A railway flat car has an artillery gun installed on it. The combined system has a mass `M` and moves with a velocity `V`. The barrel of the gun makes an angle a 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 so that it may stop after the firing is

Options

A. `(mv)/(M+m)`
B. `((Mv)/(M+m))cosalpha`
C. `(mv)/(M+m))cosalpha`
D. `(M+m))cosalpha`

Explanation

To solve the problem, we need to apply the principle of conservation of linear momentum. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Momentum
Initially, the railway flat car with the artillery gun (combined mass \( M \)) is moving with a velocity \( V \). The total initial momentum \( P_i \) of the system can be expressed as:
\[ P_i = (M + m) \cdot V \]
where \( m \) is the mass of the shell.

### Step 2: Analyze the Final Momentum
After the shell is fired, the flat car comes to a stop, meaning its final velocity is \( 0 \). The shell leaves the barrel at a speed \( v \) relative to the barrel, making an angle \( \alpha \) with the horizontal. The velocity of the shell can be resolved into two components:
- Horizontal (x-direction): \( v_x = v \cdot \cos(\alpha) \)
- Vertical (y-direction): \( v_y = v \cdot \sin(\alpha) \)

The final momentum \( P_f \) of the system after firing is:
\[ P_f = 0 + m \cdot v_x = m \cdot (v \cdot \cos(\alpha)) \]

### Step 3: Apply Conservation of Momentum
According to the conservation of momentum, the initial momentum must equal the final momentum:
\[ (M + m) \cdot V = m \cdot (v \cdot \cos(\alpha)) \]

### Step 4: Solve for the Velocity \( V \)
Rearranging the equation to solve for \( V \):
\[ V = \frac{m \cdot (v \cdot \cos(\alpha))}{M + m} \]

### Step 5: Final Expression
This gives us the speed of the flat car so that it may stop after firing:
\[ V = \frac{m \cdot v \cdot \cos(\alpha)}{M + m} \]

### Conclusion
Thus, the speed of the flat car so that it may stop after firing is given by the formula above.

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