A bullet moving with a speed of 400 m/s penetrates a sand bag and drops dead inside it. The masses of the bullet and the sand bag are 0.25 kg and 4.75 kg respectively. If the bag is free to move, its velocity shall be (in m/s)

Options

A. 10
B. 15
C. 21
D. 50

Answer

To solve the problem, we will use the principle of conservation of momentum. The momentum before the bullet hits the sandbag must equal the momentum after the bullet becomes embedded in the sandbag.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the bullet (m1) = 0.25 kg
- Speed of the bullet (u1) = 400 m/s
- Mass of the sandbag (m2) = 4.75 kg
- Initial speed of the sandbag (u2) = 0 m/s (since it is at rest)

2. **Write the conservation of momentum equation:**
The total momentum before the collision is equal to the total momentum after the collision.
\[ m_1 \cdot u_1 + m_2 \cdot u_2 = (m_1 + m_2) \cdot v \]
where \(v\) is the final velocity of the combined system (bullet + sandbag).

3. **Substitute the known values into the equation:**
\[ (0.25 \, \text{kg}) \cdot (400 \, \text{m/s}) + (4.75 \, \text{kg}) \cdot (0 \, \text{m/s}) = (0.25 \, \text{kg} + 4.75 \, \text{kg}) \cdot v \]

4. **Calculate the left side of the equation:**
\[ 0.25 \cdot 400 + 0 = 100 \, \text{kg m/s} \]

5. **Combine the masses on the right side:**
\[ (0.25 + 4.75) \cdot v = 5 \cdot v \]

6. **Set the two sides of the equation equal to each other:**
\[ 100 = 5 \cdot v \]

7. **Solve for \(v\):**
\[ v = \frac{100}{5} = 20 \, \text{m/s} \]

### Final Answer:
The velocity of the sandbag after the bullet penetrates it is **20 m/s**.