Two rectangular blocks \(A\) and \(B\) of masses \(2 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) respectively are connected by a spring of spring constant \(10.8 \mathrm{Nm}^{-1}\) and are placed on a frictionless horizontal surface. The block A was given an initial velocity of \(0.15 \mathrm{~ms}^{-1}\) in the direction shown in the figure. The maximum compression of the spring during the motion is

As the block \(A\) moves with velocity \(0.15 \mathrm{~ms}^{-1}\), it compresses thespring which pushes \(B\) towards right. \(A\) goes on compressing the spring till the velocity acquired by \(B\) becomes equal to the velocity of \(A\), ie, \(0.15 \mathrm{~ms}^{-1}\). 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
or
\(\mathrm{m}_{\mathrm{A}} \mathrm{u} =\left(\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{v} \)
\( \mathrm{v} =\frac{\mathrm{m}_{\mathrm{A}} \mathrm{u}}{\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}} \)
\( =\frac{2 \times 0.15}{2+3}=0.06 \mathrm{~ms}^{-1}\)
According to the law of conservation of energy.
\(\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{u}^{2}=\frac{1}{2}\left(\mathrm{~m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{v}^{2}+\frac{1}{2} \mathrm{kx}^{2} \)
\( \frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{u}^{2}-\frac{1}{2}\left(\mathrm{~m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{v}^{2}=\frac{1}{2} \mathrm{kx}^{2} \)
\( \frac{1}{2} \times 2 \times(0.15)^{2}-\frac{1}{2}(2+3)(0.06)^{2}=\frac{1}{2} \mathrm{kx}^{2} \)
\( 0.0225-0.009=\frac{1}{2} \mathrm{kx}^{2} \text { or } 0.0135=\frac{1}{2} \mathrm{kx}^{2} \)
\( \text {or } \quad \mathrm{x}=\sqrt{\frac{0.027}{\mathrm{k}}}=\sqrt{\frac{0.027}{10.8}}=0.05 \mathrm{~m}\)