Two toy trains each of mass 'M' are moving in opposite directions with velocities `v_1` and `v_2` over two smooth rails. Two stuntmen of mass 'm' each are also moving with the trains (at rest w.r.t. trains). When trains are opposite to each other the stuntmen interchange their positions, then find the final velocities of the trains.

Explanation

To solve the problem of two toy trains with stuntmen exchanging positions, we will use the principle of conservation of linear momentum. Here’s a step-by-step solution:

### Step 1: Understand the Initial Setup
We have two toy trains, each with mass \( M \):
- Train 1 moving with velocity \( v_1 \) to the right.
- Train 2 moving with velocity \( v_2 \) to the left.

Each train has a stuntman of mass \( m \) moving with it. When the trains are opposite each other, the stuntmen interchange their positions.

### Step 2: Define the Initial Momentum
The initial momentum of the system can be calculated as follows:
- Momentum of Train 1: \( P_1 = M v_1 \) (to the right)
- Momentum of Train 2: \( P_2 = -M v_2 \) (to the left, hence negative)

The total initial momentum \( P_{initial} \) is:
\[ P_{initial} = M v_1 - M v_2 \]

### Step 3: Define the Final Momentum After the Exchange
After the stuntmen interchange their positions, the final velocities of the trains will change. Let:
- Final velocity of Train 1 be \( V_1 \)
- Final velocity of Train 2 be \( V_2 \)

The momentum of the system after the stuntmen interchange positions will be:
- Momentum of Train 1: \( M V_1 \)
- Momentum of Train 2: \( -M V_2 \)

The total final momentum \( P_{final} \) is:
\[ P_{final} = M V_1 - M V_2 \]

### Step 4: Apply Conservation of Momentum
According to the conservation of momentum:
\[ P_{initial} = P_{final} \]
Substituting the expressions we derived:
\[ M v_1 - M v_2 = M V_1 - M V_2 \]

### Step 5: Simplify the Equation
We can divide the entire equation by \( M \) (assuming \( M \neq 0 \)):
\[ v_1 - v_2 = V_1 - V_2 \]
Rearranging gives:
\[ V_1 = v_1 - v_2 + V_2 \]

### Step 6: Analyze the Stuntmen's Momentum
Now, we consider the momentum of the stuntmen. Initially, the momentum of the stuntmen is:
- Stuntman in Train 1: \( m v_1 \)
- Stuntman in Train 2: \( -m v_2 \)

After they switch positions, their momentum becomes:
- Stuntman in Train 1 (now in Train 2): \( -m V_2 \)
- Stuntman in Train 2 (now in Train 1): \( m V_1 \)

The total initial momentum of the stuntmen is:
\[ P_{stuntmen, initial} = m v_1 - m v_2 \]
The total final momentum of the stuntmen is:
\[ P_{stuntmen, final} = m V_1 - m V_2 \]

### Step 7: Set Up the Equation for Stuntmen
Using conservation of momentum for the stuntmen:
\[ m v_1 - m v_2 = m V_1 - m V_2 \]
Dividing by \( m \) (assuming \( m \neq 0 \)):
\[ v_1 - v_2 = V_1 - V_2 \]
This is the same equation we derived earlier.

### Step 8: Solve for Final Velocities
From the two equations derived, we can express \( V_1 \) and \( V_2 \) in terms of \( v_1 \) and \( v_2 \):
1. \( V_1 = \frac{M v_1 - m v_2}{M + m} \)
2. \( V_2 = \frac{M v_2 - m v_1}{M + m} \)

### Final Answer
The final velocities of the trains after the stuntmen interchange positions are:
\[ V_1 = \frac{M v_1 - m v_2}{M + m} \]
\[ V_2 = \frac{M v_2 - m v_1}{M + m} \]