A-Level Mathematics MM0 Mechanics Final Mark scheme 6360 June 07 Version/Stae: v.0
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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 = 0.4469....4 θ = 0.97 OR. m = m 9.8.4 cosθ cos 0 0.7 cosθ = cos 4.0 0 θ = 0.9 OR.4. (.4 ) 0 = θ θ =.4 = 0.97.4 0. 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
MARK SCHEME A-LEVEL MATHEMATICS MM0 JUNE 07 (a) T =.4 T 49 49 0. 0. 98d 49 = 3.9 47 98d ( 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 = 49 0. 0.4 T (.0 x 0.) 49 0.4 9.8 0. = 47.04 98x 3.9 98x 0.96 = 96x d x = 490x 0 Period = = 490 3 ( 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) 7 0 0.m s - v max = 490 0.0 = = : Use of a ω. : Correct max speed. (c) (ii) v = 490 = 0.987 v = ( 0.0 0.0 ) 0.987 = 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
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
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
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
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