A railway flat car, whose mass together with the artillery gun is `M = 2 m`, moves at a speed `V`. The gun barrel makes an angle `alpha = 60^(@)` with the horizontal. A shell of mass `m` leaves the barrel at a speed `v = 12 ms^(-1)`, relative to the barrel. Find the speed of the flat car `V` (in `ms^(-1)`) in order that it may stop after the firing.

Explanation

To solve the problem, we will apply the principle of conservation of momentum. The momentum of the system before firing must equal the momentum of the system after firing.

### Step-by-Step Solution:

1. **Identify the Initial Momentum:**
- The initial momentum of the system consists of the railway flat car (mass = 2m) and the artillery gun moving at speed V, and the shell is still in the barrel.
- Therefore, the initial momentum \( P_{initial} = (2m) \cdot V \).

2. **Identify the Final Momentum:**
- After the shell is fired, the flat car and the gun come to a stop, so their final momentum is 0.
- The shell (mass = m) is fired at a speed \( v = 12 \, \text{ms}^{-1} \) at an angle \( \alpha = 60^\circ \) with the horizontal.
- The horizontal component of the shell's velocity \( v_x = v \cdot \cos(\alpha) = 12 \cdot \cos(60^\circ) \).
- Since \( \cos(60^\circ) = \frac{1}{2} \), we have \( v_x = 12 \cdot \frac{1}{2} = 6 \, \text{ms}^{-1} \).
- Therefore, the final momentum of the shell is \( P_{final} = m \cdot v_x = m \cdot 6 \).

3. **Apply Conservation of Momentum:**
- According to the conservation of momentum:
\[ P_{initial} = P_{final} \]
\[ (2m) \cdot V = m \cdot 6 \]

4. **Simplify the Equation:**
- We can cancel the mass \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 2V = 6 \]

5. **Solve for V:**
- Dividing both sides by 2 gives:
\[ V = \frac{6}{2} = 3 \, \text{ms}^{-1} \]

### Final Answer:
The speed of the flat car \( V \) in order that it may stop after firing is \( 3 \, \text{ms}^{-1} \).

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