Explanation

(c) Block $A$ moves with velocity $0.15ms^{- 1}$ compresses the spring which pushes $B$ towards right. $A$ goes on compressing the spring till the velocity acquired by $B$ becomes equal to the velocity of $A$. Let this velocity be $v$. This state occurs when the spring is in a state of maximum compression. Let $x$ be the maximum compression in this stage
According to the law of conservation of linear momentum, we get
$m_{A}u = (m_{A} + m_{B})v$
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$\Rightarrow v = \frac{m_{A}u}{m_{A} + m_{B}} = \frac{2 \times 0.15}{2 + 3} = 0.06\ ms^{- 1}$
According to the conservation of energy, we get
$\frac{1}{2}m_{A}u^{2} = \frac{1}{2}\left( m_{A} + m_{B} \right)v^{2} + \frac{1}{2}kx^{2}$
$\frac{1}{2}m_{A}u^{2} - \frac{1}{2}\left( m_{A} + m_{B} \right)v^{2} = \frac{1}{2}kx^{2}$
$\frac{1}{2} \times 2 \times (0.15)^{2} - \frac{1}{2}(2 + 3)(0.06)^{2} = \frac{1}{2}\ kx^{2}$
$0.0225 - 0.009 = \frac{1}{2}kx^{2} \Rightarrow 0.0135 = \frac{1}{2}kx^{2}$
$\Rightarrow x = \sqrt{\frac{0.027}{k}} = \sqrt{\frac{0.027}{10.8}} = 0.05\ m$