Mark Scheme (Results) January Pearson Edexcel International A Level in Mechanics 2 (WME02/01)

Transcription

1 Mark (Results) January 05 Pearson Edexcel International A Level in Mechanics (WME0/0)

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 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: January 05 Publications Code IA All the material in this publication is copyright Pearson Education Ltd 05

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. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 PEARSON EDEXCEL IAL MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edexcel Mathematics mark schemes use the following types of marks: M marks These are marks given for a correct method or an attempt at a correct method. In Mechanics they are usually awarded for the application of some mechanical principle to produce an equation. e.g. resolving in a particular direction, taking moments about a point, applying a suvat equation, applying the conservation of momentum principle etc. The following criteria are usually applied to the equation. To earn the M mark, the equation (i) should have the correct number of terms (ii) be dimensionally correct i.e. all the terms need to be dimensionally correct e.g. in a moments equation, every term must be a force x distance term or mass x distance, if we allow them to cancel g s. For a resolution, all terms that need to be resolved (multiplied by sin or cos) must be resolved to earn the M mark. M marks are sometimes dependent (DM) on previous M marks having been earned. e.g. when two simultaneous equations have been set up by, for example, resolving in two directions and there is then an M mark for solving the equations to find a particular quantity this M mark is often dependent on the two previous M marks having been earned. A marks These are dependent accuracy (or sometimes answer) marks and can only be awarded if the previous M mark has been earned. E.g. M0 is impossible. B marks These are independent accuracy marks where there is no method (e.g. often given for a comment or for a graph) A few of the A and B marks may be f.t. follow through marks.

5 3. General 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. All A marks are correct answer only (cao.), unless shown, for example, as ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks. 5. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 6. Ignore wrong working or incorrect statements following a correct answer.

6 General Principles for Mechanics Marking (But note that specific mark schemes may sometimes override these general principles) Rules for M marks: correct no. of terms; dimensionally correct; all terms that need resolving (i.e. multiplied by cos or sin) are resolved. Omission or extra g in a resolution is an accuracy error not method error. Omission of mass from a resolution is a method error. Omission of a length from a moments equation is a method error. Omission of units or incorrect units is not (usually) counted as an accuracy error. DM indicates a dependent method mark i.e. one that can only be awarded if a previous specified method mark has been awarded. Any numerical answer which comes from use of g = 9.8 should be given to or 3 SF. Use of g = 9.8 should be penalised once per (complete) question. N.B. Over-accuracy or under-accuracy of correct answers should only be penalised once per complete question. However, premature approximation should be penalised every time it occurs. In all cases, if the candidate clearly labels their working under a particular part of a question i.e. (a) or (b) or (c), then that working can only score marks for that part of the question. Accept column vectors in all cases. Misreads if a misread does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, bearing in mind that after a misread, the subsequent A marks affected are treated as A ft Mechanics Abbreviations M(A) Taking moments about A. NL NEL HL Newton s Second Law (Equation of Motion) Newton s Experimental Law (Newton s Law of Impact) Hooke s Law SHM Simple harmonic motion PCLM Principle of conservation of linear momentum RHS, LHS Right hand side, left hand side.

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8 January 05 WME0/0 Mechanics Mark (a) Use of Imvmu M Used = 0.6i3j4ij correct unsimplified Allow for subtraction the wrong way round i3j or simplified equivalent = i j I M use of Pythagoras on their impulse = 3.3 Or better [4], condone (b) KE lost = i j Change in KE. Terms of correct structure. Subtract in either order. Correct unsimplified. Accept = =. (J) CAO [3] Allow after approximation e.g.

9 (a) Driving force = g sin M Requires both terms. Condone sign and sin/cos confusion - each error = (N) A 0 Use of 9.8 is an error Rate of work = 0 (395) M Use of P Fv = 7.9 kw or 7900 (W) [5] If they use.87 this leads to inaccurate answers. Allow all the marks apart from the final in each part, but watch out for those who tell you.87 and actually use the correct value and score full marks. (b) The question specifies work-energy 50d 500gsin dtheir 395d M Energy equation. Requires all 3 terms (of correct form), with no duplication, but condone sign errors and sin/cos confusion their 395 Correct unsimplified equation in d. - each error d g d d A 0 MA available for correct work leading to -ve d d 50 (m) accept 53 (answer must be +ve) [4]

10 3(a) t t v i t j 4 0 when t 4, t t M Differentiate r. Attempt seen for one or both components. (See at least one power going down) i component correct j component correct Allow if M earned here but j component not seen in (a) but then seen correct in (b) DM Dependent on the first M. Set i component of their v equal to zero Allow with no j component or incorrect j component *Given answer* Allow with no j component or incorrect j component [5] 3(b) t 4, speed = 38 (m s - ) B Must be a scalar, not a vector. [] 3(c) 3t a 4i0j M Differentiate v = 6i + 0j ISW CSO [] 3(d) t 0, r 5i B t 4, r 7i7j B Distance = 3 7 (m) M Use Pythagoras to find r4 r 0 for r 0 0, r (m) , 8 97 [4]

11 4(a) Use total height A to C B a acos seen the centres of mass of the rhombuses lie on a straight line passing through the centre of mass. 0.9a a acos M A using the symmetry of the figure. Condone sin/cos confusion. - each error cos 0.8 *Given answer* alt B Distance from A to centre of rhombus = acos. acos.cos. acos cos 0.9a cos 0.9 cos cos 0.8 M A Taking moments about an axis through A parallel to FB. Condone sin/cos confusion. - each error *Given answer* From exact working alt B Distance from A to centre of rhombus = acos the centres of mass of the rhombuses lie on a straight line passing through the centre of mass. acos cos 0.9a M A using the symmetry of the figure. Condone sin/cos confusion. - each error cos 0.9 cos cos 0.8 *Given answer* From exact working Working backwards from cos 0.8 to deduce that the distance is 0.9a is acceptable for 5/5

12 alt 3 EFBC EDC FAB lamina B Correct division of lamina with correct mass ratios a sin a sin cos a sin cos a sin a a cos a a cos cos 3 3 a 0.9a a a sin acos a sin cos acos a acos Moments equation addition and subtraction of terms M 3 3 consistent with their division. A a sin 0.9a - each error cos 0.8 *Given answer* alt 4 EDF & BCD AFD & ADB lamina a cos 3 a sin sin a a sin cos 3 a 0.9a a cos a sin a cos a sin 3 3 a sin 0.9a cos 0.8 *Given answer* B M A Correct division of lamina with correct mass ratios Moments equation addition and subtraction of terms consistent with their division. - each error Alt5 AM acos DM a acos M 0.9a acos a acos A 0.4a acos, cos 0.8 [5] B Let M be the midpoint of FB. Centre of mass lies at the midpoint of DM *Given answer*

13 Alt6 C Using the symmetry of the figure, the centres of mass of the rhombuses lie on a straight line passing through the centre of mass. D 0.a 0.4a θ 0.5a B 0.5a θ 0.5a A B 0.4a M cos 0.5a A 0.8 [5] 0.4a seen or implied Trig ratio for. Condone sin/cos confusion. Correct unsimplified expression 4(b) M Taking moments about B Must have both terms. Condone trig & sign errors kw a cos W 0.9a acos A - each error 0.8kW 0.W, k 8 4(b) alt Centre of mass is on BF. M Taking moments about A 0.9aW k W a cos k 0.8aW A - each error 8 k [4]

14 5(a) 5a Moments about A: F mgacos kmgacos 4 M Requires all 3 terms. Condone trig & sign errors A - each error 4mg cos 6 F k mg k 5 5 Substitute for cos and obtain GIVEN ANSWER [4] sin 6 sin 5 M 48 5 mg k 5(b) H F mg k V mg k Fcos M 64 mg mg k mg k k Resolve horizontally Resolve vertically. Need all three terms. Condone trig & sign errors Correct unsimplified Correct unsimplified with X and cos substituted. Accept 5(b) alt H V mg kmg 4H 3V 3mg k M Resolve parallel to the rod or perpendicular to the rod V F H mg kmg V H mgk mgk 4 5 M Obtain second equation in H and V and solve for H or V Both equations correct unsimplified 48 H mg k 5 Or equivalent 64 mg V mg k mg k 63k 5 5 Or equivalent. Accept [5]

15 4 mg L mgk H L R, V R L (b) alt R F mgk R 4k mg 3mg 48mg H k 4 k k mg 3 3mg mg V 4 k k 63k M M [5] Component R perpendicular to the rod at A and L parallel to the rod. Attempt to find both. Express V and H in terms of R and L Correct unsimplified Correct unsimplified. Accept 5(c) Use of H V to form equation in k M 48 mg k mg k mg k 5 5 DM Correct for their H, V and solve for k. 3 k or better 0.3 [3]

16 6(a) Conservation of energy: m mg mv A - each error v 5.7 m s - Accept 6 Not M Equation must include all three terms. (a)alt Find horizontal and vertical components of speed at B M Use of suvat for both components and combine v 7cos55 x vy 7sin 55 0g v 5.7 m s - Accept 6 Not 5.6 [4] NB: Use of 7 in v u as scores 0/4 in (a) but allow ft marks in subsequent parts if that work follows correctly 6(b) 7cos55 cos, their v cos 7cos55, 49 0g their v tan y M Correct trig to form equation in a relevant angle 7cos55 7sin55 0g tan 7cos to the horizontal (75) (75. from 5.7) [3] 4.9 to the vertical (direction seen or implied) A0 if magnitude and direction contradict. 6(c) Vertical distance: 0 7sin 55 t 4.9t M Use of suvat - condone sign errors A - each error 7sin55 7sin t 9.8 DM Solve for t. Incorrect answers must be supported by working. =.3 (s) Accept.

17 (c)alt M Complete strategy to find t. Vertical component of speed at B their 5.7 sin their 75. : v u at sin55 gt DM Use suvat. Condone sign error(s) t =.3 (s) (.) [5] Accept. 6(c) alt 6(c) alt 3 M Time to top: 07sin55 gt, t Distance to top: 07sin55 9.8s, s DM Time to fall.68:.68 gt t Total time tt.3 (s) M 7sin557sin55 gt, t.7... Time to fall 0 m: t Total time tt.3 (s) [5] 0 7sin 55t 9.8t DM Complete strategy for time to top + Top to ground Complete strategy for time to level + time to fall 0 m

18 3u 7(a)(i) P m Q m v CLM: 3mu mv mw 3u v w w M Requires all three terms, but condone sign errors Impact law applied the right way round, but condone sign errors A0 here if the signs in the two equations are not consistent. wu e DM Dependent on the previous M marks. Solve for w or v. Impact: wv 3eu M 3w3 u( e), (ii) v w 3eu u ue speed = u e Both speeds correct. Must both be positive. 7(b) Change in direction e 0 M Correct inequality for reversal of direction (for their v) e CWO. e not required. [] [6]

19 7(c) u v and u 5 w u B 3 u Q m R 3m q m w 3mu mq 3mr 6u rqewu 9 r M CLM & impact equations Both correct u 6u q 3r, 3r3q 3 3 DM Dependent on previous M. Solve for q 5q 5u, q u u u therefore Q will collide with P a second time 3 Given answer [6]

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