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    Edexcel - M1 - June 2013 (R) - Q4
    Erm. Well. How on earth do you do that question?
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    (Original post by HarrisonGCSE)
    Edexcel - M1 - June 2013 (R) - Q4
    Erm. Well. How on earth do you do that question?
    Can you link the question paper?
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    https://ce0f1d00673c9b9787bbeaa1aa08...Kinematics.pdf

    Third question from the bottom!
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    (Original post by HarrisonGCSE)
    https://ce0f1d00673c9b9787bbeaa1aa08...Kinematics.pdf

    Third question from the bottom!
    When A and B are at the same height, the vertical displacement that object B will have moved through will be the
    height, h, but the vertical displacement that A will have moved through will h-50
    Draw a diagram if you can't quite see why this is.
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    simultaneously solve.

    1. diagram
    2. look for an appropriate equation
    3. simultaneously solve.

    You have 2 things the same. They are both at h at time T.

    (I think, without actually trying the questions.)

    Do you have the answers? Madness to do these kind of questions if you cant check the answer after.
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    (Original post by HarrisonGCSE)
    https://ce0f1d00673c9b9787bbeaa1aa08...Kinematics.pdf

    Third question from the bottom!
    Okay, so you have two ball A:

    S = h-50 (because it starts off 50 metres above the ground so it's displacement at height h is h-50 metres)
    U = 2 ms^(-1) (initial speed)
    V =
    A = -9.81 = -g
    T = T

    That gives us h - 50 = 2T - \frac{1}{2}gT^2

    Ball B:

    S = h
    U = 20
    V =
    A = -g = -9.81
    T=T

    That gives us h = 20T - \frac{1}{2}gT^2

    You know how two equations in h and T... simultaneous equations.
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    (Original post by Zacken)
    Okay,
    don't hand him the answer Zac

    let him fight it out - stays in your brain then.

    you have now given him a gcse question.

    The hard bit is to produce that equation.
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    (Original post by RMIM)

    The hard bit is to produce that equation.
    There isn't a hard bit.
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    (Original post by Zacken)
    There isn't a hard bit.
    ok, the main bit.

    I dare say u might get even half marks or more just by producing the equations without even solving them

    but anyway - Harrison, u have seen how it's done now. This type of question comes up all the time and the method stays the same.
 
 
 
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