JEE Advanced 2021 Paper 2 Physics: Paragraph 1

Just before the collision, $\omega=\sqrt{k/m}=10\ \mathrm{rad\,s^{-1}}$ and at $x=5$ cm the speed is $v=\omega\sqrt{A^2-x^2}=10\times0.0866=0.866\ \mathrm{m\,s^{-1}}$. The putty has no horizontal momentum, so horizontal momentum conservation gives $v'=\tfrac{2}{4}v=0.433\ \mathrm{m\,s^{-1}}$, while the position $x=5$ cm is unchanged. The new angular frequency is $\omega'=\sqrt{200/4}=7.07\ \mathrm{rad\,s^{-1}}$. The new total energy is $\tfrac12kx^2+\tfrac12Mv'^2=0.25+0.375=0.625\ \mathrm{J}$, and equating to $\tfrac12kA'^2$ gives $A'=\sqrt{2\times0.625/200}=0.0791\ \mathrm{m}=7.91\ \mathrm{cm}$. The value $7.07$ cm results from wrongly assuming the collision happens at the mean position, and $8.66$ cm is the pre-collision speed-based amplitude $\sqrt{A^2-x^2}$.

After the collision the system is a mass $M=4\ \mathrm{kg}$ on the same spring, with total mechanical energy $E=\tfrac12kx^2+\tfrac12Mv'^2$ where $x=0.05\ \mathrm{m}$ and $v'=0.433\ \mathrm{m\,s^{-1}}$. Thus $E=\tfrac12(200)(0.0025)+\tfrac12(4)(0.1875)=0.25+0.375=0.625\ \mathrm{J}$. The block is fastest at the equilibrium position, where all this energy is kinetic: $\tfrac12Mv_{max}^2=0.625$, giving $v_{max}=\sqrt{2\times0.625/4}=\sqrt{0.3125}=0.559\ \mathrm{m\,s^{-1}}$. Equivalently $v_{max}=\omega'A'=7.07\times0.0791=0.559\ \mathrm{m\,s^{-1}}$. The value $0.43$ is the speed immediately after the collision, which is not the maximum, and $0.87$ is the speed before sticking. Hence option (C) is the only one consistent with energy conservation after the sticking collision; the energy lost in the inelastic collision is irrelevant for what follows.