Mark Scheme (Results) January Pearson Edexcel International A Level in Mechanics 3 (WME03/01)

Transcription

1 Mark (Results) January 06 Pearson Edexcel International A Level in Mechanics 3 (WME03/0)

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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 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.

4 General Instructions for Marking EDEXCEL IAL MATHEMATICS. 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. 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. 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.

5 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. MARKS MUST BE ENTERED IN THE SAME ORDER AS THEY APPEAR ON THE MARK SCHEME. 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.

6 Jan 06 WME03/0 M3 Mark Question. Rsin 60 mg Rcos60 tan 60 mr g r g g 3 r 3 * ddcso 3r [6] dd cso Resolve vertically, allow with 60 or Correct equation, with 60 or Equation of motion along radius, acceleration in either form, 60 or Correct equation, with 60 or acceleration to be r now Eliminate R, substitute a numerical value for and solve to... or... Depends on both previous M marks Complete to the given answer 6

7 (a) dv a 6 t dt v 6t t c t 0, v8 c 8 v t t 6 8 (3) (b) v6tt 8 0 t t 0 t, t s 6tt 8dt 3t t 3 8t 3 c 3 3t t 8t t t 8t (a) (b) ft Total distance = 8 m ft (5) [8] dv Using a and attempting the integration Use of dt Correct integration w/wo constant Find constant and correct statement for the velocity a dv v scores M0 d x Find the times when P is at rest (Usual rules for factorising or formula) Integrate v to obtain an expression for s (c not needed) Correct displacement for either time interval Correct displacement for second time interval Add (positive) distances NB: Using v 8 Is NOT a misread. 7

8 3 R( ) R cos0 Fcos70 800g v NL( ) Rcos70Fcos F R 0.5R Rcos00.5Rcos70 800g R 800g cos00.5cos70 800g cos700.5cos0 v 0 cos00.5cos70 OR v cos700.5cos0 0g cos00.5cos70 dd v or m s - dddcao [9] Resolve vertically if equation completely correct; A0 one error; A0A0 two or more errors Equation of motion along the radius, acceleration in either form LHS correct RHS correct inc acceleration in form r m or 800 for these marks dd Use F R and eliminate R to obtain v... Depends on both previous M marks ddd Complete to a numerical value for v or v All previous M marks needed. Correct value of v or 3 sf only v 8

9 (a) R A a B C Tsin mg 3 sin or cos B T mg 6 a T 3a mg mg * 6 cso (5) (b) a a EPE at D at C 3a 3a (or equivalents with half strings) B (either) a a 3 mv mg a 6a 6a v ag d 7 v 7ag (5) [0] (a) Resolve vertically Must have T. Correct equation B A correct trig value for an angle - seen implicitly or used Hooke's law inc attempting the extension in terms of a cso Obtain the given value of from correct working (b) B Obtain the correct EPE at either start or finish of the motion. May be those shown (full strings) or half of these (half strings) (May have already sub their ) x An energy equation with the correct number of terms. EPE terms to be of the form k l A fully correct equation Solve to v... A correct expression for v Any equivalent form 9

10 5 (a) (b) (c) 3 T mgsin mg 5 B 5 l 3 T mg l 5 3mg * cso (3) mg sin T mx 3 3mg 5 l x mg mx 5 l 3g x x l d SHM () 3g 3g x a l ft () max l 3g (d) Time from D to B: l lsin t l (a)b cso (b) d (c) ft (d) dd l t 6 3g Time from B to nat. length: 3g l lsin t 5 l l t sin 3g 5 l l dd,cao Total time sin, 0.5 k g g () [3] 3 T 5 mg shown explicitly or used Attempt Hooke's law Obtain the given result Equation of motion along the plane inc T in terms of l and x. Allow with acceleration as a Correct equation, acceleration can still be a. Can still have Re-arrange to the required form. Acceleration must be x oe now Correct result as shown or x x and SHM stated. If sub made for g, answer must be ml or 3 sf. But may be possible to award earlier. Use x a max Correct answer follow through their Must be positive.7 or 5 allowed Find time from D to B or from C to D Find time from B to natural length or from C to natural length Add or subtract (as appropriate) the two times obtained Correct result for k Must be sf but need not be shown explicitly. 0

11 r xy d x 0 6(a) r x r x x d 0 r x xr 0 r Mx xy dx 3 r r 3 x r* (5) 3 8 (b) Mass m M (m+m) Dist from O 3 8 r r 3 r x B 35 rm 3rM m M x 8 ft x 35m M r 8 m M () (c) x cos OA r cos OA B OA x r x 35m M r 7r 8 m M r 35mM 3mM m M * 0 (5) []

12 (a) Using xy dx and/or may be missing limits need not be shown Correct integration., and limits need not be shown cso Use limits to obtain r Use Mx xy dx with their previous result. and must be seen in both sides or neither. Obtain 3 8 r (b) B Correct distances from O or centre of plane face (can both be positive). Construct a moments equation with their distances using their masses (which may be volumes) ft Correct moments equation, follow through their distances (but not their masses). If working from centre of plane face one term must be negative now. Correct result for x (any equivalent) (inc fractions within fractions) (c) Attempting an inequality with x and OA (Sign either way round or =) B A correct trig function connecting the angle used and OA OA Obtain x r Use their expression for x in their inequality/equality. cso Obtain the given result. ALT for (c) Use distance from centre of common face and tangents x r r M is min when tan r r

13 7(a) For complete circles there must be a speed at the top. (or equiv statement) B mgl 5 mu gl u * () 5 (b) NL at bottom v Tmax mg m l mg 9 5 Energy to bottom mv mu mgl cos m8gl Tmax mg u l 5 v NL at top Tmin mg m l m l 5 Tmin u gl mg m m8gl u glmg mg u l 5 l 5 dddd 8gl u glgl gl u 5 5 u 5gl 7gl * 5 5 cso () [5] (a) B Statement (oe) shown seen in working. Use energy (and above statement) to obtain an inequality for u Correct inequality Deduce the given result ALT B, Energy equation including speed at top State v > 0 and use in energy equation to obtain an inequality As above 3

14 (b) Attempt NL at bottom. Acceleration in either form. v Correct equation inc acceleration in r form Energy equation from A to lowest point Correct energy equation Eliminate v to obtain the max tension in terms of m, g, l u as shown oe Attempt NL at top. Acceleration in either form. v Correct equation inc acceleration in r form Use energy to top to obtain an expression for least tension in terms of m, g, l u Correct expression as shown oe dddd Connect the two tensions using on either side cso Obtain the given result.

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