JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2019
- Hard
- Numerical
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 ................ \(\mathrm{cm}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
v = \sqrt {2 \times 10} \times 100\\
\,\,\,\, = 20\sqrt 5 \\
COLM \to \,{v_p} = \frac{{1 \times 20\sqrt 5 }}{4}\\
COTME \to \\
\frac{1}{2} \times 4 \times {\left( {5\sqrt 5 } \right)^2} + \frac{1}{2}1.25 \times {10^6}{\left( {\frac{{30}}{k}} \right)^2}
\end{array}\)
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}\)
\(\begin{array}{l}
= - 4 \times 10 \times \left( {n - \frac{{30}}{k}} \right) + \frac{1}{2}k{n^2}\\
\Rightarrow \,\,250 + \frac{{900}}{{2 \times 1.25 \times {{10}^6}}}\\
\,\, = - 40x + \frac{{1200}}{k} + \frac{1}{2}k{x^2}\\
\Rightarrow \,x = 2\,cm
\end{array}\)
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