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Transfer of momentum problem?

  1. Nov 6, 2012 #1
    Here's an interesting puzzle:

    The whole scenario takes place in one dimension of space.

    Ball B is at rest. Ball A moves with momentum [tex]p_a[/tex]

    Ball A makes a perfectly elastic head-on collision with Ball B. Ball B moves off with momentum [tex]p_b[/tex]

    Prove that [tex]|\frac{p_b}{p_a}|<2[/tex]
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  3. Nov 6, 2012 #2


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    Momentum conservation :
    [tex]{p_a} = {p_a}' + {p_b}[/tex]
    [tex]1- \frac{{p_a}'}{p_a} = \frac{p_b}{p_a}[/tex]

    Energy conservation (Ball B had intially no KE, so speed of ball A cannot increase):
    [tex]|\frac{{p_a}'}{p_a}| < 1[/tex]
    Last edited: Nov 6, 2012
  4. Nov 7, 2012 #3


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    All true, but I don't see how that gets you to the answer required.
    Using energy conservation more completely:
    Combining with momentum eqn. we get
  5. Nov 7, 2012 #4


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    You have to combine the two of course:
    [tex]1- \frac{{p_a}'}{p_a} = \frac{p_b}{p_a}[/tex]
    [tex]|\frac{{p_a}'}{p_a}| < 1[/tex]
    This leads to:
    [tex]0 < \frac{p_b}{p_a} < 2[/tex]

    Yes, the clean & complete way is to derive the ratio as function of the masses. Mine was just showing that the ratio is < 2.
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