to conserve momentum,
(1.93*10^4)(3.12) + (2)(1.93*10^4)(1.04) = 3(1.93*10^4)(v)
3.12+2.08 = 5.79v
v = 0.898 m/s
Old KE = 1/2 m*3.12^2 + 1/2 * 2m * 1.04^2 = 5.949m
new KE = 1/2 * 3m * 0.898^2 = 1.209m
loss of KE = 4.739m = 9.147*10^4 J
A railroad car with a mass of 1.93 ✕ 104 kg moving at 3.12 m/s joins with two railroad cars already joined together, each with the same mass as the single car and initially moving in the same direction at 1.04 m/s.
What is the decrease in kinetic energy during the collision?
J
1 answer
1 answer