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 glven as `f(1/2mv^2)` then `f=(m/(M+m))`
Statement II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

Options

A. Statement - I is true, Statement - II is true, Statement - II is not a correct explanation of Statement - I.
B. Statement – I is true, Statement – II is false.
C. Statement – I is false, Statement – II is true
D. Statement – I is true, Statement – II is true, Statement – II is a correct explanation of Statement – I.

Answer

To solve the problem, we need to analyze both statements regarding the collision of two particles and determine their validity.

### Step-by-Step Solution:

1. **Understanding the Collision**:
- We have a point particle of mass \( m \) moving with speed \( v \) colliding with a stationary point particle of mass \( M \).

2. **Conservation of Momentum**:
- Before the collision, the momentum of the system is given by:
\[ p_{\text{initial}} = mv + 0 = mv \]
- After the collision, if the two particles stick together (perfectly inelastic collision), they will move with a common velocity \( v' \). The total mass after the collision is \( m + M \), so the momentum after the collision is:
\[ p_{\text{final}} = (m + M)v' \]
- By conservation of momentum:
\[ mv = (m + M)v' \]
- Solving for \( v' \):
\[ v' = \frac{mv}{m + M} \]

3. **Calculating Initial and Final Kinetic Energy**:
- The initial kinetic energy \( KE_{\text{initial}} \) of the system is:
\[ KE_{\text{initial}} = \frac{1}{2}mv^2 \]
- The final kinetic energy \( KE_{\text{final}} \) after the collision is:
\[ KE_{\text{final}} = \frac{1}{2}(m + M)(v')^2 = \frac{1}{2}(m + M)\left(\frac{mv}{m + M}\right)^2 \]
- Simplifying \( KE_{\text{final}} \):
\[ KE_{\text{final}} = \frac{1}{2}(m + M)\frac{m^2v^2}{(m + M)^2} = \frac{1}{2}\frac{m^2v^2}{m + M} \]

4. **Calculating the Energy Loss**:
- The loss in kinetic energy \( \Delta KE \) is given by:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2}mv^2 - \frac{1}{2}\frac{m^2v^2}{m + M} \]
- Factor out \( \frac{1}{2}v^2 \):
\[ \Delta KE = \frac{1}{2}v^2\left(m - \frac{m^2}{m + M}\right) = \frac{1}{2}v^2\left(\frac{m(m + M) - m^2}{m + M}\right) = \frac{1}{2}v^2\left(\frac{mM}{m + M}\right) \]

5. **Finding the Value of \( f \)**:
- The maximum energy loss can be expressed as:
\[ \Delta KE = f\left(\frac{1}{2}mv^2\right) \]
- Setting the two expressions equal gives:
\[ f\left(\frac{1}{2}mv^2\right) = \frac{1}{2}v^2\left(\frac{mM}{m + M}\right) \]
- Solving for \( f \):
\[ f = \frac{mM}{m + M} \]

### Conclusion:
- **Statement I**: The statement claims that \( f = \frac{m}{M + m} \), which is incorrect because we found \( f = \frac{mM}{m + M} \).
- **Statement II**: This statement is true because maximum energy loss occurs during a perfectly inelastic collision when the particles stick together.

### Final Answer:
- Statement I is false, and Statement II is true.