A railway flat car, whose mass together with the artillery gun is M, moves at a speed V. The gun barrel makes an angle with the horizontal. A shell of mass m leaves the barrel at a speed v, relative to the barrel. The speed of the flat car in order that it may stop after the firing is 

Options

A. `(mv cos alpha)/(M - m)`
B. `(mv)/(M + m)`
C. `(M + m) v cos alpha`
D. `(Mv cos alpha)/(M + m)`

Answer

To solve the problem, we will use the principle of conservation of linear momentum. The steps are as follows:

### Step-by-Step Solution:

1. **Identify the Initial Momentum:**
The initial momentum of the system (railway flat car + artillery gun + shell) before firing is given by:
\[ P_i = (M) \cdot V \]
where \( M \) is the mass of the flat car and the artillery gun combined, and \( V \) is their initial speed.

2. **Determine the Final Momentum:**
After the shell is fired, the flat car comes to rest, so its final momentum is:
\[ P_f = 0 \]
The shell is fired at a speed \( v \) relative to the barrel at an angle \( \theta \) with the horizontal. The horizontal component of the shell's velocity is:
\[ v_x = v \cdot \cos(\theta) \]
Therefore, the momentum of the shell after firing is:
\[ P_{shell} = m \cdot v_x = m \cdot (v \cdot \cos(\theta)) \]

3. **Apply Conservation of Momentum:**
According to the conservation of momentum, the initial momentum must equal the final momentum:
\[ P_i = P_f + P_{shell} \]
Substituting the expressions we have:
\[ (M) \cdot V = 0 + m \cdot (v \cdot \cos(\theta)) \]

4. **Rearranging the Equation:**
Rearranging the equation gives:
\[ M \cdot V = m \cdot v \cdot \cos(\theta) \]

5. **Solve for the Speed \( V \):**
To find the speed \( V \) of the flat car such that it stops after firing, we rearrange the equation:
\[ V = \frac{m \cdot v \cdot \cos(\theta)}{M} \]

### Final Answer:
The speed of the flat car in order that it may stop after firing is:
\[ V = \frac{m \cdot v \cdot \cos(\theta)}{M} \]