A 1 215 kg car traveling initially with a speed of 25.0 m/s in an easterly direc

Question

A 1 215 kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the back of a 9 200 kg truck moving in the same direction at 20.0 m/s. 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?
20.92 m/s east

(b) What is the change in mechanical energy of the car–truck system in the collision? (Use input values with an adequate number of significant figures to calculate this answer.)
? J

Explanation / Answer

The mass of the car m = 1215kg the mass of the truck M = 9200 kg the speed of the car v1i = 25m/s the speed of the truck v2i = 20 m/s after collision the speed of the car v1f = 18m/s then from law of conservation of momentum        m1v1i + m2v2i = m1v1f + m2v2f then the speed of the truck after collsion       v2f = (m v1i + M v2i - mv1f) / M                           = [(1215)(25) + (9200) (20)] - (1215)(18) / 9200              = 20.92 m/s (b) The kinetic energy before collision        E = 1/2mv1i^2 + 1/2M v2i^2             = (0.5) (1215)(25)^2 + (0.5)(9200)(20)^2          = 379687.5 + 1840000 = 2219687.5 the kinetic energy after collision         E = 1/2mv1i^2 + 1/2M v2i^2             = (0.5) (1215)(18)^2 + (0.5)(9200)(20.92)^2          = 196830 + 2013173.44 = 2210003.5 therefore the difference           E = 9684 J the kinetic energy after collision         E = 1/2mv1i^2 + 1/2M v2i^2             = (0.5) (1215)(18)^2 + (0.5)(9200)(20.92)^2          = 196830 + 2013173.44 = 2210003.5 therefore the difference           E = 9684 J therefore the difference           E = 9684 J

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