a. M1*V1 + M2*V2 + M3*V3=M1V+M2V+M3V
21500*3.15+21500*1.2+21500*1.2 = 64,500V.
116,100 = 64,500V, V = 1.6 m/s.
A railroad car of mass 2.15 ✕ 104 kg moving at 3.15 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s.
(a) What is the speed of the three coupled cars after the collision?
____________ m/s
(b) How much kinetic energy is lost in the collision?
____________ J
2 answers
b. KE1 = 0.5M*V1^2+0.5M*V2^2+0.5M*V3^2.
KE1 = M(0.5V1^2+0.5V2^2+0.5V3^2).
KE1 = 21,500(4.96 + 0.72 + 0.72) = 137,600 J.
KE2 = 0.5M*V^2 = 32,250*1.8^2 = 105,300 J.
KE!-KE2 = 137,600 - 105,300 = 32,300 J. lost.
KE1 = M(0.5V1^2+0.5V2^2+0.5V3^2).
KE1 = 21,500(4.96 + 0.72 + 0.72) = 137,600 J.
KE2 = 0.5M*V^2 = 32,250*1.8^2 = 105,300 J.
KE!-KE2 = 137,600 - 105,300 = 32,300 J. lost.
1 answer