A `40kg` shell is flying at a speed of `72km//h` It explodes into two pieces One piece of mass `15kg` just stops . What is the speed of the other ?

Explanation

To solve the problem, we will use the principle of conservation of momentum. The momentum before the explosion must equal the total momentum after the explosion.

### Step-by-Step Solution:

1. **Convert the initial speed from km/h to m/s**:
\[ \text{Speed} = 72 \text{ km/h} = \frac{72}{3.6} = 20 \text{ m/s} \]

2. **Identify the masses and velocities**:
- Total mass of the shell, \( M = 40 \text{ kg} \)
- Mass of the first piece after the explosion, \( m_1 = 15 \text{ kg} \)
- Mass of the second piece, \( m_2 = M - m_1 = 40 \text{ kg} - 15 \text{ kg} = 25 \text{ kg} \)
- Initial velocity of the shell, \( u = 20 \text{ m/s} \)
- Final velocity of the first piece after the explosion, \( v_1 = 0 \text{ m/s} \) (it just stops)
- Final velocity of the second piece after the explosion, \( v_2 \) (to be determined)

3. **Apply the conservation of momentum**:
\[ \text{Initial momentum} = \text{Final momentum} \]
\[ M \cdot u = m_1 \cdot v_1 + m_2 \cdot v_2 \]
Substituting the known values:
\[ 40 \cdot 20 = 15 \cdot 0 + 25 \cdot v_2 \]
\[ 800 = 0 + 25 \cdot v_2 \]

4. **Solve for \( v_2 \)**:
\[ 800 = 25 \cdot v_2 \]
\[ v_2 = \frac{800}{25} = 32 \text{ m/s} \]

### Final Answer:
The speed of the other piece is \( 32 \text{ m/s} \).

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