. Four railroad cars, each of mass 2.50 104 kg, are coupled together and coasting along horizontal tracks at a speed of 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 4.00 m/s southward. The remaining three cars continue moving toward the south, now at 2.00 m/s. (a) Find the initial speed of the cars. (b) How much work did the actor do

a)     v₀ = 2.5 m / s,   heading south.

b)  W = 1,219 10⁵ J

Explanation:

a) For this exercise we can use the conservation of the moment, we create a system formed by all the 4 cars, in this case when the last one separates the forces are intense and the moment is conserved

initial instant. Before separation

        p₀ = M v₀

final instant. When uncoupling the last car

        p_f = 3m v₁ + m v₂

where they indicate that the speed of the wagons is v₁ = 2.00 m / s and the speed of the last wagon is v₂ = 4.00 m / s

the total mass is M = 4m

how the moment is preserved

           p₀ = p_f

         4m v₀ = 3m v₁ + mv₂

         v₀ = ¾ v₁ + v₂ / 4

let's calculate

           v₀ = ¾ 2 + ¼ 4

           v₀ = 2.5 m / s

heading south.

b) work is equal to the change in kinetic energy

              W = ΔK = K_f -K_o

              W = ½ m v_f² - ½ m v₀²

               W = ½ m (v_f² -v₀²)

               W = ½ 2.50 10⁴ (4² - 2.5²2)

               W = 1,219 10⁵ J

it may be wrong but i think its a

double the distance = divide strength by 4