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Consider two particles with mass $m_1$ and $m_2$ travelling both with the same speed $v=v_1=v_2$ towards each other. Assume that they collide in a perfectly inelastic way, so they merge in a unique particle of mass $m_1+m_2$. By a simple calculation, the variation of Kinetic energy in this case is given by

$$\Delta K=\frac{m_1m_2}{m_1+m_2}(v_1-v_2)^2$$

Therefore, in our specific case $\Delta K=0$, therefore the collision is indeed elastic. It seems a contradiction.

Moreover, why does the energy loss depends on the relative velocity before the impact?

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Let $v_1$ and $v_2$ be components of the velocities in a given direction.

If $v_1=v_2$ (the masses are travelling in the same direction) then indeed $\Delta K=0$ because they never collide. Relative velocity before "collision" is zero.

If $\hat {v_1}=-\hat {v_2}$ (the masses are travelling in opposite directions) then $\Delta K=\frac{m_1m_2}{m_1+m_2}(v_1+v_2)^2$ and the collision is inelastic.
Relative velocity before collision is $v_1+v_2$.

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    $\begingroup$ I suppose that one can reasonably talk about an "equal-velocity collision" by saying something like "so long as $v_1 \neq v_2$, the bodies collide, and in the limit of $v_1 \to v_2$, $\Delta K \to 0$." $\endgroup$ Commented Nov 21, 2022 at 14:09
  • $\begingroup$ @MichaelSeifert A very nice way of analysing the situation. $\endgroup$
    – Farcher
    Commented Nov 21, 2022 at 14:18

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