Four railroad cars, each of mass `2.50 xx 10^4` kg, are coupled together and coasting along horizontal tracks at speed `v_(I)`towards the south. A very strong movie actor, riding on the second car, uncouples the front car and gives it a big push, increasing its speed to `4.00 m//s` southward. The remaining three cars continue moving south, now at `2.00 m//s`. (a) Find the initial speed of the cars. (b) How much work did the actor do?

Explanation

To solve the problem step by step, we will use the principles of conservation of momentum and the work-energy theorem.

### Step-by-Step Solution

**Step 1: Understand the system and given data.**
- Mass of each railroad car, \( m = 2.50 \times 10^4 \) kg
- Total number of cars = 4
- Speed of the front car after being pushed, \( v_2 = 4.00 \) m/s (south)
- Speed of the remaining three cars after the push, \( v_1 = 2.00 \) m/s (south)

**Step 2: Apply the conservation of momentum.**
The total momentum before the push must equal the total momentum after the push.

- Let the initial speed of all cars be \( V_i \).
- Initial momentum of the system (all four cars):
\[ P_{\text{initial}} = 4mV_i \]

- Final momentum after the push:
\[ P_{\text{final}} = m v_2 + 3m v_1 = m(4.00) + 3m(2.00) = m(4 + 6) = 10m \]

**Step 3: Set the initial momentum equal to the final momentum.**
\[ 4mV_i = 10m \]

**Step 4: Solve for \( V_i \).**
We can cancel \( m \) from both sides since it is non-zero:
\[ 4V_i = 10 \]
\[ V_i = \frac{10}{4} = 2.5 \text{ m/s} \]

**Step 5: Calculate the work done by the actor.**
The work done is equal to the change in kinetic energy of the system.

- Initial kinetic energy of the system (all four cars):
\[ KE_{\text{initial}} = \frac{1}{2} \times 4mV_i^2 = \frac{1}{2} \times 4m(2.5)^2 \]
\[ = 2m(6.25) = 12.5m \]

- Final kinetic energy of the system (after the push):
\[ KE_{\text{final}} = \frac{1}{2} mv_2^2 + \frac{1}{2} \times 3m v_1^2 \]
\[ = \frac{1}{2} m(4)^2 + \frac{3}{2} m(2)^2 \]
\[ = \frac{1}{2} m(16) + \frac{3}{2} m(4) = 8m + 6m = 14m \]

**Step 6: Calculate the change in kinetic energy.**
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 14m - 12.5m = 1.5m \]

**Step 7: Substitute the mass to find the work done.**
\[ \Delta KE = 1.5 \times (2.50 \times 10^4) = 3.75 \times 10^4 \text{ J} \]

### Final Answers
(a) The initial speed of the cars, \( V_i = 2.5 \) m/s.
(b) The work done by the actor, \( W = 3.75 \times 10^4 \) J.