Question

A block of mass $m\text{kg}$ is suspended by a string attached with a $2m\text{kg}$ mass block. The $2m$ mass block is suspended by a spring as shown in the figure. The system is initially at equilibrium and rest. The spring and string are massless. Now the string is cut. The accelerations of mass $2m$ and $m$ just after the string is cut will be

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

Answer 1

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

Initially, when system is at equilibrium and rest,
F.B.D of $m\text{kg}$ and $2m\text{kg}$ block:


From the FBDs,
$T=mg$ ... $\left(1\right)$
and $2mg+T=kx$ ... $\left(2\right)$

Putting (1) in (2)
$kx=3mg$ ... $\left(3\right)$

As we know, force due to spring does not change instantanously. (After cutting the string, initially it will exert the same force)
But after cutting the string, $T=0$

F.B.D for $m\text{kg}$ and $2m\text{kg}$ block after cutting the string:

For $m\text{kg}$ block, only gravitational force is acting. Hence, acceleration of block $m$ is $g$ downward after cutting the string.

Assume acceleration of $2m$ block, $a$ upwards
Now, $kx-2mg=2ma$
$\because (kx=3mg)$ (from (3))
$\Rightarrow 3mg-2mg=2ma$
$\Rightarrow a=\frac{g}{2}$
It is positive means assumed direction of acceleration is correct.
$\therefore$ Acceleration of $2m$ block is $\frac{g}{2}$ upwards.