Mark Scheme (Results) June GCE Mechanics M4 (6680) Paper 1

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Mark (Results) June 011 GCE Mechanics M4 (6680) Paper 1

Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edexcel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on 0844 576 005 or visit our website at www.edexcel.com. If you have any subject specific questions about the content of this Mark that require the help of a subject specialist, you may find our Ask The Expert email service helpful. Ask The Expert can be accessed online at the following link: http://www.edexcel.com/aboutus/contact-us/ June 011 Publications Code UA08446 All the material in this publication is copyright Edexcel Ltd 011

EDEXCEL GCE MATHEMATICS 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) should not be subdivided. 3. Abbreviations These 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. 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

1. June 011 6680 Mechanics M4 Mark v 1 ( 3i 4 j) m s A(m) a B(3m) b 1 ( i+ 3 j) m s w a = 3 & b= b Conservation of linear momentum : 4 + 3 3 = v 3w = 1 ( ) ( ) Restitution : v+ w= e 7 = 3 Solve the simultaneous equations DM1 giving v= and w= 1 1 1 KE lost = m ((16 + 9) (4 9)) + 3 m ((9 + 4) (1 4)) = 4 m J ( ) 10 GCE Mechanics M4 (6680) June 011 1

. 5 m 5-x X 100-x x C 4 m u B α v 7.5 m β Y At X: usinα = vsin β vcos β = eucosα 4vcos β = 3ucosα Eliminate u & v by dividing: tan α tan β = 3 4 M1 Substitute for the trig ratios: 5 x x = 3 4 4 7.5 D Solve for x: 37.5 7.5x = 3x DM1 x = 3.57 ( m) or better, 5 7 9 3. (a) Velocity of C relative to S = ( 8 + u ) ( 1 + 16 ) = + ( u ) i j i j M1 1 ( 4 16 )( m s ) i j (b) (i) C intercepts S relative velocity is parallel to i. u 16 = 0, u = 16 (ii) 1 10 km at 4 km h takes.5 hours, so.30pm () () () GCE Mechanics M4 (6680) June 011

(c) u = 8, relative velocity = 4i 8 j. S A 10 θ C d 4i 8 j m s 1 Correct distance identified 8 Using velocity: tanθ = = sinθ = 4 5 d Using distance: sinθ = =, 10 5 0 (5) d= = 4 5 = 8.9 ( km) 5 11 GCE Mechanics M4 (6680) June 011 3

4. (a) W rel H 5 o W 40 o 5 o H 5 5 o 40 o 5 W rel H H 40 o vector triangles with a common side M1..correct and drawn on a single diagram Wind is from bearing 05, (N 5 E) (3) (b) 5 W = o o sin 5 sin 40 5 o ) o 5 sin40 W = = 7.6 km h sin 5 1 ( ) (ft on their ft (4) GCE Mechanics M4 (6680) June 011 4

5. (a) Need an equation linking speed and displacement, so dv mv ( a bv ) dx = + 6v M1 Separating the variables: dv = 1dx a+ bv Integrating : 3 ln( a + bv ) = x + ( C ) b 3 3 X ln( a bu ) ln( a) ln 1 bu = b + = + b a ** as required (6) (b) dv Equation connecting v and t: 6 (1 3 v ) dt = + M1 6 M1, Separate the variables: dv = 1dt 1 + 3v 0 U M1 dv = v = T U + v d 4 0 4 + v tan 1U tan 1U T = = ( s ) (5) 11 GCE Mechanics M4 (6680) June 011 5

6. (a) Using F = ma: d x 4 dt = 9x 1v = 9x 1 dx dt M1 d x dx Hence 4 + 1 + 9x = 0 ** dt dt (4) (b) Auxiliary eqn : 4m + 1m+ 9= 0, m+ 3 = 0, m= 3/, λ = 3/ ( ) t = 0, x= 4 B= 4 λt t = 0, x = e λ At+ B + A = 0 6 + A= 0, A= 6 ( ( ) ) 3 3 (c) t 3 x = e ( (6t+ 4) + 6) = 9te 3 t 3 x = e ( 9 ( 9 t) ), so acceleration = 0 when t = /3 1 at which time, v 6e t 1 =, so max speed 6 / e.1 m s ( 3sf ) M1 =, (4) (5) 13 GCE Mechanics M4 (6680) June 011 6

7. (a) θ a B a θ a C mg a 4mg θ θ R a A BR = acosθ = 4acosθ (4a cos θ ) EPE = 3mg a M1 = 1mga cos θ = 6mga + 6mga cos θ (b) GPE: taking AR as the level of zero GPE, GPE = GPE of AB + GPE of BC M1+M1 = 4mg a sin θ + mg a sin θ a / cos θ ( ) = 8mgasin θ mga cos θ Total V = 8mgasin θ + 5mgacos θ + constant, as required. ** dv = 16mga cos θ 10mga sin θ dθ M1 dv 0 10sinθ 16cosθ dθ M1 8 tan θ = θ = 0.51 radians (9.0 ) 5 Or: 8mga sin θ + 5mga cos θ = 8 89mga cos( θ α), tanα = 5 t. pts when θ α = nπ θ = 0.51 rads. (7) (4) (c) dv = 3mga sin θ 0mga cos θ dθ M1 d V θ = 0.51 < 0, equilibrium is unstable. dθ cso Or: θ α = 0 cos(θ -α) = 1 Max value equilibrium is unstable (3) 14 GCE Mechanics M4 (6680) June 011 7

Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN Telephone 0163 467467 Fax 0163 450481 Email publication.orders@edexcel.com Order Code UA08446 June 011 For more information on Edexcel qualifications, please visit www.edexcel.com/quals Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Essex CM0 JE