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 subject teaches. This mak scheme includes any amendments made at the standadisation meeting attended by all examines and is the scheme which was used by them in this examination. The standadisation meeting ensues that the mak scheme coes the candidates esponses to questions and that eey examine undestands and applies it in the same coect way. As pepaation fo the standadisation meeting each examine analyses a numbe of candidates scipts: altenatie answes not aleady coeed by the mak scheme ae discussed at the meeting and legislated fo. If, afte this meeting, examines encounte unusual answes which hae not been discussed at the meeting 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 deeloped 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 aoided; 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 aailable to download fom the AQA Website: www.aqa.og.uk Copyight 00 AQA and its licensos. All ights eseed. COPYRIGHT AQA etains the copyight on all its publications. Howee, egisteed centes fo AQA ae pemitted to copy mateial fom this booklet fo thei own intenal use, with the following impotant exception: AQA cannot gie pemission to centes to photocopy any mateial that is acknowledged to a thid paty een 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 67) and a egisteed chaity (egisteed chaity numbe 07). Registeed addess: AQA, Deas Steet, Mancheste M 6EX
Key to mak scheme and abbeiations used in making M m o dm A B E mak is fo method mak is dependent on one o moe M maks and is fo method mak is dependent on M o m maks and is fo accuacy mak is independent of M o m maks and is fo method and accuacy mak is fo explanation o ft o F follow though fom peious incoect esult MC mis-copy CAO coect answe only MR mis-ead CSO coect solution only RA equied accuacy AWFW anything which falls within FW futhe wok AWRT anything which ounds to ISW ignoe subsequent wok ACF any coect fom FIW fom incoect wok AG answe gien BOD gien benefit of doubt SC special case WR wok eplaced by candidate OE o equialent FB fomulae book A, o (o 0) accuacy maks NOS not on scheme x EE deduct x maks fo each eo G gaph NMS no method shown c candidate PI possibly implied sf significant figue(s) SCA substantially coect appoach dp decimal place(s) No Method Shown Whee the question specifically equies a paticula method to be used, we must usually see eidence of use of this method fo any maks to be awaded. Howee, thee ae situations in some units whee pat maks would be appopiate, paticulaly when simila techniques ae inoled. You Pincipal Examine will alet you to these and details will be poided on the mak scheme. Whee the answe can be easonably obtained without showing woking and it is ey unlikely that the coect answe can be obtained by using an incoect method, we must awad full maks. Howee, the obious penalty to candidates showing no woking is that incoect answes, howee 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 gien to less than the degee of accuacy accepted in the mak scheme, when it gains no maks. Othewise we equie eidence of a coect method fo any maks to be awaded.
MMB (a) = d ds t M = 0t sin t AA Total Kinetic enegy = M M fo eithe d ds t = (J) A (b) PE lost is = g M = g o 99. = 00 J A Accept 99, g o of tems coect (ignoe signs) (c)(i) KE is + g M M (a) + (b) =. = 0 J A (if done (c)(i) in (b) 0 maks; if done (b) and then (c)(i) in (b) M only) (ii) Using KE = m = 0.6 M Speed of stone is.9 ms A Accept.8 fom 0 (d) eg Stone is a paticle E No ai esistance If use constant acceleation fomulae in D, possible maks in (c) BUT no maks if initial speed is teated as being etical Not no esistance; accept no wind esistance Total 9 (a) Symmety E Only accept symmety (b) Moments about B: 0. + 0. 8 = 0. x MA M tems, coect x =. 0. =.8 cm A Total
MMB (cont) (a) Using F = ma, 00 cos π t i + 600t j = 00 a M a = cos π t i + t j A (b) = a dt M M fo eithe a dto of tems coect = sin π ti + t j + c π Am m fo + c When t =, = i + 6j, 6j + c = i + 6j m c = i 8j = ( sin π π t )i + (t 8)j A Do not accept π Accept.7 fo π (c) When paticle is moing due west, nothely component is zeo M t 8= 0 A t = A (d) When t =, = i + 0j B Speed of paticle is m s B B fo change to + Total d λ = M dt d = λ dt m Condone one of +, λd t, λ = λt + c AA m t = 0, = u c = = u t λ u A m fo + c = u λt A 7 Total 7
MMB (cont) 6(a) Using powe = foce elocity Powe = (0 8) 8 M = 690 watts A AG (b) When speed is 0 m s, max foce exeted is 690 0 = 78 N B Acceleating foce is 78 00 N M Using F = ma: 8 = 00a m a = 0. m s A (c) Foce exeted by engine is 690 B Foce exeted by the engine = 0 mg sin M (Use of cos delete A,A of A tems) 0 6.7 (o 00gsin )= 690 A All tems coect AA A Two tems coect 0 6.7 690 = 0 A SC fo 0 + 6.7 690 = 0 7(a) 6.7 ± 6.7 + 0 690 = M 0 Speed is 9. m s A 7 Total R F S C B B fo S and 6g (in coect place) B fo R and F o combined etical foce at C A 6g (b) Moments about C: S cos 0 = 6g cos 0 MA M tems, tem coect S = 9.6 N o g A R, F not coect 0 maks in (c)(i) and (c)(ii) (c)(i) Moments about A: O 6g cos 0 = R MA Moments about mid-point of od: R = 6.8 N A S cos 0 = P cos 0 (o esoling, R = 6g cos0 S cos0 P = 9. N o g = g cos0) (O esoling etically P = g) (ii) Resole paallel to AB: R = P cos 0 M A S cos 70 + F = 6g cos 70 M = 6.8 N A F = g cos 70 F = P sin 0 M =. N A =. N A (o F = 6g sin0 S sin0 = g sin0) (d) Using F = μ R : M M (c)(ii) = μ (c)(i). = μ 6.8 μ = 0.6 o tan0 A (condone ) Total 6
MMB (cont) 8(a) Using conseation of enegy: m mg( cos θ ) M m mgh = 6 g( cos) m = (6 g[ cos]) =. A SC:. (b) When paticle is at est, esole adially T = mg cos MA M T mgcos= m o T = mgsin = mg cos m =. A Total 7 m o 9 As paticle moes, T = M using unknown as extension: If adius is, extension is. B If extension is x, adius is. + x B λx λx Using T = : Using T = : l l 9(.) T =. M T = 9 x. M = 60(.) A = 60x A m 8 T = 60(.) = T = m 8 60x = M. + x M M 60 9 = 7 ( o 9 0. = 86. ) A 9x+ 60x = 7 A 0 9 = 0 0x + x 9 = 0 (0+ )( ) = 0 M (0x )( x+ ) = 0 M =. o 0. x = 0. o. Radius is. A 8 Radius is. A Total 8 TOTAL 7 7