Question

confused with the problem and how to do the math can you continue the math where I left off thank you

A railroad car of mass 2.75 × 104 kg moving at 2.50 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? (b) How much kinetic energy is lost in the collision? (a) What is the speed of the three coupled cars after the collision? (mixy) + (m) (リ= (mi) (y) + (m2) (vi) (m,)(vi) + (m) (v) = (mi+ m 2) (v) but there are two railroad cars, so m = 2 m)+(2m)(+2m) (v) (2.75 x 10t kg) (2.50 m/s)+ 2(2.75 x 10 kg) (1.20 m/s) (2.75 x 10 kg) 2(2.75 x 10 kg) (v) 68,750 kg* m/s + 66,000 kg* m/s = 2.75 x 104 kg + 55,000 kg (v) 134,760 kg*m/s 82500 kg (v) 134,760 kg*m/s=82500 kg (v.) 82500 kg 82500 kg 1.63 =y

Answer 1

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