A-LEVEL Mathematics Mechanics 5 MM05 Mark scheme 6360 June 015 Version/Stage: 1.0: Final
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MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE 015 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 through from previous incorrect result CAO correct answer only CSO correct solution only AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given 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 significant figure(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 showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working 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 working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded. 3 of 8
MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE 015 Q Solution Mark Tot al Comment 1.(a 0.8 = 0.ω : Use of 0.8 = 0.ω or 0. = 0.8ω ω = 4 : Correct ω : Correct period (1.57 π π Period = = 4 3 1.(b v = 4 (0. - 0.1 : Seeing (0. 0.1 or v = 3-1 3 (0.4 0.3 in an equation. 0.48 = = 0.693 m s 5 : Correct equation. : Correct speed. 1. (c s = 0. 0.cos(4t 3 : Seeing 0.-... : Seeing 0.cos... : Seeing 4t. 1. (d 0.1 = 0. 0.cos(4t : Using 0.1 and answer to (c0 to form cos(4t = 0.5 an equation. t = 0.6 s : Correct equation. 3 : Correct time. Total 1 1 1. : Obtaining two times and adding. T1 = π = 0.54966 4 g : One correct time. : Second correct time. 1 0.7 T = π = 0.41981 : Correct total time. 4 g T + T = 0.54966 + 0.41981 = 0.969 s (to 3sf 4 1 Total 4 4 of 8
MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE 015 3. (a θ = : Seeing or clear use of θ =. θ = t : Correct expression for r in terms of t θ = 0 : Correct expression for r in terms of t : Use or radial formula. 1 aθ at r = ae θ = ae : Correct radial acceleration. : Use or transverse formula. 1 aθ 1 aθ at r = a e θ + ae θ = a e : Correct transverse acceleration. Radial : r r θ Transverse: = a = e at r θ + r θ = ae at e ( a at 1 4 e 1 = 4ae at at 7 3. (b e at = 1 (( a 1 + (4a = 0 4 4a + 8a + 4 = 400 4 a + a 99 = 0 ( a + 11( a 9 = 0 a = ± 3 5 Total 1 : Using initial value to get expression to eliminate t. : Finding magnitude. : Correct magnitude in terms of a. : Solving for a : Correct values of a 5 of 8
MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE 015 4. (a VAB = mga sinθ : Two GPE terms found. a : Both correct. VBC = mg cosθ : EPE for one string. 1 mg V = (5a a a sinθ : Correct EPE A a : EPE for other string. = 4mga( sinθ : Correct EPE = 4mga(4 4sinθ + sin θ : Correct total from correct V 1 8mg = (5a a a cosθ working. B a = 4mga(4 cosθ = 4mga(16 8cosθ + cos θ mga V = (4sinθ + cosθ + 3 3sinθ + 8sin θ + 18 64cosθ mga = (168 63cosθ 8sinθ 7 4. (b 4. (c dv mga = (63sinθ 8cosθ dθ mga (63sinθ 8cosθ = 0 8 4 tanθ = = 63 9 θ = 0.418 d V mga = (63cosθ + 8sinθ dθ d V mga θ = 0.418 = (68.9 > 0 dθ Stable 5 3 : Differentiating energy expression. : Correct derivative. : Solving to obtain tan θ : Correct value for tan θ : Correct θ. (Accept 3.96 : Finding second derivative. : Correct derivative. : Correct conclusion. Total 15 6 of 8
MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE 015 5. (a g = 4e : Using Hookes Law. : Correct extension. g e = : Correct distance. 4 g 49 1 7 x = + 0.35 = + = 4 60 60 6 3 Accept 1.17 5. (b (i d x dx : Suitable four term equation. = g 14 4( x 0.35 : Three terms correct. dt dt : All terms correct. d x dx + 7 + 1x = 9.8 + 1 0.35 dt dt d x dx + 7 + 1x = 14 dt dt 4 5. (b (ii PI 7 x = 6 CF λ + 7λ + 1 = 0 ( λ + 3( λ + 4 = 0 λ = -3 or λ = -4 x = Ae x = Ae -3t 3t x = 1.8, t = 0 + Be + Be -4t 4t + 7 6 7 1.8 = A + B + 6 19 A + B = 30 dx 3t = 3Ae 4Be 4 dt x = 0, t = 0 0 = 3A 4B 4B A = 3 38 19 A =, B = 15 10 38 3t 19 x = e e 15 10 4t + 7 6 t d d 10 5. (b (iii Heavy damping 1 : Correct statement. Total 18 : Correct PI. : Quadratic equation for λ. : correct values for λ. d: Using initial values to obtain an equation containing A and B. : Correct equation. d: Using derivative to obtain a second equation for A and B. : Correct equation. : Value of A correct. : Value of B correct. : Correct final expression. 7 of 8
MARK SCHEME A-LEVEL MATHEMATICS MM05 JUNE 015 6. (a dv : Correct equation for block. T = M dt : Impulse-momentum equation with ( mg T dt = ( m + dm( v + dv dmv mv = mdv mg T or T mg. : Correct equation. dv dv mg M = m : Correct differential equation. dt dt : Use of expression for mass and it's dv mg = ( m + M derivative to complete expression dt : Correct final expression. m = M λt d dv mg = dt m + M dv ( M λt g = dt M λt 6 6. (b dv ( M l t g = dt M lt Mg = g M lt Mg v = gt + ln(m lt + c l Mg v = 0, t = 0 c = ln(m l Mg M lt v = gt + ln l M d d 5 : Splitting RHS into two terms, with one a constant. d: Integrating to obtain linear and ln terms.. : Correct integral. d: Use of initial conditions. : Correct final expression. 6. (c M t = l Mg Mg 1 v = + ln l l Mg = (1 ln l 3 Total 14 TOTAL 75 : Use of correct value of t. : Correct substitution into correct result. : Simplified correct time. 8 of 8