This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement - I: A point particle of mass $m$ moving with speed $v$ collides with stationary point particle of mass $M$. If the maximum energy loss possible is given as $f(\frac{1}{2}mv^2)$ then $f=(m/(M+m))$. Statement - II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

Step 1: Analyze Statement II and the condition for maximum energy loss.

The maximum loss of kinetic energy in a collision occurs when the collision is perfectly inelastic, meaning the particles stick together after the collision.

Thus, Statement II is true.

Step 2: Apply conservation of linear momentum for a perfectly inelastic collision.

Let $u$ be the common velocity of the combined mass $(m + M)$ after the collision.

$$mv = (m + M)u$$ $$u = \frac{mv}{m + M}$$

Step 3: Calculate the final kinetic energy.

Substitute the expression for $u$ into the kinetic energy formula.

$$KE_f = \frac{1}{2}(m + M)u^2$$ $$KE_f = \frac{1}{2}(m + M)\left(\frac{mv}{m + M}\right)^2$$ $$KE_f = \left(\frac{1}{2}mv^2\right)\left(\frac{m}{m + M}\right)$$

Step 4: Calculate the maximum energy loss.

The maximum energy loss $\Delta KE_{\max}$ is the difference between the initial and final kinetic energies.

$$\Delta KE_{\max} = KE_i - KE_f$$ $$\Delta KE_{\max} = \frac{1}{2}mv^2 - \left(\frac{1}{2}mv^2\right)\left(\frac{m}{m + M}\right)$$ $$\Delta KE_{\max} = \left(\frac{1}{2}mv^2\right)\left(1 - \frac{m}{m + M}\right)$$ $$\Delta KE_{\max} = \left(\frac{1}{2}mv^2\right)\left(\frac{M}{m + M}\right)$$

Step 5: Determine the fraction $f$ and evaluate Statement I.

Comparing the maximum energy loss to the given expression $f\left(\frac{1}{2}mv^2\right)$ yields the value of $f$.

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

Since Statement I claims $f = \frac{m}{M+m}$, Statement I is false.

Answer: Statement - I is false, Statement - II is true.