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(B) The velocity of the centre of mass $(V_{cm})$ of a system of particles is given by the formula:$V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$In th

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$\frac{m_1 v_1 - m_2 v_2}{m_1 + m_2}$

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(B) The velocity of the centre of mass $(V_{cm})$ of a system of particles is given by the formula:$V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$In this problem,the particles are moving towards each other. Let us define the direction of $m_1$ as positive and $m_2$ as negative (or vice versa) to account for their relative motion.If $v_1$ is the velocity of $m_1$ in the positive direction,then $v_2$ (moving towards $m_1$) must be in the negative direction,i.e.,$-v_2$.Substituting these into the formula:$V_{cm} = \frac{m_1 v_1 + m_2 (-v_2)}{m_1 + m_2} = \frac{m_1 v_1 - m_2 v_2}{m_1 + m_2}$Thus,the correct option is $B$.

$A$ system consists of two particles of masses $m_1$ and $m_2$ separated by a distance $r$. If the particles move towards each other with velocities $v_1$ and $v_2$ respectively,the velocity of the centre of mass of the system is:

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