A railroad car of mass 2.65 104 kg moving at 2.60 m/s collides and couples with

Question

A railroad car of mass 2.65 104 kg moving at 2.60 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 A railroad car of mass 2.65 104 kg moving at 2.60 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 A railroad car of mass 2.65 104 kg moving at 2.60 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

Explanation / Answer

Total momentum of system
Momentum initial = momentum final

(2.650 x 10^4 kg)(2.6 m/s) + (5.3 x 10^4 kg)(1.20 m/s) = 7.95x10^4kgV
V = 1.67 m/s <=========== answer to A

Total kinetic energy:
Initial: (1/2)(2.65 x 10^4)(2.60)^2 + (1/2)(5.30 x 10^4)(1.20)^2 = 127730 J

Final: (1/2)(7.95x 10^3)(1.67)^2 = 11085.88 J

Difference: (242080 - 185641) = 116644.12 J lost

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