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    (Original post by jelmes96)
    Ok it's really annoting me, people thinking that 7.65s (my answer) was too long. I'm sorry, but let's think about it.

    It's already going at 8ms^-1. There's an acceleration against motion, at -gSin15. so, it's got to go at 0ms^-1, which would take about 3. odd seconds. Then, it's got to travel downwards, at 13m. From the first 3 seconds, it's going to go slightly further than 13m. So to you, who put 7.67, WELL DONE! :P
    thanks for that my rounding was a little off i think but yeah 7.67 ish 7.65 is probably exactly right
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    S = 321 N
    T = 80.4 N

    if angle alpha is 5.71, Tan5.71 = roughly 0.1 = a / 30. Therefore a= 30 x 0.1 = 3m. Voila!
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    (Original post by JG1027)
    did you do pythagoras to find T and do tan to find the angle and - it from 90 degrees?
    yeah pythag for 2 but sine rule to show everything else

    cant remember though
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    (Original post by jg1027)
    did you do pythagoras to find t and do tan to find the angle and - it from 90 degrees?
    sqrt (250^2 + 25^2) = 251 n
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    (Original post by jelmes96)
    S = 321 N
    T = 80.4 N

    if angle alpha is 5.71, Tan5.71 = roughly 0.1 = a / 30. Therefore a= 30 x 0.1 = 3m. Voila!
    yeah i got those two forces.

    Howeverm I did sine rule to find it was 3m
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    (Original post by jelmes96)
    sqrt (250^2 + 25^2) = 251 n
    so basically we found c ?
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    (Original post by JG1027)
    so basically we found c ?
    Pretty much. That's why they wanted us to draw a triangle of forces.

    For the first question, did people get 16g as the weight and reaction force?
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    I have a paper I can post, just to confirm, am I allowed to?
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    (Original post by stefanocattaneo)
    I have a paper I can post, just to confirm, am I allowed to?
    Yes you can. MEI M1 is only taken at 9.00 UK time.
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    Going by TSR guidelines: "Exams from the examination boards AQA, OCR, CCEA, WJEC and CIE (other than orals and practicals) are fine to talk about straight after the exams. In any situations where exams from these boards are taken internationally, the exam board is satisfied that appropriate controls are already in place."

    Here is the paper If someone could get going on an unofficial Mark Scheme that would be fab x

    Since its too big for TSR, you can download it here: https://dl.dropboxusercontent.com/u/...une%202013.pdf
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    (Original post by stefanocattaneo)
    Going by TSR guidelines: "Exams from the examination boards AQA, OCR, CCEA, WJEC and CIE (other than orals and practicals) are fine to talk about straight after the exams. In any situations where exams from these boards are taken internationally, the exam board is satisfied that appropriate controls are already in place."

    Here is the paper If someone could get going on an unofficial Mark Scheme that would be fab x

    Since its too big for TSR, you can download it here: https://dl.dropboxusercontent.com/u/...une%202013.pdf
    Cheers for this on another note has anyone got the D1 paper, haven't seen that one surface yet
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    I'm on it. It will take about 45 minutes
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    (Original post by jelmes96)
    Pretty much. That's why they wanted us to draw a triangle of forces.

    For the first question, did people get 16g as the weight and reaction force?
    Up : 9g and 7g

    Down 16g

    You had to write the numerical value for each (i.e. substitute g for 9.8)
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    what do we reckon grade boundaries will be like??
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    (Original post by rachel1508)
    what do we reckon grade boundaries will be like??
    Judging by the general reaction to the paper, I reckon somewhere along the lines of June 2006:

    A - 55
    B - 47
    C - 40


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    (Original post by jelmes96)
    S = 321 N
    T = 80.4 N

    if angle alpha is 5.71, Tan5.71 = roughly 0.1 = a / 30. Therefore a= 30 x 0.1 = 3m. Voila!
    i put t = 80.66 obviosuly due to rounding etc do u think this will be correct
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    Give me 5 minutes to upload
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    (Original post by jelmes96)
    Give me 5 minutes to upload
    Cheers mate


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    (Original post by jelmes96)
    Give me 5 minutes to upload
    thanks
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    (Original post by Axion)
    Up : 9g and 7g

    Down 16g

    You had to write the numerical value for each (i.e. substitute g for 9.8)
    Well, g is a constant, as 9.8, so having put 16g, won't matter. 16g is a number, it's just neat On the papers they use g as a constant, so why can't we? Anyway, I've had some problems, but it's uploading now.
 
 
 
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