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Elastic collision of blocks

  1. May 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Two blocks collide on a perfectly elastic collision, after sliding on a frictionless surface.
    va = 1m/s
    vb = 1m/s
    ma = 0.05kg
    mb = 0.03kb

    I need to find the speed of both blocks after the crash.

    2. Relevant equations
    Conservation of momentum: mava+mbvb = mava' + mbvb'
    Conservation of energy: 1/2mava^2 + 1/2mbvb^2 = 1/2mava'^2 + 1/2mbvb'^2

    General solution for these two equations: -(vb−va)=vb'-va'

    Refer to this thread for cool formatting of these equations (i dont really know how to do that):

    https://www.physicsforums.com/threads/elastic-collision.753780/

    I dont really understand where that last equation comes from or how to get there.


    3. The attempt at a solution

    I solved for the conservation of momentum:

    mava+mbvb = mava' + mbvb'
    0.05kg.1m/s + 0.03kg.1m/s = 0.05kg.va' + 0.03kg.vb'
    0.08ms = 0.05va' + 0.03vb' (*)

    Then on the third equation:

    -(vb−va)=vb'-va'
    -(-1m/s - 1m/s) = vb'-va'
    2m/s = vb'-va'
    2m/s +va' = vb'

    Then i plug this on (*) and i get

    0.08m/s = 0.05va' + 0.03(2m/s + va')
    0.08m/s = 0.05va' + 0.06m/s + 0.03va'
    0.02m/s = 0.08va'
    0.25m/s = va'

    then vb' = 2.25m/s

    The expected solution is 0.5m/s and 2m/s .

    So my two questions are, am i doing something wrong here? And how do i get to the third equation?
     
  2. jcsd
  3. May 17, 2015 #2

    ehild

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    Gold Member

    If both blocks have the same velocity they never collide. Check the problem text.
    It is not true.
    That third equation is wrong. You get the correct one by arranging both the first and second equations that quantities labelled by "a" are on one side and those labelled by "b" are on the other side.

    ##m_a (v_a-v_a' ) = - m_b(v_b-v_b') ## *
    ##m_a(v_a^2-v_a'^2)= - m_b(v_b^2-v_b'^2) ## **
    Factorize the second equation and divide it by the first one: you get a third equation
    [tex]v_a+v_a' =v_b+v_b'[/tex]
     
  4. May 17, 2015 #3
    Thanks for the answer, my bad on that veocity, vb should be -1 m/s.

    Then technically va + va' = vb + vb'
    is the same as -(vb − va)=vb' - va'.

    I see how to find that equation now.
     
  5. May 17, 2015 #4

    ehild

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    Gold Member

    Sorry, I misread your equation, they are the same
     
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