A man fires a bullet of mas 20g at a speed of 100 m/s. The gun is 2 kg mass. By what velocity the gun rebounds backward?

Options

A. 2 m/s
B. 1 m/s
C. 5 m/s
D. 6 m/s

Explanation

To solve the problem of the gun's recoil velocity when a bullet is fired, we will use the principle of conservation of momentum. Here are the steps to find the solution:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the bullet (m_bullet) = 20 g = 0.02 kg (conversion from grams to kilograms)
- Speed of the bullet (v_bullet) = 100 m/s
- Mass of the gun (m_gun) = 2 kg

2. **Understand the Conservation of Momentum:**
According to the law of conservation of momentum, the total momentum before firing the bullet is equal to the total momentum after firing the bullet. Initially, both the gun and bullet are at rest, so the initial momentum is zero.

\[ \text{Initial momentum} = 0 \]

After firing, the momentum of the bullet and the gun can be expressed as:
\[ \text{Final momentum} = m_bullet \cdot v_bullet + m_gun \cdot v_gun \]

Where \( v_gun \) is the recoil velocity of the gun that we need to find.

3. **Set Up the Equation:**
Since the initial momentum is zero, we can write:
\[ 0 = m_bullet \cdot v_bullet + m_gun \cdot v_gun \]

Rearranging the equation gives:
\[ m_gun \cdot v_gun = - m_bullet \cdot v_bullet \]

4. **Substitute the Known Values:**
Plugging in the values we have:
\[ 2 \cdot v_gun = - (0.02 \cdot 100) \]

This simplifies to:
\[ 2 \cdot v_gun = -2 \]

5. **Solve for \( v_gun \):**
Dividing both sides by 2:
\[ v_gun = -1 \text{ m/s} \]

The negative sign indicates that the gun moves in the opposite direction to the bullet.

### Final Answer:
The recoil velocity of the gun is **1 m/s backward**.

---