A-LEVEL Mathematics. MM05 Mechanics 5 Mark scheme June Version 1.0: Final

Transcription

1 A-LEVEL Mathematics MM05 Mechanics 5 Mark scheme 6360 June 016 Version 1.0: Final

2 Mark schemes are prepared by the Lead Assessment Writer and considered, toether with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students scripts. Alternative answers not already covered by the mark scheme are discussed and leislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a workin document, in many cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the uidin principles of assessment remain constant, details will chane, dependin on the content of a particular examination paper. Further copies of this mark scheme are available from aqa.or.uk Copyriht 016 AQA and its licensors. All rihts reserved. AQA retains the copyriht on all its publications. However, reistered schools/collees for AQA are permitted to copy material from this booklet for their own internal use, with the followin important exception: AQA cannot ive permission to schools/collees to photocopy any material that is acknowleded to a third party even for internal use within the centre.

3 MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE 016 Key to mark scheme abbreviations M mark is for method m or dm mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for explanation or ft or F follow throuh from previous incorrect result CAO correct answer only CSO correct solution only AWFW anythin which falls within AWRT anythin which rounds to ACF any correct form AG answer iven SC special case OE or equivalent A,1 or 1 (or 0) accuracy marks x EE deduct x marks for each error NMS no method shown PI possibly implied SCA substantially correct approach c candidate sf sinificant fiure(s) dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showin workin and it is very unlikely that the correct answer can be obtained by usin an incorrect method, we must award full marks. However, the obvious penalty to candidates showin no workin is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without workin earns full marks, unless it is iven to less than the deree of accuracy accepted in the mark scheme, when it ains no marks. Otherwise we require evidence of a correct method for any marks to be awarded. 3 of 8

4 MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE (a) π x : Equation with no more than one error. = π : Correct equation : Correct distance. 9.8 x = = (b) π 7 : Uses ω = 3.5 θ = cos t : Seein ±π/0 used. 0 : Correct equation. π π 7 = cos t cos t = d d: Solvin equation for t. 7 π t = 3 : Correct time. π 5 t = = s 1 Total 8. (a) TA = ( d 0.5) 0.5 : One correct tension. 3 TB = ( d 0.) 0. : Other correct tension..6 + T = T :Equation with correct terms. B A : Correct equation d = 8d : Correct distance. d = = (b) (i) TA = 8(1. + x 0.5) F TB = 7.5(0.8 x 0.) F.6 x =.6 + TB TA = x 5.6 8x = 15.5x 15.5 x = x.6 5. (b) (ii).6 Period = π = 0.8 s (b) (iii) v max = 0. = 1.53 m s.6 Total 1 : One correct tension in terms of x. : Other correct tension in terms of x.. : equation with correct terms. : Correct equation. : Correct expression. : Usin their ω to find period. : Correct period. : Usin their ω to find max speed. : Correct max speed. of 8

5 MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE (a) m a V = m a tanθ + a : Correct GPE a cosθ : Correct EPE = ma + tanθ cos θ cosθ 3 : Correct total ACF = ma((secθ 1) tanθ ) 3. (b) dv sinθ sinθ 1 : Attempts to differentiate. = ma 3 : Correct derivative. dθ cos θ cos θ cos θ θ = 0.7 dv d d: Substitution of 0.7. = 0.355ma dθ θ = 0.8 d d: Substitution of 0.8. dv : Correct values obtained. = 0.513ma dθ As there is a sin chane there must be a F: Correct conclusion based on their zero between 0.7 and 0.8. Hence there is a F 6 values. position of equilibrium. 3. (c) Stable, as the enery is a minimum : Stable dv because chanes from neative to dθ : Correct justification. positive. Total 11 5 of 8

6 MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE 016. (a) No transverse acceleration 1 d : Uses transverse component equal to ( r θ ) = 0 zero. r r θ = c. (b). (c) c θ = r Where c is a constant. km m( r r θ ) = r c k r = 3 r r r = 0, r = a c c r r r ak k r = 3 r r k = r = ak ak k rr = r r 3 r r 1 ak k r = + + d r r r = U, r = a 1 k k U = + + d a a 1 d = U r = a 1 k k 1 r = + + U a a k r = + U a d d 5 5 Total 1 : Correct conclusion from correct workin. : Uses Newton's second law with radial component. : Correct equation : Substitutes for θ. : Correct result from substitution. : Correct expression for r. : Multiplyin by r. d: Interatin. : Correct constant. d: Substitutin and simplifyin correctly. : Correct final answer. 6 of 8

7 MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE (a) mx = m cmx mx x + cx + x = For critical dampin c 16 = 0 c = d : Equation of motion with correct terms. : Correct equation. d: Equation to find c from discriminant. : Obtainin c = from correct workin. 5. (b) CF λ + λ + = 0 λ = t x = e ( A + Bt) PI x = AG : Findin λ : Correct form for CF. : Correct PI. 5. (c) 5. (d) General Solution t x = e ( A + Bt) + x = 0, t = 0 A = t x = e ( A + Bt) + Be t = e ( B A + Bt) x = 0, t = 0 B = A B = t e x = (1 + t) t x = ( e e t = te t t x = ( e te ) 1 x = 0 t = x max = e t t (1 + t)) 8 : Findin A. : A correct. : Findin B. : B correct. 1 : CAO Total 17 : Correct final answer : Correct x. : Obtains second derivative. : Correct time. : Correct final answer. 7 of 8

8 MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE (a) ( m + dm)( v + dv) mv = m : Equation with correct terms. mdv + vdm = m : Correct equation. dv dm m + v = m dv m + vλm = m dv = λv 6. (b) 1 dv= lv+ : Correct equation with derivatives. : Correct DE from correct workin. : Separation of variables. 6. (c) 1 ln( lv+ ) = + t c l t = 0, v= U 1 c= ln( lu + ) l v = ln( ) = + t ln( lu + ) l l 1 lu t = ln + 1 l dm = λm λt m = Me λu m = M Total 13 : Correct interation. : Findin constant. : Correct constant. : Use of v = 0 : Correct time. : Use of DE to find m. : Correct expression for m. : Correct mass. 8 of 8