A-Level Mathematics. MM05 Mechanics 5 Final Mark scheme June Version/Stage: v1.0

Transcription

1 A-Level Mathematics MM0 Mechanics Final Mark scheme 6360 June 07 Version/Stae: v.0

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 07 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 MM0 JUNE 07 (a).4 Period = : Uses formula with correct 9.8 lenth. = (b) = 3.4 s.4 Averae Speed = 0 - = 0.98 m s 3 : Correct period. : Correct distance found. : Correct expression for averae speed. : Correct averae speed. (c) θ = cos(t) 0.4 v = sin(t).4. = sin(t). t = sin = θ = 0.97 OR. m = m cosθ cos cosθ = cos θ = 0.9 OR.4. (.4 ) 0 = θ θ =.4 = d () () () (d) () () () () () (d) () () Total 0 : Correct expression for θ. : Correct expression for velocity. d: Formin equation to find t. : Correct time. : Correct θ. : Enery equation with two terms correct. : Correct terms but allow sin errors. : Correct equation. d: Solvin for θ. : Correct value of θ. : Use of v = ω ( a x ) with consistent terms. : Correct terms but possible sin errors. : Correct terms. d: Solvin for θ. : Correct value of θ. 3 of 8

4 MARK SCHEME A-LEVEL MATHEMATICS MM0 JUNE 07 (a) T =.4 T d 49 = d ( d 0.) = 0.4 ( d 0.) 96d = 99.9 d =.0 0 d 4 : Three force equation with at least two terms correct. : Correct equation. d: Solvin for d. : Correct d. (b) d x 0.4 = T = T (.0 x 0.) = x x 0.96 = 96x d x = 490x 0 Period = = ( x.0 0.) : Equation of motion with at least two terms correct. : Correct terms but possible sin errors. : Correct equation. : Correct simplified differential equation. : Correct period from correct workin. (c) (i) m s - v max = = = : Use of a ω. : Correct max speed. (c) (ii) v = 490 = v = ( ) = 0.99 m s 47 = 60-3 : Use of v = ω ( a x ) with correct ω. : Correct equation. : Correct speed. Total 4 4 of 8

5 MARK SCHEME A-LEVEL MATHEMATICS MM0 JUNE 07 4m EPE = x y : EPE in terms of x and y. 3 (a) ( ) 4 = m( x 6 8x x ) 4 = m( 6 7x x ) = m GPE = my = m(4 4 ( 7 7x x 6 7x x 4 4 ( 38 x x 4 6 7x x ) V = m ) B : EPE in terms of x. : EPE expanded correctly. B: Correct GPE. : Correct final result from correct workin. 3 (b) dv = m 30x 8x 3 8x 3 6 7x 8x x 4 : Differentiates with no more than one error. : Correct derivative. dv x =, = 0 So there is a position of equilibrium when x =. d 4 d: Substitutes x =. : Obtains correct conclusion. 3 (c) At x =.9 At x =. dv dv =.99m = 3.94m As increasin this corresponds to a minimum value of V and hence is a position of stable equilibrium. 3 : Substitutes a value of x just less than. : Substitutes a value of x just reater than. : Uses values to reach the correct conclusion. OR 4x 30x 3 3 d V ( 8x 8x )( x 7) = m. 4 ( 6 7x x ) ( 8-4x ) 4 6 7x x d V At x =, = 34m This corresponds to a minimum value of V and hence is a position of stable equilibrium. () () () (3) : Second derivative of the correct format. : Correct second derivative. : Correct value at x = and correct conclusion. Total of 8

6 MARK SCHEME A-LEVEL MATHEMATICS MM0 JUNE 07 4 (a) = sin t : Usin r = to form an equation. t = n 4 : Findin t. r = cos t B B: Correct θ. θ = θ = 0 r θ r θ = 8cos t : Expression transverse component. cos n = 0 : Obtainin zero from correct workin. OR = sinθ cosθ = 0 r = cosθ θ = θ = 0 r θ r θ = 8cosθ = 0 () () (B) () () () : Usin r = to form an equation. : Findin cos θ. B: Correct θ. : Expression transverse component. : Obtainin zero from correct workin. 4 (b) 4 (c) r = 4sin t r r θ = 4sin t 4sin t = 8sin t 0 = 8sin t n t = 0 r θ = sin t sin( n ) = 0 r θ = 0 4 : Findin radial component. : Correct radial component. : Formin equation to find t. : Correct time(s). : Findin transverse component of the velocity. : Correct conclusion from correct workin. Total 6 of 8

7 MARK SCHEME A-LEVEL MATHEMATICS MM0 JUNE 07 d x.mx : Formin equation of motion, with at m = m m least two terms correct. 0. : Correct differential equation. d x x= PI x = B B: Correct PI. CF λ λ = 0 4 ± λ = = ± i x= e ( Asin t Bcos t) t = 0, x= 0 0 = B B= / x = e ( Asin t Bcos t) e (Acos t Bsin t) x = e (( A B)sin t t = 0, x = 0 0= B A ( B A)cos t) A = 0 sin t cos t x = e 0 sin t x = e 8 x = 0 t = sin cos x = e 0 Max Lenth of Strin = e 0. =.87 m : Solvin auxiliary equation. : Correct roots. : Correct eneral solution. : Findin one constant. : Correct constant. : Findin derivative. : Correct derivative. : Second constant correct. : Findin time for zero speed. : Correct time. : Usin time to find max lenth. : Correct maximum lenth. Total 7 of 8

8 MARK SCHEME A-LEVEL MATHEMATICS MM0 JUNE 07 6 (a) mδt = ( m δm)( v δv) : Use of momentum-impulse equation. ( δm)( v δv U ) mv (Must have vu) : Correct equation. mδt = mδv Uδm dv dm m = m U d d: Formin differential equation. dm = λ, m = M λt : Expression for M in terms of t. dv m = m λu dv λu = M λt : Correct result from correct workin. 6 (b) dv t = 0, m = M, < 0 λu < 0 M d M U > λ 3 : Statement that acceleration is less than zero. d: Use of time as zero. : Correct result from correct workin. 6 (c) λu dv = M λt v = t U ln( M λt) c U U t = 0, v = c = U ln M 0 0 M λt U v = t U ln M 0 M t = 0λ M 3M 9 3M v = ln 0λ λ 0 40λ 3M 9 7M = ln λ 0 40λ : Interatin to obtain linear and ln terms. : Correct interal. : Correct constant. : Correct time. : Correct velocity. Total 3 8