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

Question

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 (A) g/2 upwards, g downwards (B) g upwards, g/2 downwards (C) g upwards, g downwards (D) 2g upwards, g downwards

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 upwards and downwards on mass 2 m ∴ am=(mg/m)=g (downwards) and a2m=(mg/2m)=(g/2) (upwards)

Q. 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

$g /2$ upwards, $g$ downwards

$g$ upwards, $g /2$ downwards

$g$ upwards, $g$ downwards

$2 g$ upwards, $g$ downwards

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 $2 m$
$∴ a_{m} = \frac{m g}{m} = g$ (downwards) and
$a_{2 m} = \frac{m g}{2 m} = \frac{g}{2}$ (upwards)

Q. 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 2m and m just after the string is cut will be

g/2 upwards, g downwards

g upwards, g/2 downwards

g upwards, g downwards

2g upwards, g downwards