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    I thought this was a challenging, but quite a good paper in all. I especially liked the part of the last question asking you to find the max velocity. Range of approaches acceptable and nice link with the 'Core' modules (of which I think there should be more).
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    im sure youll get the b or above, seems like most of our answers come out the same (:
    ive seen easier papers before, this one was quite hard
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    (Original post by ajay_ganger)
    i missed out that complete the square ? at the end time run out!!!!! i got t = 1 for that minimum acceleration ?
    I got t=1 as well.

    Edit: I didn't think this was much harder than the past paper Qs except for the final question. I knew if that (the 6 marker) type of question came up I hadn't learnt it, so that was a shame. The vertical circular motion was a nice easy one though, because they can be really tough. If you have access, try Jan '05; the last question is a killer.
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    I got t=0 for minimum acceleration. I differentiated acceleration, which came to 24ti, and put it equal to zero to find the minimum point of the acceleration, which was 0.
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    what did you guys get for the max distance on end question? 54 or something like that?

    for the 31.8 one,
    i just differentiated it and equaled it to zero to get the max value
    i hate completing the square

    i also got t=1 for the acceleration q
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    (Original post by maxfire)
    I got t=0 for minimum acceleration. I differentiated acceleration, which came to 24ti, and put it equal to zero to find the minimum point of the acceleration, which was 0.
    t could = 0 or 1, so I took 1 because 0 was a given.

    (Original post by jahan218)
    what did you guys get for the max distance on end question? 54 or something like that?

    i just differentiated it and equaled it to zero to get the max value
    i hate completing the square

    i also got t=1 for the acceleration q
    I think I got 54 ish as well. I just used my equation solver on my calc. I wonder if I'll get all the method marks therefore. I just wrote "use quadratic formula" next to it, lol :|
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    i just pretended to look like i used the quadratic formula when i used my calc aswell, youll be alright if you stated it i would think
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    What did people get for the maximum velocity of the jumper? I think I ****** that up and got 2800 or something, which doesn't sound remotely correct.
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    (Original post by maxfire)
    What did people get for the maximum velocity of the jumper? I think I ****** that up and got 2800 or something, which doesn't sound remotely correct.
    i just subbed in 31.8 into the equation i think, i got 29 or someting like that
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    I did the same. hmmmmmm
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    we have to COMPLETE THE SQUARE on the last question??? whaaat... i just used the quadratic equation .....and i managed to get (u^2+2ag)/(a^2+2ag) for the angle pheta question

    this is bad...... i thought this paper was alot harder than the past papers, especially with the amount of core maths involved, and i totally ignored core cos our exam is in june, now im totally regretting it
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    (Original post by bunzy)
    we have to COMPLETE THE SQUARE on the last question??? whaaat... i just used the quadratic equation .....and i managed to get (u^2+2ag)/(a^2+2ag) for the angle pheta question

    this is bad...... i thought this paper was alot harder than the past papers, especially with the amount of core maths involved, and i totally ignored core cos our exam is in june, now im totally regretting it
    i dont think we had to complete the square, it didnt say so
    which question are we talking about here, i cant really remember,

    there was, find the maximum x, where you could just use the quadratic formula

    then there was, show that the maxspeed is at the point x=31.8 where you could complete the square or use calculus

    and then it was, find this speed, just subbing 31.8 into the equatoin
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    (Original post by jahan218)
    i dont think we had to complete the square, it didnt say so
    which question are we talking about here, i cant really remember,

    there was, find the maximum x, where you could just use the quadratic formula

    then there was, show that the maxspeed is at the point x=31.8 where you could complete the square or use calculus

    and then it was, find this speed, just subbing 31.8 into the equatoin
    oh right okay, i think i just differentiated for the 31.8 one, ah i can't remember any questions or my answers which is making me panic slightly, not that there's anything i can do except to hope until march time

    coming on tsr after exams is definetely adding to the feeling of dooom and failure :o:
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    In case anyone is interested:

    mga(1-cosx) = 0.5mv^2 - 0.5mu^2
    mga(1-cosx) + 0.5mu^2 = 0.5mv^2
    2ga(1-cosx) + u^2 = v^2
    v = sqrt[2ga(1-cosx) + u^2]


    mgcosx - R = (mv^2)/r R = 0, r = a
    gacosx = 2ga(1-cosx) + u ^2
    gacosx = 2ga - 2gacosx + u^2
    3gacosx = 2ga + u^2
    cosx = (2ga + u^2)/3ga
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    dats what i got
 
 
 
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