. 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

d) the number of neutrons plus the number of protons

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)