Question

Mass of block $A$ is $m$ and that of block $B$ is $2m$. Spring constant is $k$ and there is no friction. System is released from rest with the spring unstretched. Pulley is massless. Identify the correct statement(s).

• A. Maximum extension of the spring is $\frac{4mg}{k}$
• B. Acceleration of block B is $\frac{g}{3}$ downwards, when extension in the spring is $\frac{mg}{k}$
• C. Maximum extension in the spring is $\frac{2mg}{k}$
• D. Acceleration of block $B$ is $\frac{g}{3}$ downwards, when extension in spring is $\frac{4mg}{k}$

Answer 1

Answer: (A), (B)

The correct options are
A Maximum extension of the spring is $\frac{4mg}{k}$
B Acceleration of block B is $\frac{g}{3}$ downwards, when extension in the spring is $\frac{mg}{k}$

$m_A=m$, $m_B=2m$
System is released from rest, so initial velocity of both blocks is $0\text{m/s}$.
Let the maximum extension in spring be $x_m$.


At the maximum externsion in spring, both the blocks will be at rest momentarily.
Applying mechanical energy conservation,
$\text{Loss in gravitational PE of B}=\text{Gain in spring PE}$
(K.E is zero initially and finally)
$\Rightarrow 2mg{x}_m=\frac{1}{2}k{x}_m^2$
$\therefore x_m=\frac{4mg}{k}$
Let the acceleration of block $B$ be $a$. Then acceleration of block $A$ will also be $a$ (due to string constraint).


FBD of block $A$:


$T_1-T_2=ma$
If extension in spring is $x=\frac{mg}{k}$,
$T_2=kx=mg$
$\Rightarrow T_1-mg=ma...\left(i\right)$
FBD of block $B$:


$2mg-T_1=2ma...\left(ii\right)$
Adding $\left(i\right)\&\left(ii\right)$:
$a=\frac{g}{3}$ (downwards)
Similarly, when extension in spring is $\frac{4mg}{k}$, then:
$T_1-4mg=ma...\left(i\right)$ for block $A$.
$[\because T_2=kx=4mg]$
For block $B$
$2mg-T_1=2ma...\left(ii\right)$
Adding $\left(i\right)\&\left(ii\right)$:
$a=\frac{-2g}{3}$
$\therefore$ when $x=\frac{4mg}{k}$, block $B$ will move upwards with acceleration $\frac{2g}{3}$.
Only option A and B are correct.