Question

A 1,165-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 8,300-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? (Round your answer to at least three decimal places.)

(b) How much mechanical energy is lost in the collision?

account for this loss in energy

Answer 1

a) from conservation of momentum we have

m1u1+m2u2=m1v1+m2v2

1165kg*25.0m/s+8300 kg*20 m/s=1165*18-m2v2

v2=174155/8300=20.983 m/s

so answer is 20.983 m/s or 21.0 m/s

b) the kinetic energy before collision

K.Ei=1/2 (m1u1²+m2u2²)=1/2(1165*25²+8300*20²)=2024063 J

after collsion

K.Ef=1/2(m1v1²+m2v2²)=1/2(1165*18²+8300*20.983²)=2015836 J

so mechanical energy lost=K.Ei-K.Ef=8226 J

so answer is 8226 J or 8226.254 J or 8230 J or 8.23*10³ J

this loss in energy is lost as heat to the surrounding , energy lost due to friction between tires and road , energy lost as sound energy etc.