A 1,240-kg car traveling initially with a speed of 25.0 m/s in an easterly direc
ID: 2033856 • Letter: A
Question
A 1,240-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 9,200-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. +25.0 m/s +20.0 m/s +18.0 m/s Before After (a) What is the velocity of the truck right after the collision? (Round your answer to at least three decimal places.) m/s (east) (b) How much mechanical energy is lost in the collision? Account for this loss in energyExplanation / Answer
Given :-
u1 = initial velocity of car
u2 = initial velocity of truck
v1 = final velocity of car
v2 = final velocity of truck
m1 = mass of the car
m2 = mass of the truck
According to conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2
(1240 kg x 25 m/s) + (9200 kg x 20 m/s) = (1240 kg x 18 m/s) + (9200 kg x v2)
v2 = 20.94 m/s
b)
The energy before collision is
Ei = 1/2m1u1^2 + 1/2m2u2^2
Ei = [0.5 x 1240 x 25^2] + [0.5 x 9200 x 20^2]
Ei = 2227500 J
The energy after collision is
Ef = 1/2m1v1^2 + 1/2m2v2^2
Ef = (0.5 x 1240 x 18)^2 + (0.5 x 9200 x 20.94^2)
Ef = 2217904.56 J
Energy lost in collision is
= Ef - Ei
= 2217904.56 - 2227500 J
= -9595.5 J
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