# Can non-relativistic elastic collision have mass exchange?

I noticed that all the problems about elastic collision have "strict mass conservation" that is, if two masses $$m_1$$ and $$m_2$$ collides and make $$m_3$$, $$m_4$$, atleast one of them is gonna equal $$m_1$$ and the other is gonna equal $$m_2$$.

Curious, I tried to prove that the objects' mass between the collision has to be conserved, that the only way to satisfy conservation of momentum and conservation of energy at the same time in an elastic collision is to have one of resulting mass equal to $$m_1$$, but I couldn't really come up with a satisfactory answer.

I also couldn't make a counterexample, where given a configuration of $$m_1, m_2$$ and initial velocities $$v_1, v_2$$ I could make some masss $$m_3$$ that is $$m_3\neq m_1$$ and $$m_3 \neq m_2$$ with the momentum and energy conserved.

I have some heuristics, in the center of momentum frame of reference, I could prove $$m_1|\vec{u_1}| = m_3|\vec{u_3}|, |\vec{u_1}|+|\vec{u_2}| = |\vec{u_3}| + |\vec{u_4}|$$ where $$\vec{u_1}, \vec{u_2}$$ are velocities before the collision and $$\vec{u_3}, \vec{u_4}$$ are velocities after (inside the center of momentum frame). But I haven't got far from this.

Is there any way to prove this?

I had to come up with a counterexample. The key was that $$m_1, m_2, v_1, v_2$$ essentially didn't matter, since all you could find out in the conservation law was about momentum and energy. So if $$m_1 = 1\ \mathrm{kg}, m_2 = 1\ \mathrm{kg}, v_1 = 1\ \mathrm{m/s}, v_2 = -1\ \mathrm{m/s}$$. You could essentially give the condition that momentum is zero and energy is 1 joules. Since that's all the information you'll be able to extract from that.
With that, I picked a case of $$m_3 = 3/2\ \mathrm{kg}, m_4 = 1/2\ \mathrm{kg}$$, some substitution and polynomial solving later, you would get $$v_3 = \pm1/\sqrt{3}\ \ \mathrm{and}\ \ v_4 = \mp\sqrt{3}$$