A ball of mass `m_(1)` is moving with velocity 3v. It collides head on elastically with a stationary ball of mass `m_(2)` . The velocity of both the balls become v after collision. Then the value of the ratio `(m_(2))/( m_(1))` is

Options

A. (a)1
B. (b)2
C. (c)3
D. (d)4

Answer

To solve the problem, we will use the principles of conservation of momentum and the fact that the collision is elastic.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Mass of the first ball: \( m_1 \)
- Initial velocity of the first ball: \( 3v \)
- Mass of the second ball: \( m_2 \)
- Initial velocity of the second ball: \( 0 \) (stationary)

2. **Identify the Final Conditions**:
- After the collision, both balls have a final velocity of \( v \).

3. **Apply the Conservation of Momentum**:
The total momentum before the collision must equal the total momentum after the collision.
\[ \text{Initial Momentum} = \text{Final Momentum} \]
\[ m_1 \cdot 3v + m_2 \cdot 0 = (m_1 + m_2) \cdot v \]
Simplifying this, we have:
\[ 3m_1 v = (m_1 + m_2) v \]

4. **Cancel out the common factor \( v \)** (assuming \( v \neq 0 \)):
\[ 3m_1 = m_1 + m_2 \]

5. **Rearrange the equation to find \( m_2 \)**:
\[ 3m_1 - m_1 = m_2 \]
\[ 2m_1 = m_2 \]

6. **Calculate the ratio \( \frac{m_2}{m_1} \)**:
\[ \frac{m_2}{m_1} = \frac{2m_1}{m_1} = 2 \]

Thus, the ratio \( \frac{m_2}{m_1} \) is \( 2 \).

### Final Answer:
\[ \frac{m_2}{m_1} = 2 \]