A rail road car of mass M is at rest on frictionless rails when a man of mass m starts moving on the car towards the engine. If the car recoils with a speed v backward on the rails, with what velocity is the man aproaching the engine?

Explanation

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

### Step 1: Understand the system
- We have a railroad car of mass \( M \) at rest on frictionless rails.
- A man of mass \( m \) starts moving towards the engine, causing the car to recoil backward with speed \( v \).

### Step 2: Set up the momentum conservation equation
- Initially, the total momentum of the system (man + car) is zero since both are at rest.
- According to the conservation of momentum, the final momentum of the system must also be zero.

### Step 3: Define the directions
- Let's assume the direction towards the engine is positive.
- Therefore, the backward direction (where the car is moving) is negative.

### Step 4: Write the final momentum equation
- The final momentum can be expressed as:
\[ \text{Final Momentum} = m \cdot u - M \cdot v \]
where \( u \) is the velocity of the man relative to the ground.

### Step 5: Set the initial momentum equal to the final momentum
- Since the initial momentum is zero, we have:
\[ 0 = m \cdot u - M \cdot v \]

### Step 6: Solve for the man’s velocity
- Rearranging the equation gives:
\[ m \cdot u = M \cdot v \]
\[ u = \frac{M \cdot v}{m} \]

### Step 7: Adjust for the relative motion
- The velocity \( u \) calculated is the velocity of the man relative to the ground. However, since the car is also moving backward, we need to account for this.
- The actual velocity of the man relative to the car is:
\[ u' = u + v = \frac{M \cdot v}{m} + v \]
\[ u' = v \left( \frac{M}{m} + 1 \right) \]

### Final Result
- Therefore, the velocity of the man approaching the engine, relative to the ground, is:
\[ u' = v \left( \frac{M + m}{m} \right) \]