A man fires a bullet of mass 200 g at a speed of 5 `m //s` . The gun is of one kg mass. By what velocity the gun rebounds backwards

Options

A. `0.1m//s`
B. `10m//s`
C. `1m//s`
D. `0.01m//s`

Answer

To solve the problem of the gun's recoil when a bullet is fired, we can use the principle of conservation of momentum. Here’s a step-by-step solution:

### Step 1: Understand the System
When the bullet is fired, the system consists of the bullet and the gun. Initially, both the bullet and the gun are at rest, so the total initial momentum of the system is zero.

### Step 2: Define the Masses
- Mass of the bullet, \( m_b = 200 \, \text{g} = 0.2 \, \text{kg} \) (since 1 g = 0.001 kg)
- Mass of the gun, \( m_g = 1 \, \text{kg} \)

### Step 3: Define the Speeds
- Speed of the bullet after firing, \( v_b = 5 \, \text{m/s} \)
- Let the recoil speed of the gun be \( v_g \) (this will be a negative value since it moves in the opposite direction of the bullet).

### Step 4: Apply Conservation of Momentum
According to the conservation of momentum:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
Since the initial momentum is zero, we have:
\[ 0 = m_b \cdot v_b + m_g \cdot v_g \]

### Step 5: Substitute the Values
Substituting the known values into the equation:
\[ 0 = (0.2 \, \text{kg}) \cdot (5 \, \text{m/s}) + (1 \, \text{kg}) \cdot v_g \]

### Step 6: Solve for \( v_g \)
Rearranging the equation gives:
\[ (1 \, \text{kg}) \cdot v_g = - (0.2 \, \text{kg}) \cdot (5 \, \text{m/s}) \]
\[ v_g = - \frac{(0.2 \cdot 5)}{1} \]
\[ v_g = -1 \, \text{m/s} \]

### Conclusion
The gun rebounds backwards with a velocity of \( 1 \, \text{m/s} \).

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