Question

Find the total kinetic energy of the two particles in the reference frame fixed to their centre of inertia.
Here $\mu =\frac{m_1{m}_2}{m_1+m_2}$.

• A. $T=\frac{1}{2}\mu (v_1^2+v_2^2)$.
• B. $T=\frac{1}{2}\mu (v_1^2+2v_2^2)$
• C. $T=\frac{1}{3}\mu (v_1^2+v_2^2)$
• D. $T=\frac{1}{2}\mu (2v_1^2+v_2^2)$

Answer 1

Answer: (A)

$T=\frac{1}{2}\mu (v_1^2+v_2^2)$.

Velocity of center of inertia frame=$\frac{m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}}{m_1+m_2}$

Velocity of $m_1$ relative to this frame=$\overrightarrow{v_{r1}}=\overrightarrow{v_1}-$ $\frac{m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}}{m_1+m_2}=\frac{m_2(\overrightarrow{v_1}-\overrightarrow{v_2})}{m_1+m_2}$

Similarly velocity of $m_2$ relative to this frame=$\overrightarrow{v_{r2}}=\overrightarrow{v_2}-$ $\frac{m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}}{m_1+m_2}=\frac{m_1(\overrightarrow{v_2}-\overrightarrow{v_1})}{m_1+m_2}$

Hence the kinetic energy with respect to this frame=$\frac{1}{2}{m}_1{v}_{r1}^2++\frac{1}{2}m{v}_{r2}^2=\frac{1}{2}\mu {|\overrightarrow{v_2}-\overrightarrow{v_1}|}^2=\frac{1}{2}\mu (v_1^2+v_2^2)$