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GCEEdexcel GCEMathematicsMechanics 3 M3 (6679June 2008MathematicsEdexcel GCEMark Scheme (Final)
General Marking Guidance All candidates must receive the same treatment. Examiners must mark the firstcandidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what theyhave shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perceptionof where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be usedappropriately. All the marks on the mark scheme are designed to be awarded. Examiners should alwaysaward full marks if deserved, i.e. if the answer matches the mark scheme. Examinersshould also be prepared to award zero marks if the candidate’s response is not worthyof credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by whichmarks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to acandidate’s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with analternative response.
June 20086679 Mechanics M3Mark SchemeSchemeQuestionNumberQ1(a)1 λ 1 L 2 L 2 EPE stored 1m 2gL ( mgL)2λL mg LEPE KE 8Marks2B1 λL 8 B1KE gained (b)*i.e. λ 8mgM1A1cso(4)M1EPE GPE KE28mgL1 8mg 1 L 1 mg mu 2 L 2 L 2 82 2mgL2 m 2u2 u gLA1A1M1A1 (5)9 Marks
SchemeQuestionNumberQ2 (a)MarksOAT 3 2πB2π3) ; a 0.12 , u 2 a 2 ω 2 , u 0.12 ω 0.251 ms 1 (0.25 m s-1) ω ωu ω 2 (a 2 - x 22M1A1M1A1(b)Time from O A O 1.5s t 0.5 π OP 0.12 sin 3 x a sin ωt(v 2 ω2 a2 - x 2v 2π3)0.12 2 0.104. 2 B1M1A1Distance from B is 0.12 – OP 0.12 – 0.104 0.016m(c)(4)M1A1(5)M12π 0.05983 0.13 ms 1A1(2)11 Marks
SchemeQuestionNumberMarksQ3 (a)M1A1 θlhrsin θ T cos θ N Mg(1)T sin θ mrω 2rlsub into (1)from (2)M1A1(2)T mlω 2M1ml cos θ ω 2 N mgA1N mg mhω 2Since in contact with table N 0 ω 2 „gh*M1A1 cso(8)(b)r : h :l 3: 4:5 extension mg2mg h h425mh 22gT mlω 2 ω ω 45hT h4B1M1A1M1A1(5)13 marks
QuestionNumberSchemeQ4 (a) Mass a 32π :3C of M from O:216 Moment :Marks 2168208271263 6a83 2a8M1A1xUse of3r82a 36a 3 8 208 x88480a 30a 20813x M1M1*A1 cso(5)(b) Massπa 3 :C of M:Moments :416330a13 320ay (c) S24 48839a 216a 201a61B1B1yM1488y3*A1 cso(4)201a612atan θ .12a tan θ 2a12a 20161 aθ 12.93.so critical angle 12.93 if θ 12 it will NOT topple.M1M1A1A1 (4)13 marks
SchemeQuestionNumberMarksQ5(a)T – mg cos θ F maSub forv2:aM1A11mv 2 mga cos θ2v 2 2ga cos θEnergyT mg cos θ 2mg cos θ : θ 60 T mv 2aM1A13mg2M1A1(6)(b)Speed of P before impact 2 gaPCLM :2 ga u 0 m3mB1 u 2 ga 4ga8*M1A1cso4m(3)(c) (i)At A v 0 so conservation of energy gives:14mu 2 4m ga (1- cos θ )2ga ga (1- cos θ )16cos θ 15, θ 20o16M1A1M1A1(ii)At AT 4mg cos θ 15mg4(accept 3.75mg)M1A1 (6)15 Marks
SchemeQuestionNumberQ6 (a)3F ma ( )(x 1)3 ( x 1) 33 x(c)dxv dt dvdxM1A1dx 0.5 v dv32(x 1) 2x 0, v 0 c –(b) 0.5a 0.5 vMarks1 2v ( c)432 1v 2 6 1 2 ( x 1) *26 ( x 1) 2 1x 1 x 1dx ( x 1) 2 1( x 1) 2 1M1A1 cso(6)B1(1)M16 dt6 t c′ M1A1(Q ( x 1) always 0 )v 6 v2 6Separate and M1M1 A1t 0, x 0 c′ 0M1t 2 ( x 1) 1 (2 6 ) 22(x 1)2 25 x 4 ( c′ need not have been found)M1A1 cao(7)14 Marks
Aug 07, 2008 · Mathematics Mechanics 3 M3 (6679 Edexcel GCE Mathematics. General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be appl
