A stone 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

Answer

To solve the problem, we will use the principle of conservation of momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the scenario
We have a stone of mass \( m_1 \) moving with a uniform speed \( v \). It suddenly explodes into two fragments: one fragment has mass \( m_2 \) and is at rest after the explosion, while the other fragment has mass \( m_1 - m_2 \).

### Step 2: Write down the conservation of momentum
According to the law of conservation of momentum, the total momentum before the explosion must equal the total momentum after the explosion.

Before the explosion, the momentum \( P_{\text{initial}} \) is:
\[ P_{\text{initial}} = m_1 \cdot v \]

After the explosion, the momentum \( P_{\text{final}} \) can be expressed as:
\[ P_{\text{final}} = m_2 \cdot 0 + (m_1 - m_2) \cdot v' \]
where \( v' \) is the speed of the fragment with mass \( m_1 - m_2 \).

### Step 3: Set up the equation
Setting the initial momentum equal to the final momentum, we have:
\[ m_1 \cdot v = (m_1 - m_2) \cdot v' \]

### Step 4: Solve for \( v' \)
To find the speed \( v' \) of the fragment with mass \( m_1 - m_2 \), we rearrange the equation:
\[ v' = \frac{m_1 \cdot v}{m_1 - m_2} \]

### Conclusion
Thus, the speed of the other fragment (the one with mass \( m_1 - m_2 \)) after the explosion is:
\[ v' = \frac{m_1 \cdot v}{m_1 - m_2} \]