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A particle $m_1$ is traveling with velocity $v$ toward a stationary particle $m_2$. The velocity of the center of mass is given as $v_c=\frac{m_1}{m_1+m_2}v$. Changing to a moving coordinate system, the Center-of-mass Coordinate System (CMCS), we now have the two particles heading toward each other, $m_1$ with speed $v-v_c$ and $m_2$ with speed $v_c$. The total momentum is found to be 0 in this new coordinate system.

Depending on how $m_1$ collides with $m_2$, it may leave the collision in any direction. After the collision $m_2$ will have the same magnitude of momentum but opposite direction.

Now the assertion is made that in an elastic collision, $m_1$ and $m_2$ have the same speeds before and after the collision. In other words, the speed of $m_1$ is $v-v_c$ and the speed of $m_2$ is $v_c$ after the collision.

I don't see why this must be the case. Apparently momentum conservation and energy conservation lead to only one solution for these speeds. How?

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    $\begingroup$ Just try conserving kinetic energy (because the collision is elastic) and momentum (always!) at the same time. There is only one solution. By trying it I mean write down the equations. It's a system of two equations in two unknowns, so it is solvable. $\endgroup$ – dmckee Oct 9 '14 at 3:56
  • $\begingroup$ It is the definition of "elastic " after all. If the kinetic energies of the particles change ( deformation for a classical example) it is called inelastic. $\endgroup$ – anna v Jun 26 '15 at 3:35
  • $\begingroup$ Greg: "Changing to a moving coordinate system, the Center-of-mass Coordinate System (CMCS), we now have [...]" -- The equations and the speed values $v$ and $v_c$ given in your question don't seem to involve any coordinates at all. So instead of referring to coordinate systems, perhaps you mean a comparison of two reference systems; i.e. first considering the inertial reference system of which particle $m_2$ was a member, and then changing to considering the Center-of-mass Reference System (CMRS). $\endgroup$ – user12262 Jul 29 '15 at 21:55
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Aug 17 '16 at 10:56
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The math is almost trivial for someone beyond algebra 1. Write the kinetic energy of each particle as $p_n^2/2m_n$. Then converse momentum and kinetic energy in the center-of-momentum. You will see that the magnitude of the momentum each particle does not change.

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Let $v_1$ and $v_2$ be the particle's velocities in the center of mass coordinate system after the collision. by conservation of momentum and energy we have \begin{gather}\tag{1}m_1v_1+m_2v_2=0 \\ \tag{2}m_1v_1^2+m_2v_2^2=m_1(v-v_c)^2+m_2v_c^2 \end{gather} Isolating $v_1$ in $(1)$ and substituting in $(2)$: $$m_1\left(-\frac{m_2v_2}{m_1}\right)^2+m_2v_2^2=m_1\left(\frac{m_2v}{m_1+m_2}\right)^2+m_2\left(\frac{m_1v}{m_1+m_2}\right)^2 \implies\\ \frac{m_2(m_1+m_2)}{m_1}v_2^2=\frac{m_1m_2}{m_1+m_2}v^2 \implies \\ v_2=\frac{m_1}{m_1+m_2}v=v_c,$$ which is exactly the initial velocity. Substituting the result, yields also the same velocity for $v_1$.

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    $\begingroup$ The speeds will be the same, but the velocities cannot be the same if there is a collision. The directions of the particles must change is there is a collision. One of the mathematical solutions which satisfies the conservation equations is that the velocities are the same, but there must be a 2nd solution in which the velocity vectors are different. $\endgroup$ – Bill N Jun 25 '15 at 20:38

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