Ans.

Let us consider two particles of masses $m_1$ and $m_2$ having position vector $\vec{r_1}$ and $\vec{r_2}$ with respect to the origin O. Let $\vec{R}$ be the position vector of centre of mass C with respect to the origin O.

Therefore, $\displaystyle{\vec{R}=\frac{m_1\vec{r_1}+m_2\vec{r_2}}{m_1+m_2}}$

or, $\displaystyle{\dot{\vec{R}}=\frac{m_1\dot{\vec{r_1}}+m_2\dot{\vec{r_2}}}{m_1+m_2}}$

or, $\displaystyle{\dot{\vec{R}}=\frac{m_1\vec{v_1}+m_2\vec{v_2}}{m_1+m_2}}$

or, $\displaystyle{\vec{V}=\frac{m_1\vec{v_1}+m_2\vec{v_2}}{m_1+m_2}}$

or, $(m_1\vec{v_1}+m_2\vec{v_2})=(m_1+m_2)\vec{V}\tag{1}$

where, $\vec{v_1}$ and $\vec{v_2}$ are the velocities of the masses $m_1$ and $m_2$ respectively and $\vec{V}$ is the velocity of the centre of mass.

If $\vec{v}$ be the velocity of $m_1$ relative to $m_2$ then

$\vec{v}=\frac{d}{dt}(\vec{r_1}-\vec{r_2})\\=\dot{\vec{r_1}}-\dot{\vec{r_2}}$

or, $\vec{v}=\vec{v_1}-\vec{v_2}\tag{2}$

Solving equations (1) and (2) we get

$\displaystyle{\vec{v_1}=\vec{V}+\frac{m_2\vec{v}}{m_1+m_2}}$

and, $\displaystyle{\vec{v_2}=\vec{V}-\frac{m_1\vec{v}}{m_1+m_2}}$

Therefore the total kinetic energy is given by,

$T=\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2$

$=\frac{1}{2}m_1{\left(\vec{V}+\frac{m_2\vec{v}}{m_1+m_2}\right)}^2+\frac{1}{2}m_2{\left(\vec{V}-\frac{m_1\vec{v}}{m_1+m_2}\right)}^2$

$=\frac{1}{2}(m_1+m_2)V^2+\frac{1}{2}\frac{m_1m_2}{m_1+m_2}v^2$

or, $T=\frac{1}{2}MV^2+\frac{1}{2}{\mu}v^2$

where, $M=(m_1+m_2)$ is the total mass of the system and $\mu=\frac{m_1m_2}{m_1+m_2}$ is the reduced mass of the system.