A body of mass `M_(1)` collides elastically with another mass `M_(2)` at rest. There is maximum transfer of energy when :

Options

A. `M_(1)gtM_(2)`
B. `M_(1)ltM_(2)`
C. `M_(1)=M_(2)`
D. same for all values of `M_(1)` and `M_(2)`

Explanation

To determine when there is maximum transfer of energy during an elastic collision between two bodies, we can analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding Elastic Collision**: In an elastic collision, both momentum and kinetic energy are conserved. We have two masses, \( M_1 \) (moving) and \( M_2 \) (at rest).

2. **Initial Conditions**: Let the initial velocity of mass \( M_1 \) be \( V \) and mass \( M_2 \) be at rest, so its initial velocity is \( 0 \).

3. **Final Velocities After Collision**: After the elastic collision, let the final velocities of \( M_1 \) and \( M_2 \) be \( V_1' \) and \( V_2' \) respectively.

4. **Conservation of Momentum**: The total momentum before and after the collision must be equal:
\[ M_1 \cdot V = M_1 \cdot V_1' + M_2 \cdot V_2' \]

5. **Conservation of Kinetic Energy**: The total kinetic energy before and after the collision must also be equal:
\[ \frac{1}{2} M_1 V^2 = \frac{1}{2} M_1 (V_1')^2 + \frac{1}{2} M_2 (V_2')^2 \]

6. **Maximum Transfer of Energy**: For maximum transfer of kinetic energy from \( M_1 \) to \( M_2 \), we want \( M_1 \) to stop completely after the collision, meaning \( V_1' = 0 \). This scenario occurs when the masses are equal:
\[ M_1 = M_2 \]

7. **Conclusion**: When \( M_1 \) and \( M_2 \) are equal, the entire velocity \( V \) of \( M_1 \) is transferred to \( M_2 \), resulting in maximum kinetic energy transfer. Thus, the condition for maximum energy transfer is:
\[ M_1 = M_2 \]

### Final Answer:
The maximum transfer of energy occurs when \( M_1 = M_2 \).