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In a reference frame K, two particles move along the X-axis, one of mass $m_{1}$ with velocity ${\vec{v}}_{1}$ and the other of mass $m_{2}$ with velocity ${\vec{v}}_{2}$ . There exists another reference frame $K_{0}$ in which the total kinetic energy of the two masses $m_{1}$ and $m_{2}$ will be the minimum. The instantaneous velocity $\vec{v}$ of the reference frame $K_{0}$ in the reference frame K will be

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$\vec{v} = {\vec{v}}_{1} + {\vec{v}}_{2}$

$\vec{v} = {\vec{v}}_{1}$

$\vec{v} = {\vec{v}}_{2}$

$\vec{v} = \frac{m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}}{m_{1} + m_{2}}$

answer is D.

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Detailed Solution

The kinetic energy is minimum in centre of mass reference frame.

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