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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\vec{v}_1 + m_2\vec{v}_2}{m_1 +

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$\frac{m_1v_1 - m_2v_2}{m_1 + m_2}$

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(B) The velocity of the centre of mass $(V_{cm})$ of a system of particles is given by the formula:$V_{cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}$In this problem,the particles move towards each other. Let the direction of $m_1$ be positive. Then the velocity of $m_1$ is $\vec{v}_1 = v_1$ and the velocity of $m_2$ is $\vec{v}_2 = -v_2$ (since it moves in the opposite direction).Substituting these values into the formula:$V_{cm} = \frac{m_1(v_1) + m_2(-v_2)}{m_1 + m_2}$$V_{cm} = \frac{m_1v_1 - m_2v_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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