Mark Scheme (Results) Summer GCE Mechanics 2 (6678/01)

Transcription

1 Mark Scheme (Results) Summer 013 GCE Mechanics (6678/01)

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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 e applied positively. Candidates must e 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 oundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should e used appropriately. All the marks on the mark scheme are designed to e awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also e 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 y which marks will e awarded and exemplification may e limited. Crossed out work should e marked UNLESS the candidate has replaced it with an alternative response.

4 EDEXCEL GCE MATHEMATICS General Instructions for Marking 1. The total numer 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 e awarded if the relevant method (M) marks have een earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not e sudivided. 3. Areviations These are some of the traditional marking areviations that will appear in the mark schemes: od enefit of dout ft follow through the symol will e used for correct ft cao correct answer only cso - correct solution only. There must e no errors in this part of the question to otain this mark isw ignore susequent 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 e followed through. After a misread however, the susequent A marks affected are treated as A ft, ut manifestly asurd answers should never e awarded A marks. 5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question: If all ut 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. 7. Ignore wrong working or incorrect statements following a correct answer. 8. In some instances, the mark distriutions (e.g., and ) printed on the candidate s response may differ from the final mark scheme

5 General Rules for Marking Mechanics Usual rules for M marks: correct no. of terms; dim correct; all terms that need resolving (i.e. multiplied y cos or sin) are resolved. Omission or extra g in a resolution is accuracy error not method error. Omission of mass from a resolution is 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 e awarded if a previous specified method mark has een awarded. Any numerical answer which comes from use of g = 9.8 should e given to or 3 SF. Use of g = 9.81 should e penalised once per (complete) question. N.B. Over-accuracy or under-accuracy of correct answers should only e penalised ONCE per complete question. In all cases, if the candidate clearly laels their working under a particular part of a question i.e. (a) or () 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, earing in mind that after a misread, the susequent A marks affected are treated as A ft.

6 1. Use of I = mv-mu Must e sutracting. Condone sutraction in the wrong order v = (3i + 6j) +(i 4j) Correct unsimplified equation (= 5i j) v =.5i - j Speed = (.69(m s -1 )) Use of correct Pythagoras with their v Exact form or s.f. or etter. Watch out for fortuitous answers from.5i + j.

7 a Work done = 15 R g cos0 = 18g cos0 166 (J) [3] Fmax 3g cos 0 (11.05). R must e resolved ut condone trig confusion. 15 their F max. Independent M 15 F max... is M0 or 170 (J) Energy: WD against F + GPE + final KE = initial KE Must include all four correct terms (including resolving). Condone sign errors and trig confusion. Any sign errors in the KE terms count as a single error. Follow their WD their WD 3gsin v 3 0 Aft -1ee Follow their WD v 13.7 (m s -1 ) or 14 Or 3a 0.4 3g cos 0 3g sin 0 and use of v u as ft [4] Complete method. Their F max +component of weight A correct equation with their F max. Allow for a acting down the slope a v 0 a 15( ) ft Correct equation for their a v 13.7 (m s -1 ) [4] or 14 (m s -1 )

8 3a v 0 t 14t 0 Set v 0 t t 5 Solve for t t or t 5 [3] There are many different approaches to part (). The allocation of the two M marks is : A method to find the time when the velocity is a minimum : Evaluate the speed at that time e.g. t 0, v 0 (m s -1 ) a 4t 14 = t, v Must see 4.5 Clearly stated & correct conclusion. Max speed = 0 ms -1 Depends on the two M marks. From correct solution only. alt1 t 0, v 0 (m s -1 ) Sketch with symmetry aout their t 3.5 v(their 3.5) Evaluate v at min Correct work Max speed = 0 ms -1 alt t 0, v 0 (m s -1 ) Justification of minimum or taulate sufficient values to confirm location Clearly stated & correct conclusion. Depends on the two M marks. From correct solution only. Evaluate v at min. Correct work Correct conclusion. Depends on the two M marks Clearly stated & from correct solution only.

9 alt 3 t 0, v 0 (m s -1 ) Complete the square as far as 7 t 7 9 t Max speed = 0 ms -1 Clearly stated & correct conclusion. Depends on the two M marks. From correct solution only. c 3 3 t 14t 0 dt t 7t 0 t( C ) Integration. Need to see majority of powers going up All correct. Condone C missing Distance = t 7t 0t t 7t 0t Correct method to find the distance, for their Correct unsimplified 3 3 t 7t 0t t 7t 0t (m) 3 3 4

10 4a F A E m p O s m B q r D For a valid division into asic elements: e.g. pair of rhomuses m m C AOCB OCDE whole 1 1 1/ 1/ y Correct mass ratios for parts and the arrow shape Correct vertical distances from a horizontal axis 1 1 y 1 1 y 0.5 (m) Moments equation aout a horizontal axis Correct equation for their axis a alt AOB OBCD DOE whole y Rhomus + two triangles 4y 1 Moments equation y 0.5 (m)

11 a alt 3 Hexagon AOEF whole y 4y 0 1 y 0.5 (m) a alt 4 h = height of each triangle = 3 Distances of c of m from horizontal through O p q r s whole cos30 3 h cos30 3 h 0 y y 1 cos y 0.5 (m) Hexagon rhomus 4 triangles

12 4 In 4() the first two marks are : Indentify a triangle, with one angle correct, and attempt to find the lengths of two sides ft: sides correct, follow their answer to (a) D: Work sufficient to e ale to go on to find the required angle. Dependent on the preceding ft: follow their answer to (a) D: Find the required angle. Dependent on the preceding Correct answer. for example F A θ E O 0.5 cos30 3, "0.5" sin ft Their 0.5 & their 3 tan their 1.5 their 3 Required angle = tan D D ft D [6] Use of tan in a right angled triangle. Accept the reciprocal Correct for their angle. Ft their 0.5 Correct strategy to find required angle e.g. " " 30 or " " Accept 11, 10.9 or etter

13 4alt E m θ O 0.5 m 10 d G SAS in a relevant triangle ft Their 0.5 d cos D Correct cosine rule. ft Correct equation. Their 0.5 sin sin D 10.9 [6]

14 5a F B V C mg A θ H Moments aout A: mg F acos mg acos mg( 3a cos mg ) A -1 each error Moments aout A. Requires all three terms and terms of correct structure (force x distance). Condone consistent trig confusion F 5 : 3amg cos H Fsin *Answer given* 3amg cos sin [4] Resolve horizontally. Condone trig confusion RHS correct. Or equivalent. : mg V F cos Resolve vertically. Condone sign error and trig confusion Correct equation 3amg cos 3amg cos V mg cos mg RHS correct. Or equivalent

15 5c 3amg cos mg 3amg cos sin 3acos sin 3a cos sin cos tan 3a cos 3asin 3a, a 3 D [4] Use of tan, either way up. V, H, F sustituted. Correct for their components in θ only Simplify to otain the ratio of a and, or equivalent 5c alt The centre of mass of the comined rod + particle is 3 a from A F R mg alt c 3 3 forces in equilirium must e concurrent 3 a a 3 R acts along the rod, so resolve forces perpendicular to the rod. F mg cos mg cos mg cos 3amg cos D a 3 [4] [4] Not on the spec, ut you might see it. Resolve and sustitute for F Eliminate θ

16 alt c 4 c alt 5 R acts along the rod. Take moments aout C mg cos a mg cos a a a, a Resultant parallel to the rod R mg sin And V H R 3 3amg cos sin 3amg cos mg sin mg Eliminate θ a 3 D [4] D [4] Moments aout B gives a F amg cos and sustitute for F Sustitute for V, H and R in terms of θ

17 6a Conservation of energy: 1 1 mu 16mg 4mu mu mg 8 m u 16 3 mg mu, A -1ee 16g u D 3 Energy equation must contain the correct terms, ut condone sign error. Correct unsimplified Solve for u 16g u 7. Accept 7.3. Accept 3 6 g Condone sign errors or trig error. u must e Vertical distance: 8 u sin 4 resolved. A -1ee Correct equation for their u. g 8 sin u or [4] 6c Min speed at max height, i.e. u cos Condone consistent trig confusion with part () = 4.3 (m s -1 or 4.3 (ms -1 ) ) []

18 7a CLM: mu mv 3mw All three terms required, ut condone sign errors Impact: w v eu Condone sign error, ut must e sutracting and e must e used correctly. Penalise inconsistent signs here. Sust v w eu: u w eu 3w 5w eu D Solve for w. Requires the two preceding M marks w 1 e u *Answer Given* 5 (6) 7 7u w 10 CLM: 3mw 3mx 4my and Impact: Sust: 3w 3x 4 x 3 w 4 x 0, y 3 1 w u 4 40 y x 3w 4 D (6) Seen, or implied y correct speeds. Both needed Solve for x or y. Dependent on the preceding M mark 0.55u, 7c u v 0 u 1u 3u Speed of separation = (3) [15] Correct velocity of P Correct use of their values and sustitute for e. Check directions carefully 0.575u

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