Question

Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively:

Answer 1

The free body diagram of the blocks A and B of masses 3m and m respectively are given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(20,0){2}{\circle*{1}}\multiput(0,0)(20,0){2}{\vector(0,-1){10}}\multiput(0,0)(20,0){2}{\vector(0,1){10}}\put(-3.5,0){ \sf{A} }\put(16.5,0){ \sf{B} }\put(-14,-12){ \sf{3mg+T} }\put(14.5,-12){ \sf{mg} }\put(-2,10){ \sf{F} }\put(17.5,10){ \sf{T} }\end{picture}$

Here T is the tension in the string and F is the restoring force in the spring.

From the FBD of the block A, we get,

$\displaystyle\longrightarrow\sf{F=3mg+T\quad\quad\dots(1)}$

And from the FBD of the block B, we get,

$\displaystyle\longrightarrow\sf{T=mg\quad\quad\dots(2)}$

On solving (1) and (2) we get the restoring force,

$\displaystyle\longrightarrow\sf{F=4mg}$

After cutting the string, the tension in the string becomes zero. So the block A moves upwards due to the restoring force in the spring and the block B moves downwards as it loses contact with the system.

Hence the new free body diagrams of the blocks are as follows.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(20,0){2}{\circle*{1}}\multiput(0,0)(20,0){2}{\vector(0,-1){10}}\put(0,0){\vector(0,1){10}}\put(-3.5,0){ \sf{A} }\put(16.5,0){ \sf{B} }\put(-7,-12){ \sf{3mg} }\put(14.5,-12){ \sf{mg} }\put(-2,10){ \sf{F} }\put(24,5){\vector(0,-1){10}}\put(4,-5){\vector(0,1){10}}\put(5,0){ \sf{a_A} }\put(25,0){ \sf{a_B} }\end{picture}$

where $\sf{a_A}$ and $\sf{a_B}$ are accelerations of blocks A and B respectively.

From FBD of block A, the net force is,

$\displaystyle\longrightarrow\sf{F-3mg=3ma_A}$

$\displaystyle\longrightarrow\sf{4mg-3mg=3ma_A}$

$\displaystyle\longrightarrow\sf{mg=3ma_A}$

$\displaystyle\Longrightarrow\sf{\underline{\underline{a_A=\dfrac{g}{3}\ m\,s^{-2}}}}$

And from FBD of the block B, the net force is,

$\displaystyle\longrightarrow\sf{mg=ma_B}$

$\displaystyle\Longrightarrow\sf{\underline{\underline{a_B=g\ m\,s^{-2}}}}$

Hence the acceleration of A and B are $\sf{\dfrac{g}{3}}$ upwards and $\sf{g}$ downwards respectively.