B Alternative elastic collision formula / physical interpretation

[bobdavis]

Standard formula for final velocities $v_1$, $v_2$ in elastic collision with masses $m_1$, $m_2$ and initial velocities $u_1$, $u_2$ is given by $$v_1 = \frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2$$$$v_2 = \frac{2m_1}{m_1+m_2}u_1+\frac{m_2-m_1}{m_1+m_2}u_2$$.

By rearranging terms this seems to be equivalent to $$v_1 = \frac{p}{\bar{m}}-u_1$$$$v_2 = \frac{p}{\bar{m}}-u_2$$ where $p = m_1u_1+m_2u_2$ is total momentum and $\bar{m} = \frac{m_1+m_2}{2}$ is average mass.

The term $\frac{p}{\bar{m}}$ seems to be the same as $2v_c$ where $v_c$ is the velocity of the center of mass of the system. By substituting this into the formula and rearranging it seems the formula is equivalent to $$\bar{v}_1=v_c$$$$\bar{v}_2=v_c$$ where $\bar{v}_i=\frac{u_i+v_i}{2}$ is the average of the initial and final velocities of particle $i$

Are these alternative formulas correct and if so is there a good way to physically interpret these alternative formulas and the "momentum over average mass" term? If these formulas are correct is there a reason the seemingly more complex standard formula is used instead?

 

[anuttarasammyak]

Are these alternative formulas correct and if so is there a good way to physically interpret these alternative formulas and the "momentum over average mass" term?

It is interpreted as Galilean transformed event that in Center of Mass system velocity of each particle does not changes its magnitude but signature by collision.

 

[bobdavis]

I see so in the center of mass system $v_c = 0$ and the formula becomes $$\bar{v}_1=\frac{u_1+v_1}{2}=0$$$$\bar{v}_2=\frac{u_2+v_2}{2}=0$$ so $$v_1 = -u_1$$$$v_2=-u_2$$ ?

 

[anuttarasammyak]

Yes, for 1D system as if a time reverse takes place.
For 3D system in COM system,
m_1\mathbf{u_1}+m_2\mathbf{u_2}=m_1\mathbf{v_1}+m_2\mathbf{v_2}=0
|\mathbf{u_1}|=|\mathbf{v_1}|
|\mathbf{u_2}|=|\mathbf{v_2}|