# Energy loss after an inelastic collision at the same speed

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?

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 $$v_1+v_2$$.
• 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 $v_1 \to v_2$, $\Delta K \to 0$." Commented Nov 21, 2022 at 14:09