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Homework Help: Elastic Collision and speed of a ball

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A 120 g ball moving to the right at 4.5 m/s catches up and collides with a 420 g ball that is moving to the right at 1.2 m/s.

    If the collision is perfectly elastic, what is the speed of the 420 g ball after the collision?

    2. Relevant equations

    [tex]m_1v_{1i}+}m_2v_{2i} = m_1v_{1f}+m_2v_{2f} [/tex]

    3. The attempt at a solution

    I manipulated the formula a little bit to solve for V2_f and came up with this:
    [tex](m_1v_{1i}+}m_2v_{2i} - m_1v_{1f}) / m_2[/tex]

    then I plugged in my numbers:
    (.12*4.5 + .42*1.2 - .12*.63) / .42

    simplified to this:
    (.54 + .504 - .0756) / .42

    and got 2.3057. Mastering physics said that is wrong however and I'm not sure what I'm doing wrong. Is the conservation of momentum the correct equation for this problem? I found the velocity of the 120 gram ball fine, which was .63 m/s, but am stuck on this one. Any help would be appreciated.
  2. jcsd
  3. Nov 20, 2009 #2


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    Homework Helper

    Hi Atlos, welcome to PF.
    How did you get v1f = 0.63 m/s?
  4. Nov 20, 2009 #3
    I came to find v1f by using the following equation:

    [tex][(m_1-m_2) / (m_1+m_2) ]v_{1i} + [2m_2 / (m_1+m_2]v_{2i}[/tex]
  5. Nov 21, 2009 #4
    A useful fact for straight line elastic collision is that

    v_2b - v_1b = - (v_2a - v_1a)

    v_1a = velocity of m1 before collision
    v_1b = velocity of m1 after collision etc.

    Relative velocity before the collision is the same as after, just opposite in sign.
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