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    Does anyone happen to have a mark scheme for this paper? Rep on offer if anyone does!

    Thanks very much

    Does anyone have any past papers short of the specimen and the january 2005 one too? (I don't actually think there are any...)
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    Hey , I dont have an official mark scheme, but i have a good set of model answers from my head of maths, its on paper though ! if u want i can type in a few answers
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    If you have a minute could you possibly type me question 7 (the one with the inclined plane and the 2 particles)? I'd be really grateful lol

    Thanks
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    I wasn't aware that there was a new spec for January 2005. The specimen and syllabus on the OCR site are both for 2000 spec. Could you tell me what's changed in the new specification?
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    Hi, this is what we were given for qu 7:

    i) particle a R=0.3g and F=0.4*0.3g=0.12g

    Applying F=ma to both particles gives:

    T-0.12g=0.3a
    0.2gsin70-T=0.2a

    so

    0.2gsin70-0.12g=0.5a
    a=1.33
    T=1.57

    ii) F=ma
    -0.12g=0.3a
    a=-0.4g

    a=10.4g
    u=1.5
    v=0
    s=?

    v^2=u^2+2as

    0=1.5^2-2(0.4g)s
    s=0.247

    iii) For a
    a=-0.4g
    u=1.5
    v=1.5
    t=?

    v=u+at
    0=1.5-0.4gt
    t=0.383s

    a=gsin70
    u=1.5
    t=0.383
    s=?

    s=ut+0.5at^2
    s=2.42m

    Hope this helps.

    Joanne
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    HERES WHAT I WAS GIVEN:

    APPLY F=MA UP THE SLOPE FOR A

    T-MJU R - 0.1G SIN ALPHA = MA
    T-0.25 X 0.784 - 0.1 X 9.8 X 0.6 = 0.1A
    T= 0.1A + 0.784

    APPLY F=MA DOWNWARDS TO B

    0.32G - T = 0.32A
    T=3.136-0.32A

    COMBINING BOTH EQUATIONS

    0.1A + 0.784 = 3.136 - 0.32A
    0.42A = 2.352
    A=2.352/0.42

    A = 5.6 MS-2

    ii) U=0,V=2.8,A=5.6,S=?

    V^2=U^2 + 2AS
    7.84= 11.2 S
    S= 0.7 M

    WHEN NO STRING APPLY F=MA DOWN THE SLOPE TO FIND THE DECELLERATION

    MJU R + 0.1G SIN ALPHA = 0.1A
    0.25 X 0.784 + 0.1 X 9.8 X 0.6 = 0.1A
    0.784 = 0.1A
    A = (-)7.84 MS^-2

    U=2.8,V=0,A=-7.84,S=?

    V^2=U^2+ 2AS

    7.84 = 15.68S

    S = 0.5 M

    TOTAL DISTANCE = 0.7M + THE EXTRA 0.5

    =1.2M

    HOPE THIS HELPS

    ANY REP IS APPRECIATED
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    Ps, Method By Joanne 88 Is Completely Wrong And Mabey For A Different Paper, If It Is For This Paper I Have No Clue What It Is On About, I Have Done This Question And My Marks Scheme Agrees With My Answers
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    (Original post by simxp)
    I wasn't aware that there was a new spec for January 2005. The specimen and syllabus on the OCR site are both for 2000 spec. Could you tell me what's changed in the new specification?
    I don't know that it was new for Jan 2005 but it's referred to on the OCR site under GCE AS Mathematics (New) somewhere I believe

    Thanks all for your help
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    http://www.ocr.org.uk/OCR/WebSite/do...1997&x=41&y=47

    Under New GCE.
    However i believe the other can still be followed to a certain point, I just know that I'm doing this one.
 
 
 
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