Mark Scheme (Results) Summer Pearson Edexcel GCE In Mechanics M4 (6680/01)

Transcription

1 Mark Scheme (Results) Summer 07 Pearson Edexcel GCE In Mechanics M4 (6680/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at or Alternatively, you can get in touch with us using the details on our contact us page at Pearson: helping people progress, everywhere Pearson aspires to be the world s leading learning company. 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 50 years, and by working across 70 countries, in 00 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 07 Publications Code 6680_0_706_MS All the material in this publication is copyright Pearson Education Ltd 07

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 Q Scheme Marks Notes a Position vectors after t hours: 4t 5 t and t 3t M Use position vectors to find position of one ship relative to the other a alt a alt 7 5t 3 t A d 7 5t 3 t M Correct method for magnitude 6t 76t 58 Differentiate: 5t 76 0 DM or complete the square 76 t.46 hrs, A (5) Position vectors after t hours: 4t 5 t and Use position vectors to find position M t 3t of one ship relative to the other 7 5t 3 t 5 7 5t At closest point: 0, 3 t 5 7 5t 3 t 0 A M Scalar product of relative velocity and relative position 35 5t 3t 0 DM Scalar product = 0 38 t.46 hrs, A (5) Angle between initial positions of A and B and relative velocity 3 tan tan MA d 58 cos M d Time taken DM 6 t.46 hrs, 3.8 A (5) A θ α B d 4

5 b balt Distance : d 6t 76t 58 4, (condone equality) M 6t 76t 54 0 Time interval: M Difference between roots hrs (9 mins) A 0.49 or better (3) sin sin , d 8.695, d 6.4 M Find at least one distance t d t.705, t.9 6 M Interval hrs (9 mins) A 0.49 or better (3) [8] 5

6 w (5i - j) m s - A 3m kg 5i B m kg 3i (3i + 4j) m s - v a For A, component perpendicular to loc = 5 B For B, component perpendicular to loc = 3 B 85 3 m5 m3 v 00 M v, v 4 4 A Equation for kinetic energy of B For their "3" 6m 4m 3mw mv 3mw 3.5m w 0.5 M Aft CLM parallel to loc. No missing/additional terms Condone sign error(s) Correct unsimplified equation for CLM (with their values if substituted) Select correct root and state velocities: DM v 3 i 3.5 j (m s- ) A One correct B v 5 i 0.5 j (m s- ) A Both correct A (9) b v w e 4 M e Aft e A 3 (3) [] Impact law parallel to loc. Used the right way round. Condone sign error(s) Correct unsimplified or with their values 6

7 3a 3a alt dv 75v 50 3v d 75v dv dx 50 3v M x A Differential equation in v and x No additional/missing terms. Condone sign error(s) M Separate variables 5 x ln 50 v 5 50 v ln 50 x v ln, v 5 50 v 50e x 5 v 0 50e x 5 DM A Integrate and use limits DM Change the subject to v or v A (7) dv v 4 dx 5 MA 5 Integrating factor: x x x v e 4e dx 50e C v 0, x 0 C 50 v x e v 50e x 5 e x M MA Integrate and use limits DM Change the subject to v or v A (7) 7

8 3b d d v Differential equation in v and t v t M No additional/missing terms v 5 t dv 50 v 5 dv v 50 v Condone sign error(s) dv dt M Separate variables and integrate v v 5 ln 50 ln v 5 50 t ln ln v A With or without constant of integration 5 v Or arc tanh DM Use limits 0 and v 5 50 v ln v A (5) [] 5 50 v ln 4 50 v or equivalent 8

9 -7j 4a -ai+j -4i - 3j Components parallel to the plane unchanged: Use of scalar product. 0 4 a 4 M Do not need to see a 3 a 6 A *Given Answer* () 4a alt -ai+j α θ -7j θ -4i - 3j 4b θ Components parallel to the plane: 7sin vcos 7sin v cos cos sin sin, M 7tan a tan 8tan a 6 A () Component of -7j parallel to the plane M 4i 3j 4. A M A Both correct 9.36 e 3.36 DM e A (6) Equate components and form an equation in a and Scalar product of -7j and unit vector parallel to plane Use Pythagoras to find components perpendicular to the plane Use if impact law Dependent on preceding M mark 9

10 Alt4 b Alt 4b Alt 4b Component of -7j perpendicular to the 0 3 plane Component of aijperpendicular to 6 3 the plane 5 4 Impact law: e Components perpendicular to the e7cos vsin plane: e v MA MA DM A 7cos sin cos cos sin A , v 37 Substitute for : 7e 6tan Solve for e : 3 7e 6, 4 e 4 Components perpendicular to the e7cos vsin plane: e v M (6) MA DM A M 7cos sin cos cos sin A Divide and substitute for : tan tan M ecot tan tantan A (6) Solve for e : 3 e 4 4 DM A (6) 0

11 Q Scheme Marks Notes Alt Parallel: 7sin 37 cos MA Pair of equations 4b Perpendicular: e7 cos 37 sin 49sin 37 cos 49 sin sin 37 M 49e cos 37 sin sin A e , e Square and substitute to eliminate MA Substitute for θ to obtain e. (6) [8]

12 Q Scheme Marks Notes 45 x 5 5 alt B B One correct triangle Two triangles combined using their common w. (seen or implied) cos30 x cos 45 M Horizontal components equal x A w x x cos 45 M Cosine rule w 0.5 (km h - ) A sin sin 45 EITHER: Sine Rule: x w sin sin 60 w M 8. A Direction 6 A OR: wcos cos30 wsin cos30 cos30 tan, cos 30 (M) 8. (A) Direction 6 (A) (9) xcos 45 w x cos 45 B w expressed as a vector cos 30 w sin 30 y B Second expression of w as a vector cos30 x cos 45 M Horizontal components equal x cos A Or x 6 6 θ w 60 y w M Use of Pythagoras w 0.5 (km h - ) A 6 3 tan M A Direction 6 A (9) [9] Correct method for direction of w

13 Q Scheme Marks Notes 7e 6a 0.g B T mg at equilibrium (e = 0.4) 0.8 Equation of motion 7x e No missing/additional terms 0.x 0.g v M 0.8 Condone sign error(s) Allow with m and l At most one error A m and l substituted 7x 0.x x Correct unsimplified equation A 0.8 m and l substituted x 0x 43.75x 0 A *given answer* (5) AE: m 0m , 6b 5 3 M Form and solve AE m 5 i 5t x e Acos t Bsin t A t 0, x 0. A 0. B 6c 6c alt 5t x 5e Acost B sin t 5t e A sint B cost 0 5A B B, e x cos sin 5 t t 3 5t instantaneous rest when t M B M 5 3 A 0.73 A 5 3 () 5t x 5e cost sin t t e sint cost Use of t 0, x 0 to find B (6) (s) M Use of periodic time 5t 5 3 3e sin t 0 M Solve x t, t 0.73 s A 5 3 () [3] 3

14 Q Scheme Marks Notes 7a Relative to the fixed point A, PE of mass at C 3mg 4acos B PE of rods mg acos mg 3a cos B Extension in the spring 4acos a B 7mg 4a cos a 0mga cos M Total PE 4a 7mga 4cos 4cos 0mga cos 8mga cos 48mga cos constant (5) A *given answer* 7b 7c dv M Differentiate d 56mga cos sin 48mga sin A dv 8sin7cos 6 0 M 0 and solve for d cos 3, 0.54r A 7 (4) 6 Differentiate to obtain second M derivative d V 56mga cos sin d A 48mga cos Find value of second derivative mga 48mga DM when cos Dependent on preceding M mga, stable A (4.8...mga) (4) [3] 4

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