Y543/01 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 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. Y543/01 Mark Scheme June 2023 6 A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words “Determine” or “Show that”, or some other indication that the method must be given explicitly. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. d. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep*’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. e. The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only – differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. If this is not the case please, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question. f. We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so. • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value. Y543/01 Mark Scheme June 2023 7 • 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. Question Answer Marks AO Guidance 1 (a) Initial KE = ½m5.32 B1 1.1 14.045m Award if substitution for v seen in energy equation after cancelling the m’s Y543/01 Mark Scheme June 2023 8 Energy at A = ½mv2 + mg0.8(1 – cos(π/3)) = ½m5.32 M1 1.1 DR CoE: equating their energy (KE + PE) at the bottom with their energy at A (KE + PE). At least one PE must be non-zero. Condone an incorrect height for M1, but do not award A0. Allow sin/ cos interchange For origin taken as 0 of PE ½mv2 - mg0.8cos(π/3)) = ½m5.32 -0.8mg v2 = 5.32 – 20.8g(1 – 0.5) = 20.25 => v = 20.25 = 4.5 so 4.5 ms–1 as required A1 1.1 CoE: equating their energies and attempting to solve for v2 or v AG Must come from a correct derivation of h [3] (b) 2 4.5 0.8 ra = M1 1.1 Use of 2v a r = awrt 25.3 ms–2... A1 1.1 Allow 405/16 ... towards O B1 1.2 Need to have a clear indication that it is towards the centre, which can rely on a clearly stated angle. Allow “radially inwards” or “centripetally” Do not allow 𝜋 6above horizontal unless it specifies negative horizontal axis. Do not allow ‘inwards’ on it’s own. [3] (c) (−)𝑚𝑔𝑠𝑖𝑛 𝜋 3 = 𝑚𝑎𝑡 M1 1.1 Resolving weight and using F = ma in the tangential direction. Condone missing sign. Allow direct substitution into 𝑔𝑠𝑖𝑛 𝜋 3 = 𝑎𝑡 So magnitude of tangential acceleration is awrt 8.49 ms–2... A1 1.1 4.9 3. or 𝑔√3 2 Do not allow -8.49 for final magnitude [2] Question Answer Marks AO Guidance Y543/01 Mark Scheme June 2023 9 2 (a) [Strain] = [extension]/[length] = L/L = 1 or L0 or M0L0T0 or dimensionless B1 2.2a AG. Condone = 0 (or =k) or [L] provided that the ratio is correct and stated as dimensionless Do not allow use of m (metres). Do not allow L : L [1] (b) [Stress] = MLT–2 L–2 = ML–1T–2 B1 1.1 So because “Strain is dimensionless”, [E] = [Stress] = ML–1T–2 B1 2.2a oe e.g show divide by 1 when considering dimensions of strain Allow clear [stress]/[strain] = [stress] [2] (c) [V] = L3 and [] = ML–3 and [c] = LT–1 B1 3.3 All used in the solution B1 correct dimensions all used in solution c = kEV (where k is dimensionless) => LT–1 = (ML–1T–2)(L3)(ML–3) M1* 3.3 Setting up the model with their dimensions (condone missing k here). Must include ,, or equivalent M: 0 =  +  L: 1 = – + 3 – 3 T: –1 = –2 M1dep 1.1 Comparing dimensions on both sides to derive three equations in ,  and  . Allow one incorrect equation M2 for correct consideration by division with their dimensions and manipulation of indices leading to their dimensions for c  = ½ =>  = –½ =>  = 0 A1 1.1 Give this if  = 0 seen in final equation (i.e. no dependence on V) E c k  = A1 1.1 k necessary here. Allow indices also If correct answer seen from a division method using correct dimensions allow A2. If k not present A1. [5] (d) (i) 5002 ms–1 or awrt 707 ms–1 B1FT 3.4 Follow through their model from part (c) Numerical answers needed [1] Y543/01 Mark Scheme June 2023 10 (d) (ii) According to the model there is no dependency on V so 500 ms–1 B1 3.5a Some explanation involving (or calculation using) the model must be given with reference to V. Allow “volume doesn’t matter” so unchanged Needs to be an explanatory comment, not just “500” No FT in this part [1] (e) There may be other (dimensionless) parameters which affect c but have not been, or cannot be, taken into account (eg temperature) B1 3.5b Allow: can’t calculate/find value of dimensionless constant There may be additional terms that are dimensionally consistent Do not allow “there could be a dimensionless constant” B1 BOD can’t identify [1] Question Answer Marks AO Guidance Y543/01 Mark Scheme June 2023 11 3 vAy = uAy = 4sin60 B1 1.1 Correct perpendicular component of velocity of A before collision, which is unchanged by the collision = 23 = 3.46... may be seen on the diagram 54cos60 + 3–6 = 5vAx + 3vBx M1 3.4 Conservation of momentum condone sin/cos need to be mv for all terms 5vAx + 3vBx = –8 A1 1.1 Correct rearrangement to a 3 term equation Check diagram for directions 3 4 4cos60 6 Bx Ax v v − = −− M1 3.4 Restitution – needs to be consistent 6 Bx Ax v v − = oe A1 1.1 Correct rearrangement to a 3 term equation vAx = –13/4 (= –3.25) A1 1.1 vBx = 11/4 (= 2.75) 1 2 3 tan 46.8 3.25  − = =  (1 dp) A1 2.2a Could be 133.2 or –46.8 or + 0.817 rad or -0.817 rad or 2.32 rad [7] Y543/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 4 (a) (BD is a line of) symmetry (of the kite) B1 2.1 [1] (b) 2 2 0.37 0.35 0.12 ABC h = − = so area = 0.042 B1 3.1b ALT (considering the CoM of triangle ABD) with X as the point where the diagonals meet: ℎ𝐴𝐵𝑋= 0.12 so area = 0.021 2 2 0.91 0.35 0.84 ADC h = − = so area = 0.294 B1 1.1 ℎ𝐴𝑋𝐶= 0.84 so area = 0.147 Measuring from B: (0.042 0.294) 0.042 0.08 0.294 (0.12 0.28) x + =  +  + M1 1.1 Attempt to balance moments about any point with at least two non-zero terms each comprising the product of a force or mass with an appropriate distance. May be one error in the distance. Measuring from AC (0.042 + 0.294)𝑥̄ = −0.04 × 0.042 + 0.28 × 0.294 Gives 𝑥̅ = 0.08064 0.336 = 0.24 Final step: add 0.12 to measure from B Measuring from B: (0.021 + 0.147)𝑥̅ = 0.021 × 2 3 × 0.12 + 0.147 × (0.12 + 1 3 × 0.84) 0.12096 0.36 0.336 x = = A1 1.1 AG. Some intermediate working must be seen. 𝑥̅ = 0.06048 0.168 = 0.36 AG [4] (b) Alternative method Y543/01 Mark Scheme June 2023 13 E.g. Taking B as (0,0) 2 2 0.37 0.35 0.12 ABC h = − = so A (0.12, 0.35) B1 1.1 Need to see clear coordinates, may be on diagram. Award if average of trianlge coords method used 2 2 0.91 0.35 0.84 ADC h = − = so D (0.96,0) B1 1.1 From B, 𝑥= 1 3 (0 + 0.12 + 0.96) M1 1.1 = 1.08 3 = 0.36 as triangle BAD and triangle BCD are symmetrical A1 1.1 AG. Intermediate working and mention of symmetry along BD must be seen. [4] (c) ( ) 0.24 tan 0.35 CAG  = B1 1.1 34.4 (3 sf) CAG  =  ∠𝐴𝐺𝐵= 55.6° (3 sf) 1 0.84 180 90 "34.4 " tan 0.35  − = − − − M1 3.1a ∠𝐴𝐶𝐷= 𝑡𝑎𝑛−1 0.84 0.35 = 𝑡𝑎𝑛−1 12 5 , 67.4° (3 sf) Allow a sign error. May see ∠𝐵𝐷𝐶= 22.6°=𝑡𝑎𝑛−1 0.35 0.84 Allow for method to gain 101.8 degrees with the vertical seen |180° −"55.6°" −𝑡𝑎𝑛−1 0.35 0.84| Allow for combination of one of (∠𝐶𝐴𝐺 or ∠𝐴𝐺𝐵) and one of (∠𝐴𝐶𝐷 or ∠𝐵𝐷𝐶) to find angle with vertical or horizontal. 11.8 (3 sf) =  A1 1.1 or –11.8 C is above D A1 2.2a Needs to come from correct working, may have found angle from horizontal or vertical [4] Question Answer Marks AO Guidance Y543/01 Mark Scheme June 2023 14 5 (a) ( ) 2 d 2 1 d v v v x = − + M1 3.3 Using F = ma to derive a differential equation in v and x only. Allow a single sign error Condone m seen in equation (not 2). Do not allow 𝑑𝑥 𝑑𝑣 2 2 d 1 v v x c v = −+ +  M1 1.1 Correctly separating the variables and integrating at least one side Condone m seen in equation (not 2). Can be given if m substituted later 2 ln( 1) v x c  + = −+ A1 1.1 Correct relationship between v and x (condone missing + c). Must see brackets for this A mark, and correct numerical m. 2 0, 3 ln( 1) ln10 x v c u = =  = + = M1 3.4 Using initial condition to find c for an expression of correct format ln(𝑓(𝑣)) = ±𝑥+ 𝑐 May see use of limits [ln (𝑣2 + 1)]3 𝑣= [−x]0 𝑥 ⇒ln(𝑣2 + 1) −ln(10) = −𝑥 2 1 e 10e 10e 1 x c x x v v −+ − −  + = = = − A1 1.1 The –1 must be clearly under the root sign oe [5] (b) v = 2 => ln(22 + 1) = –x + ln10 M1 3.4 Substituting v = 2 into a solution of the correct form, which includes a numerical value for the constant of integration Or integrating between the correct limits. [ln (𝑣2 + 1)]3 2 = [−x]0 𝑥 ln(5) −ln(10) = −𝑥 x = ln10 – ln5 = ln2 so distance is ln2 m or awrt 0.693 m A1 1.1 Award SC1 if answer seen with no working (need to see v=2) [2] Y543/01 Mark Scheme June 2023 15 (c) For Q, F = –1 => a = –0.5 => v = u – 0.5t’ or v2 = u2 – s M1 1.1 Calculating the (constant) acceleration and using it in a suvat equation ALT method 1: (using F = ma to set up a differential equation): 𝐹= −1 = 𝑚𝑑𝑣 𝑑𝑡= 2 𝑑𝑣 𝑑𝑡 ⟹−𝑡+ 𝑐= 2𝑣 At 𝑡= 0, 𝑣= 3 ⟹𝑐= 6 Hence 𝑣= 1 2 (6 −𝑡) ALT method 2: 2𝑣𝑑𝑣 𝑑𝑥= −1 ⇒𝑣2 = −𝑥+ 𝑐 At 𝑥= 0, 𝑣= 3 ⇒𝑐= 9 Hence 𝑣2 = 9 −𝑥 oe Q stops when vQ = 0. v = u + at’ => –0.5t’ = –3 => t’ = 6 v2 = u2 + 2as => s = 9 A1 2.1 AG Need to see v = 0 used. ALT method 1 vQ = 0 when t’ = 6 −𝑡’ + 6 = 2 𝑑𝑥 𝑑𝑡’ ⟹−𝑡’2 2 + 6𝑡’(+𝑘) = 2𝑥 (and k = 0) so 𝑥= 1 2 (6𝑡’ − 𝑡’2 2 ) At t’ = 6, x = 9 ALT method 2 Q stops when v=0, therefore 9-x = 0, therefore x = 9. They will also need to use one of the other listed methods to justify the limit on t for this method mark Y543/01 Mark Scheme June 2023 16 But for P, F = 1 + v2 > 1 (so aP > 0.5 for all v) so P will slow down more quickly than Q (and be travelling more slowly at any comparable time) so it must stop before time t = 6 and it cannot reach x = 9.  t < 6 and x < 9 A1 2.2a AG Include: • F = 1 + v2 > 1 (explicit comparison) • A reference to P coming to rest more quickly than Q (comparison). Needs to make the link – a larger resistive force will decelerate P more quickly or a larger resistive force will mean P is travelling more slowly than Q (from starting point of 3) • Final conclusion: So it comes to rest when t < 6 and x < 9 Be generous with t and t’ confusion [3] (d) B1 3.1b Graph in 1st quadrant only. Strictly decreasing from positive value on vertical (v) axis to the horizontal (t) axis. Negative gradient at vertical axis intercept. Allow concave. Do not allow linear No values required. Should come to rest [1] (e) v = 0 => ln1 = –x + ln10 M1 3.4 Substituting v = 0 into a solution of the DE needs to be correct form Or considering 10e–x – 1 ≥ 0 x = ln10 so maximum displacement is ln10 m or awrt 2.30 m A1 1.1 Allow -ln(1/10) Award SC1 if answer seen with no working (need to see v=0) [2] Y543/01 Mark Scheme June 2023 17 Question Answer Marks AO Guidance 6 (a) (i) (0) (32sinh0) (32cosh0 257) 225 = + − = − v i j j M1 3.1b DR required (determine) Attempt to find v at t = 0 Alternative using calculus: 𝒂= (64 𝑐𝑜𝑠ℎ2𝑡 64 𝑠𝑖𝑛ℎ2𝑡) so 𝑭= 3𝒂= 192 (𝑐𝑜𝑠ℎ2𝑡 𝑠𝑖𝑛ℎ2𝑡) seen used in F.v (ln 2) (32sinh(2ln 2)) (32cosh(2ln 2) 257) 60 189 = + − = − v i j i j M1 1.1 Attempt to find v at t = ln2 Uses 𝑊𝐷= ∫𝐹∙𝑣 𝑑𝑡 𝑊𝐷= 192 ∫ (𝑐𝑜𝑠ℎ2𝑡 𝑠𝑖𝑛ℎ2𝑡) ∙( 32 𝑠𝑖𝑛ℎ2𝑡 32 𝑐𝑜𝑠ℎ2𝑡−257) 𝑑𝑡 𝑙𝑛2 𝑡=0 ( ) 2 2 2 (ln2) 60 189 39321 v = + − = M1 1.1 Attempt to use v.v to find v2 at t = ln2 (v = 198.295...) may see KE = 58981.5J Attempts the dot product: = 192 ∫ 32 𝑠𝑖𝑛ℎ4𝑡−257 𝑠𝑖𝑛ℎ2𝑡 𝑑𝑡 𝑙𝑛2 0 ( ) 2 1 3 39321 ( 225) 2 WD =  −− J M1 1.1 Using WD = change in KE 1 =Δ 2 mv.v Integrates and applies limits: = 192 [8 𝑐𝑜𝑠ℎ4𝑡−257 2 𝑐𝑜𝑠ℎ2𝑡] 0 𝑙𝑛2 awrt –17000 J A1 1.1 -16956 J ISW if replaced with magnitude [5] (ii) It is negative because over the interval P ends up moving slower than when it started B1 2.4 Condone eg “Because P is slowing down”, “particle is decelerating”, “P is opposing the motion of the particle” Or “(the overall effect of F over this interval is such that) the force F is acting in the opposite direction to the motion of P” “The particle does work against the force” Allow “F is a resistive force” [1] (b) P moving parallel to x-axis => vy = 0 => cosh2t = 257/32 M1 2.2a Deriving equation for t for P to be moving parallel to x-axis => 2t = ln16 but t > 0 => t = ln4 (or awrt 1.39) A1 2.3 May go straight to t = ln 4 (or 2t = ln 16). May see t=0.5arcosh (257/32) 1.386… t = ln4 => (32sinh(2ln 4)) 255 = = v i i A1 1.1 or just vx. If shown, vy must be 0 Y543/01 Mark Scheme June 2023 18 d 3 (192cosh 2 ) (192sinh 2 ) d m t t t = = = + v F a i j M1* 3.1b Attempt to differentiate v (implied by either (64cosh2t)i or (64sinh2t)j) and multiply by m (may just see the i-component) May be seen in part (a), must be used in part (b) to gain this method mark t = ln4 => 192cosh(2ln 4) 1542 x F = = M1dep 1.1 Attempting to find (i- component) of F at their time. Ignore attempt to find Fy P = F.v => Power is 1542255 = awrt 393000 W A1 1.1 3 x 514 x 255=393210 [6] Y543/01 Mark Scheme June 2023 19 Question Answer Marks AO Guidance 7 (a) For A: 2 1.25 0.9 A T  =  M1 3.3 NII for A using a = r2 or v2/r or v and correct values for m and r For B: ↕ Tcos = 2g B1 3.4 Correctly balancing forces in the vertical for B T = 24.5 = 2.5g 𝑐𝑜𝑠𝜃= 4 5 , 𝑠𝑖𝑛𝜃= 3 5 rB = (1.26 – 0.9)sin B1 2.2a Correct expression or value for the radius of B’s circle used 0.36sin or 0.216  Tsin = 2rBB2 M1 3.4 NII for B using a = r2 or v2/r or v and correct value for m, 𝑠𝑖𝑛𝜃 and their calculated rB, T 𝑠𝑖𝑛𝜃=1.5g May be seen as one equation for B divided by the other in which case give 1st B1 2nd M1 if correct 𝑡𝑎𝑛𝜃= 2𝑟𝐵𝜔𝐵 2 2𝑔 T = 24.5 => B = 35/6 or awrt 5.83 and A = 14/3 or awrt 4.67 A1 1.1 [5] (b)  for strings to realign need (35/6)t – (14/3)t = k M1FT 3.1b Condone 180 or any integer multiple of  rads or 180 . Condone, 𝑇= 2𝜋 Δ𝜔, Need to see a calculation of difference of angular speed Do not award for partially complete common multiple methods. SC2 for fully correct common multiple or ratio method, finding 4 half turns for one object and 5 half turns for the other leading to t = 6 /7 oe. Must come from correct . 7t/6 =  => t = 6 /7 so time is awrt 2.69 s A1 1.1 cao [2] Y543/01 Mark Scheme June 2023 20 Question Answer Marks AO Guidance 8 Initial KE = ½1.752.42 J B1 3.4 5.04 J Suppose the distance is d m. Initial PE = 1.75gdsin J M1 3.1b 1.05gd or 10.29d. Attempt to use mgh with h different from d 𝑐𝑜𝑠𝜃= 4 5 , 𝑠𝑖𝑛𝜃= 3 5 d is the distance moved along the plane. h is the vertical distance between O and the point where P stops. EPE/Total energy when P stops = 4.8(d – 2.1)2/(22.1) J M1 3.4 Correct use of 2 2 x l  (but could be in terms of x rather than d). 8(d – 2.1)2 / 7 or 8x2 / 7 C = 1.75gcos => Fr = 0.7321.75gcos B1 1.1 Using law of friction to derive an expression for the frictional force WD against friction = Fr d J M1 3.4 Using WD = Fd to find an expression for the energy lost 1.0248gd or 10.04304d Do not allow sin cos interchange Y543/01 Mark Scheme June 2023 21 5.04 + 10.29d = 8(d – 2.1)2/7 + 10.04304d M1 3.4 Energy budget equation involving initial energy, final energy and energy loss on the correct side, signs correct, in terms of one correct unknown distance. Do not allow sin/cos interchange Equation could be in terms of extension, x: 5.04 + 10.29(x + 2.1)= 8x2/7 + 10.04304(x +2.1) ALT method Candidates may consider interim energy at 2.1m (giving 𝑣2 =6.3527). Can award B1 B1 for the initial KE and frictional force as per the main scheme. All method marks awarded as per the main scheme for energy consideration involving a correct algebraic distance by considering motion from 2.1m to the point the particle comes to rest. KE (d=2.1) +PE(d=2.1) = EPE stored + WD against Fr 5.558616 + 10.29x= 8x2/7 + 10.04304x Or 5.558616 = 8x2/7 – 0.24696x d(d – 4.41609) = 0 => (d = 0) or d = awrt 4.42 so distance travelled must be awrt 4.42 m A1 1.1 Could be BC (at least one correct root seen) 8x2 – 1.72872x – 38.910312 = 0 (x + 2.1)(8x – 18.52872) = 0 x = –2.1 or awrt 2.32 So: awrt 2.1 + 2.32 = 4.42 m Rejection of d = 0 (as start point) A1 3.2a explicit rejection must be seen, for correct roots only But x > 0 do reject x =-2.1m [8] 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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