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. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we will use the principles of conservation of momentum and conservation of kinetic energy, which hold true for elastic collisions.

### Step-by-Step Solution:

1. **Identify 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. **Write the Equation for Conservation of Momentum**:
The total momentum before the collision must equal the total momentum after the collision.
\[ \text{Initial Momentum} = \text{Final Momentum} \]
\[ m_1 \cdot 3v + m_2 \cdot 0 = m_1 \cdot v + m_2 \cdot v \]
Simplifying this gives:
\[ 3m_1v = m_1v + m_2v \]

3. **Rearranging the Equation**:
We can rearrange the equation to isolate \( m_2 \):
\[ 3m_1v - m_1v = m_2v \]
\[ 2m_1v = m_2v \]

4. **Canceling Out the Common Terms**:
Since \( v \) is not zero, we can divide both sides by \( v \):
\[ 2m_1 = m_2 \]

5. **Finding the Ratio \( \frac{m_2}{m_1} \)**:
To find the ratio of \( m_2 \) to \( m_1 \):
\[ \frac{m_2}{m_1} = 2 \]

### Final Answer:
The value of the ratio \( \frac{m_2}{m_1} \) is \( 2 \).

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