Statement-I: A point particle of mass m moving with speed v...

Initial kinetic energy (K_i=$\frac{1}{2}$m v²) If after collision the two masses stick (perfectly inelastic), common velocity by momentum conservation is (V=$\frac{m v}{m+M}$ Final kinetic energy then is (K_f=$\frac{1}{2}($m+M) V²=$\frac{1}{2}($m+M) $\frac{m^2 v^2}{(m+M)^2}=\frac{1}{2} \frac{m^2 v^2}{m+M}$ Energy lost: ($\Delta K=K_i-K_f=$\frac{1}{2}$m v²-$\frac{1}{2} \frac{m^2 v^2}{m+M}=\frac{1}{2}$m v²$\left(1-\frac{m}{m+M}\right)=\frac{1}{2}m v² $\cdot \frac{M}{m+M}$ So the maximum fractional loss $f$ (such that ($\Delta K=f$\cdot \frac{1}{2}$m v²)) is (f=$\frac{M}{m+M}$Thus Statement-I (which asserts (f=$\frac{m}{m+M}) is wrong. Statement-II is correct because the maximum kinetic energy loss occurs for the perfectly inelastic case (particles stick, $e=0$).