Question

Two blocks $A$ and $B$ of same mass $m$ attached with a light spring are suspended by a string as shown in the figure. Find the acceleration of block $A$ and $B$ respectively, just after the string is cut.

• A. $g$ upwards, $g$ downwards
• B. $g$ downwards, zero
• C. $2g$ upwards, $g$ upwards
• D. $2g$ downwards, zero

Answer 1

The correct option is D $2g$ downwards, zero

When block $A$ and $B$ are in equilibrium position


From FBD of B, $T_0=mg--\left(i\right)$

From FBD of A,
$T=mg+T_0$
$T=2mg$

When string is cut, tension $T$ becomes zero. But spring does not change its shape after cutting. So same spring force acts on mass B. Again drawing FBD of blocks A and B as shown.


From FBD of B,
$T_0-mg=m{a}_B$
$\Rightarrow m{a}_B=0$
$\Rightarrow a_B=0$ (from (i))


From FBD of A,
$mg+T_0=m{a}_A$
$\Rightarrow 2mg=m{a}_A$ (from (i))
$\Rightarrow a_A=2g$ (downwards)