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 apply the principle of conservation of momentum. Here's the step-by-step solution:

### Step 1: Understand the Initial Conditions
- The mass of the railroad car is \( M \).
- The initial velocity of the car is \( v_0 \).
- The rain adds mass at a rate of \( m \) kg/s.

### Step 2: Determine the Initial Momentum
The initial momentum \( P_{\text{initial}} \) of the car can be calculated as:
\[ P_{\text{initial}} = M \cdot v_0 \]

### Step 3: Calculate the Mass Added After Time \( t \)
After \( t \) seconds, the mass of water added to the car is:
\[ \text{mass of water} = m \cdot t \]
Thus, the total mass of the car after \( t \) seconds becomes:
\[ \text{Total mass} = M + m \cdot t \]

### Step 4: Define the Final Velocity
Let the final velocity of the car after \( t \) seconds be \( V \). The final momentum \( P_{\text{final}} \) can be expressed as:
\[ P_{\text{final}} = (M + m \cdot t) \cdot V \]

### Step 5: Apply Conservation of Momentum
According to the conservation of momentum:
\[ P_{\text{initial}} = P_{\text{final}} \]
Substituting the expressions we derived:
\[ M \cdot v_0 = (M + m \cdot t) \cdot V \]

### Step 6: Solve for Final Velocity \( V \)
Rearranging the equation to solve for \( V \):
\[ V = \frac{M \cdot v_0}{M + m \cdot t} \]

### Final Answer
The velocity of the car after \( t \) seconds is:
\[ V = \frac{M \cdot v_0}{M + m \cdot t} \]

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