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Conservation of Momentum Elastic Collision Problem

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data
    I am basically being asked to use the law of conservation of momentum and kinetic energy to derive two equations for the final velocity of the two masses involved in the system. (i.e. [tex]\acute{v_{1}}[/tex] and [tex]\acute{v_{2}}[/tex])

    2. Relevant equations
    Conservation of Momentum: [tex]m_{1}v_{1}+m_{2}v_{2} = m_{1}\acute{v_{1}}+m_{2}\acute{v_{2}}[/tex] (Simply refered to as Therom 1)

    Conservation of Kinetic Energy: [tex]\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} = \frac{1}{2}m_{1}\acute{v_{1}}^{2}+\frac{1}{2}m_{2}\acute{v_{2}}^{2}[/tex] (Simply refered to as Therom 2)

    For both equations: [tex]m_{1}\neq m_{2} , v_{1}>v_{2}[/tex]

    3. The attempt at a solution
    I know that we can get rid of the fractions from Therom 2 by simple factoring and multiplication:
    [tex]m_{1}v_{1}^{2}+m_{2}v_{2}^{2} = m_{1}\acute{v_{1}}^{2}+m_{2}\acute{v_{2}}^{2}[/tex]

    Then we can solve for an arbitrary final velocity, say [tex]\acute{v_{1}}[/tex], with Theorm 1:
    [tex]m_{1}v_{1}+m_{2}v_{2} = m_{1}\acute{v_{1}}+m_{2}\acute{v_{2}}[/tex]

    a.[tex]m_{1}\acute{v_{1}} = m_{1}v_{1}+m_{2}v_{2}-m_{2}\acute{v_{2}}[/tex]

    b.[tex]\acute{v_{1}} = \frac{m_{1}v_{1}+m_{2}v_{2}-m_{2}\acute{v_{2}}}{m_{1}}[/tex]

    c.[tex]\acute{v_{1}} = \frac{m_{1}v_{1}+m_{2}\left(v_{2}-\acute{v_{2}}\right)}{m_{1}}[/tex]

    Now we solve for [tex]\acute{v_{2}}[/tex] by substituing in for [tex]\acute{v_{1}}[/tex] the equation that we just solved for:
    a.[tex]m_{1}v_{1}^{2}+m_{2}v_{2}^{2} = m_{1}\acute{v_{1}}^{2}+m_{2}\acute{v_{2}}^{2}[/tex]

    b.[tex]m_{2}\acute{v_{2}}^{2} = m_{1}v_{1}^{2}+m_{2}v_{2}^{2}-m_{1}\acute{v_{1}}^{2}[/tex]

    c.[tex]m_{2}\acute{v_{2}}^{2} = m_{1}v_{1}^{2}+m_{2}v_{2}^{2}-m_{1}\left(\frac{m_{1}v_{1}+m_{2}\left(v_{2}-\acute{v_{2}}\right)}{m_{1}}\right)^{2}[/tex]
    (the [tex]\frac{1}{m_{1}^{2}}[/tex] and the [tex]m_{1}[/tex] will cancel out one of the [tex]\frac{1}{m_{1}^{2}}[/tex]'s to get...)

    d.[tex]m_{2}\acute{v_{2}}^{2} = m_{1}v_{1}^{2}+m_{2}v_{2}^{2}-\frac{m_{1}^{2}v_{1}^{2}+m_{2}^{2}\left(v_{2}-\acute{v_{2}}\right)^{2}}{m_{1}}[/tex]

    Basically its at this point were my brain explodes from all of the m's and v's and their appropriate subscripts. I still have a [tex]\acute{v_{2}}[/tex] on the wrong side of the equals sign and have no idea how to get it on the right side other than voodoo magic and satanic rituals (not really but I'm almost at that point). Any help would be appreciated. :smile:
  2. jcsd
  3. Nov 12, 2009 #2


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    There is a very nice method to derive the formulas for v1' and v2': Arrange both equations so that the quantities with index "1" are on one side and those with index "2" on the other side.

    m1v1-m1v1'= m2v2'-m2v2


    m1v12-m1v1'2= m2v2'2-m2v22

    Factor the second equation and divide by the first one., you get a very simple relation between the velocities. Try.

  4. Nov 12, 2009 #3


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    From theorem I you get
    m1v1 - m1v1' = m2v2' - m2v2-------(1)
    From theorem 2 you get
    m1v1^2 - m1v1'^2 = m1v2'^2 - m2v2^2-------(2)
    Dividing 1 by 2 you get
    (v1-v1')/(v1^2 - v1'^2) = (v2' - v2)/(v2'^2 - v2^2)
    On simplification you get
    v1 + v1' = v2 + v2'----(3)
    Substitute the value of v2' in eq. no.1 to get v1'.
  5. Nov 13, 2009 #4
    Ok I understand the whole dividing the two equations part but I am stuck trying to solve for the general equation. This is what I have:


    I am not sure where to go from here. I tried to cross multiply but it looked like it would get messy so i gave that up and now I don't know where to go.
  6. Nov 13, 2009 #5
    However complicated it may seem, it's only a quadratic in [itex] v_2^' [/itex]. just multiply everything out and group the powers of [itex] v_2^' [/itex].

    A more elegant way to do this is to use the reference frame of the center of mass of m1 and m2

    the speed of the center of mass is:

    [tex] u = \frac {m_1 v_1 + m_2 v_2} {m_1 + m_2} [/tex]

    now the new speeds [itex] w_1 [/itex] and [itex] w_2 [/itex] are [itex] v_1 - u [/itex] and [itex] v_2 - u [/itex].

    In this new reference frame, total momentum is 0, and it is much easier to find [itex] w_1^' [itex] and [itex] w_2^' [itex]
  7. Nov 13, 2009 #6


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    Now factorize the denominator.
    v1^2 - v1'^2 = (v1 + v1')*(v1 - v1')
    On simplification you get
    v1 +v1' = v2 + v2'
    Write v2' =........... and substitute in the eq. 1 to get v1'
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