MEI M1 exam discussion - Friday 1st June


    Rep:
    Hi all - couldn't find a thread on this exam so I thought I'd start one. Just wanted to know the general feelings towards it - I take A level physics and find there's quite a bit of overlap, so I'm pretty confident...though I've been confident in the past and it's turned out not so well...anyway, pre/post discussion, revision help, questions and the like here

    Rep:
    any1 got the june11/jan12 paper for MEI mechanics 1? Thanks

    Rep:
    took me ages to find a thread i thought there must be one im dreading it although i took physics and mechanics in that is fine with maths its so much harder because i just get confused by the questions! june 11 should be on the ocr website and i have jan 12 but i dont know how to put it on here if you want it

    Rep:
    (Original post by firebolt)
    took me ages to find a thread i thought there must be one im dreading it although i took physics and mechanics in that is fine with maths its so much harder because i just get confused by the questions! june 11 should be on the ocr website and i have jan 12 but i dont know how to put it on here if you want it
    Some of the questions are quite complicated - half of the trick is to work out what the question's actually asking...what type of question are you having trouble with? Do you have jan 12 as a pdf file? If so you can save it and attach it in a post

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    I'm doing this exam on Friday and I am so prepared to fail it. In all the mock papers I've done I've been getting 40% ish, which isn't going to get me the grade I need :/

    Rep:
    (Original post by DavidH20)
    Some of the questions are quite complicated - half of the trick is to work out what the question's actually asking...what type of question are you having trouble with? Do you have jan 12 as a pdf file? If so you can save it and attach it in a post
    It's not really a specific type of question i think i'll have to spend loads of time spending time understanding the question, no i don't i just remembered it was my M1 mock my teacher took in and didnt let us take home although i can get the worked out answers off my colleges vle

    Rep:
    Here is a post with the pictures of m1 jan 2012 paper

    http://www.thestudentroom.co.uk/show...1901129&page=3

    Rep:
    Could someone please help me with q2ii on the Jan 09 paper. I'm trying to use SUVAT but its not working

    http://www.mei.org.uk/files/papers/m109jn_86dj.pdf

    Rep:
    (Original post by nju)
    Could someone please help me with q2ii on the Jan 09 paper. I'm trying to use SUVAT but its not working

    http://www.mei.org.uk/files/papers/m109jn_86dj.pdf
    You can work it out like this:

    In part i) you should have found the velocity to be
    6+4*2=14m/s

    Therefore in part ii) you can use this information.
    u=14m/s
    v=-6m/s(travelling in negative direction)
    a=-8m/s^2
    t=?(what we want to find)

    therefore, using v=u+at
    -6=14-8t (just sub in values)
    -20=-8t (take 14 over)
    2.5=t (divide by -8)
    then just add the extra 2 seconds, which gives 4.5s.(correct in mark scheme)

    Rep:
    Can someone please help me out!
    I don't understand question 7 v) bits of June 2009 paper
    How can the sphere move upwards when the string is broken? Surely, it should move downwards...?
    I am mega confused....:confused:

    Rep:
    (Original post by tissue_26)
    Can someone please help me out!
    I don't understand question 7 v) bits of June 2009 paper
    How can the sphere move upwards when the string is broken? Surely, it should move downwards...?
    I am mega confused....:confused:
    The sphere was moved 1.8m above P then it breaks so then it falls past P again

    Rep:
    good luck to anyone doing this exam I took it in January and got an A, i'm doing mei m2 on 1st june as well not looking forward to it.

    Rep:
    (Original post by firebolt)
    The sphere was moved 1.8m above P then it breaks so then it falls past P again
    but it says at 1.8m above P, the string breaks leaving the sphere moving upwards...how??

    Rep:
    (Original post by tissue_26)
    Can someone please help me out!
    I don't understand question 7 v) bits of June 2009 paper
    How can the sphere move upwards when the string is broken? Surely, it should move downwards...?
    I am mega confused....:confused:
    "The system is stationary when the sphere is at point P. When the sphere is 1.8m above P the string
    breaks, leaving the sphere moving upwards at a speed of 3ms−1."

    The wording is a little dubious, but basically what it means is:
    The sphere starts at P, with initial velocity of zero, then accelerates to 1.8m above P where the velocity is 3m/s.

    So for part:
    A)
    As soon as the string breaks, the only force acting upon it is gravity. Therefore, the acceleration must be 9.8m/s^2

    B)
    we know the following things:
    a=9.8m/s^2(from part A)
    s=1.8m(to P)
    u=3(this is at 1.8m above P, just after the string breaks)
    t=?

    use equation s=ut + (at^2)/2
    1.8=3t -4.9t^2(simply sub in values)
    4.9T^2-3T+1.8=0(rearrange equation)

    C)
    First of all, we need the time taken to get from P to 1.8m above P
    u=0
    s=1.8m
    v=3m/s
    t=?

    use equation s=t(u+v)/2
    1.8=1.5t
    t=1.2

    Then the rest is simply the root of the equation:

    (using the quadratic equation, or via calculator) = 0.985

    Total time = 1.2 + 0.985 = 2.19 (3.s.f)


    Hopefully that makes sense!

    Rep:
    (Original post by NJam)
    "The system is stationary when the sphere is at point P. When the sphere is 1.8m above P the string
    breaks, leaving the sphere moving upwards at a speed of 3ms−1."

    The wording is a little dubious, but basically what it means is:
    The sphere starts at P, with initial velocity of zero, then accelerates to 1.8m above P where the velocity is 3m/s.

    So for part:
    A)
    As soon as the string breaks, the only force acting upon it is gravity. Therefore, the acceleration must be 9.8m/s^2

    B)
    we know the following things:
    a=9.8m/s^2(from part A)
    s=1.8m(to P)
    u=3(this is at 1.8m above P, just after the string breaks)
    t=?

    use equation s=ut + (at^2)/2
    1.8=3t -4.9t^2(simply sub in values)
    4.9T^2-3T+1.8=0(rearrange equation)

    C)
    First of all, we need the time taken to get from P to 1.8m above P
    u=0
    s=1.8m
    v=3m/s
    t=?

    use equation s=t(u+v)/2
    1.8=1.5t
    t=1.2

    Then the rest is simply the root of the equation:

    (using the quadratic equation, or via calculator) = 0.985

    Total time = 1.2 + 0.985 = 2.19 (3.s.f)


    Hopefully that makes sense!
    Ohhhh.....I see....omg thanks so much!
    The wording is..very strange...or is it just me

    Rep:
    (Original post by tissue_26)
    but it says at 1.8m above P, the string breaks leaving the sphere moving upwards...how??
    Imagine the sphere is a tennis ball and the string is your hand. If you move your hand up, you accelerate the tennis ball, moving it up, lets say 1.8m. If you then let go (equivalent to the string snapping) the ball still moves upwards, but it then accelerates in the opposite direction.

    Does it make sense now?

    Rep:
    (Original post by NJam)
    Imagine the sphere is a tennis ball and the string is your hand. If you move your hand up, you accelerate the tennis ball, moving it up, lets say 1.8m. If you then let go (equivalent to the string snapping) the ball still moves upwards, but it then accelerates in the opposite direction.

    Does it make sense now?
    Yes! I get it! Ah thank you so much

    Rep:
    (Original post by tissue_26)
    Yes! I get it! Ah thank you so much
    No problem, glad i could help.

    Rep:
    Can we use C4 Cos(Theta)=a.b/|a||b| formula?

    Rep:
    (Original post by ngnav)
    Can we use C4 Cos(Theta)=a.b/|a||b| formula?
    I think so but I don't think you will need it as it is not on the m1 specification

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