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A body of mass $1 \mathrm{~kg}$ falls freely from a height of $100 \mathrm{~m}$, on a platform of mass $3 \mathrm{~kg}$ which is mounted on a spring having spring constant $\mathrm{k}=1.25 \times 10^{6} \mathrm{~N} / \mathrm{m}$. The bodysticks to the platform and the spring's maximum compression is found to be $x$. Given that $g=10 \mathrm{~ms}^{-2},$ the value of $\mathrm{x}$ will be close to :
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The correct answer is:
$4 \mathrm{~cm}$
Velocity of $1 \mathrm{~kg}$ block just before it collides with $3 \mathrm{~kg}$
block $=\sqrt{2 \mathrm{gh}}=\sqrt{2000} \mathrm{~m} / \mathrm{s}$
Using principle of conservation of linear momentum just before and just after collision, we get
$1 \times \sqrt{2000}=4 \mathrm{v} \Rightarrow \mathrm{v}=\frac{\sqrt{2000}}{4} \mathrm{~m} / \mathrm{s}$
Initial compression of spring $1.25 \times 10^{6} \mathrm{x}_{0}=30 \Rightarrow \mathrm{x}_{0} \approx 0$
using work energy theorem, $\mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{sp}}=\Delta \mathrm{KE}$
$\Rightarrow 40 \times x+\frac{1}{2} \times 1.25 \times 10^{6}\left(0^{2}-x^{2}\right)$
$=0-\frac{1}{2} \times 4 \times v^{2}$
solving $x \approx 4 \mathrm{~cm}$
block $=\sqrt{2 \mathrm{gh}}=\sqrt{2000} \mathrm{~m} / \mathrm{s}$
Using principle of conservation of linear momentum just before and just after collision, we get
$1 \times \sqrt{2000}=4 \mathrm{v} \Rightarrow \mathrm{v}=\frac{\sqrt{2000}}{4} \mathrm{~m} / \mathrm{s}$
Initial compression of spring $1.25 \times 10^{6} \mathrm{x}_{0}=30 \Rightarrow \mathrm{x}_{0} \approx 0$
using work energy theorem, $\mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{sp}}=\Delta \mathrm{KE}$
$\Rightarrow 40 \times x+\frac{1}{2} \times 1.25 \times 10^{6}\left(0^{2}-x^{2}\right)$
$=0-\frac{1}{2} \times 4 \times v^{2}$
solving $x \approx 4 \mathrm{~cm}$
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