2 (4 4 47 30 170 0) t t t t + - - + = 1 2 :1 2 0 t t − = ⇒= i 2 : 2 13 0 2.3117..., 2.8117... t t t + − = ⇒= − j No value of t is common to both components, so P is never stationary A1 [2] 2.2a BC – need not see the negative value of t or for substituting t = 0.5 into quadratic expression for j and showing this gives a non-zero answer (oe) with correct working and conclusion A1 for the correct quartic equation with roots stated as 2.3 0.35i, 2.8 0.65i ± - ± + correct conclusion 12 (b) ( ) 2 4 1 t ⇒ = − + + a i j (m s–2) B1 [1] 1.1 Correct derivative – or as a column vector Brackets must be around the 4t + 1 H240/03 Mark Scheme June 2022 23 12 (c) ( ) ( ) 2 2 2 13 1 1 2 t t t − + − = − M1* 3.1b Setting up a quadratic equation in t only – allow sign errors (including on the 1 and 2) and the 1 and –2 on the wrong side Or multiples of 1 and –2 2 4 25 0 t − = 2.5 t ⇒= M1dep* 1.1 Solves their (two or three term) quadratic and selects their positive value of t Check unsupported solutions if incorrect quadratic equation ( ) { } 0.5 2 4 1 m t = ⇒ = − + + F a F i j M1* 3.4 Substitute their a into F = 0.5a or their a into 0.5 . = F a If F not stated in terms of t then one component must be correct following through from their a (and possibly t) Must use correct value of 0.5 for m but can be in terms of t ( ) 2 2 1 5.5 = − + F M1dep* 1.1 Dependent on previous M mark only From a value of t > 0 5.59 = F (N) A1 [5] 1.1 awrt 5.59 (exact: 5 5 2 ) 5.590169… H240/03 Mark Scheme June 2022 24 12 (d) ( ) ( ) 2 3 2 2 1 13 3 2 t t t t t   = − + + − +     s i j c M1* 1.1 Integrates v wrt t – at least three terms correct Allow without +c ( ) 1 1, 0 12 6 t = = ⇒ = + s j c i j A1 3.4 Uses given conditions to find correct c – dependent on a completely correct integrated expression for s www When t = 1.5, 3 33 4 8 = − − s i j M1dep* 1.1 Substitute t = 1.5 into their s 𝑡𝑡𝑡𝑡𝑡𝑡−1 ቀ ±3/4 ±33/8ቁ or 𝑡𝑡𝑡𝑡𝑡𝑡−1 ቀ ±33/8 ±3/4 ቁ M1 3.1b Attempt to find a relevant angle using the components of their s (allow use of sin/cos with the magnitude of s) Dependent on both previous M marks. Written in terms of arctan is sufficient Bearing = 1 3/ 4 180 tan 33/ 8 −  +     = 190° A1 3.2a awrt 190 (or from 270 −𝑡𝑡𝑡𝑡𝑡𝑡−1 ቀ 33/8 3/4 ቁ) 190.3048465… [5] H240/03 Mark Scheme June 2022 25 13 (a) x Ut = 2 1 2 y Vt gt = − M1* 3.3 Setting up expressions for x and y using 2 1 2 s ut at = + with a = 0 in x and g ± in y oe. M0 if using U instead of V vertically Allow sign errors. May use t = x / U in 2 1 2 y Vt at = + 2 1 2 x x y V g U U     = −         M1dep* 3.4 Eliminating both t terms in y to get an equation in y, x, U, V (and possibly g) 2 2 2 2 2 2 2 Vx gx y U y UVx gx U U = − ⇒ = − A1 2.2a AG so sufficient working must be shown www [3] 13 (b) B passes through ( ) 2 2 1 2 , 2 2 2 a a a U UVa ga   ⇒ = −     ( ) 2 2 U UV ga ⇒ = − B1 3.4 Substituting ( ) 1 2 ,a a into given result from (a) B passes through ( ) ( ) ( ) 2 4 , 0 2 4 4 0 a UV a g a ⇒ − = ( ) 2 0 UV ga ⇒ − = B1 3.1b Substituting ( ) 4 , 0 a into given result from (a) – note that using (3 ,0) a is not a MR 2 3 , 3 ga U ga V = = or 2 3 U V = M1* 2.1 Solve simultaneously (oe) to find either U or V (or their squares) in terms of a and g only, or for a linear equation (oe) in V and U only, if correct 2U = 3V oe e.g., if correct 2 4 3 V ga = and 2 3 U ga = ( ) 2 3 tan tan 3 ga V U ga θ θ = ⇒ = M1dep* 3.1b Using tan V U θ = with their U and their V 2 tan 33.7 3 θ θ = ⇒ = ° (3 sf) A1 2.2a awrt 33.7 (an answer of 36.9 from using (3 ,0) a scores (if from correct working) B1 B0 M1 M1 A0) [5] H240/03 Mark Scheme June 2022 26 13 (c) 4 3 54.6 3 ga ga + = M1 3.4 Using 2 2 54.6 U V + = to set up an equation in a (and possibly g) a = 70.2 A1 1.1 www awrt 70.2 an answer of 97.344 (awrt 97.3) from using (3a, 0) in (b) scores M1 A0 [2] Alternative 1 54.6cos(33.7...) 3ga = or 4 3 54.6sin(33.7...) ga = M1 Setting either the horizontal component equal to their U (from (b)) or the vertical component equal to their V (from (b)). Allow sin/cos confusion M0 for an unsupported value of θ (if used) a = 70.2 A1 www awrt 70.2 [2] Alternative 2 54.6cos(33.7...) 54.6sin(33.7...) 2 2 UV a g g ´ = = M1 Using their expression for a (possibly seen in (b)) in terms of U and V with 54.6 and their θ . M0 for an unsupported value of θ Allow sin/cos confusion a = 70.2 A1 www awrt 70.2 [2] H240/03 Mark Scheme June 2022 27 13 (d) 2 2 0 2 2 V V gH H g   = − ⇒ =       M1* 3.3 Setting up the model using 2 2 2 v u as = + with 0 v = and a g = - H is the maximum height of B 1 4 2 2 3 3 ga H a g   = =     M1dep* 3.4 Using their expression for V from (b) to get an expression for H in terms of a (oe) e.g., 2 where 54.6sin 2 V H V g θ = = (allow cosθ ) with their value of θ M0 for an unsupported value of θ ( ) 2 70.2 46.8 3 H = = (m) A1 2.2a awrt 46.8 an answer of 54.756 (awrt 54.8) from using (3a, 0) in (b) scores M1M1A0 [3] Alternative 1 0 V gt = - 54.6sin(33.7...) t g Þ = M1* Find t at maximum height with v = 0, a g = - and u = their V 54.6 ¹ (allow sin/cos confusion). M0 for an unsupported value of θ t = 3.090472522… 2 1 2 2 1 2 (54.6sin(33.7...)) (54.6sin(33.7...))(3.09...) (3.09...) H t gt g = - = - M1dep* Substituting their t into 2 1 2 s Vt gt = - V 54.6 ¹ (allow sin/cos confusion) H = 46.8 (m) A1 awrt 46.8 [3] Alternative 2 2 x a = 2 2 2 4 4 U y UVa ga Þ = - M1* Setting x = 2a or 1.5a and substituting into path equation from (a) 2 2 2(54.6cos(33.7)) 4(54.6cos(33.7))(54.6sin(33.7))(70.2) 4 (70.2) H g = - M1dep* Substituting their U, V and a to form an equation in H (and possibly g) only. M0 for an unsupported value of θ U and V 54.6 ¹ (allow sin/cos confusion) H = 46.8 (m) A1 awrt 46.8 [3] H240/03 Mark Scheme June 2022 28 13 (e) examples of possible refinements include • taking into account the size of B • taking into account that B is not a particle • taking into account the spin of B • taking into account the dimensions of B • taking into account the wind/weather • taking into account the thickness of the wall (which is assumed to be ‘thin’) • taking into account the clearance of the ball at the top of the wall B1 3.5c Allow any correct refinement, including use of a more accurate value of g rather than the assumed 9.8 B0 if referring to • the mass or weight or shape of B • the ground is unlikely to be horizontal • modelling the problem as three dimensional rather than two dimensional (unless specific detail given) • air resistance (only) If multiple refinements given, then all must be valid to score B1 [1] H240/03 Mark Scheme June 2022 29 Method marks for solving/factorising quadratics of the form 2 ax bx c + + - however note that candidates may solve quadratics on their calculators without showing any working so (unless the scheme says otherwise) unsupported answers from an incorrect quadratic equation need to be checked by examiners to award the corresponding M mark(s) 1. Factorisation: The modulus of the product of the first two terms in the brackets must equal 2. a x The modulus of the product of the second two terms in the bracket must equal , c e.g. 2 2 6 x x + - any of the following would score M1: 2 ( 2) 3( 2) x x x ± ± ± ± ( 2 3)( 2) x x ± ± ± ± 2. Formula: If the formula is written down correctly, then allow a maximum of one sign error in the substituted values for M1 3. Completing the square: Must get as far as halving the coefficient of x and dealing with the square of that term as well as moving the constant terms to one side ready to take the square root. Condone 1 sign error. e.g. 𝑥𝑥2 + 5𝑥𝑥−3 = 0 ⇒ቀ𝑥𝑥+ 5 2ቁ 2 = 3 ± ቀ 5 2ቁ 2 would get M1 e.g. 𝑥𝑥2 + 5𝑥𝑥−3 = 0 ⇒ቀ𝑥𝑥+ 5 2ቁ 2 = −3 + ቀ 5 2ቁ 2 would get M1 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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