Mark (Results) January 013 GCE Mechanics M3 (6679/01)
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General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. Unless indicated in the mark scheme a correct answer with no working should gain full marks for that part of the question.
General Instructions for Marking EDEXCEL GCE MATHEMATICS 1. The total number of marks for the paper is 75.. The Edexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) should not be subdivided. In some instances, the mark distributions (e.g. M1, B1 and ) printed on the candidate s response may differ from the final mark scheme. 3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark. isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures or AG: The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for example, as ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but incorrect answers should never be awarded A marks. 5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt.
7. Ignore wrong working or incorrect statements following a correct answer. 8. The maximum mark allocation for each question/part question(item) is set out in the marking grid and you should allocate a score of 0 or 1 for each mark, or trait, as shown: am aa bm1 b bb bm ba 0 1
January 013 6679 M3 Mark Question 1. dv v 9x dx = M1 1 ( ) = + 9 v x c v 9x c = + M1dep x= v= 6 36 = 9 4+ c c= 0 v = 9x
Question (a) Mass: 1 π r kπ r 3 3 3 3 k + k B1 Dist from O: 3 1 r kr x B1 8 4 3 kr r+ = x( + k) M1ft 4 4 x = ( k 3) ( k + ) 4 r * (b) ( k 3) r tanθ = r 4 ( k + ) M1 ( k ) 3 11 = 4 14 ( k + ) 14k 4 = 44k+ 88 7k k 65= 0 ( k )( k ) 7 + 13 5 = 0 k = 5 M1dep
Question 3 (a) ( t ) 1 0.6a = M1 + 1 0.6 dv= dt ( t + ) 1 0.6v= + ( t + ) ( c) M1dep t = 0 v= 15 0.6 15= 6+ c c= 3 M1dep 1 0 0.6 3 5 5 4 v= + v= + = + 1 t+ t+ t+ ( ) ( ) * (b) dx 4 = 5 + 1 dt t+ 4 x = 5 + 1 dt t + M1 ( ( ) ) ( ) x = 54lnt+ + t + c M1dep t = 0, x= 0 c = 0ln ( ) t = 5 x= 5 4ln7+ 5 0ln = 50.05... = 50.1 or better 7 or 0ln + 5 M1dep
Question 4 a θ T O P (a) ( ) mg R Tcosθ = mg M1 a T = mg ( a+ x) Hooke s Law: T = 6mgx 3mgx a = a M1 3mgx a = mg a a x ( + ) M1dep 6x = a+ x x = a * 5 (b) mv T sinθ = M1 r mv 3mg sinθ = 5 1a sinθ 5 M1dep 6 1a sin v = g θ 5 5 4a 11 sin θ = 1 = 1a 36 5 7ag 11 ag v = = M1dep 5 36 5
Question 5 (a) x = asinωt 0.15 = 0.5sin 0.1ω M1 1 sin 0.1ω = π 0.1ω = 6 π 10π ω = = M1dep 0.6 6 π 6 Period = = ( = 1.) ω 5 (B1 on e-pen) (b) 5 x = 0.5sin πt 3 t = 5 x= 0.5sin π 3 M1 x = 0.165... Dist from B = 0.5 + x = 0.033 m ft (c) Max accel π = aω = 0.5 = 6.853... = 6.85 3 5 M1 (d) Max speed 5π aω = 0.5 = 1.308... = 1.31 3 M1
Question 6 B v mg 150 A O u (a) At B mg cos 60 ( R) mv + = M1 a 1 v ag g = v= * a (b) Energy A to B: 1 1 ag mu m = mga sin 30 M1 u ag 1 = + ag 3ag u = (c) Horiz speed = ag 1 ag cos 60 = M1 ag ag Initial vert speed = ( ) sin 60 = ( ) 1 3 M1 1 3ag a v = + g M1 4 11ag v = 8 vert 11ag 8 tanθ = = = 11 M1 horiz 8 ag θ = 73... = 73
Question 7 (a) T λx λ 18 = 40 = M1 l 30 λ = 400 (b) 4 cm T 7 cm θ T 1.5g Extension = 10 cm or 0 cm (used in (b) or (c)) B1 T 400 10 800 = = 15 3 M1ft ( ) θ ( ) R Tcos 1.5g = ± 1.5a M1 1600 7 1.5 9.8 = ( ± ) 1.5a 3 5 a = 89.75... a= 90 m s - or 89.8 (positive) (c) 1 0. E.P.E. = 400 0.3 B1ft (any correct EPE) 1 0. 00 0.18 + = M1 0.3 0.3 1.5g 0.07 1.5v 00 1 0. 00 0.18 v = 00 1.5g 0.07 0.75 0.3 0.3 M1dep v 1 =.3... =.3 m s
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