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Velocity of COM

Two masses m ...

Question

Two masses $m_{1}$ and $m_{2}$ are moving with velocities $v_{1}$ and $v_{2}$ respectively. Find their total kinetic energy in the reference frame of centre of mass.

A

\frac{m_{1} m_{2}}{m_{1} + m_{2}} (v_{1} - v_{2})^{2}

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B

\frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} (v_{1} - v_{2})^{2}

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C

\frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} (v_{1} + v_{2})^{2}

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D

\frac{1}{2} (m_{1} m_{2}) (v_{1} - v_{2})^{2}

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Solution

The correct option is B $\frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} (v_{1} - v_{2})^{2}$
$K = \frac{1}{2} m_{1} v_{1 c}^{2} + \frac{1}{2} m_{2} v_{2 c}^{2}$ …… (i)
Where $v_{1 c}$ and $v_{2 c}$ are velocities relative to the CM.
$v_{1 c} = v_{1} - v_{C M} = v_{1} - (\frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}) = (\frac{m_{2} (v_{1} - v_{2})}{m_{1} + m_{2}})$
$v_{2 c} = v_{2} - v_{C M} = v_{2} - (\frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}) = (\frac{m_{1} (v_{2} - v_{1})}{m_{1} + m_{2}})$
Putting these in Equation (i), $K = \frac{1}{2} \frac{m_{1} m_{2}}{(m_{1} + m_{2})} (v_{1} - v_{2})^{2}$
Note: For a system of two particles of masses $m_{1}$ and $m_{2}$ , the total kinetic energy is $k = (1 / 2) μ v^{2} + (1 / 2) (m_{1} + m_{2}) v_{C M}^{2}$ where $μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}$ (known as reduced mass of the system and
$\to v = {\to v}_{1} - {\to v}_{2}$ (relative velocity).

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