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Conservation of linear momentum and energy in a collision

  1. Sep 19, 2011 #1
    Consider an ideal case of a mass m1 moving at constant velocity v1 on a frictionless surface, colliding with another masss m2 at rest. After collision, can someone tell me if it is possible for m1 to move off at -v1 while conserving the momentum and energy of the entire system? This is quite a simple scenario depicted in many text book but is this kind of collision physically possible at all, without m2 being infinitely large to absorb the momentum change?

    Before the collision the linear momentum of the entire system is p1 = m1v1, and kinetic energy k1 = (p1)^2/2m1. But after the collision, m1 moved off with -p1 = -m1v1, carrying the same kinetic energy k1 as before. However due to conservation of linear momentum, m2must move in the opposite direction to m1 after collision with momentum of 2m1v1. Hence, m2 will carry with it the kinetic energy of 4k1*m1/m2. There is an energy surplus after collision just by conserving the linear momentum. Where does this energy come from? If this is not a physical possible collision (be it elastic or inelastic), why some of the text books keep depicting collision like this? Or is there any implicit assumption made without explicitely expressed? I even read similar depiction in one of the text book talking about electromagnetic radiation pressure when shining on an object, talking about the momentum transfer but without mentioning about the energy change in the system.

    Appreciate enlightenment. Thanks.
  2. jcsd
  3. Sep 19, 2011 #2


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    You are quite right in your analysis. Textbooks often use this example because it is relatively easy to work through in order to introduce students to the concept of collisions and because it represents a reasonable approximation when m2 >> m1. If we denote the ratio, [itex]\epsilon = m_1/m_2[/itex], then [itex]0<\epsilon<<1[/itex]. For a head on collision between the two masses, the velocities afterwards are,

    [tex]v_1 = \frac{\epsilon - 1}{\epsilon+1}v_\text{initial}[/tex]


    [tex]v_2 = \frac{2\epsilon}{\epsilon+1}v_\text{initial}\;.[/tex]

    We can then expand

    [tex]v_1 \sim -(1 - 2\epsilon)v_\text{initial} + \mathcal{O}(\epsilon^2)[/tex]


    [tex]v_2 \sim 2\epsilon v_\text{initial} + \mathcal{O}(\epsilon^2)[/tex]

    Suppose we now take the leading order approximation:

    [tex]v_1\sim - 1v_\text{initial} \text{ and } v_2 \sim 0[/tex]

    So, for the case when m2 is a hundred times m1 you are looking at a relative error of around 0.02, in the velocities.

    However, strictly speaking you are correct. The case when the second mass remains stationary does indeed violate conservation of momentum and therefore cannot occur. That said, it is a relatively good approximation for a large contrast in masses. For example, if you imagine throwing a ball at a wall, you would assume in your calculations that the wall doesn't move - however, in actuality is does a little bit, and the earth moves a little bit.
    Last edited: Sep 19, 2011
  4. Sep 19, 2011 #3
    Thank you very much, appreciate the clear explanation. I hope all text books could be more rigorous and state the necessary assumptions much more clearer and explicitly.
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