Maximum energy transfer for an elastic collision will occur if one body is at rest when

Options

A. `m_(1)=m_(2)`
B. `m_(2)=(1)/(2) m_(1)`
C. `m_(1) gt gt m_(2)`
D. `m_(2) gt gt m_(1)`

Explanation

To solve the problem of maximum energy transfer during an elastic collision when one body is at rest, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Scenario**:
- We have two bodies: Body 1 (mass \( m_1 \)) is moving with velocity \( V_1 \) and Body 2 (mass \( m_2 \)) is at rest (velocity = 0).

2. **Apply the Concept of Elastic Collision**:
- In an elastic collision, both momentum and kinetic energy are conserved.
- The coefficient of restitution \( e \) for an elastic collision is equal to 1.

3. **Set Up the Equations**:
- The coefficient of restitution can be expressed as:
\[ e = \frac{V_1' - V_2'}{V_1 - V_2} \]
where \( V_1' \) and \( V_2' \) are the velocities after the collision.
- Since Body 2 is initially at rest, this simplifies to:
\[ 1 = \frac{V_1' - 0}{V_1 - 0} \implies V_1' = V_1 \]
- This indicates that if Body 2 moves after the collision, it can only do so if \( m_1 \) and \( m_2 \) are equal.

4. **Conservation of Momentum**:
- The principle of conservation of momentum states:
\[ m_1 V_1 + m_2 \cdot 0 = m_1 V_1' + m_2 V_2' \]
- Rearranging gives:
\[ m_1 V_1 = m_1 V_1' + m_2 V_2' \]

5. **Substituting Velocities**:
- If we assume maximum energy transfer occurs when \( m_1 = m_2 \), we can substitute \( m_2 \) with \( m_1 \):
\[ m_1 V_1 = m_1 V_1' + m_1 V_2' \]
- Dividing through by \( m_1 \) (assuming \( m_1 \neq 0 \)):
\[ V_1 = V_1' + V_2' \]

6. **Conclusion**:
- The maximum energy transfer occurs when the masses are equal, i.e., \( m_1 = m_2 \). Thus, the correct answer is that maximum energy transfer for an elastic collision will occur if one body is at rest when the masses are equal.

### Final Answer:
The maximum energy transfer for an elastic collision will occur if the two bodies have equal mass.