Question

Four railroad cars, each of mass 2.70 104 kg, are coupled together and coasting along horizontal tracks at speed vi toward the south. A very strong but foolish movie actor, riding on the second car, uncouples the front car and gives it a big push, increasing its speed to 3.85 m/s southward. The remaining three cars continue moving south, now at 2.00 m/s.
(a) Find the initial speed of the four cars.

m/s

(b) By how much did the potential energy within the body of the actor change?

kJ

(c) State the relationship between the process described here and the process in which a railroad car collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 2.00 m/s.

Answer 1

(a)

From conservation of momentum,

4mvi = 3mv1 + mv2

4vi = 3*2 + 3*3.85

vi = 2.46 m/s

(b)

Change in potential energy = change in KE

PE = Kf - Ki = (1/2)3m*v1^2 + (1/2)mv2^2 - (1/2)4mvi^2

PE = 1.5*2.7*10⁴*2² + 0.5*2.7*10⁴*3.85² - 2*2.7*10⁴*(2.46)²

PE = 3.53*10^4 J

(c)

ln both the situations momentum is conserved and energy is also conserved.

The change of potential energy equal to the change of kinetic energy

The change of potential energy equal to the change of kinetic energy