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. `(m_(A)+m_(B))//m_(A)`
D. `v_(A)//v_(B)`

Answer

To solve the problem, we need to analyze the elastic collision of two masses \( m_A \) and \( m_B \) moving in opposite directions with velocities \( v_A \) and \( v_B \). After the collision, the masses switch their velocities. We need to find the ratio \( \frac{m_A}{m_B} \).

### Step-by-Step Solution:

1. **Understand the Momentum Conservation**:
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.

2. **Write the Initial Momentum**:
The initial momentum of the system before the collision can be expressed as:
\[ P_{\text{initial}} = m_A v_A - m_B v_B \]
(Here, we consider \( v_A \) as positive and \( v_B \) as negative since they are in opposite directions.)

3. **Write the Final Momentum**:
After the collision, the momentum of the system is:
\[ P_{\text{final}} = m_A v_B + m_B v_A \]
(Here, \( m_A \) moves with \( v_B \) and \( m_B \) moves with \( v_A \).)

4. **Set Initial Momentum Equal to Final Momentum**:
According to the conservation of momentum:
\[ m_A v_A - m_B v_B = m_A v_B + m_B v_A \]

5. **Rearrange the Equation**:
Rearranging the equation gives:
\[ m_A v_A - m_A v_B = m_B v_A + m_B v_B \]
\[ m_A (v_A - v_B) = m_B (v_A + v_B) \]

6. **Solve for the Ratio \( \frac{m_A}{m_B} \)**:
Dividing both sides by \( m_B (v_A - v_B) \) gives:
\[ \frac{m_A}{m_B} = \frac{v_A + v_B}{v_A - v_B} \]

7. **Simplifying the Ratio**:
Since we know that after the collision \( m_A \) moves with \( v_B \) and \( m_B \) moves with \( v_A \), we can conclude that:
\[ \frac{m_A}{m_B} = 1 \]

### Final Answer:
Thus, the ratio \( \frac{m_A}{m_B} \) is:
\[ \frac{m_A}{m_B} = 1 \]