A 1,210 -kg car traveling initially with a speed of 25.0 m/s in an easterly dire
ID: 1903501 • Letter: A
Question
A1,210-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a9,800-kg truck moving in the same direction at 20.0 m/s (see figure below). 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? m/s (east)
(b) How much mechanical energy is lost in the collision? (Use input values with an adequate number of significant figures to calculate this answer.) kJ
Account for this loss in energy.
Explanation / Answer
Q: A 1 290 kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the back of a 8 200 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? answer = ? m/s east (b) What is the change in mechanical energy of the car–truck system in the collision? answer = J ? (c) Account for this change in mechanical energy? ANS: (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.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.