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    (Original post by aishamarie)
    can anyone help with the qs that I've posted on this threaad?>>>>>>>>>>>>>>> http://www.thestudentroom.co.uk/show...970&highlight=

    thank you! I will rep!
    Q5 solution. Hope it helps!
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    (Original post by Axion)
    it hangs vertically downwards with the 2.3m length slack. you can get the tensions from that
    so one of the 0.5m feels all the tension?
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    (Original post by bluekipper)
    Q5 solution. Hope it helps!
    it really did, thanks a lot! - I've reached my rep limit for today :mad: but I'll defo do it tomorrow !
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    hi does anyone want solutions for june 2010 jan 2012
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    (Original post by rohan11)
    so one of the 0.5m feels all the tension?
    yup!
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    (Original post by Economics247)
    hi does anyone want solutions for june 2010 jan 2012
    yes please!
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    (Original post by Axion)
    Good luck everyone tomorrow by the way

    Who is AS level and who is A2 Level?
    I'm A2. Did S1 at AS which I hated but I don't exactly like M1 either!
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    (Original post by aishamarie)
    yes please!
    they not that great im just scanning them now
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    (Original post by Economics247)
    they not that great im just scanning them now
    all solutions are great solutions
    - thanks
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    (Original post by aishamarie)
    haha omg sorry, good luck tomorrow!
    & btw, I love your blog, will defo follow after tomorrow is over!
    you too! ah just did that paper and I can't believe how much better my mech has gotten after 4 months! and thanks!
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    (Original post by aishamarie)
    all solutions are great solutions
    - thanks
    ahh man it exceed the amount
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    (Original post by Economics247)
    ahh man it exceed the amount
    uhh could you only send the last question of the june 2010 paper then please?
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    http://www.mei.org.uk/files/papers/2012_Jan_m1.pdf

    Can anyone explain the first part of the last question to me pleas, the projectile one, it seems impossible!
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    (Original post by CWE)
    http://www.mei.org.uk/files/papers/2012_Jan_m1.pdf

    Can anyone explain the first part of the last question to me pleas, the projectile one, it seems impossible!
    Okay to work out T, think about where it will be at it's halfway point, use suvat and consider the vertical bits:

    v = 0 ms^-1 as objects are instantaneously at rest at highest point

    u = 40 x sin a ms^-1 , vertical component of initial velocity

    a = -g , as gravity acts downwards in opposite direction to the motion

    t = ?

    Now use v= u+at, remembering that this is just for half the time (T/2):

    0 = 40sin a - g x (T/2)

    T/2 = 40sin a / g

    T = 80sin a / g

    For R, just think about the horizontal component of the velocity (using cos a) and use suvat, remembering that there isn't any acceleration in this direction as air resistance is negligible. Sub in T = 80sin a / g for time t and you'll get the answer
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    (Original post by aishamarie)
    uhh could you only send the last question of the june 2010 paper then please?
    Pm me your email
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    Hi guys need help with question 3 june 07 http://www.mei.org.uk/files/papers/m107ju_8gxe4.pdf
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    (Original post by Economics247)
    Hi guys need help with question 3 june 07 http://www.mei.org.uk/files/papers/m107ju_8gxe4.pdf
    1) The pulley is smooth so there is no friction
    2) The system is in equilibrium.
    Right side = Left side
    20g + rod = 15g
    So the force in the rod must be -5g
    It is a thrust as the force is -ve
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    (Original post by loua96)
    Okay to work out T, think about where it will be at it's halfway point, use suvat and consider the vertical bits:

    v = 0 ms^-1 as objects are instantaneously at rest at highest point

    u = 40 x sin a ms^-1 , vertical component of initial velocity

    a = -g , as gravity acts downwards in opposite direction to the motion

    t = ?

    Now use v= u+at, remembering that this is just for half the time (T/2):

    0 = 40sin a - g x (T/2)

    T/2 = 40sin a / g

    T = 80sin a / g

    For R, just think about the horizontal component of the velocity (using cos a) and use suvat, remembering that there isn't any acceleration in this direction as air resistance is negligible. Sub in T = 80sin a / g for time t and you'll get the answer
    Perfect, thank you!
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    (Original post by Alotties)
    1) The pulley is smooth so there is no friction
    2) The system is in equilibrium.
    Right side = Left side
    20g + rod = 15g
    So the force in the rod must be -5g
    It is a thrust as the force is -ve
    Thanks really appreciated i dont even get question 5
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    (Original post by JG1027)
    same so do you think past papers for 5 days would be enough to get at least a C ?

    5 days is more than enough! I wish I had that much time! I only started revising yesterday...I'm in A2 and resitting...need an A in mechanics...not had much time to revise though due to synoptic papers for my other subjects!


    have a feeling C4 isn't going to go too well :eek:
 
 
 
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