A gun weighing 10 kg fires a bullet of 30 g with a velocity of 330 `ms^(-1)` . With what velocity does the gun recoil? What is the combined momentum of the gun and bullet before firing and after firing?

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the gun, \( m_g = 10 \, \text{kg} \)
- Mass of the bullet, \( m_b = 30 \, \text{g} = 0.03 \, \text{kg} \)
- Velocity of the bullet after firing, \( v_b = 330 \, \text{m/s} \)

2. **Calculate Initial Momentum:**
- Before firing, both the gun and bullet are at rest, so the initial momentum \( p_{initial} \) is:
\[ p_{initial} = m_g \cdot 0 + m_b \cdot 0 = 0 \]

3. **Set Up the Conservation of Momentum Equation:**
- According to the conservation of momentum:
\[ p_{initial} = p_{final} \]
- The final momentum \( p_{final} \) can be expressed as:
\[ p_{final} = m_g \cdot v_g + m_b \cdot v_b \]
- Where \( v_g \) is the recoil velocity of the gun.

4. **Substituting Values into the Equation:**
- Since \( p_{initial} = 0 \):
\[ 0 = m_g \cdot v_g + m_b \cdot v_b \]
- Substituting the known values:
\[ 0 = 10 \cdot v_g + 0.03 \cdot 330 \]

5. **Solving for the Recoil Velocity \( v_g \):**
- Calculate the momentum of the bullet:
\[ 0.03 \cdot 330 = 9.9 \, \text{kg m/s} \]
- Now, substitute this back into the equation:
\[ 0 = 10 \cdot v_g + 9.9 \]
- Rearranging gives:
\[ 10 \cdot v_g = -9.9 \]
- Therefore, solving for \( v_g \):
\[ v_g = -\frac{9.9}{10} = -0.99 \, \text{m/s} \]

6. **Conclusion for Recoil Velocity:**
- The recoil velocity of the gun is \( v_g = -0.99 \, \text{m/s} \). The negative sign indicates that the gun recoils in the opposite direction to the bullet.

7. **Calculate Combined Momentum After Firing:**
- The combined momentum after firing is still equal to the initial momentum, which is:
\[ p_{final} = 0 \, \text{kg m/s} \]

### Final Answers:
- Recoil velocity of the gun: \( -0.99 \, \text{m/s} \)
- Combined momentum before firing: \( 0 \, \text{kg m/s} \)
- Combined momentum after firing: \( 0 \, \text{kg m/s} \)