Question

Two blocks ‘$A$’ and ‘$B$’ of same mass '$m$' connected by a light spring are suspended by a string as shown in figure. Find the acceleration of block ‘$A$’ and ‘$B$’ just after the string is cut.

• A. $0,2g$
• B. $0,0$
• C. $g,g$
• D. $2g,0$

Answer 1

The correct option is D $2g,0$

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


For block B,
$kx=mg$
For block A,
$T=kx+mg=2mg$
When string is cut, tension $T$ becomes zero. But spring does not change its shape just after cutting. So spring force acting on mass $B$ remains the same. Again drawing F.B.D. of block $A$ and $B$ as shown in figure.

Let $a_A$ and $a_B$ be the acceleration of the blocks A and B.
For block B,
$kx-mg=m{a}_B$
Since $kx=mg$
$\Rightarrow a_B=0$
For block A,
$mg+kx=m{a}_A$
$2mg=m{a}_A$
$\Rightarrow a_A=2g$ (Downwards)