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Homework Help: Elastic Collision

  1. Jan 2, 2012 #1
    1. The problem statement, all variables and given/known data

    A ball of mass 5.0kg moving at a speed of 5.0m/s has a head on collision with a stationary bal of mass 6.0kg. If the collision were perfectly elastic what would be the speeds of the two balls after the collision?

    2. Relevant equations

    P = mv
    KE = 0.5mv2
    3. The attempt at a solution

    Using together kinetic energy and momentum equations, I can solve for final velocities.

    http://img855.imageshack.us/img855/6519/centralkootenayj2012010.jpg [Broken]

    Is this correct?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 2, 2012 #2
    The formulas for elastic collisions are:
    v1 = u1(m1-m2)/(m1+m2)
    v2 = 2m1u1/(m1+m2)

    v1 = 5(5-6)/(6+5) = -5/11 m/s = -.45 m/s
    v2 = (2*5*5)/(5+6) = 50/11 m/s = 4.5 m/s

    So you got the right answers if you add a negative sign to v1 since it bounces backwards after the collision.
     
    Last edited by a moderator: Jan 3, 2012
  4. Jan 2, 2012 #3

    How do you know that the first ball will bounce backwards?
     
    Last edited by a moderator: Jan 3, 2012
  5. Jan 2, 2012 #4
    My advice would be not to rely on these formulas and use conservation of KE and momentum conservation and the equation of restitution ... with these 3 things you can solve nearly all collision problems.

    PS:
    KE Coservation: [itex]\frac{1}{2}{m_1u_1}^2 + \frac{1}{2}{m_2u_2}^2 = \frac{1}{2}{m_1v_1}^2 + \frac{1}{2}{m_2v_2}^2[/itex] - valid only when e=1

    Momentum Conservation: [itex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/itex] - valid for all [itex]e\in[0,1][/itex]

    Eqn of coefficient of restitution: [itex](v_2 - v_1) = e(u_1 - u_2)[/itex]
     
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