A block $m_{2}$ is loaded onto another block $m_{1}$ placed on a spring of stiffness $k$ as shown in the figure. If block $m_{2}$ is suddenly removed, then magnitude of acceleration of block $m_{1}$ just after the removal of $m_{2}$ will be:

(\frac{m_{2}}{m_{1}} g)

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\frac{\sqrt{m_{1} m_{2}}}{m_{1} + m_{2} g}

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\frac{m_{1} + m_{2}}{m_{1} - m_{2}}

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(\frac{m_{1}}{m_{2}} g)

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Solution

The correct option is A $(\frac{m_{2}}{m_{1}} g)$
Considering the equilibrium of system, just before removal of $m_{2} .$

FBD before removal of $m_{2}$ :

Here, $x =$ compression in spring

$k x = (m_{1} + m_{2}) g \dots (i)$
$∴ x = \frac{(m_{1} + m_{2}) g}{k}$

Spring force will remain as it is after removal of $m_{2},$ because spring will not change its length instantaneously.

FBD after removal of $m_{2}$ :

Applying Newton's 2nd law on block $m_{1}$ in direction of acceleration:

$k x - m_{1} g = m_{1} a$
$\Rightarrow (m_{1} + m_{2}) g - m_{1} g = m_{1} a$
(from (i))

$∴ a = (\frac{m_{2}}{m_{1}} g)$

Hence, block $m_{1}$ will move with acceleration $(\frac{m_{2}}{m_{1}} g)$ upwards.

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