Question

Two mass $\text{A}$ and $\text{B}$, each of mass $m$, are initially in equilibrium as shown in figure. The acceleration of block $A$ just after string between $\text{A}$ and $B$ is cut will be :
( $g=$ acceleration due to gravity).

• A. $2g$ downward
• B. $2g$ upward
• C. $g$ downward
• D. $g$ upward

Answer 1

The correct option is D $g$ upward

Under equilibrium, the forces have been shown in the figure below.


From FBD of system we have,

$2T=2mg$

or, $T=mg$

Also $T$ is equal to the spring force.

Let $x$ be extension in spring at equilibrium.

i.e. $T=kx$

After the string between $\text{A}$ and $\text{B}$ is cut, the spring force will retain its value at that instant.

So, the F.B.D of $\text{A}$ :

$2T-mg=ma$

$2mg-mg=ma$

or,

$mg=ma$

$\therefore$ $a=g$ (upwards)

$\begin{array}{c}\text{Why this question?}\\ \text{Tip:- If a string is cut, tension becomes zero instantly. However spring force does not instantly become zero}\end{array}$