A 1 265 kg car traveling initially with a speed of 25.0 m/s in an easterly direc

Question

A 1 265 kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the back of a 8 400 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? m/s

(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.) in J

(c) Account for this change in mechanical energy.

Explanation / Answer

a)

using Law of conservation of momentum

M1V1 +M2V2 = M1U1 + M2U2

(1265*18) + (8400*V2) = (1265*25)+(8400*20)

8400*V2 = 176855

V2 = 21.054 m/s

b)

Final KE of System = 0.5M1V12 + 0.5M2V22 = 2066.667KJ

Initial KE of the System = 0.5M1U12 + 0.5M2U22 = 2075.312KJ

Change = Final KE - Initial KE = 2066.667 - 2075.312 = -8.645 KJ

c)

The loss of energy indicate that this collision is a non-elastic collision

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