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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
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The correct answer is:
$0.05 \mathrm{~m}$
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
$\begin{aligned}
\mathrm{m}_{\mathrm{A}} \mathrm{u} &=\left(\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{v} \\
\mathrm{v} &=\frac{\mathrm{m}_{\mathrm{A}} \mathrm{u}}{\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}} \\
&=\frac{2 \times 0.15}{2+3}=0.06 \mathrm{~ms}^{-1}
\end{aligned}$
According to the law of conservation of energy.
$\begin{gathered}
\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{u}^{2}=\frac{1}{2}\left(\mathrm{~m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{v}^{2}+\frac{1}{2} \mathrm{kx}^{2} \\
\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{u}^{2}-\frac{1}{2}\left(\mathrm{~m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{v}^{2}=\frac{1}{2} \mathrm{kx}^{2} \\
\frac{1}{2} \times 2 \times(0.15)^{2}-\frac{1}{2}(2+3)(0.06)^{2}=\frac{1}{2} \mathrm{kx}^{2} \\
0.0225-0.009=\frac{1}{2} \mathrm{kx}^{2} \text { or } 0.0135=\frac{1}{2} \mathrm{kx}^{2} \\
\text { or } \quad \mathrm{x}=\sqrt{\frac{0.027}{\mathrm{k}}}=\sqrt{\frac{0.027}{10.8}}=0.05 \mathrm{~m}
\end{gathered}$
According to the law of conservation of linear momentum, we get
or
$\begin{aligned}
\mathrm{m}_{\mathrm{A}} \mathrm{u} &=\left(\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{v} \\
\mathrm{v} &=\frac{\mathrm{m}_{\mathrm{A}} \mathrm{u}}{\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}} \\
&=\frac{2 \times 0.15}{2+3}=0.06 \mathrm{~ms}^{-1}
\end{aligned}$
According to the law of conservation of energy.
$\begin{gathered}
\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{u}^{2}=\frac{1}{2}\left(\mathrm{~m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{v}^{2}+\frac{1}{2} \mathrm{kx}^{2} \\
\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{u}^{2}-\frac{1}{2}\left(\mathrm{~m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{v}^{2}=\frac{1}{2} \mathrm{kx}^{2} \\
\frac{1}{2} \times 2 \times(0.15)^{2}-\frac{1}{2}(2+3)(0.06)^{2}=\frac{1}{2} \mathrm{kx}^{2} \\
0.0225-0.009=\frac{1}{2} \mathrm{kx}^{2} \text { or } 0.0135=\frac{1}{2} \mathrm{kx}^{2} \\
\text { or } \quad \mathrm{x}=\sqrt{\frac{0.027}{\mathrm{k}}}=\sqrt{\frac{0.027}{10.8}}=0.05 \mathrm{~m}
\end{gathered}$
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