The system shown in the 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

Question

The system shown in the 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 <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-ziqfmqskr0y6enk1.jpg" /> (A) (g/2) upward, g downward (B) g upward, g downward (C) g upward, 2g downward (D) 2g upward, g downward

Tardigrade Admin · Accepted Answer

Initially under equilibrium of mass m T=mg Now, the string is cut. Therefore, T=mg force is decreased on mass m upward and downwards on mass 2 m. ∴ am=(m g/m)=g (d o w n w a r d s) and a2 m=(m g/2 m)=(g/2) (.upwards.)

Q. The system shown in the 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

$\frac{g}{2}$ upward, $g$ downward

$g$ upward, $g$ downward

$g$ upward, $2 g$ downward

$2 g$ upward, $g$ downward

Initially under equilibrium of mass $m$
$T = m g$
Now, the string is cut. Therefore, $T = m g$ force is decreased on mass $m$ upward and downwards on mass 2 $m .$
$∴ a_{m} = \frac{m g}{m} = g (d o w n w a r d s)$
and $a_{2 m} = \frac{m g}{2 m} = \frac{g}{2} (u pw a r d s)$

Q. The system shown in the figure is in equilibrium and at rest. The spring and string are massless, now the string is cut. The acceleration of mass 2m and m just after the string is cut will be

2g​ upward, g downward

g upward, g downward

g upward, 2g downward

2g upward, g downward