Two rectangular blocks A and B of masses 2 kg and 3 kg respectively...

As the block $A$ moves with velocity $0.15 \mathrm{~ms}^{-1}$, 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., $0.15 \mathrm{~ms}^{-1}$. Let this velocity be $v$. Now, spring is in a state of maximum compression. Let $x$ be the maximum compression at this stage. According to the law of conservation of linear momentum, we get $m_A u=\left(m_A+m_B\right) v$ or $v=\frac{m_A u}{m_A+m_B}$ $=\frac{2 \times 0.15}{2+3}=0.06 \mathrm{~ms}^{-1}$ According to the law of conservation of energy. $\frac{1}{2} m_A u^2=\frac{1}{2}\left(m_A+m_B\right) v^2+\frac{1}{2} k x^2$ $\frac{1}{2} m_A u^2-\frac{1}{2}\left(m_A+m_B\right) v^2=\frac{1}{2} k x^2$ $\frac{1}{2} \times 2 \times(0.15)^2-\frac{1}{2}(2+3)(0.06)^2=\frac{1}{2} k x^2$ $0.0225-0.009=\frac{1}{2} k x^2$ or $\quad 0.0135=\frac{1}{2} k x^2$ or $\quad x=\sqrt{\frac{0.027}{k}}=\sqrt{\frac{0.027}{10.8}}=0.05 \mathrm{~m}$