A 1,110-kg car traveling initially with a speed of 25.0 m/s in an easterly direc

Question

A 1,110-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a8,100-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?
_______________________J
Account for this loss in energy.

_________________________________________________________________________

Explanation / Answer

mc = mass of car = 1110 kg

mt = mass of truck = 8100 kg

Vic = initial velocity of car = 25 m/s

Vit = initial velocity of truck = 20 m/s

Vfc = final velocity of car = 18 m/s

Vft = final velocity of truck

a)

Using conservation of momentum

mc Vic + mt Vit = mc Vfc + mt Vft

1110 x 25 + 8100 x 20 = 1110 x 18 + 8100 Vft

Vft = 21 m/s

b)

initial Total kinetic energy = KEi = (0.5) (mc Vic2 + mt Vit2 ) = (0.5) (1110 x 252 + 8100 x 202 ) = 1966875 J

final Total kinetic energy = KEf = (0.5) (mc Vfc2 + mt Vft2 ) = (0.5) (1110 x 182 + 8100 x 212 ) = 1965870 J

loss of mechanical energy = KEf - KEi = 1965870 - 1966875 = - 1005 J

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