A 1,260-kg car traveling initally with a speed of 25.0 m/s in an easterly direct

Question

A 1,260-kg car traveling initally with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 9,600-kg truck moving in the same direction at 20.0 m/s. The velocity of the car right after the 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. (Use input values with adequate number of significant figures to calculate this answer.) ___________kJ

C) Account for this loss in energy.

Explanation / Answer

The law of conservation of momentum says that the total momentum before the collision is equal to the total momentum after the collision. So you may write an equation like this: P(i) = P(f) m?v?(i) + m?v?(i) = m?v?(f) + m?v?(f) (1200kg)(25.0m/s) + (9000kg)(20.0m/s) = (1200kg)(18.0m/s) + (9000kg)v?(f) 210,000kg?m/s = 21,600kg?m/s + (9000kg)v?(f) v?(f) = (210,000kg?m/s - 21,600kg?m/s) / 9000kg = 20.9m/s

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