Laws Of Motion Ques 56
- System shown in figure is in equilibrium and at rest. The spring and string are massless, now the string is cut. The acceleration of mass $2 m$ and $m$ just after the string is cut will be
(2006, 3M)
(a) $g / 2$ upwards, $g$ downwards
(b) $g$ upwards, $g / 2$ downwards
(c) $g$ upwards, $2 g$ downwards
(d) $2 g$ upwards, $g$ downwards
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Answer:
Correct Answer: 56.(a)
Solution:
Formula:
- Initially under equilibrium of mass $m$
$$ T=m g $$
Now, the string is cut. Therefore, $T=m g$ force is decreased on mass $m$ upwards and downwards on mass $2 m$.
$$ \begin{array}{lll} \therefore & a_{m}=\frac{m g}{m}=g & \text { (downwards) } \\ \text { and } & a_{2 m}=\frac{m g}{2 m}=\frac{g}{2} & \text { (upwards) } \end{array} $$
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