Question

A railroad car of mass 3.10 ✕ 10^4 kg moving at 3.50 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?

__m/s

(b) How much kinetic energy is lost in the collision?

__J

Answer 1

A railroad car of mass 3.1*10^4 kg moving at 3.500 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.

“couples” means all 3 cars are coupled together after the collision

Total mass after collision = 3 * 3.1 * 10^4 kg

Momentum is always conserved!

Initial momentum = 3.1 * 10^4 * 3.500 + 2 * 3.1 * 10^4 * 1.20

Final momentum = 3 * 3.1 * 10^4 * v

(a) What is the speed of the three coupled cars after the collision?

Final momentum = Initial momentum

3 * 3.1* 10^4 * v = 3.1 * 10^4 * 3.500 + 2 * 3.1* 10^4 * 1.20

solve

v = (5.9÷ 3) m/s is the velocity of the 3 coupled cars after the collision!

(b) How much kinetic energy is lost in the collision?

Initial KE = (½ * 3.1 * 10^4 * 3.500^2) + (½ * 2 * 3.1 * 10^4 * 1.20^2) =234515 j

Final KE = ½ * 3 * 3.1 * 10^4 * (5.9/3)^2 =179851.6667 j

Initial KE – Final KE = kinetic energy is lost in the collision

=54663.33333 j