A srtone of mass `m_(1)` moving at uniform speed v suddenly explodes into two fragments. If the fragment of mass `m_(2)` is at rest, the speed of the other fragment is

Options

A. `(m_(1)v)/((m_(1)-m_(2)))`
B. `(m_(2)v)/(m_(1)-m_(2))`
C. `(m_(1)v)/((m_(1)+m_(2)))`
D. `(m_(1)v)/(m_(2))`

Explanation

To solve the problem of a stone of mass \( m_1 \) moving at a uniform speed \( v \) that suddenly explodes into two fragments, where one fragment of mass \( m_2 \) comes to rest, we will use the principle of conservation of momentum.

### Step-by-Step Solution:

1. **Understand the Initial Momentum**:
The initial momentum of the stone before the explosion can be calculated using the formula:
\[ p_{\text{initial}} = m_1 \cdot v \]

2. **Define the Masses After Explosion**:
After the explosion, the stone splits into two fragments:
- Fragment 1 with mass \( m_2 \) (which is at rest, so its velocity \( v_2 = 0 \))
- Fragment 2 with mass \( m_1 - m_2 \) (let's denote its velocity as \( v' \))

3. **Calculate the Final Momentum**:
The total momentum after the explosion must equal the total momentum before the explosion (conservation of momentum). Therefore, the final momentum can be expressed as:
\[ p_{\text{final}} = m_2 \cdot 0 + (m_1 - m_2) \cdot v' = (m_1 - m_2) \cdot v' \]

4. **Set Up the Conservation of Momentum Equation**:
According to the conservation of momentum:
\[ p_{\text{initial}} = p_{\text{final}} \]
Substituting the expressions we derived:
\[ m_1 \cdot v = (m_1 - m_2) \cdot v' \]

5. **Solve for the Velocity of the Other Fragment**:
Rearranging the equation to solve for \( v' \):
\[ v' = \frac{m_1 \cdot v}{m_1 - m_2} \]

### Final Answer:
The speed of the other fragment (mass \( m_1 - m_2 \)) after the explosion is:
\[ v' = \frac{m_1 \cdot v}{m_1 - m_2} \]