A 1,290-kg car traveling initially with a speed of 25.0 m/s in an easterly direc

Question

A 1,290-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 8,100-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 +25.0 m/s +18.0 m/s +20.0 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.) X m/s (east) (b) How much mechanical energy is lost in the collision? 874596 362 Account for this loss in energy This answer has not been graded yet

Explanation / Answer

Using momentum conservation

Pi = Pf

m1u1 + m2u2 = m1v1 + m2v2

v2 = (m1u1 + m2u2 - m1v1)/m2

Using given values:

v2 = (1290*25 + 8100*20 - 1290*18)/8100

v2 = +21.115 m/sec (east)

Part B

Loss in mechanical energy will be

dKE = KEi - KEf

dKE = 0.5*m1u1^2 + 0.5m2u2^2 - (0.5*m1v1^2 + 0.5*m2v2^2)

dKE = 0.5*1290*25^2 + 0.5*8100*20^2 - (0.5*1290*18^2 + 0.5*8100*21.115^2)

dKE = 8479.939 J

Part C

Energy lost due to sound energy or heat energy.

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