Mark Scheme (Results) June GCE Mechanics M3 (6679) Paper 1

Transcription

1 Mark (Results) June 011 GCE Mechanics M (6679) Paper 1

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3 EDEXCEL GCE MAHEMAICS General Instructions for Marking 1. he total number of marks for the paper is 75.. he 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. marks are unconditional accuracy marks (independent of M marks) should not be subdivided.. Abbreviations hese are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. here 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 he answer is printed on the paper he second mark is dependent on gaining the first mark

4 1. (a) O 0.75x June 011 Mechanics M 6679 Mark P 0.5 kg x x dv 0.5v = 0.75x dx v x c t = 0, v=, x= = + c c = 10 1 v = x + 60 * (b) v= 5 x = x = 7.77 (4) () 6 GCE Mechanics M (6679) June 011 1

5 . ( ) 4 V = π 9 x dx= π ( 81 18x + x ) dx x 648 = π 81x 6x + = π OR: π ( 9 x ) xdx 5 π ( + ) x 18x x dx π ( 9 x ) = = π 6 x x + x π = 0+ ( 9) = π + 6 = 4 π 4 π 4 15 x = = ( accept 0.94) 5 (a) Mass ratio ( l) l ( l) Dist. from O Moments equation: π 5 ρ π ρ 81π l ρ l l x l 4 l = 9x 8 ft 8 x = l 9 (9) 9 (5) GCE Mechanics M (6679) June 011

6 (b) 8 7 GX = 5l l = l 9 9 1ft l 7 tanθ = = 7 l 7 9 ft θ = 6.1 accept 6, 0.6 or 0.60 rad or better (4) 9 GCE Mechanics M (6679) June 011

7 4. (a) A 4a C a A P a mg (b) 4 cos θ = or sinθ = R vert cos 45 + mg = cosθ ( ) mg = 5 A ( ) θ m aω A R horiz sin + cos 45 = = A 1 A + = maω 5 4 A mg = maω a = maω + mg 5 A 5 A = m( aω + g) 7 * (8) 4 b = a mg 5 = 4 ( ) 7 m aω g mg + = ( 4 ) 7 m aω g oe () GCE Mechanics M (6679) June 011 4

8 (c) b 0 4aω g g ω 4a 1 g ω * a (Allow strict inequalities in (c).) () 5. (a) O 7 6 l R A F 1 mg (b) mg 1 1 = l = mg l 6 1 R ( ) R = mg R ( ) 1 F = = mg F µ R 1 mg mg 1 µ * 1 mg 1 mgl E.P.E. lost = l = l 8 1 l Work done by friction = mg mgl 1 1 l = mv + mg 8 gl v = 4 1 v= gl 1 1 ft (4) GCE Mechanics M (6679) June (5)

9 (c) mgl 1 = mgx 8 ft l x = 4 () 1 6. (a) Energy to : 1 ( ) 1 m ag mv = mag 9ag V = ag V = 7ag NL along radius at : V + mg = m a + mg = 7mg = 6mg > 0 particle reaches (6) GCE Mechanics M (6679) June 011 6

10 (b) Energy to C: 1 1 mu m( ag ) = mag U = ag+ 9ag U = 11ga () (c) θ ag 1 Energy from C to rest: 1 5 m 11 ag = mga( 1 cos θ ) ag = ga ( 1 cos θ ) cosθ = 144 θ = θ = 87 (or 1.5 rad) or better (4) 1 GCE Mechanics M (6679) June 011 7

11 7. x x (a) otal extn. = 0.6 λ ext 0. ( x) b = = = ( 10 x) * l (b) a ( x ) + 0. = = ( 10 x+ ) (c) b a = 0.5x ( 10x) ( 10x+ ) = 0.5x ft 7 7 0x = 0.5x 7 40 x = x S.H.M. (d) (e) ( ) π Period = = π = π * ω v A 1 m a 80 7 O 1 max = aω = 0. o.e. or a.w.r.t m s 80 x = acosωt = 0.cos t x = 0.1 = cos t cos t = 80 x ( ) b (1 x) 7 π π 7 t = = o.e. (accept a.w.r.t. 0.6) s () (1) (6) () (4) 15 GCE Mechanics M (6679) June 011 8

12 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN elephone Fax Order Code UA0844 June 011 For more information on Edexcel qualifications, please visit Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Essex CM0 JE