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.