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In the system shown in figure $m_{1} > m_{2}$ . System is held at rest by thread BC. Just after the thread BC is burnt.

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acceleration of $m_{1}$ will be equal to zero

acceleration of $m_{2}$ will be downwards

magnitude of acceleration of two blocks will be non-zero and unequal

magnitude of acceleration of both the blocks will be $(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g$

answer is A.

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Detailed Solution

$m_{1} g = k x$

$k x = m_{2} g + T$ in equilibrium

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