A bullet leaves the rifle of mass one kg and rifle recoils thereby with velocity 30 cm/s. If the mass of the bullet is 3 g, the velocity of the bullet is

Options

A. `10^(4)" cm"//"s"`
B. `3xx10^(2)" cm"//"s"`
C. `10^(2)" cm"//"s"`
D. `10" cm"//"s"`

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. **Identify the Masses and Velocities**:
- Mass of the rifle (M_r) = 1 kg
- Mass of the bullet (M_b) = 3 g = 3 x 10^-3 kg (conversion from grams to kilograms)
- Velocity of the rifle after firing (V_r) = 30 cm/s = 0.3 m/s (conversion from cm/s to m/s)
- Velocity of the bullet (V_b) = ? (this is what we need to find)

2. **Apply the Conservation of Momentum**:
The total momentum before firing is equal to the total momentum after firing. Initially, both the rifle and bullet are at rest, so the initial momentum is 0.

\[ \text{Initial Momentum} = 0 \]
\[ \text{Final Momentum} = \text{Momentum of bullet} + \text{Momentum of rifle} \]
\[ 0 = M_b \cdot V_b + M_r \cdot (-V_r) \]
(The negative sign for the rifle's momentum indicates it recoils in the opposite direction to the bullet.)

3. **Substitute the Known Values**:
\[ 0 = (3 \times 10^{-3}) \cdot V_b + (1) \cdot (-0.3) \]
\[ 0 = (3 \times 10^{-3}) \cdot V_b - 0.3 \]

4. **Rearranging the Equation**:
\[ (3 \times 10^{-3}) \cdot V_b = 0.3 \]

5. **Solve for V_b**:
\[ V_b = \frac{0.3}{3 \times 10^{-3}} = \frac{0.3}{0.003} = 100 \text{ m/s} \]

6. **Convert to cm/s**:
Since the question asks for the velocity in cm/s:
\[ V_b = 100 \text{ m/s} = 10000 \text{ cm/s} \]

### Final Answer:
The velocity of the bullet is **10000 cm/s**.