A body of mass 1 ~kg falls freely from a height of 100 ~m , on a...

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}