A bullet of mass `20g` is fired horizontly with a velocity of `150ms^(-1)` from a pistol of maass `2kg`. What is the recoil velocity of the pistol?

Options

A. 1.5 m/s
B. 4 m/s
C. -1.5 m/s
D. -4 m/s

Explanation

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

### Step-by-Step Solution:

1. **Identify the masses and velocities:**
- Mass of the bullet, \( m_1 = 20 \, \text{g} = 20 \times 10^{-3} \, \text{kg} = 0.02 \, \text{kg} \)
- Mass of the pistol, \( m_2 = 2 \, \text{kg} \)
- Velocity of the bullet, \( v_1 = 150 \, \text{m/s} \)
- Recoil velocity of the pistol, \( v_2 = ? \)

2. **Apply the conservation of momentum:**
The total momentum before firing is zero (since both the bullet and the pistol are at rest). After firing, the momentum of the bullet and the pistol must still sum to zero:
\[ m_1 v_1 + m_2 v_2 = 0 \]

3. **Rearranging the equation to solve for \( v_2 \):**
\[ m_1 v_1 = -m_2 v_2 \]
\[ v_2 = -\frac{m_1 v_1}{m_2} \]

4. **Substituting the known values:**
\[ v_2 = -\frac{(0.02 \, \text{kg}) \times (150 \, \text{m/s})}{2 \, \text{kg}} \]

5. **Calculating the value:**
\[ v_2 = -\frac{3}{2} = -1.5 \, \text{m/s} \]

6. **Conclusion:**
The negative sign indicates that the direction of the recoil velocity is opposite to the direction of the bullet's velocity. Therefore, the recoil velocity of the pistol is \( 1.5 \, \text{m/s} \) in the opposite direction to that of the bullet.

### Final Answer:
The recoil velocity of the pistol is \( 1.5 \, \text{m/s} \) (opposite direction to the bullet).

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