A 1200-kg car traveling initially with a speed of 25.0 m/s in an easterly direct

Question

A 1200-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 9000-kg truck moving in the same direction at 20.0 m/s. The velocity of the car right after the collision is 18.0 m/s to the east. (a) What is the velocity of the truck right after the collision? (b) How much mechanical energy is lost in the collision? Account for this loss in energy

Velocity of truck after collision = ____________________________________

Loss in Mechanical Energy = ________________________________________

Energy Loss “Shows Up As” = _______________________________________

Explanation / Answer

applying momentum conservation

m1u1 + m2u2 = m1v1 + m2v2

(1200 x 25) + (9000 x 20) = (1200 x 18) + (9000 v2)

v2 = 20.93 m/s (velocity of truck after collision)

Energy before collision

Ebc = KE of car + KE of truck = 1/2 m1u1^2 + 1/2 m2u2^2 = 1/2 x 1200 x (25)^2 + 1/2 x 9000 x (20)^2

Ebc = 2175000 J

Eac = KE of car + KE of truck = 1/2 m1v1^2 + 1/2 m2v2^2 = 1/2 x 1200 x (18)^2 + 1/2 x 9000 x (20.93)^2

Eac = 2165692.05 J

Loss of energy = Ebc - Eac = 2175000 - 2165692.05 = 9307.95 J

This loss done in the form of friction between cat and truck, not perfact elastic collision, sound energy etc

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