Equations for Perfect Collisions 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.
 A: 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
A: 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. 
