The bullet of 5 g is shot from a gun of mass 5 kg. The muzzle velocity of the bullet is 500 `m//s`.The recoil velocity of the gun is

Options

A. `0.5m//s`
B. `0.25m//s`
C. `1m//s`
D. Data is insufficient

Answer

To find the recoil velocity of the gun when a bullet is fired, we can use the principle of conservation of momentum. According to this principle, the total momentum before firing must equal the total momentum after firing.

### Step-by-Step Solution:

1. **Identify the masses and velocities:**
- Mass of the bullet, \( m_b = 5 \, \text{g} = 5 \times 10^{-3} \, \text{kg} \)
- Mass of the gun, \( m_g = 5 \, \text{kg} \)
- Muzzle velocity of the bullet, \( v_b = 500 \, \text{m/s} \)
- Recoil velocity of the gun, \( v_g \) (this is what we need to find).

2. **Apply the conservation of momentum:**
The total momentum before firing (when both the gun and bullet are at rest) is zero. After firing, the momentum of the bullet and the gun must still equal zero:
\[ m_b \cdot v_b + m_g \cdot v_g = 0 \]

3. **Rearranging the equation:**
Since the gun recoils in the opposite direction to the bullet, we can express the equation as:
\[ m_b \cdot v_b = - m_g \cdot v_g \]
This implies:
\[ v_g = -\frac{m_b \cdot v_b}{m_g} \]

4. **Substituting the known values:**
\[ v_g = -\frac{(5 \times 10^{-3} \, \text{kg}) \cdot (500 \, \text{m/s})}{5 \, \text{kg}} \]

5. **Calculating the recoil velocity:**
\[ v_g = -\frac{(5 \times 10^{-3}) \cdot 500}{5} \]
\[ v_g = -\frac{2.5}{5} \]
\[ v_g = -0.5 \, \text{m/s} \]

6. **Conclusion:**
The recoil velocity of the gun is \( 0.5 \, \text{m/s} \) in the opposite direction to the bullet.