Asked by Meagan

In 1896 in Waco, Texas, William Crush of the “ Katy” railroad parked two locomotives at opposite ends of a 6.4 km track, fired them up, tied their throttles open, and then allowed them to crash head-on at full speed in front of 30,000 spectators. Hundreds were hurt by flying debris; several were killed. Assuming the weight of each locomotive was 1.2 x 10^6 N and its acceleration prior to the collision was a constant 0.26 m/s^2, what was the total
kinetic energy of the two locomotives just before the collision?
Answered by Henry
V1 = V2.
d1 = d2 = 6.4km/2 = 3.2 km before collision.
M*g = 1.2*10^6, M = 1.2*10^6/9.8 = 1.22*10^5 kg.

V^2 = Vo^2 + 2a*d.
V^2 = 0 + 0.52*3200
V = 40.8 m/s. = Velocity before collision.

KE = 2(M*V^2/2) = M*V^2.
Answered by gabriel
good, very good.Danke