A gun of mass M(including the carriage) fires a shot of mass `m`. The gun along with the carriage is kept on a smooth horizontal surface. The muzzle speed of the bullet `v_r` is constant. Find
(a) The elevation of the gun with horizontal at which maximum range of bullet with respect to the ground is obtained.
(b) The maximum range of the bullet.

Explanation

To solve the problem, we will break it down into two parts: finding the elevation angle for maximum range and calculating the maximum range itself.

### Part (a): Finding the elevation angle for maximum range

1. **Understanding the System**:
- We have a gun of mass \( M \) (including the carriage) that fires a bullet of mass \( m \).
- The bullet has a muzzle speed \( v_r \) and is fired at an angle \( \theta \) with respect to the horizontal.

2. **Initial Conditions**:
- Before firing, the entire system (gun + bullet) is at rest. Therefore, the initial momentum of the system is zero.

3. **Applying Conservation of Momentum**:
- When the bullet is fired, the gun recoils. By conservation of momentum:
\[ 0 = m v + M V_g \]
where \( v \) is the velocity of the bullet and \( V_g \) is the velocity of the gun after firing.
- The velocity of the bullet can be expressed as:
\[ v = v_r \cos \theta + V_g \]

4. **Finding the Velocity of the Gun**:
- Rearranging the momentum equation gives:
\[ V_g = -\frac{m v_r \cos \theta}{M} \]

5. **Velocity of the Bullet**:
- Substituting \( V_g \) back into the bullet's velocity equation:
\[ v = v_r \cos \theta - \frac{m v_r \cos \theta}{M} \]
- Simplifying gives:
\[ v = v_r \cos \theta \left(1 - \frac{m}{M}\right) \]

6. **Time of Flight**:
- The time of flight \( T \) for the bullet can be calculated using the vertical component of the bullet's velocity:
\[ T = \frac{2 v_r \sin \theta}{g} \]

7. **Finding the Elevation Angle for Maximum Range**:
- The horizontal range \( R \) is given by:
\[ R = V_g T \]
- Substituting \( V_g \) and \( T \):
\[ R = -\frac{m v_r \cos \theta}{M} \cdot \frac{2 v_r \sin \theta}{g} \]
- To maximize \( R \), we need to maximize \( \sin 2\theta \). This occurs at \( \theta = 45^\circ \).

### Part (b): Calculating the Maximum Range

1. **Substituting \( \theta = 45^\circ \)**:
- At \( \theta = 45^\circ \), \( \sin 2\theta = 1 \).
- The range becomes:
\[ R = \frac{m v_r^2}{g (M + m)} \]

2. **Final Expression for Maximum Range**:
- Thus, the maximum range \( R \) of the bullet is:
\[ R = \frac{m v_r^2}{g (M + m)} \]

### Summary of Results:
- **(a)** The elevation angle \( \theta \) for maximum range is \( 45^\circ \).
- **(b)** The maximum range of the bullet is \( R = \frac{m v_r^2}{g (M + m)} \).