The two bodies of mass $m_{1}$ and $m_{2}$ respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and then released. Then acceleration of the centre of mass of the system is

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

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

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Solution

The correct option is A ${(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} g$
For such a pulley system eq.1 $m_{1} g - T = m_{2} a$ and eq.2 $T - m_{2} g = m_{2} a$ . from both these equations we get $a = \frac{(m_{1} - m_{2}) g}{(m_{1} + m_{2})}$ . Now for such a system $a_{c m} = \sum m_{i} a_{i} / \sum m_{i} = a_{c m} = a (m_{1} - m_{2}) / (m_{1} + m_{2})$ Now putting the value of a into $a_{c m}$ we get $a_{c m} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}})^{2} g$

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The two bodies of mass m1 and m2 respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and then released. Then acceleration of the centre of mass of the system is

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The two bodies of mass $m_{1}$ and $m_{2}$ respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and then released. Then acceleration of the centre of mass of the system is