Transfer of momentum problem?

  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]
     
  2. jcsd
  3. A.T.

    A.T. 5,302
    Gold Member

    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. haruspex

    haruspex 12,529
    Science Advisor
    Homework Helper
    Gold Member
    2014 Award

    All true, but I don't see how that gets you to the answer required.
    Using energy conservation more completely:
    [tex]\frac{{p_a}'^2}{m_a}+\frac{{p_b}^2}{m_b}=\frac{{p_a}^2}{m_a}[/tex]
    Combining with momentum eqn. we get
    [tex]\frac{p_b}{p_a}=\frac{2m_b}{m_a+m_b}<2[/tex]
     
  5. A.T.

    A.T. 5,302
    Gold Member

    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.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook