H240/03 Mark Scheme June 2023 5 5. Subject Specific Marking Instructions 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 2023 6 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 2023 7 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 2023 8 Question Answer Marks AO Guidance ENSURE THAT PAGE 20 (WHICH APPEARS ABOVE QUESTION 1) IS EITHER LINKED TO THE CORRESPONDING QUESTION OR ANNOTATED AS ‘BP’ OR ‘SEEN’. NOTE 5(b), 7(b) AND 9(b) ARE OVER TWO PAGES AND THEREORE THE 2nd PAGE MUST BE CHECKED AND MARKED IF USED, OR ANNOTATED AS ‘SEEN’ IF NOT USED 1 2 1 4 5 (2 1)log4 log5 x x x x M1* 1.1 Take logs of both sides (any base) correctly and use power law correctly at least once. Common correct answers that score M1 are: 4 2 1 log 5 , x x 2 1 5 log 4 , x x (2 1)ln4 ln(5 ) x x Condone lack of bracket on the 2x + 1 term (2log4 log5) log4 x M1dep* 1.1 Re-arrange to get an equation with a single term in x – condone sign errors only e.g. the following score the first two M marks 1 log5 2 , log 4 x log5 2 1, log 4 x 4 (log 5 2) 1, x 5 5 (1 2log 4) log 4 x (2ln4 ln5) ln4 0 x This mark can be implied by a correct answer provided the first M mark was awarded (that is, we must see logs being taken and the power law used at least once) log 4 1.19 log5 2log 4 x A1 1.1 awrt –1.19 –1.19184404… Correct answer with no working scores no marks H240/03 Mark Scheme June 2023 9 [3] 1 ALTERNATIVE 2 1 5 4 5 4 16 x x x B1 Correctly re-writes the given equation in the form x a b with a and b correct (check carefully for other equivalent correct equations) 5 16 log 4 x or 5 log log4 16 x M1 Taking logs correctly of their , xa b where a and b are both positive, to obtain either log log x a b (any base) or loga x b (for their a and b) Not dependent on the B mark 1.19 x A1 awrt –1.19 –1.19184404… Correct answer with no working scores no marks [3] H240/03 Mark Scheme June 2023 10 Question Answer Marks AO Guidance 2 (a) 5 R B1 1.1 B0 for 5, 25 R etc. unless replaced with 5 No working required for this mark. Ignore working 4 cos 3 tan 3 sin 4 R R    M1 1.1 M1 for tan k  where 3 4 , 4 3 k or equivalent e.g. 3 cos , R  4 sin R  with their value of R (but not just R and do not allow reciprocals for this mark). 53.1 (or better) with no working implies M1 SC If cos 3, sin 4   4 tan 3  explicitly seen then this scores M1 A0 but do not penalise again in (b) (if correct answer seen) 53.13  A1 1.1 www awrt 53.13 (at least 4 sf required) so 53.1 (or 53) is A0 (but if an awrt 53.13 seen then isw if replaced with a less accurate value) 53.13010235… - an answer in radians scores A0 53.13 from sin( ) R x  soi [3] 2 (b) 2 53.13 arcsin 5 x M1 1.1 M1 for 2 arcsin x R  or 2 arcsin x R  with their R and substituted SC B1 for 76.7 only (in the given range) from using an alternative method e.g. 2 2 9sin (2 4cos ) x x 76.7 x A1 1.1 awrt 76.7 (at least 3 sf required) – ignore any answers given outside the range 0 90 x but do not award this mark if any other values in this range are given – www but see SC in (a) Correct answer with no working seen scores SC B1 Answer in radians scores A0 [2] H240/03 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 3 (a) (i) 3 2 f( ) + f ( ) 3 x x px q x x p M1 1.1 Attempt at differentiating f(x) with at least one non-zero term correct M0 for 2 f ( ) q x x p x f (2) 13 1 p A1 1.1 Correct value for p [2] 3 (a) (ii) 3 2 2 0 p q M1 1.1a Substituting x = 2 into f(x) and equating to 0 or for correctly re- writing as 2 ( 2)( 2 4 ) x x x p (with p or their value of p from (a)(i)) Could be in terms of q only e.g., 32 2 ( ) 0 their p q Possibly seen in an attempt at long division 10 q A1 1.1 Correct value for q [2] 3 (b) 3 ( 2) ( 2) 3 y x p x q B1 B1 1.1 1.1 Substituting (x ± 2) into both x terms of y = f(x) Subtracting 3 (oe) at some stage Allow with p and q or with incorrect values of p and/or q – note that using 2 vertically and/or 3 horizontally cannot be treated as a MR 3 2 2 3 3 ( 2) 3 ( 2) ( 2) 2 3 y x x x px p q M1 1.1 Correct expansion of (x ± 2)3. Can be unsimplified for M1. If correct bracketing not seen then must be implied by later working 3 2 3 2 ( 6 12 8) 2 10 3 6 13 23 y x x x x y x x x A1 2.2a cao – condone just the expression 3 2 6 13 23 x x x (so do not need to see y = …) For reference: 6, 13, 23 a b c [4] H240/03 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 4 (a) 2 2 ( 3) ( 5) 9 25 x y k M1 3.1a Complete the square (for x and y) to obtain 2 2 ( 3) ( 5) ... x y or for 2 2 ... 3 5 k 34 0 34 k k A1 2.3 cao – www allow equivalent in either set notation e.g. { : 34} k k or interval notation e.g. ( ,34) or ( ,34] but not [ ,34) or [ ,34] unless 34 k already seen If implying a lower limit then A0 Allow 34 k [2] H240/03 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 4 (b) 2 2 6 10 46 0 x y x y For reference d d 2 2 6 10 ( 0) d d y y x y x x M1* 3.1a Attempt to differentiate the equation for C implicitly – must be four terms including a d 2 d y y x term and two other terms correct (condone either –46 or k appearing in their deriv. as a 5th term) but if the derivative of 2 2 6 10 46 0 x y x y is put equal to 1 2or d d y x (and used subsequently) then M0 or applying the chain rule to an expression of the form 2 5 ( 3) x  for some non-zero , so must be of the form 1 1 2 2 (f( )) g( ) x x where f(x) is quadratic and g(x) is linear d 6 2 3 5 2 10 d 1 2 y x x y y x M1dep* 1.1 Sets derivative equal to 1 2 or substitutes 1 2for d d y x 2 2 2 1 (1 2 ) 6 10(1 2 ) 46 ( 0) x y x x x x M1 2.1 Substitutes their linear expression into the given equation of C with 46 k (must be five terms with two quadratic terms and two linear terms in x oe if using completing the square form from (a)) to obtain an expression/equation in x (or y) only Dependent on first two M marks – if chain rule used this mark is implied by setting d d y x equal to 1 2 2 5 30 35( 0) x x M1 1.1 Simplify to a 3TQ in x (or y) (allow sign errors only when simplifying from their five term equation/expression or completing the square equation/expression) Dependent on all M marks For y if correct: 2 10 39( 0) y y (7, 13), ( 1,3) A1 2.2a BC - do not need to be stated as coordinates Two values of x and y only H240/03 Mark Scheme June 2023 14 Question Answer Marks AO Guidance [5] H240/03 Mark Scheme June 2023 15 Question Answer Marks AO Guidance Alternative Method for first two marks 4 (b) 5 3 r y m x or 5 ( 3) r y m x M1* Finding an expression for the gradient of the line segment that passes through the centre and any point (x, y) on the circumference of C Follow through their centre from (a) – gradient expression must be correct for their centre 5 2 3 y x or 5 2( 3) y x M1dep* Equating gradient with –2 (oe) H240/03 Mark Scheme June 2023 16 Question Answer Marks AO Guidance 5 (a) d 0.2 sin d x t t B1* 1.1 B1 for d 0.2 sin d x t t 2 0 sin (0.2 sin )d k t t t  2 0 (1 cos )(0.2 sin )d k t t t  M1dep* 3.1a Uses d d d x y t t and replaces 2 sin t with 2 1 cos t to obtain an expression involving 2 cos t and sint No limits required for this mark 2 2 0 (0.2 sin 0.2cos sin cos )d ( 1) k t t t t t  A1 2.2a AG (so must be checked carefully for any errors e.g. must contain relevant brackets around the 2 1 cos t term(s)) A correct expression e.g. 2 0 (1 cos )(0.2 sin )d k t t t  followed by the correct given answer can score this mark – limits and dt must be seen at least once (but need not be on the final integral) Do not need to see = 1 anywhere in their solution (and condone lack of brackets around the integrand) [3] H240/03 Mark Scheme June 2023 17 Question Answer Marks AO Guidance 5 (b) 2 1 2 (0.2 sin 0.2( )(1 cos2 ) sin cos )d t t t t t B1 1.2 Correctly applies 2 2cos 1 cos2 t t (so not just stating this identity) - implied by seeing 0.1 0.05sin2 t t after integration e.g. applies could be for an attempt to integrate 1 2(1 cos2 )t or stating this identity in an integral 0.2 cos t t B1 1.1 First two terms integrated correctly (Look out for those that have 0.1t only from combining 0.2t with –0.1t) This mark should be awarded if 0.2 is combined with another constant term and integrated correctly M1* 1.1 M1 for an answer of the form 3 sin2 cos pt q t r t or an answer of the form sin2 sin sin2 cos cos2 pt q t r t t u t t or an answer of the form cos sin2 sin sin2 cos cos2 pt q t r t u t t v t t for non-zero p, q, r (and u, v) from integrating 2 2 0.2cos sin cos t t t 3 1 0.1 0.05sin 2 cos 3 t t t A1 1.1 A1 for the correct remaining three/four terms (Alternatives: 1 1 0.1 0.05sin 2 sin sin 2 cos cos2 6 3 t t t t t t or 1 1 0.1 0.5cos 0.05sin 2 sin sin 2 cos cos2 3 6 t t t t t t t ) 3 0 3 1 0.1 cos 0.05sin 2 cos 3 1 1 (0.1 ( 1) 0 ( 1) ) (0 1 0 )3 3 t t t t   M1dep* 1.1 Uses correct limits correctly ( ) (0) F F  - condone limits the wrong way round only if the sign of their answer is subsequently changed Must be using exact values (so must include term(s) in ) 30 2 (0.1 2 ) 1 3 3 40 k k   A1 1.1 www oe e.g. 1 4 0.1 3 k  - isw once a correct expression for k seen Correct exact value must be seen at some stage for the final mark – check first A mark carefully [6] H240/03 Mark Scheme June 2023 18 Question Answer Marks AO Guidance 6 (a) 1 3 2sin , 2 3cos u a u a d   and 7 4 2 3 sin u a d  For reference 7sin 3cos 2 d   B1* 2.1 Forming a correct expression for d (or a correct equation containing d) e.g. 1 2 ( ) ( 3cos 2sin ) d  , 7 1 3 2 ( ) ( sin 2sin )( 0.5sin ) d    Can be implied e.g. 7 sin 2sin 3(...) 2   seen Can be implied by a correct equation for  7 3cos 2sin 2 sin 3cos 2 ( )     3 tan 3  M1dep* 3.1a Obtaining an equation of the form tan k  from a trigonometric equation which initially had 3 sine and 1 cosine terms or 2 sine and 2 cosine terms e.g. if correct 7 7 2 2 sin 2sin 3( sin 3cos )     5 6   A1 2.2a Condone 6  stated too but A0 if any other value given in the interval 1 2   (but ignore any values that are given outside this range) Exact answer must be seen at some stage [3] H240/03 Mark Scheme June 2023 19 Question Answer Marks AO Guidance 6 (b) 100 100[2(2sin ) (100 1) ] 2 S d  B1ft 1.2 Correct formula for the sum of an AP with 2sin a  (with either  or their value of  substituted) and either d or their value of d substituted or their expression for d Follow through their values of  and d if used provided 100[2(2sin ) (100 1) ] 2 d  implied 7 5 5 1 sin( ) 3cos( ) ( ) 2 6 6 4 d   B1ft 1.1 Correct expression for d using their  (e.g. 1 2 ( 3cos 2sin ), d   7 1 3 2( sin 2sin ) d  ) Follow through their value of  only 100 1337.5 S B1 2.2a www – must have come from 5 6   correctly derived in (a) oe (not for 1338 or 1340 unless 1337.5 seen so isw once 1337.5 (oe e.g. 2675 2 ) seen) Correct answer with no working scores all 3 marks [3] H240/03 Mark Scheme June 2023 20 Question Answer Marks AO Guidance 7 (a) DR 2 2 8 1 0 2(8 ) (7 4 ) 0 2 7 4 t t t t M1* 1.1 Setting equation for a equal to zero and removing 2t correctly from the denominator e.g. 2 1 2 8 (7 4 ) 0 t t to obtain the equivalent of a 3TQ in t only This mark can be implied by a correct 3TQ in t 2 4 16 7 0 (2 1)(2 7) 0 t t t t M1dep* 1.1 Correct method for solving their 3TQ in t If factorising: 2 ( )( ) at bt c mt n pt q where a mpand one of mq np b or c nq so note that 2 4 16 7 ( 0.5)( 3.5) t t t t is M0 but e.g. ( 2 7)(2 1) t t is M1 bod If using the formula: must apply the correct formula for their three-term quadratic in t (no errors) If completing the square: The M mark is not awarded until correctly getting to the stage 9 2 4 t for their 3TQ in t (must include so implying two roots) with no errors (so consistent with applying the formula correctly) Must see the method – the correct answers do not imply this mark therefore 2 4 16 7 0 0.5 and 3.5 t t t scores M1 M0 B1 As a minimum must see (if correct) 16 144 8 0.5 t or 3.5 t B1 1.1 This mark is not dependent on the previous M mark(s) So M1 M0 B1 is common or M0 M0 B1 if no working seen H240/03 Mark Scheme June 2023 21 Question Answer Marks AO Guidance [3] H240/03 Mark Scheme June 2023 22 Question Answer Marks AO Guidance 7 (b) 2 d 8 1 d 2 7 4 v t v t t B1* 3.1b Stating the correct differential equation Possibly implied by correct separation of variables 2 ln 1 ln(7 4 ) ( ) 2 v t t c B1dep* B1dep* 1.1 1.1 Correct lhs Correct rhs – not multiplied by v or 5 (or any other constant) Condone lack of +c for both B marks 0, 17.5 ln17.5 ln7 t v c M1* 3.1a Uses correct initial conditions to find c from an equation of the form 2 1 2 3 ln ln(7 4 ) k v k t k t c (note that e.g. + c may appear on the lhs) With non-zero values of ik - accept any equivalent form e.g. 2 0.5 (7 4 )e t v A t and then use initial conditions to find A (if correct then A = 2.5) 2 5 1 ln5 ln(7 4 ) ln 2 2 T T M1dep* 2.1 Uses , 5 t T v to obtain an equation in T only – dependent on previous M mark Condone use of t for T throughout the remainder of the question 2 2 1 5(7 4 ) 7 4 ln 2ln 2 2 5 2 T T T T A1 2.2a AG so at least one step of intermediate working from substitution of t = T and v = 5 Condone 2 7 4 2ln 2 T T [6] H240/03 Mark Scheme June 2023 23 Question Answer Marks AO Guidance 7 (c) 2 1 0 1 2 3 4 5 7 4 2ln 2 11.25 11.09523175... 11.04058716... 11.02111643... 11.01415608... 11.011665... n n T T T T T T T T B1 1.1 Uses given result and given starting value to obtain correct 1T and 2T (so the first two iterations after the initial value of 11.25) to at least 4 sf (rot) – but all stated values in these two terms must be correct 11.01 T B1 1.1 Must be stated to 4 sf only – not dependent on the first B mark – can be awarded if either of 2T and/or 3T incorrect (assume that the iterative process corrected itself or a slip in the candidate writing down an earlier value) Must be clear that T is 11.01 (and not the final term shown in the iterative process) – this mark can be awarded from using alternative iterative methods e.g. Newton-Raphson [2] 7 (d) 11.01 – 3.5 = 7.51 (s) B1 2.2a awrt 7.51 No follow through from incorrect earlier values [1] H240/03 Mark Scheme June 2023 24 Question Answer Marks AO Guidance 8 6 4(3 2 ) i u i j M1* 3.3 Applying t v u a correctly - working must imply that v and a are vectors Or for 3 2 t t v i j c and using 4, 6 t v i to find c M0 if 6 14 u i j 6 8 u i j A1 2.5 or for (3 6) ( 2 8) t t v i j and setting t = 0 to obtain correct u 2 2 ( 6) 8 u M1dep* 1.1 Correctly taking the magnitude of their u but condone 2 2 6 8 36 64 Correct answer following 6 8 i j (with no wrong working) scores full marks 1 10 (ms ) u A1 1.1 www [4] H240/03 Mark Scheme June 2023 25 Question Answer Marks AO Guidance 9 (a) B1 1.2 Correct force diagram showing the weight, normal contact, and frictional contact forces only (but ignore labelling of these forces even if labelled incorrectly) but must include all arrows pointing in the correct direction – the line of action of all forces must pass through the block (or imply passing through the block) but do not need to be attached to the block Ignore components of any of the three forces if shown – if only components shown then B0 [1] H240/03 Mark Scheme June 2023 26 Question Answer Marks AO Guidance 9 (b) M1* 3.3 Resolving parallel or perpendicular to the plane – correct number of terms - must be using 10 for the weight so no marks until this value used/applied (however, see SC in the final A mark) Allow sin/cos confusion and sign errors – but the 2 and 10 must be resolved and the R and F should not (if considering & ) 2sin 10cos R   2cos 10sin F   A1 A1 1.1 1.1 Condone or with F for this mark but withhold the final A mark Where R is the normal contact force and F is the frictional contact force 2cos 10sin 0.8(10cos 2sin )     M1dep* 3.4 Applying F R  or F R  to obtain an equation/inequality in  only – where R and F are a linear combination of the correct number of relevant resolved terms 10cos 50sin 40cos 8sin 30 15 (tan greatest value of) tan is 58 29       A1 2.2a oe e.g. 30 58 but A0 for 6 11.6 as a final answer. Allow awrt 0.517. Isw if candidates go on to work out . Allow 15 29 tan (oe) as a final answer Can use equals throughout (no justification required) But not 15 29 tan If using 10g for the weight, then SC M1* A1 A0 M1dep* A0 max (where the first A mark is for both equations correct using 10g rather than 10) [5] Alternative for first three marks M1* Resolving vertically and horizontally M1 conditions as above cos sin 10 R F   A1 cos 2 sin F R   A1 Final M1dep* and A marks as above H240/03 Mark Scheme June 2023 27 Question Answer Marks AO Guidance 10 (a) ( 4 ) (2 ) a b R i j B1 1.1 oe e.g. 4 2 a b ( 4 ) (2 ) ( 3 ) a b k i j i j M1 3.1b Sets their R equal to k times 3 i j (or k times R) where k is non- numerical/unknown Implied by a correct equation in a and b e.g. 4 1 2 3 a b but not from stating 4 1 a and 2 3 b (so must have come from a ‘gradient’ approach if k not seen) 4 k a therefore 2 3(4 ) b a so 3 10 a b A1 2.2a AG Eliminate k and derive given result Using an assumed value of k is A0 [3] 10 (b) 6 8 2 6 a b R i j B1 1.1 2 2 2 ( 6) R M1* 3.3 Calculate R or 2 R for their R 2 2 2 ( 6) (5 10) m M1dep* 3.4 N2L applied to the magnitude of their R with 5 10 40 0.4 5 10 m A1 2.2a www 0.4 (oe) but square roots cancelled [4] H240/03 Mark Scheme June 2023 28 Question Answer Marks AO Guidance 11 (a) M1 3.1b Moments about A to form an equation with the correct number of terms and both the weight and contact force at B resolved - condone 20g for the weight of the rod. All relevant values must have been substituted. May take moments about another point (and resolve) but must end up after elimination with an equation in B R only Resolved for the weight means either sin or cos with an angle of either 30 or 60 only, and for the contact force at B either sin or cos with an angle of 25 or 35 or 55 or 65 only 1.4(20cos30) (2.8cos25) B R M1 1.1 One of the two moment terms correct Must be as part of a two- term moment equation (so dimensionally correct so M0 if 20g for the weight) but the other term need not be resolved (or resolved correctly) e.g. 1.4(20cos30) 2.8 B R scores M0 M1 28cos30 9.5555330... 2.8cos25 B R = 9.56 (N) (correct to 3 significant figures) A1 1.1 AG - correct equation followed by 9.56 scores all 3 marks Allow awrt 9.56 [3] For reference in parts (a) and (b): Moments about B: (2.8cos30) (2.8sin30) 20(1.4cos30) A A R F Moments about the mid-point: (1.4cos30) (1.4sin30) (1.4cos25) A A B R F R H240/03 Mark Scheme June 2023 29 Question Answer Marks AO Guidance 11 (b) M1* 3.3 Attempt at resolving vertically or horizontally – correct number of terms with the contact force at B resolved (angle must be one of 25, 35, 55 or 65 only) – or taking moments again (see below) (same conditions for taking moments as in part (a)). M0 if using W or mg only for the weight of the rod (must substitute in the given value before awarding this mark – see second guidance column if using 20g) Allow sign errors, sin/cos confusion and 20g used as the weight for the M mark. Allow B R instead of the given value of 9.56 for the M mark 9.56sin35 20 A R 9.56cos35 A F A1 A1 1.1 1.1 Correct expression/equation for A R Correct expression/equation for A F 20 5.48... 14.5... A R 7.83... A F 2 2 (9.56cos35) (20 9.56sin35) M1dep* 1.1 Correct method for calculating the magnitude of the contact force at A where A R and A F are a linear combination of the correct number of terms with 9.56 resolved in both equations (angle must be one of 25, 35, 55 or 65 only) 16.5 (N) A1 2.2a www awrt 16.5. For reference if using 9.56 for B R then 16.494179… If using more accurate value for B R then 16.494698… [5] H240/03 Mark Scheme June 2023 30 Question Answer Marks AO Guidance For reference in parts (a) and (b): Moments about B: (2.8cos30) (2.8sin30) 20(1.4cos30) A A R F Moments about the mid-point: (1.4cos30) (1.4sin30) (1.4cos25) A A B R F R H240/03 Mark Scheme June 2023 31 Question Answer Marks AO Guidance 12 (a) 2 0 (39sin ) 2( 10)h  M1 3.3 Using 2 2 2 v u as with v = 0, a = 10 or 9.8 or , g and 39sin u  or 39cos . u  Accept any other complete method (using correct suvat equations) to find the maximum height e.g. 0 39sin 10t  and with t then substituted into 2 (39sin ) 0.5( 10) s t t  Condone g = 9.8 for full marks in (a) and (b) 2 5 0 39 2( 10) 13 h A1 1.1 A correct equation with the correct value of sin substituted or for 2 0 (39sin(22.6...)) 2( 10)h Allow using g (and not replaced with 10 or 9.8) for this A mark Max. height = 20 + h = 31.25 (m) A1 2.2a www • accept awrt 31.2 (using 22.6… as the angle and g = 10) • accept awrt 31.5 (coming from exact value of sin or 22.6… and g = 9.8) • condone 31.3 (3 sf) [3] H240/03 Mark Scheme June 2023 32 Question Answer Marks AO Guidance 12 (b) 2 1 2 20 (39sin ) ( 10) T T  M1 3.3 Applying 2 1 2 s ut at with 20, s a = 10 or 9.8 or g and 39sin u  or 39cos u  Condone g = 9.8 for full marks in (a) and (b) Accept any other complete method to find T 2 1 2 5 20 39 ( 10) 13 T T A1 1.1 With correct value of sine or sin(22.6) Allow 2 1 2 5 20 39 ( 9.8) 13 T T Allow using g (and not replaced with 10 or 9.8) for this A mark T = 4 only A1 1.1 BC • accept awrt 4.07 (using g = 9.8 and exact value of sine) • accept awrt 4.06 (using g = 9.8 and 22.6…) • a value of 4(.00…) coming from 3.998… (using g = 10 and 22.6… for sine) Condone t = 4 [3] H240/03 Mark Scheme June 2023 33 Question Answer Marks AO Guidance 12 (c) Examples of possible limitations • The ball will have dimensions/volume • The spin/rotational forces of the ball • A 1sf approx. to g was used • Other weather conditions (ignore wind and air resistance) • The ball is not a particle • g is modelled as a (universal) constant B1 3.5b Allow any correct limitation B0 if referring to • Air resistance and/or wind (only) • The ground is unlikely to be horizontal (only) • The mass or weight or shape of P (only) • The angle/heights/speeds may not be as quoted (only) • Modelling the problem as 3D rather than in 2D (only) If multiple limitations given, and any are incorrect, then B0 [1] H240/03 Mark Scheme June 2023 34 Question Answer Marks AO Guidance 12 (d) 2 3 3 12 2 a kt t M1 3.4 Differentiate given v (at least two terms correct) 12 39 ' ' ( 144) 13 BC T M1* 3.1b Applying s = ut horizontally to find distance BC with correct value of cos (or cos(22.6…)) and their value of T from (b) Allow g = 9.8 which if correct leads to 146.34… 4 3 2 3 1 4 4 2 ( ) s kt t t c M1* 2.1 Integrate given v (at least two terms correct) 4 3 2 3 1 4 4 (4) 2(4) (4) 144 k M1dep* 3.4 Puts their integrated expression for s, with t = their T from (b), equal to their distance for BC to form an equation in k only – dependent on the two previous M marks only Must not include a +c unless dealt with as part of a definite integral. However, if a +c is included then subsequently ignored/set equal to zero without justification then give bod for this and any subsequent A marks (if earned) 1 16 k A1 1.1 Correct exact value for k (oe e.g. 0.0625) Final two marks can only be awarded if g = 10 used 52.5 a 2 (m s ) A1 2.2a oe [6] H240/03 Mark Scheme June 2023 35 Question Answer Marks AO Guidance SEE APPENDIX AT THE END OF THE MS FOR IMPORTANT INFORMATION REGARDING THIS PART 13 (a) M1* 2.1 Applying N2L parallel to the plane for P – correct number of terms and weight component resolved – allow sign errors and sin/cos confusion. Allow g missing but M0 if 4ga in N2L For the first five marks condone:  used as the coefficient of friction for both P and B, or implying that 2 B P  rather than the correct 2 P B   4 sin60 4 P g F T a A1 1.1 NB 4 sin60 4 P T F g a is A1 (taking up the plane as +ve dir.) Where P F is the frictional force for P 4 cos60 P R g B1 3.3 Resolving correctly perpendicular to the plane for P Where P R is the normal contact force for P 4 sin60 (4 cos60) 4 P g g T a  2 3 2 4 P g g T a  M1dep* 3.4 Use of F R  in the attempt at N2L for P with their R which must be a component of 4g only Where P is the coefficient of friction between P and the plane (2 ) 2 B T g a  M1 3.3 Applying N2L parallel to the surface for B – correct number of terms – allow sign errors but note that (2 ) 2 B g T a  is consistent with 4 sin60 4 P T F g a and therefore gives the correct answer (and is not incorrect working) Where B is the coefficient of friction between B and the plane Allow g missing but M0 if 2ga in N2L or for 2 B T F a only 2 3 2 4 P g g T a  and 2 4 4 B T g a  with 2 P B   gives 2 3 2 g T T A1 3.3 Solving simultaneously with 2 P B  (soi) to obtain a correct equation in T only H240/03 Mark Scheme June 2023 36 Question Answer Marks AO Guidance 2 3 2 3 2 3 g T T T g A1 2.2a Must be seen in terms of g Accept awrt 1.15g [7] H240/03 Mark Scheme June 2023 37 Question Answer Marks AO Guidance 13(b) 2 1 2 1.9 2(0.5) (0.5) P a B1 3.4 Applying 2 1 2 s ut at correctly to find P a 7.2 P a B1 1.1 4 sin60 (4 cos60) 4 P P g g a  M1 3.1b Set T = 0 (or apply N2L) to obtain an expression for the acceleration of P when the string breaks Correct number of terms, dimensionally correct (so g not missing), must be using the correct mass of 4 – allow sin/cos mix and sign errors only 0.26266... 0.13(1...) P B   A1 3.4 Using their acceleration of P to correctly calculate the coefficient of friction for B. Can be implied from a correct deceleration of B. Accept 0.13 (so 2 sf) or better For reference: exact value is 72 49 3 98 (which scores A1) B B a g  deceleration is 1.29 (m s–2) A1 3.2a Accept awrt 1.29 or 1.29 1.28704895… or, for reference, exact value is 72 49 3 10 (which scores A1) [5] H240/03 Mark Scheme June 2023 38 APPENDIX (To assist with 13a) Exemplar responses for Q13(a) – see main MS for Guidance for the requirements/conditions to award each of these marks Response Max. Mark Case 1: candidates who immediately set  as the coefficient of friction for B and 2as the coefficient of friction for P (correct) 4 sin60 4 P g F T a M1 A1 4 cos60 P R g B1 4 sin60 2 (4 cos60) 4 g g T a  M1 (2 ) 2 T g a  M1 2 3 2 g T T A1 2 3 3 T g A1 Or equivalent e.g. replace 2 above with and  above with 0.5 7 Case 2: candidates who immediately set 2 as the coefficient of friction for B and as the coefficient of friction for P or set both coefficients of frictions equal to  (both of these are incorrect) 4 sin60 4 P g F T a M1 A1 4 sin60 4 P g F T a M1 A1 4 cos60 P R g B1 4 cos60 P R g B1 4 sin60 (4 cos60) 4 g g T a  M1 4 sin60 (4 cos60) 4 g g T a  M1 2 (2 ) 2 T g a  M1 (2 ) 2 T g a  M1 4 sin60 4 cos60 2 8 g g T T g   A0 A0 4 sin60 4 cos60 2 4 g g T T g   A0 A0 5 Case 3: assuming that as 2 P B   then this implies that 2 P B F F without justification (e.g. no calculation of the normal contact forces for P and B considered – so while it turns out to be correct in this case it is not true in general) – if this is assumed then 4 sin60 2 4 g F T a M1 A1 2 T F a M1 (so now allow this M mark even though in the main MS this would have been M0) leading to 2 3 2( 2 ) 4 g T a T a and therefore 2 3 3 T g no further marks (3 marks max. also if assuming that 2 P B F F ) 3 H240/03 Mark Scheme June 2023 39 Case 4: if assuming that 0 a or g or any other value then they can score the B mark only for 4 cos60 P R g 1 Need to get in touch? If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in touch with our customer support centre. Call us on 01223 553998 Alternatively, you can email us on support@ocr.org.uk For more information visit ocr.org.uk/qualifications/resource-finder ocr.org.uk Twitter/ocrexams /ocrexams /company/ocr /ocrexams OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge. For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR 2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA. Registered company number 3484466. OCR is an exempt charity. OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals. OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these resources. Though we make every effort to check our resources, there may be contradictions between published support and the specification, so it is important that you always use information in the latest specification. We indicate any specification changes within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy between the specification and a resource, please contact us. Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more information using our Expression of Interest form. Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.