Ans.

Let us consider a system of two masses \( m_1 \) and \( m_2 \) having position vectors \( \vec{r_1} \) and \( \vec{r_2} \) respectively with respect to the origin O. If \( \vec{R} \) be the position vector of the centre of mass C of the system then

\( \vec{R}=\frac{m_1\vec{r_1}+m_2\vec{r_2}}{m_1+m_2}\tag{1} \).

Let, \( \vec{R_1} \) and \( \vec{R_2} \) be the position vectors of the masses \( m_1 \) and \( m_2 \) with respect to the centre of mass C. \( \vec{r} \) be the position vector of \( m_1 \) with respect to the mass \( m_2 \).

Now from Fig. 1 we can write

\( \vec{R_1}=\vec{r_1}-\vec{R}\tag{2} \) \( \vec{R_2}=\vec{r_2}-\vec{R}\tag{3} \) \( \vec{r}=\vec{r_1}-\vec{r_2}\tag{4} \)

From equations (1) and (2) we can write,

\( \vec{R_1}=\vec{r_1}-\vec{R} \)

or, \( \vec{R_1}=\vec{r_1}-\frac{m_1\vec{r_1}+m_2\vec{r_2}}{m_1+m_2} \)

or, \( \vec{R_1}=\frac{m_2}{m_1+m_2}(\vec{r_1}-\vec{r_2}) \)

or, \( \vec{R_1}=\frac{1}{m_1}\frac{m_1m_2}{m_1+m_2}\vec{r} \)

or, \( \vec{R_1}=\frac{\mu}{m_1}\vec{r} \)

differentiating both side with respect to time

\( \displaystyle{\dot{\vec{R_1}}=\frac{\mu}{m_1}\dot{\vec{r}}} \)

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

Similarly from equations (1) and (3) we get

\( \vec{R_2}=\vec{r_2}-\vec{R} \)

or, \vec{R_2}=\vec{r_2}-\frac{m_1\vec{r_1}+m_2\vec{r_2}}{m_1+m_2}

or, \( \vec{R_2}=\frac{m_1}{m_1+m_2}(\vec{r_2}-\vec{r_1}) \)

or, \( \vec{R_2}=\frac{1}{m_2}\frac{m_1m_2}{m_1+m_2}(-\vec{r}) \)

or, \( \vec{R_2}=-\frac{\mu}{m_2}\vec{r} \)

differentiating both side

\( \displaystyle{\dot{\vec{R_2}}=-\frac{\mu}{m_2}\dot{\vec{r}}} \)

So the kinetic energy of the system in the centre of mass frame of reference is

\( K.E=\frac{1}{2}m_1{\dot{\vec{R_1}}}^2+\frac{1}{2}m_2{\dot{\vec{R_2}}}^2 \)

= \( \displaystyle{\frac{1}{2}{\mu}^2{\dot{\vec{r}}}^2\left(\frac{1}{m_1}+\frac{1}{m_2}\right)} \)

= \( \displaystyle{\frac{1}{2}\mu{\dot{\vec{r}}}^2} \)