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Momentum problem

  1. Mar 29, 2008 #1
    1. The problem statement, all variables and given/known data

    A ball of mas m1 = 2kg is moving with a speed of 5 m/s and collides with a ball of mass m2 = 2.5 kg (there are losses so the collision is inelastic). After the collision the incoming ball has deviated by 35 degrees from its origin. The struck ball has moved off at an angle of 50 degrees. Find both balls final velocities.


    2. Relevant equations

    Pxi = Pxf
    Pyi = Pyf


    3. The attempt at a solution

    I think solved it, if someone could though, I would like to have it checked to make sure I did it right. Thanks in advance.

    I found the Conservation of Momentum in the x:

    [tex] m_{1}v_{1ix} = m_{1}v_{1fx} + m_{2}v_{2fx} [/tex]
    [tex]m_{1}v_{1}cos(0) = m_{1}v_{1f}cos(35) + m_{2}v_{2f}cos(-50)[/tex]

    [tex]m_{1}v_{1} = m_{1}v_{1f}cos(35) + m_{2}v_{2f}cos(-50)[/tex]

    Then I found it in the y and solved for [tex]v_{2f} [/tex]

    [tex]m_{1}v_{1iy} = m_{1}v_{1fy} + m_{2}v_{2fy}[/tex]
    [tex]m_{1}v_{1i}sin(0) = m_{1}v_{1f}sin(35) + m_{2}V_{2f}sin(-50)[/tex]
    [tex]0 = m_{1}v_{1f}sin(35) + m_{2}v_{2f}sin(-50)[/tex]

    [tex]m_{1}v_{1f}sin(35) = -m_{2}v_{2f}sin(-50)[/tex]

    [tex]v_{2f} = \frac{-m_{1}v_{1f}sin(35)}{m_{2}v{2f}sin(-50)} [/tex]

    Then I plugged it into the x equation.

    [tex]m_{1}v_{1} = m_{1}v_{1f}cos(35) + m_{2}\frac{-m_{1}v_{1f}sin(35)}{m_{2}sin(-50)}cos(-50)}[/tex]

    Then solved for [tex] v_{1f} [/tex]

    [tex] v_{1f}[\frac{-m_{1}sin(35)cos(-50)}{sin(-50)} + m_{1}cos(35)] = m_{1}v_{1i}[/tex]

    [tex]v_{1f}[\frac{-2sin(35)cos(-50)}{sin(-50)} + 2cos(35)] = 10 [/tex]

    For v_{1f} I got 3.84. Is this the right way to solve this problem?
     
  2. jcsd
  3. Mar 29, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me! :approve:
     
  4. Mar 29, 2008 #3
    Thanks, you have helped me with a lot of problems, I really appreciate it.
     
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