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I have two bodies of known mass $m_0$ and $m_1$. $m_0$ is at constant velocity of $v_0$ on a level friction-less plane surface, and $m_1$ is moving across the same plane at constant velocity $v_1$ towards $m_0$; the assumption is that $v_1$ > $v_0$ so they will collide at some point.

When the two bodies perfectly collide (no energy transformed) I know that kinetic energy is conserved (as I believe is momentum), but I am at a loss as to how to calculate the new velocities of $m_0$ and $m_1$ after the collision.

I think if $m_0$ and $m_1$ are the same mass and $v_0$ is initially zero, then the only solution is that $m_0$ will move at velocity $v_1$ and that $m_1$ will stop, but apart from that I can't seem to get my head around the math where $m_0$ and $m_1$ are different.

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  • $\begingroup$ Well, you can check that here. $\endgroup$
    – SarGe
    May 19, 2020 at 9:46
  • $\begingroup$ To find the velocities of two masses after an elastic collision in two dimensions, you need to know both velocities (as vectors) before the collision and at least one piece of information from after the collision. What you do know is that the magnitudes of the velocities of each mass relative to the center of mass will be the same before and after the collision. $\endgroup$
    – R.W. Bird
    May 19, 2020 at 15:19

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This is a elastic collision described above in which the kinetic energy is conserved. The momentum will always be conserved in the cases of collisions as no external force acts.

I will use these two facts - momentum conservation and kinetic energy conservation to derive the final velocities of the masses.

Let the final velocities be $v_0'$ and $v_1'$ of $m_0$ and $m_1$ respectively.

Using the first fact; $$m_0v_0 + m_1v_1 = m_0v_0' + m_1v_1' ....(i)$$

And from the second fact; $$\frac{1}{2}m_0v_0^2 + \frac{1}{2}m_1v_1^2 = \frac{1}{2}m_0v_0'^2 + \frac{1}{2}m_1v_1'^2 ...(ii)$$

We can see that we are having two variables namely $v_0'$ and $v_1'$ and two equations - $(i)$ and $(ii)$. Thus we can solve for the two variables and get final velocities, which come out to be:

$$v_0' = \frac{(m_0 - m_1)v_0 + 2m_1v_1}{m_0 + m_1}$$

and,

$$v_1' = \frac{(m_1 - m_0)v_1 + 2m_0v_0}{m_1 + m_2}$$

Note that here appropriate signs of $v_0$ and $v_1$ must be put by taking some convention, lets say direction of velocity towards right is positive and towards left is negative, in order to get right answers.

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  • $\begingroup$ Thank you - that is a great help :-) $\endgroup$ May 19, 2020 at 13:21
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You are correct to assume that the momentum will be conserved. This is a straightforward application of the Linear Momentum Conservation principle. First, we have to define a sign convention and thus, we can say that the velocities towards the right(or towards the positive x-axis, if the x-axis is defined along the ground) are positive and velocities towards the left are negative. So, according to the momentum conservation principle, $initial\:momentum = final\:momentum$. Hence, we can write the equations as $m_ov_o + m_1v_1 = m_ov_o' + m_1v_1'$, where $v_o'$ and $v_1'$ are the velocities of the bodies after the collision (All the velocities are are substituted with the appropriate sign). This is the first equation. We define the coefficient of restitution $e$ as the ratio of final to initial relative velocities, after they collide. Assuming a perfect collision (i.e. elastic collision), $e=1$,

We can say that the $final\:relative\:velocity\:between\:the\:bodies = initial\:relative\:velocity$. This is equation 2. Solve the two equations simultaneously to get the solution. I trust you to figure out the relative velocity between the bodies on your own. Cheerio!

Edit: I changed the signs of the momentum conservation equation. The momentum of each object is added, not subtracted from each other. I had made an error in the previous equation. Also, to know more about the coefficient of restitution, head to its Wikipedia Page here: https://en.wikipedia.org/wiki/Coefficient_of_restitution

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  • $\begingroup$ Note: The values that you want can be plugged into these equations. I have just given the principle behind analyzing such situations. $\endgroup$ May 19, 2020 at 11:40
  • $\begingroup$ Can you expand on your relative velocity comment please - does that only apply when the masses are equal ? $\endgroup$ May 19, 2020 at 11:59
  • $\begingroup$ See, the concept of relative velocity is independent of the masses. It's just the velocity of one body when seen from the frame of another. Another way of writing equation $2$ would be - $e$ is the ratio between the velocity of separation (After the collision) and the velocity of approach(Before the collision). For example, if $v_1$ = 5 m/s and $v_o$ = 3 m/s, their relative velocity of approach would just be $5 - 3 = 2 \: m/s$ (Since they're in the same direction). You have to make the diagram and visualize this concept to understand what I'm talking about. $\endgroup$ May 19, 2020 at 12:32
  • $\begingroup$ ok- there is no way I would have got there on my own - that relative velocity preservation seems to be counter intuitive; but thank you for your answer. $\endgroup$ May 19, 2020 at 13:29
  • $\begingroup$ No Problem!! :D $\endgroup$ May 19, 2020 at 15:15

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