Explanation

(c) As the block A moves with velocity with velocity 0.15 $ms^{- 1}$, it 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, i.e. 0.15 $ms^{- 1}$. Let this velocity be v. Now, spring is in a state of maximum compression. Let x be the maximum compression at this stage.
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According to the law of conservation of linear momentum, we get
$m_{A}u = (m_{A} + m_{B})v$
Or $v = \frac{m_{A}u}{m_{A} + m_{B}}$
$\frac{2 \times 0.15}{2 + 3} = 0.06ms^{- 1}$
According to the law of conservation of energy
$\frac{1}{2}m_{A}u^{2} = \frac{1}{2}\left( m_{A} + m_{B} \right)V^{2} + \frac{1}{2}kx^{2}$
$\frac{1}{2}m_{A}u^{2} - \frac{1}{2}\left( m_{A} + m_{B} \right)v^{2} = \frac{1}{2}kx^{2}$
$\frac{1}{2} \times 2 \times (0.15)^{2} - \frac{1}{2}(2 + 3)(0.06)^{2} = \frac{1}{2}kx^{2}$
$0.0225 - 0.009 = \frac{1}{2}kx^{2}$
$or\ 0.0135 = \frac{1}{2}kx^{2}$
Or $x = \sqrt{\frac{0.0027}{k}} = \sqrt{\frac{0.0027}{10.8}}\ = 0.05m$