Question 24 of 130

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 :

A$40 \mathrm{~cm}$B$4 \mathrm{~cm}$C$80 \mathrm{~cm}$D$8 \mathrm{~cm}$

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Answer: (B) $4 \mathrm{~cm}$

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 $x \approx 4 \mathrm{~cm}$