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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 $\vec v_s$

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1 Answer 1

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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.

If you don’t fix the relative masses, the momentum-transformation relationship is

$$ \begin{align} m_1 \vec v &= \frac{m_1 m_2}{m_1 + m_2} \vec v_s + m_1 \vec v_r \\ &= \mu \vec v_s + m_1 \vec v_r \end{align} $$

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.

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