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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?

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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 $v_3 = \pm1/\sqrt{3}\ \ \mathrm{and}\ \ v_4 = \mp\sqrt{3}$

This disprove my original conjecture that elastic collision could have some sort of symmetry I could exploit, but no, I think I need some information on the end collision to get something concrete

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  • $\begingroup$ It would appear that your example shows that a collision which exchanges mass can be elastic. I don't see that you need more information unless you want to go into two or three dimensions. $\endgroup$
    – R.W. Bird
    Jul 14, 2020 at 14:32
  • $\begingroup$ Well, I meant that elastic collision specifically would have some nice symmetry I could exploit, that the inelastic-ness would limit the mass in some way that the collision cannot exchange mass. PS: I’m also the OP, I made about 3 attempts to make a counterexample and due to some stupid mistake, I didn’t find them.So I posted this question. Then I finally made one on the fourth $\endgroup$
    – Henry Oh
    Jul 15, 2020 at 12:55
  • $\begingroup$ There is a nice symmetry in the center of mass system in that the total momentum is and must remain zero. This is also true in two or three dimensions. $\endgroup$
    – R.W. Bird
    Jul 15, 2020 at 14:36
  • $\begingroup$ I set up an Excel spreadsheet to solve 1D elastic collisions. It has one equation for momentum and another for energy. Using a tool called “Solver” you can solve for any two of the eight variables (two incoming masses with velocities and two going out). I used it to find outgoing velocities for several collisions with exchanged mass and even some with lost mass. There are some sets of numbers which yield no solution, but I have not tried to figure out why. $\endgroup$
    – R.W. Bird
    Jul 15, 2020 at 18:36
  • $\begingroup$ From what I’ve seen, completely inelastic collision doesn’t seem to work aka if their masses combine to one object. But as far as I can see, as long as there is two objects, they seem to be able to balance each other’s momentum and energy to have a valid solution $\endgroup$
    – Henry Oh
    Jul 18, 2020 at 2:34

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