Mathematics > A/As Level Mark Scheme > > GCE Further Mathematics B (MEI) Y411/01: Mechanics A Advanced Subsidiary GCE Mark Scheme for Autum (All)

> GCE Further Mathematics B (MEI) Y411/01: Mechanics A Advanced Subsidiary GCE Mark Scheme for Autumn 2021

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Mark Scheme November 2021 f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the ... units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.) We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so. • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value. • When a value is not given in the paper accept any answer that agrees with the correct value to 2 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 A the rubric specifies 3 s.f. as standard, so this statement reads “3 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. E marks are lost unless, by chance, the given results are established by equivalent working. 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” and “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.Y411/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 1 (a) [Energy M LT ] = ( −1)2 or M LT L ⋅ ⋅ −2 or ML T 2 2 − M1 1.2 Any form acceptable. This mark may be implied by the correct answer. [Specific energy ML T M L T ] = ⋅ = 2 2 1 2 2 − − − A1 1.1 [2] (b) 345 ÷ ÷ × × 4 28.35 40 4184 M1 1.1 All scale factors present. Allow one slip with × ÷ / . ≈ 509,000 J A1 2.2b [2]Y411/01 Mark Scheme November 2021 Question Answer Marks AOs Guidance 2 (a) (i) ∠OAB = ?? and ??????(∠OAB) = 3 5 = 0.6 B1 3.1a AG Need not justify why ∠OAB = ??. Allow finding angles 53.1°, 36.9° [1] (ii) cosα = 0.8 B1 1.1 [1] (b) Moments about O: 3 10sin ⋅ α = 4T2 M1 3.3 Taking moments about some point. ⇒ = T2 4.5 A1 1.1 T1 cosθ α =10sin or T1 cos 6 θ = or 3 cos T1 θ = 4T2 T T 1 sinθ α = + 10 o c s 2 or T T 1 2 sin 8 θ = + or 4T1 s n i θ =10(5) M1 A1 A1 3.3 1.1 1.1 Attempt at resolving: one equation sufficient (all terms present, but condone sin cos ↔ ) Or Taking moments about a different point Correct equation involving T1 cosθ Correct equation involving T1 sinθ M1 1.1 Correct method for finding T1 or θ (from values of T1 cosθ and T1 sinθ) Dependent on previous M1M1 12.5 tanθ θ = ⇒ ≈ ° 6 6 . 4 4 2 1 2 T = + 12.5 6 ≈13.9 A1 1.1 cao Both correct [Show More]

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