A 1290.0 kg car traveling initially with a speed of 25.000 m/s in an easterly di
ID: 2259819 • Letter: A
Question
A 1290.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 100.0 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.)
(b) What is the change in mechanical energy of the car
A 1290.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 100.0 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. What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) What is the change in mechanical energy of the car-truck system in the collision?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 etc.
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