Two masses `m_(A)` and `m_(B)` moving with velocities `v_(A)` and `v_(B)` in opposite direction collide elastically after that the masses `m_(A)` and `m_(B)` move with velocity `v_(B)` and `v_(A)` respectively. The ratio `(m_(A)//m_(B))` is

Options

A. 1
B. `(V_A - V_B)/(V_A + V_B)`
C. `(3V_A)/(2V_B)`
D. `V_A/V_B`

Answer

To solve the problem, we will use the principles of conservation of momentum and the properties of elastic collisions.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two masses, \( m_A \) and \( m_B \), moving in opposite directions with velocities \( v_A \) and \( v_B \) respectively.
- After the elastic collision, the masses switch their velocities: \( m_A \) moves with velocity \( v_B \) and \( m_B \) moves with velocity \( v_A \).

2. **Setting Up the Momentum Conservation Equation**:
- Since the collision is elastic, we can apply the conservation of linear momentum.
- The initial momentum of the system before the collision is:
\[ P_{\text{initial}} = m_A v_A - m_B v_B \]
(Note: \( v_B \) is negative because it is in the opposite direction).
- The final momentum of the system after the collision is:
\[ P_{\text{final}} = -m_A v_B + m_B v_A \]
(Here, \( -m_A v_B \) is because \( m_A \) is now moving in the opposite direction).

3. **Applying the Conservation of Momentum**:
- Setting the initial momentum equal to the final momentum:
\[ m_A v_A - m_B v_B = -m_A v_B + m_B v_A \]

4. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ m_A v_A + m_A v_B = m_B v_A + m_B v_B \]
- Factoring out \( m_A \) and \( m_B \):
\[ m_A (v_A + v_B) = m_B (v_A + v_B) \]

5. **Dividing Both Sides**:
- Assuming \( v_A + v_B \neq 0 \) (which is valid since both velocities are positive), we can divide both sides by \( (v_A + v_B) \):
\[ m_A = m_B \]

6. **Finding the Ratio**:
- Therefore, the ratio of the masses is:
\[ \frac{m_A}{m_B} = 1 \]

### Conclusion:
The ratio \( \frac{m_A}{m_B} \) is \( 1 \).