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A block of mass $m = 0.1\,kg$ is connected to a spring of unknown spring constant $k.$ It is compressed to a distance $x$ from its equilibrium. position and released from rest . After approaching half the distance $(\frac {x}{2})$ from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity $3\,ms^{-1}.$ The total initial energy of the spring is ................ $\mathrm{J}$
$0.3$
$0.6$
$0.8$
$1.5$
Solution
Applying momentum conservation
${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$
$0.1 u + m(0) = 0.1 (0) + m(3)$
$0.1 u = 3m$
$\frac{1}{2}0.1{u^2} = \frac{1}{2}m{\left( 3 \right)^2}$
Solving we get, $u = 3$
$\begin{gathered}
\frac{1}{2}k{x^2} = \frac{1}{2}K{\left( {\frac{x}{2}} \right)^2} + \frac{1}{2}\left( {0.1} \right){3^2} \hfill \\
\Rightarrow \,\,\,\,\frac{3}{4}k{x^2} = 0.9 \hfill \\
\Rightarrow \,\,\,\,\frac{3}{2} \times \frac{1}{2}k{x^2} = 0.9 \hfill \\
\therefore \,\,\frac{1}{2}K{x^2} = 0.6\,J\,\left( {total\,initial\,energy\,of\,the\,spring} \right) \hfill \\
\end{gathered} $
