Two-Body Elastic Collisions

Before specializing to two-body Coulomb collisions, it is convenient to develop a general theory of two-body elastic collisions. Consider an elastic collision between a particle of type $1$ and a particle of type $2$ . Let the mass and instantaneous velocity of the former particle be $m_1$ and ${\bf v}_1$ , respectively. Likewise, let the mass and instantaneous velocity of the latter particle be $m_2$ and ${\bf v}_2$ , respectively. The velocity of the center of mass is given by

$${\bf U} = \frac{m_1\,{\bf v}_1+m_2\,{\bf v}_2}{m_1+m_2}.$$ (3.10)

Moreover, conservation of momentum implies that ${\bf U}$ is a constant of the motion. The relative velocity is defined

$${\bf u} = {\bf v}_1-{\bf v}_2.$$ (3.11)

We can express ${\bf v}_1$ and ${\bf v}_2$ in terms of ${\bf U}$ and ${\bf u}$ as follows:

$${\bf v}_1 = {\bf U} + \frac{\mu_{12}}{m_1}\,{\bf u},$$ (3.12)

$${\bf v}_2 = {\bf U} - \frac{\mu_{12}}{m_2}\,{\bf u}.$$ (3.13)

Here,

$$\mu_{12} = \frac{m_1\,m_2}{m_1+m_2}$$ (3.14)

is the reduced mass. The total kinetic energy of the system is written

$$K = \frac{1}{2}\,m_1\,v_1^{\,2} + \frac{1}{2}\,m_2\,v_2^{\,2} = \frac{1}{2}\,(m_1+m_2)\,U^{\,2} + \frac{1}{2}\,\mu_{12}\,u^2.$$ (3.15)

Now, the kinetic energy is the same before and after an elastic collision. Hence, given that $U$ is constant, we deduce that the magnitude of the relative velocity, $u$ , is also the same before and after such a collision. Thus, it is only the direction of the relative velocity vector, rather than its length, that changes during an elastic collision.