# CoM-and-relative velocity

In our scrip we are considering the elastic collision between two particles, one with inital velocity $$\vec v$$ and the other $$\vec w$$. We also consider that the particles have the same mass.

Conservation of momentum : $$\vec v + \vec w = \vec v' + \vec w'$$

Conservation of energy: $$\vec v^2 + \vec w^2 = \vec v'^2 + \vec w'^2$$

According to the professor, we have the following relationship between the individual particle velocities and the center of mass and relative velocity:

$$\vec v = \frac 1 2 \vec v_s + \vec v_r$$.

$$\vec w = \frac 1 2 \vec v_s - \vec v_r$$.

My question is:

Shouldn't the $$\frac 1 2$$ multiply the relative velocity $$\vec v_r$$ and not the CoM velocity $$\vec v_s$$

Suppose you are already in the center-of-mass frame, so $$\vec v_s = 0$$. Your equations as written give $$\vec v = \vec v_r$$ and $$\vec w = -\vec v_r = -\vec v$$ in this limiting case. Moving the factor of one-half to the other term would make this “trivial” transformation incorrect.
\begin{align} m_1 \vec v &= \frac{m_1 m_2}{m_1 + m_2} \vec v_s + m_1 \vec v_r \\ &= \mu \vec v_s + m_1 \vec v_r \end{align}
and similarly for $$m_2 \vec w$$. You should prove this for yourself; it takes about a page of algebra the first time you do it. The quantity $$\mu = \frac{m_1 m_2}{m_1 + m_2}$$ is called the reduced mass. In this formulation, each term has units of momentum; the $$\mu$$ and $$\vec v_s$$ appear together because they both describe the system as a whole rather than just one or the other of the particles.
It’s more physically meaningful to think of the factor multiplying $$\vec v_s$$ as $$\mu/m_1$$ (or $$\mu/m_2$$ for the other particle’s transformation) than as a magic number $$1/2$$ to be moved around.