A 30 g bullet leaves a rifle with a velocity of 300 `ms^(-1)` and the rifle recoils with a velocity of `0.60ms^(-1)`. Find the mass of the rifle.

Answer

To solve the problem, we will use the principle of conservation of momentum. The momentum before the bullet is fired is equal to the momentum after the bullet is fired.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Mass of the bullet, \( m_b = 30 \, \text{g} = 0.030 \, \text{kg} \) (conversion from grams to kilograms)
- Velocity of the bullet, \( v_b = 300 \, \text{m/s} \)
- Recoil velocity of the rifle, \( v_r = 0.60 \, \text{m/s} \)

2. **Set Up the Conservation of Momentum Equation:**
- The total momentum before firing is zero (since both the bullet and rifle are at rest).
- The total momentum after firing can be expressed as:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
\[ 0 = m_b \cdot v_b + m_r \cdot (-v_r) \]
Here, \( m_r \) is the mass of the rifle, and we take the recoil velocity as negative because it is in the opposite direction to the bullet.

3. **Substituting the Known Values:**
\[ 0 = (0.030 \, \text{kg}) \cdot (300 \, \text{m/s}) + m_r \cdot (-0.60 \, \text{m/s}) \]
This simplifies to:
\[ 0 = 9 \, \text{kg m/s} - 0.60 \, m_r \]

4. **Rearranging the Equation to Solve for \( m_r \):**
\[ 0.60 \, m_r = 9 \, \text{kg m/s} \]
\[ m_r = \frac{9 \, \text{kg m/s}}{0.60 \, \text{m/s}} = 15 \, \text{kg} \]

5. **Conclusion:**
The mass of the rifle is \( m_r = 15 \, \text{kg} \).