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A body of mass $1\,kg$ falls freely from a height of $100\,m,$ on a platform of mass $3\,kg$ which is mounted on a spring having spring constant $k = 1.25 \times 10^6\, N/m.$ The body sticks to the platform and the spring’s maximum compression is found to be $x.$ Given that $g = 10\,ms^{-2},$ the value of $x$ will be close to ................ $\mathrm{cm}$
$40$
$4$
$80$
$2$
Solution
$\begin{array}{l}
v = \sqrt {2 \times 10} \times 100\\
\,\,\,\, = 20\sqrt 5 \\
COLM \to \,{v_p} = \frac{{1 \times 20\sqrt 5 }}{4}\\
COTME \to \\
\frac{1}{2} \times 4 \times {\left( {5\sqrt 5 } \right)^2} + \frac{1}{2}1.25 \times {10^6}{\left( {\frac{{30}}{k}} \right)^2}
\end{array}$
$\begin{array}{l}
= – 4 \times 10 \times \left( {n – \frac{{30}}{k}} \right) + \frac{1}{2}k{n^2}\\
\Rightarrow \,\,250 + \frac{{900}}{{2 \times 1.25 \times {{10}^6}}}\\
\,\, = – 40x + \frac{{1200}}{k} + \frac{1}{2}k{x^2}\\
\Rightarrow \,x = 2\,cm
\end{array}$
