A loaded spring gun of mass M fires a bullet of mass m with a velocity v at an angle of elevation `theta`. The gun is initially at rest on a horizontal smooth surface. After firing, the centre of mass of the gun and bullet system

Options

A. moves with a velocity `v(m)/(M)`
B. moves with a velocity `(vm)/(M) cos theta` in the horizontal direction
C. remains at rest
D. moves with a velocity `(v(M-m))/((M+m))` in the horizontal direction

Answer

To solve the problem of the loaded spring gun firing a bullet, we can analyze the situation using the principles of conservation of momentum and the concept of the center of mass.

### Step-by-Step Solution:

1. **Identify the System**:
We have a system consisting of a gun of mass \( M \) and a bullet of mass \( m \). Initially, the system is at rest.

2. **Initial Momentum**:
Since the gun and bullet are at rest, the initial momentum of the system is:
\[ P_{\text{initial}} = 0 \]

3. **Final Momentum After Firing**:
When the bullet is fired with a velocity \( v \) at an angle \( \theta \), the bullet has two components of velocity:
- Horizontal component: \( v_x = v \cos \theta \)
- Vertical component: \( v_y = v \sin \theta \)

The momentum of the bullet after firing is:
\[ P_{\text{bullet}} = m v_x = m v \cos \theta \]

By conservation of momentum, the momentum of the gun must be equal and opposite to that of the bullet. Let \( V_g \) be the velocity of the gun after firing. The momentum of the gun is:
\[ P_{\text{gun}} = -M V_g \]

According to the conservation of momentum:
\[ P_{\text{initial}} = P_{\text{bullet}} + P_{\text{gun}} \]
\[ 0 = m v \cos \theta - M V_g \]

4. **Solve for Gun's Velocity**:
Rearranging the equation gives us the velocity of the gun:
\[ M V_g = m v \cos \theta \]
\[ V_g = \frac{m v \cos \theta}{M} \]

5. **Center of Mass Velocity**:
The center of mass (CM) velocity of the system can be calculated using the formula:
\[ V_{\text{CM}} = \frac{m v + M (-V_g)}{m + M} \]
Substituting \( V_g \):
\[ V_{\text{CM}} = \frac{m v - M \left(\frac{m v \cos \theta}{M}\right)}{m + M} \]
Simplifying this:
\[ V_{\text{CM}} = \frac{m v (1 - \cos \theta)}{m + M} \]

6. **Conclusion**:
The center of mass of the gun and bullet system moves with a velocity given by:
\[ V_{\text{CM}} = \frac{m v (1 - \cos \theta)}{m + M} \]