Mark Scheme (Results) January 2011

Transcription

1 Mark Scheme (Results) January 011 GCE GCE Mechanics M (6678) Paper 1 Edexcel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WC1V 7BH

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3 General Instructions for Marking 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) Marks should not be subdivided. 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 The answer is printed on the paper The second mark is dependent on gaining the first mark

4 January 011 Mechanics M 6678 Mark Scheme Question Number 1. Scheme (a) Constant speed Driving force = resistance, F = 3. B1 P = F v= 3v= v = 1 ms Marks (b) 384 P= F v 384 = F 9, F = 9 Their F 3 = 10 a, a = ms. I= -6i+8j = v 5i+j ( ) 3i+ 4 j= v 5 i j v= i+ 5j 1 KE = v = ( + 5 ) = 9 ( J) [6] [5] 3. (a) 3 a= 4t 1t Convincing attempt to integrate 4 v= t 6t + c Use initial condition to get 4 v t t ( 1 ) = ms. (b) Convincing attempt to integrate 5 t 3 s = t + 8t( + 0) Integral of their v 5 ft () (c) Set their v = 0 Solve a quadratic in t D ( t )( t 4) = 0 at rest when t =, t = [8] GCE Mechanics M (6678) January 011 1

5 Question Number 4. Scheme (a) Work done against friction = 50 µ R = 50 ¼ 30cos Marks Gain in GPE = sin 0 Total work done = WD against Friction + gain in GPE D = 8480( J ), 8500( J) (6) (b) Loss in GPE = WD against friction + gain in KE 3 terms sin0 = 50 ¼ cos0 + ½ 30 v -1 ee A,1,0 ½ v sin0 ¼cos0 =, D -1 v = 10. m s. (5) [11] GCE Mechanics M (6678) January 011

6 5. (a) A cm F 36 cm 18 cm 18 cm E 9 cm B D cm Divide the shape into usable areas, e.g.: C Shape C of mass Units of mass Rectangle 7 x 9 (13.5,4.5) 43 (6) Right hand triangle (30,3) 40.5 (1) Top triangle (3,30) 40.5 (1) Rectangle 9 x 18 (4.5,18) 16 (4) Mass ratios B1 Centres of mass B1 (b) Take moments about AB: = 13 = 1x, A(,1,0) x = 11(cm) solve for x (or y) co-ord y = 11(cm) using the symmetry B1ft Alternative: Shape C of mass Units of mass Small triangle (1,1).5 x 18x 18 Large triangle (15,15).5 x 36 x = ( ) x etc θ 11 x tanθ = 36 y 11 ft tanθ = = θ = 4 (7) [10] GCE Mechanics M (6678) January 011 3

7 6. (a) 1 Using s = ut + at Method must be clear r = 3t i t 4.9 t j Answer given (b) j component = 0 : t 4.9t 5 ± ± 1 D quadratic formula: t = = T =.03(s),.0 (s) positive solution only. (c) Differentiating the position vector (or working from first principles) v = 3i + (5 9.8t)j (ms -1 ) () (d) At B the j component of the velocity is the negative of the i component: 5 9.8t = -3, 8 = 9.8t, () (e) t = 0.8 v = 3i 3j, speed = = 18 = 4. 4 (m s -1 ) () [1] GCE Mechanics M (6678) January 011 4

8 Question Number 7. Scheme Marks 1 m S α C 1 m B R m α 100 N A x F Taking moments about A: 3 S = 100 cosα Resolving vertically: R + S cos α = 100 Resolving horizontally: S sin α = F (Most alternative methods need 3 independent equations, each one worth. Can be done in e.g. if they resolve horizontally and take moments about X then R cosα = S (3 cos α) scores MA) Substitute trig values to obtain correct values for F and R (exact or decimal D equivalent) S = 9, R = 100 =, F = F µ R, 00 8 µ 1100, µ = Least possible µ is sf, or exact. [10] GCE Mechanics M (6678) January 011 5

9 Question Number 8. (a) 1 1 KE lost : m 36 m v = 64 Restitution: v = 1/3 x 6 = Scheme 1 1 Substitute and solve for m: m 36 m 4 = 64 = 16m m = 4 answer given Marks D (6) (b) 3 m/s m/s kg 4 kg v w Conservation of momentum: 6 8 = 4 w v their "" ft Restitution: v+ w= ⅓ + 3 their "" ft 5 v = w D Solve for w: = 4w ( w) = 6w w = 3 1 w = 418= 9m s w > 0 will collide with the wall again (7) [13] GCE Mechanics M (6678) January 011 6

10 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN Telephone Fax Order Code UA06580 January 011 For more information on Edexcel qualifications, please visit Edexcel Limited. Registered in England and Wales no Registered Office: One90 High Holborn, London, WC1V 7BH