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2D Collision problem, need help please

  1. Feb 12, 2006 #1
    Ok, I have tried to work this out for about an hour now and I haven't gotten very far. In the question there are two equal masses. One of them is moving at a given velocity and the other is not moving at all. After the collision one disk goes off on a given angle while the other disk goes off on an angle perpendicular to that of the second disk. The question wants me to solve for the final velocities.

    So this is whats given:

    Mass one Mass two

    Initial velocity = 0 Initial velocity = vi1
    Final velocity = ? Final velocity = ?
    Angle = 90-A2 (Not sure if thats right) Angle = A2
    Mass = Mass two Mass = Mass one

    Oh ya and the collision is perfectly elastic. So I think I am supposed to have like 4 unknowns and then I am suposed to have 4 different equations. But the only equation I am aware of right now is
    vi1 = v1f + v2f

    This can be split up to two other equations like so:

    vi1x = v1fx + v2fx
    vi1y = v1fy + v2fy

    Ok, so I think thats right so far, then what I did was I made four other equations like so:

    v1fy = v1fy(sinA2-90)
    v1fx = v1fx(cosA2-90)
    v2fy = v2fy(sinA2)
    v2fx = v2fx(CosA2)

    Now I plugged those into the original two equations like so:

    v1ix = v1fx(cosA2-90) + v2fx(cosA2)
    v1iy = v1fy(sinA2-90) + v2fy(sinA2)

    Now this is where I am stuck. I am not sure if I am on the right track right now but I hit a dead end. I have too many unknowns and not enough equations. Either the answer is really easy and I just can't see it, or I messed up somewhere, or theres another equation I am supposed to use. But right now I am really stuck and would appreciate any help given.
  2. jcsd
  3. Feb 13, 2006 #2


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    you have two relationships that you can use here. conservation of momentum and conservation of energy. i think that conservation of energy will help you with some of your variables

    prior to the collision

    [tex] m_1 v = p_i [/tex]


    [tex] \frac{1}{2} m_1 v^2 = KE_i [/tex]

    initially there is only an x component so the y components of momentum sum to zero afterwards

    [tex] m_1 v_{1y} + m_2 v_{2y} = 0 [/tex]

    also the initial x momentum is the final x momentum and kinetic initial is kinetic final

    [tex] m_1 v_{1x} + m_2 v_{2x} = m_1 v_x [/tex]

    [tex] \frac{1}{2} m_1 v^2_1 + \frac{1}{2} m_2 v^2_2 = \frac{1}{2} m_1 v^2 [/tex]

    these should be all of the relevant equation which lead to the solving of the problem
  4. Feb 14, 2006 #3


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    Homework Helper

    Since the masses are equal one can work with the velocities on their own for momentum conservation. Thus the vector sum of the two required velocities after the collision should be equal tot the original velocity. Since the required velocities are perpendicular to each other they can therefore be considered to be the two components of the original velocity being [itex]v\cos(\theta)[/itex] and [itex]v\sin(\theta)[/itex].
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