Question

The system shown in the figure is in equilibrium at rest and the spring and string are massless. Now the string is cut. The acceleration of masses $2\text{m}$ and $\text{m}$ just after the string is cut will be:

• A. $\frac{3g}{2}$ upwards, $g$ downwards
• B. $\frac{g}{2}$ upwards, $g$ downwards
• C. $g$ upwards, $2g$ downwards
• D. $2g$ upwards, $g$ downwards

Answer 1

The correct option is B $\frac{g}{2}$ upwards, $g$ downwards

Initially, when the system is in equilibrium:
FBD of system:


By applying equilibrium condition on system in vertical direction
$\Rightarrow F_{\text{Spring}}=3mg$

Now, after the string is cut, $T=0$ i.e tension in string becomes zero instantaneously, while spring force will act at its initial value $F_{\text{Spring}}$, due to inertia of spring.

FBD of individual blocks after string is cut, represented below:


Acceleration of block of mass $m$
$a_m=\frac{mg}{m}$
$\therefore a_m=g$ (Downwards)

Acceleration of block of mass $2m$
$a_{2m}=\frac{F_{\text{Spring}}-2mg}{2m}=\frac{mg}{2m}$

$\therefore a_{2m}=\frac{g}{2}$ (Upwards)