Physics - Center of Mass Question with Solution | TestHub

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 $x\approx 4\mathrm{cm}$