Question

Two blocks $\text{A}$ and $\text{B}$ of mass $M$ and $2M$ respectively are hanging from a ceiling by means of a light spring and a light string as shown in figure.


If string between the blocks is suddenly cut, then accelerations of block $\text{A}$ and $\text{B}$ will be:

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

Answer 1

The correct option is C $a_{\text{A}}$$=$ $2g$ $\text{upwards}$ , $a_{\text{B}}$ $=$ $g$ $\text{downwards}$

At equilibrium :

From F.B.D of both blocks,

$\Rightarrow T=2Mg$..........(1)

$F_S=T+Mg$................(2)

Here $F_S$ is the spring force on block $\text{A}$,

On adding Eq. (1) and (2) we have :

$\Rightarrow F_S$ $=3Mg$

When string is cut, the only external force acting on block $\text{B}$ is its weight


So, $a_B=\frac{2Mg}{2M}$ $=$ $g$ $\text{downwards}$

F.B.D of $\text{A}$ after string is cut :

Spring force doesn't change instantly after string is cut.

Since $F_S>Mg$, block $A$ will move upwards.

$\Rightarrow a_B=\frac{F_S-Mg}{M}=\frac{3Mg-Mg}{M}$

$\therefore a_A=2g$ $\text{upwards}$

$\begin{array}{c}\text{Why this question}?\\ \text{Always be cautious that due to inertial}\\ \text{properties the spring force do not change}\\ \text{instantly, even if some forces disappear or}\\ \text{extension at that instant.}\end{array}$