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. (a) moves with velocity `v/Mm`
B. (b) moves with velocity `(vm)/(mcostheta)` in the horizontal direction
C. (c) does not move in horizontal direction
D. (d) moves with velocity `(v(M-m))/(M+m)` in the horizontal direction

Explanation

To solve the problem, we will analyze the situation using the principles of conservation of momentum and the definition of the center of mass.

### Step-by-Step Solution:

1. **Understand the System**:
- We have a spring gun of mass \( M \) that fires a bullet of mass \( m \) at an angle \( \theta \) with an initial velocity \( v \).
- The gun is initially at rest on a smooth horizontal surface.

2. **Identify the Directions**:
- The bullet is fired at an angle \( \theta \), which means it has both horizontal and vertical components of velocity.
- The horizontal component of the bullet's velocity is \( v \cos \theta \).

3. **Conservation of Momentum**:
- Since there are no external horizontal forces acting on the system, the horizontal component of momentum before and after firing must be conserved.
- Initially, the momentum of the system (gun + bullet) is zero because both are at rest.

4. **Calculate the Momentum After Firing**:
- After firing, the momentum of the bullet in the horizontal direction is \( m(v \cos \theta) \).
- Let \( V_d \) be the recoil velocity of the gun in the horizontal direction. The momentum of the gun after firing is \( -M V_d \) (the negative sign indicates it moves in the opposite direction to the bullet).

5. **Set Up the Conservation Equation**:
- According to the conservation of momentum:
\[ 0 = m(v \cos \theta) - M V_d \]
- Rearranging gives:
\[ M V_d = m(v \cos \theta) \]
- Thus, the recoil velocity of the gun is:
\[ V_d = \frac{m(v \cos \theta)}{M} \]

6. **Determine the Center of Mass Velocity**:
- The velocity of the center of mass \( V_{cm} \) of the system (gun + bullet) can be calculated using the formula:
\[ V_{cm} = \frac{M V_d + m(v \cos \theta)}{M + m} \]
- Substituting \( V_d \):
\[ V_{cm} = \frac{M \left(\frac{m(v \cos \theta)}{M}\right) + m(v \cos \theta)}{M + m} \]
- Simplifying:
\[ V_{cm} = \frac{m(v \cos \theta) + m(v \cos \theta)}{M + m} = \frac{2m(v \cos \theta)}{M + m} \]

7. **Conclusion**:
- The center of mass of the gun and bullet system moves with a velocity of \( \frac{m(v \cos \theta)}{M + m} \) in the horizontal direction.

### Final Answer:
The center of mass of the gun and bullet system moves with a velocity of \( \frac{m(v \cos \theta)}{M + m} \) in the horizontal direction.