An open rail road car of mass M is moving with initial velocity `v_(0)` along a straight horizontla and friction less track.It suddenly starts raining at time t=0. The rain drops fall vertically with velocity with velocity u and add a mass of m `kg //sec` of water . The velocity of car after t second is

Options

A. `(M+m)/( MV_(0)t)`
B. `(MV_(0)t)/(M+v)`
C. `(Mv_(0))/(M+mt)`
D. `(M)/((M+m)v_(0))`

Answer

To solve the problem, we will use the principle of conservation of momentum. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Conditions
At time \( t = 0 \):
- The mass of the railroad car is \( M \).
- The initial velocity of the car is \( v_0 \).
- The initial momentum of the system (car) is given by:
\[ \text{Initial Momentum} = M \cdot v_0 \]

### Step 2: Analyze the Situation After Time \( t \)
After time \( t \):
- The rain adds mass to the car at a rate of \( m \) kg/s.
- The total mass added by the rain after time \( t \) is \( m \cdot t \).
- Therefore, the total mass of the car plus the rain after time \( t \) is:
\[ \text{Total Mass} = M + m \cdot t \]

### Step 3: Apply the Conservation of Momentum
Since there are no external horizontal forces acting on the system, we can apply the conservation of momentum:
- The initial momentum of the system (car) is \( M \cdot v_0 \).
- The final momentum of the system after time \( t \) is given by the total mass multiplied by the final velocity \( v \):
\[ \text{Final Momentum} = (M + m \cdot t) \cdot v \]

Setting the initial momentum equal to the final momentum:
\[ M \cdot v_0 = (M + m \cdot t) \cdot v \]

### Step 4: Solve for the Final Velocity \( v \)
Rearranging the equation to solve for \( v \):
\[ v = \frac{M \cdot v_0}{M + m \cdot t} \]

### Conclusion
The velocity of the car after time \( t \) is:
\[ v = \frac{M \cdot v_0}{M + m \cdot t} \]