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(B) Let the masses be $m_1 = 3m$ and $m_2 = m$. The acceleration of the blocks is given by $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(3m - m)g}{3m +

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$g/4$

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(B) Let the masses be $m_1 = 3m$ and $m_2 = m$. The acceleration of the blocks is given by $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(3m - m)g}{3m + m} = \frac{2mg}{4m} = \frac{g}{2}$.The acceleration of the centre of mass is $a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$.Taking the downward direction as positive for the heavier mass $(3m)$,its acceleration is $a_1 = g/2$. The lighter mass $(m)$ moves upward,so its acceleration is $a_2 = -g/2$.Substituting these values: $a_{cm} = \frac{3m(g/2) + m(-g/2)}{3m + m} = \frac{1.5mg - 0.5mg}{4m} = \frac{mg}{4m} = \frac{g}{4}$.

$A$ pulley fixed to the ceiling carries a string with blocks of mass $m$ and $3m$ attached to its ends. The masses of the string and pulley are negligible. When the system is released,what is the acceleration of its centre of mass?

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