Mark Scheme (Results) Summer Home. GCE Mechanics M2 6678/01 Original Paper

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1 Mark Scheme (Results) Summer Home GCE Mechanics M 6678/01 Original Paper

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 150 years, and by working across 70 countries, in 100 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: Summer 013 Publications Code N/A All the material in this publication is copyright Pearson Education Ltd 013

3 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. 3

4 June 013 M 6678/01 breach paper Mark Scheme Question 1. (a) a m = ( m+ 5) m = ( m + 5) M1 Moments equation for y coordinate. Terms of correct structure but condone sign slips A -1 each error ( 7 = m), m = 3. 5 (b) a+ 9 m= 0 ( ( m+ 5) ) M1 Moments equation for x coordinate. Terms of correct structure but condone sign slips a + 9 7= 0 use their m M1 Substitute their m and solve for a. a = 1 [8]. (a) (b) Use of P F V = M = 800 V V = 0 (3) 3000 M1 Find the driving force F = = gsinθ = 100a M1 Equation of motion for their F. Requires all terms. Condone sign errors and trig confusion 100g Correct equation (allow with sinθ ) = 100a 0 a = 0.91(m s - ), Accept 0.91

5 Question [7] 3. (a) Momentum = 0.5 (( 1) i + ( 9 + 6) j) M1 Substitution of m = 0.5 and t = 3 = -.5i j 5 15 or simplified equivalent i + j () (b) a = -i + (t+)j M1 Differentiate v A for each term Use of F = ma M1 Use of Pythagoras on F or a M1 F = (6 + ) = 5. or better (c) 3 t i + t + t j (+ r 0 ) 3 s = ( t ) M1 A Use of r 0 and t = 3 M1 Position vector of A is -10i + 1j (6) (5) [13] Integrate to find displacement for each component. r 0 not required. 5

6 Question. (a) Energy at A = Energy at B M1 Equation requires all 3 terms. Condone sign errors 1 1 m 15 = m 1 + mgh 7 3 Allow.13 h =.1 g 05 is A0 98 (b) 1 M1 or equivalent complete method to an equation in Use of s = ut + gt α (c) g Correct unsimplied equation = 1sinα ( sin α = ) α = Or 57 (3) 1cosα M1 Use of horizontal component and speed to find cosθ = 15 direction, or equivalent. ft follow their α 6.606, 13.7, 15 θ = from calculator values, 6. from 57 (3) [10] 6

7 Question 5. (a) (b) (c) Moments equation. Needs to include both rectangles and the whole shape. -1 each error. or equivalent Or simplified equivalent. a M1 Moments about OE: 1+ ( a + ) 1 = x A 3 = a 3 a x +, x = + 1 a M1 Moments about OA: 1+ 1 = y o.e. A 1 a y = + o.e. 3a + 1 x Hanging in Equilibrium: tanα = = M1 Condone reciprocal y a 1+ ft For their x, y 3a + 1 a + = 3 DM1 =, ( + a ) = 3(3a + ) Their tanα = and solve for a 3 a + a a = 9a + 1, = 5a, Or 0.8 a = 5 [1] Moments equation. Needs to include both rectangles and the whole shape. -1 each error. or equivalent Or simplified equivalent. 7

8 Question 6. (a) N B F R A 15 d m 1.5 m 30 N P 5 cm Moments about A: cos15 = d N M1 Requires correct terms of correct structure. Condone with value of d not yet found cos15 sin15 N A -1 each error. Requires a value for d. Accept unsimplified. d = N = 100 cos15sin15 = 5 (N) *AG* Watch out for given answer (b) Moments about P: R dcos15 = 30 ( d 1.5) cos15 M1 Terms of correct structure. Condone sign errors & trig confusion. Aft their d. Correct unsimplified. -1 each error 30( d 1.5).1 or better R = = d (c) Resolving horizontally at P: Ncos 75 = Fcos15 M1 Condone trig confusion N cos 75 F = cos15 F cos 75 M1 Use of F µ N µ : µ N cos15 µ or better [1] 8

9 Question 7. (a) u 3u P m Q m 3u u u u v w 1 M1 Correct use of impact law Q collides with wall: v = 3u = u towards P 3 P and Q collide. M1 Requires all terms. Condone sign errors Conservation of momentum: m u m u = m w m v 0 = m w m v, w = v *AG* Watch out (5) (b) Impact law: v + w = e( u + u) M1 Correct use of impact law to find w 3 w = 3eu, w = eu Kinetic energy: m u + m u = m v + m w M1 A Requires all terms. Each term of correct structure -1 each error M1 Solve for w 3u = w + w = 3w w = 1 u e = or better (8) [13] 9

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