Given

• A particle of mass $m$ moves with speed $v$.
• It collides with a stationary particle of mass $M$.
• Initial kinetic energy: $K_i=\frac12 mv^2$

When is energy loss maximum? For a given initial momentum, the final kinetic energy is minimum when both particles move together with the same velocity after collision, i.e. in a perfectly inelastic collision.

So Statement II is true.

Find the common velocity after sticking By conservation of momentum, $mv=(m+M)V$ so $V=\frac{mv}{m+M}$

Final kinetic energy in this case $K_f=\frac12 (m+M)V^2$ Substitute $V$: $K_f=\frac12 (m+M)\left(\frac{mv}{m+M}\right)^2$ $K_f=\frac12 \frac{m^2v^2}{m+M}$

Maximum loss of kinetic energy $\Delta K_{\max}=K_i-K_f$ $=\frac12 mv^2-\frac12 \frac{m^2v^2}{m+M}$ $=\frac12 mv^2\left(1-\frac{m}{m+M}\right)$ $=\frac12 mv^2\left(\frac{M}{m+M}\right)$

Hence, $f=\frac{M}{M+m}$

Check Statement I Statement I claims: $f=\frac{m}{M+m}$ But we found: $f=\frac{M}{M+m}$ Therefore Statement I is false.

Evaluate the options

• Statement I: False
• Statement II: True

Therefore the correct option is: $\boxed{\text{D}}$

Comparison with stored answer Stored correct answer: D

My derived answer matches the stored answer.