A railroad car of mass 2.30e+4 kg moving with a speed of 3.00 m/s collides and couples with two other coupled railroad cars each of the same mass that are already moving in the same direction at a speed of 1.10 m/s. what is the speed (in m/s) of the three coupled cars after the collision?

Question

Hanna Richardson

Physics

25 November, 21:17

A railroad car of mass 2.30e+4 kg moving with a speed of 3.00 m/s collides and couples with two other coupled railroad cars each of the same mass that are already moving in the same direction at a speed of 1.10 m/s. what is the speed (in m/s) of the three coupled cars after the collision?

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Ulises Wells · Answer 1

This is a case of conservation of momentum in which;

mv1 + 2 (mv2) = mfvf

Where m = mass of first railroad cars = 2.30e+4 kg = 23000 kg; v2 = velocity of first railroad car = 3.00 m/s; v2 = velocity of the two other railroad cars = 1.10 m/s; mf = total mass of the three cars after coupling = 23000*3 = 69000 kg; vf = velocity of the coupled cars

Substituting;

23000*3 + 2 (23000*1.1) = 69000*vf

69000vf = 119600

vf = 119600/69000 = 1.73 m/s

The velocity the three cars couple will be 1.73 m/s.