A 1 160.0 kg car traveling initially with a speed of 25.000 m/s in an easterly d

Question

A 1 160.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 900.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east. (a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) m/s east (b) What is the change in mechanical energy of the car

Explanation / Answer

(a)by the law of momentum conservation:- m1u1+m2u2=m1v1+m2v2 =>1290 x 25 + 8200 x 20 = 1290 x 18 + 8200 x v2 =>v2 = 21.10 m/s [East] (b) KE(initial) = 1/2 x m1 x u1^2 + 1/2 x m2 x u2^2 =>KE(i) = 1/2 x 1290 x (25)^2 + 1/2 x 8200 x (20)^2 =>KE(i) = 2043125 J KE(final) = 1/2 x m1 x v1^2 + 1/2 x m2 x v2^2 =>KE(f) = 1/2 x 1290 x (18)^2 + 1/2 x 8200 x (21.10)^2 =>KE(f) = 2034341 J Thus the Loss in KE = KE(i) - KE(f) = 8784 J (c) Converted in to other forms of energy i.e. sound,heat e

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