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

Options

A. Statement-1 is True, Statement-2 is True and Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True and Statement-2 is NOT a correct explanation for Stateme
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

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

### Step 1: Understand the scenario
We have two point particles:
- Particle 1: Mass = \( m \), moving with speed = \( v \)
- Particle 2: Mass = \( M \), stationary (speed = 0)

### Step 2: Conservation of Momentum
In a collision, the total momentum before the collision equals the total momentum after the collision. Thus, we can write:
\[ \text{Initial momentum} = \text{Final momentum} \]
\[ mv + 0 = (m + M)v_1 \]
where \( v_1 \) is the final velocity of both particles after the collision (assuming they stick together).

### Step 3: Solve for \( v_1 \)
From the conservation of momentum equation:
\[ mv = (m + M)v_1 \]
Rearranging gives:
\[ v_1 = \frac{mv}{m + M} \]

### Step 4: Calculate Initial and Final Kinetic Energy
**Initial Kinetic Energy (KE_initial)**:
\[ KE_{\text{initial}} = \frac{1}{2} mv^2 \]

**Final Kinetic Energy (KE_final)** (after they stick together):
\[ KE_{\text{final}} = \frac{1}{2}(m + M)v_1^2 = \frac{1}{2}(m + M)\left(\frac{mv}{m + M}\right)^2 \]
Calculating \( KE_{\text{final}} \):
\[ KE_{\text{final}} = \frac{1}{2}(m + M) \cdot \frac{m^2v^2}{(m + M)^2} = \frac{1}{2} \cdot \frac{m^2v^2}{m + M} \]

### Step 5: Calculate the Change in Kinetic Energy
The change in kinetic energy (energy loss) is:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
Substituting the expressions we derived:
\[ \Delta KE = \frac{1}{2} \cdot \frac{m^2v^2}{m + M} - \frac{1}{2} mv^2 \]
Factoring out \( \frac{1}{2}v^2 \):
\[ \Delta KE = \frac{1}{2}v^2\left(\frac{m^2}{m + M} - m\right) \]
Finding a common denominator:
\[ \Delta KE = \frac{1}{2}v^2\left(\frac{m^2 - m(m + M)}{m + M}\right) = \frac{1}{2}v^2\left(\frac{m^2 - m^2 - mM}{m + M}\right) = -\frac{1}{2}\frac{mM}{m + M}v^2 \]

### Step 6: Identify Maximum Energy Loss
The maximum energy loss occurs when the collision is perfectly inelastic. The expression for the maximum energy loss can be written as:
\[ F = \frac{1}{2} \frac{mM}{m + M} v^2 \]
From the problem statement, if \( F = \frac{1}{2} mf v^2 \), we can equate:
\[ f = \frac{M}{m + M} \]

### Step 7: Analyze the Statements
- **Statement 1**: \( f = \frac{m}{M + m} \) is incorrect because we derived \( f = \frac{M}{m + M} \).
- **Statement 2**: The maximum energy loss occurs when the particles stick together, which is true for perfectly inelastic collisions.

### Conclusion
- **Statement 1** is false.
- **Statement 2** is true.

### Final Answer
The correct option is that Statement 1 is false and Statement 2 is true.

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