Version 1.0. General Certificate of Education (A-level) June Mathematics MM04. (Specification 6360) Mechanics 4. Final.

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1 Vesion 1.0 Geneal Cetificate of Education (A-level) June 011 Mathematics MM04 (Specification 660) Mechanics 4 Final Mak Scheme

2 Mak schemes ae pepaed by the Pincipal Examine and consideed, togethe with the elevant questions, by a panel of subject teaches. This mak scheme includes any amendments made at the standadisation events which all examines paticipate in and is the scheme which was used by them in this examination. The standadisation pocess ensues that the mak scheme coves the candidates esponses to questions and that evey examine undestands and applies it in the same coect way. As pepaation fo standadisation each examine analyses a numbe of candidates scipts: altenative answes not aleady coveed by the mak scheme ae discussed and legislated fo. If, afte the standadisation pocess, examines encounte unusual answes which have not been aised they ae equied to efe these to the Pincipal Examine. It must be stessed that a mak scheme is a woking document, in many cases futhe developed and expanded on the basis of candidates eactions to a paticula pape. Assumptions about futue mak schemes on the basis of one yea s document should be avoided; whilst the guiding pinciples of assessment emain constant, details will change, depending on the content of a paticula examination pape. Futhe copies of this Mak Scheme ae available fom: aqa.og.uk Copyight 011 AQA and its licensos. All ights eseved. Copyight AQA etains the copyight on all its publications. Howeve, egisteed centes fo AQA ae pemitted to copy mateial fom this booklet fo thei own intenal use, with the following impotant exception: AQA cannot give pemission to centes to photocopy any mateial that is acknowledged to a thid paty even fo intenal use within the cente. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guaantee egisteed in England and Wales (company numbe 6447) and a egisteed chaity (egisteed chaity numbe 1074). Registeed addess: AQA, Devas Steet, Mancheste M1 6EX.

3 Key to mak scheme abbeviations M mak is fo method m o dm mak is dependent on one o moe M maks and is fo method A mak is dependent on M o m maks and is fo accuacy B mak is independent of M o m maks and is fo method and accuacy E mak is fo explanation o ft o F follow though fom pevious incoect esult CAO coect answe only CSO coect solution only AWFW anything which falls within AWRT anything which ounds to ACF any coect fom AG answe given SC special case OE o equivalent A,1 o 1 (o 0) accuacy maks x EE deduct x maks fo each eo NMS no method shown PI possibly implied SCA substantially coect appoach c candidate sf significant figue(s) dp decimal place(s) No Method Shown Whee the question specifically equies a paticula method to be used, we must usually see evidence of use of this method fo any maks to be awaded. Whee the answe can be easonably obtained without showing woking and it is vey unlikely that the coect answe can be obtained by using an incoect method, we must awad full maks. Howeve, the obvious penalty to candidates showing no woking is that incoect answes, howeve close, ean no maks. Whee a question asks the candidate to state o wite down a esult, no method need be shown fo full maks. Whee the pemitted calculato has functions which easonably allow the solution of the question diectly, the coect answe without woking eans full maks, unless it is given to less than the degee of accuacy accepted in the mak scheme, when it gains no maks. Othewise we equie evidence of a coect method fo any maks to be awaded.

4 MM04 1(a) = B1 1 Clea use of Fi = 0 (b) i F = j 1 = 1 1 k 1 0 i 4 F = j 1 0 = 6 k 0 4 i 4 F = j 0 1 = 7 k 4 M1 Any attempt at F o F A,,1 1 each type of eo = m1 A1F 6 Summing vecto poduct expessions Follow though thei thee answes fo F F loses final A1 (c) 0 Resultant foce = 0 and moment about 0 E,1 O is non zeo couple Total 9 E1 F = 0 E1 Moment 0

5 1 8h π (a) Volume of cone = πh= B1 Coect volume Altenative: 8h 1 8h π x dx= π 0 8 (B1) 8h 8 π d π h x x = d = π d h 4 xy x x x x M1 Use of x = = 16h π xy dx 4 16h π π A1 xy d x Coect integation of x = = M1 Attempt at fomula fo x π y dx 8h = 6h A1 AG kx (b)(i) R B1 foces θ F B1 Though a common point at base of cone W (ii) G θ d θ G above point of otation tanθ = d M1 tanθ seen anywhee = h 1 8 h 6h A1 Ratio coect θ = tan = 6.6 A1F Use of tan to get θ ; ft incoect atio (c) Resolve pep to plane N = Wcosα paallel to plane F= Wsinα B1 Both coect 6 Limiting fiction F = μn tanα = 1 M1 Eliminate W, obtain tanα α = 4.8 < 6.6 o 6 1 < 1 A1 Compaison slides fist A1F 4 Conclusion; ft (b)(ii) Total 14

6 (a) System is symmetical about vetical line though B, hence eactions ae equal E1 1 (b) Resolve vetically R = 0 M1 Attempt to esolve R = 1 N A1 (c) N T AE T AB At A esolve vetically T sin = 0 M1 Attempt at equation involving AE AE 100 T AE = = 11.4(...) = 11N sin 60 A1 AG 00 o seen At A esolve hoizontally TAB + TAE cos 60 = 0 M1 Attempt at equation involving AB 100 T AB = cos60 = 7.7(...) = 7.7N A1 4 AG sin o seen (d) By symmety, TBE = TBD = T M1 Use of symmety Resolve vetically at B:T cos0 + 0= 0 A1 T BE = = 8.86 = 8.9 N cos0 Altenative: At E esolve vetically A1 Accept 0 T AE R 60 T BE T ED R+ T cos0 + T cos0 = 0 (M1) tem equation involving R, T, T AE BE cos0 + T BE cos0 = 0 (A1) Coect equation, +/ signs and values sin 60 coect AE BE T BE = 8.86 = 8.9N (A1) () Accept 0 (e) At E esolve hoizontally T + T cos60 = T cos60 M1 tem equation including T, T, T ED BE AE 100 T ED = cos60 cos60 = cos0 sin 60 4.(...) = 4.N A1 ED BE AE Coect equation, +/ signs and values coect = 4.N A1F Follow though T AE ; accept 7 Total 1

7 4(a) MI = m = DISC ( 1.)( 0.) M1 Use of m = 0.0kg m A1 AG (b)(i) V = ω hence = 0.ω M1 1 ω = 10 ads A1 Attempt to use ω (ii) KE gained by pulley = 1 Iω = 1 ( 0.0)( 10) M1 Attempt at KE fo pulley = 1.J A1 Coect KE fo pulley Consevation of enegy: Gain KEPULLEY + KEBUCKET = loss in PE fo bucket ()( 1 ) = 1gd M1 tem equation d = 14 A1 4 AG (iii) g T = 0. θ M1A1 Equation of motion fo bucket mg T = mθ 0.T = 0.0 θ M1A1 Equation of motion fo pulley T = Iθ A1 fo coect substitutions in each case 0.0(9.8 T ) 0.T = 0. T = 4. N A1 CAO Altenative: a a= θ θ = 0. (M1) Connecting a, θ Fo bucket u = 0, v=, s = 14 8 using v = u + as gives a = = θ = = 8 ads 0. (A1) Eithe g T = 0. θ o 0.T = 0.0 θ (M1A1) One othe equation T = 4. N (A1) () CAO Total 1

8 (a) 8 ( 1, ) (, 6) a a 4 (, ) Moments about O: ( ) a( 6) + a( ) 4( ) + 8( ) M1 One coect F d paiing seen A,1 1 each type of eo = 6 6a+ 6 a = 10 no a independent A1 a cancels / comment about no a Magnitude = 10 A1F Must be positive value stated; ft single slip Altenative: i F1 = j 0 = 0 k i a 0 F = j 6 = 0 k a i 4 0 F = j a = 0 k 0 0 6a 0 (M1) One coect F seen (A1) Second F coect Total moment = a+ 6a 0 (A1) ( k components) = 10 a cancels / no a (A1) Comment about a Magnitude = 10 (A1F) () ft single slip (b)(i) F = = 8 0 M1 A1 Substitute a = 4 and add foces (ii) 8 (d, 0) Moments about O: 10 = d M1 y-component d = answe (a) d = A1F ft (a) Total 9

9 6(a) Let density = ρ m 4 m 4π = π ρ ρ= B1 ρ, m linked anywhee Mass of elemental disc / cylinde π x δ xρ M1 Attempt to use = ( ) MI of elemental disc = 1 π ( x ) δxρ x 1 ( ) MI of sphee = π ρ x dx x A1 Use of 1 m π hρ m 4 4 = ( + x x ) d x 8 m1 Attempt at integal, dep on fist M1 m 4 x x = x+ 8 A1F m = + 8 m 8 = = m 4 1 A1 6 AG Thei expession integated coectly coect numbe of tems only Coect limit used, ρ eplaced to obtain answe given Altenative: x δx Let density = ρ m= 4π ρ m ρ = 4π (B1) ρ, m linked anywhee Mass of elemental shell = 4πx δ xρ (M1) Attempt to use suface aea of sphee δ x ρ MI of elemental shell about diamete = ( 4π x δxρ) x (A1) Use of m 8 4 MI of sphee = π x ρ d x m x 4 d x 0 m x = 0 0 = (m1) Attempt at integal; dep on fist M1 (A1F) Thei expession integated coectly m = Coect limits used, ρ eplaced m = (A1) (6) AG

10 6(b)(i) 4 4 MIROD = ( m)( l) = 16ml B1 1 Use of ml with m = m and l = l (ii) MISPHERE ( about ) = ( m)( l) = ml B1 Use of ml with m = m and l = l G MISPHERE ( ) = ml + m( l P ) M1 Use of paallel axis theoem PENDULUM about = 17ml A1 I = 17ml + 16ml = 14ml A1 4 AG (iii) Angula momentum of clay befoe collision = mv (l) = mvl B1 Angula momentum fo pendulum afte collision = I1ω = 14ml ω B1 Coect fo pendulum Angula momentum fo clay afte collision = I ω = m( l) ω B1 Coect fo clay Consevation of angula momentum: mvl = 14ml ω + 9ml ω M1A1 Attempt at consevation of momentum mvl = 1ml ω v ω = 1l A1F 6 ft one slip Total 17 TOTAL 7

Version 1.0. klm. General Certificate of Education June Mathematics. Mechanics 2B. Mark Scheme

Version 1.0. klm. General Certificate of Education June Mathematics. Mechanics 2B. Mark Scheme Vesion.0 klm Geneal Cetificate of Education June 00 Mathematics MMB Mechanics B Mak Scheme Mak schemes ae pepaed by the Pincipal Examine and consideed, togethe with the eleant questions, by a panel of