A 1155.0 kg car traveling initially with a speed of 25.000 m/s in an easterly di
ID: 1457914 • Letter: A
Question
A 1155.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 8 600.0 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 1 155.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 8 600.0 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. VTi Cf Tf Before After (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-truck system in the collision? (c) Account for this change in mechanical energy.Explanation / Answer
(a)by the law of momentum conservation:-
m1u1+m2u2=m1v1+m2v2
=>1155 x 25 + 8600 x 20 = 1155 x 18 + 8200 x v2
=>v2 = 21.96 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 1155 x (25)^2 + 1/2 x 8600 x (20)^2
=>KE(i) = 2080937.5 J
KE(final) = 1/2 x m1 x v1^2 + 1/2 x m2 x v2^2
=>KE(f) = 1/2 x 1155 x (18)^2 + 1/2 x 8600 x (21.96)^2
=>KE(f) = 2260748.9 J
Thus the Loss in KE = KE(i) - KE(f) = 179811.38 J
(c) Converted in to other forms of energy i.e. sound,heat etc.
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