A 1,145-kg car traveling initially with a speed of 25.0 m/s in an easterly direc
ID: 2197933 • Letter: A
Question
A 1,145-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 8,000-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? (Use input values with an adequate number of significant figures to calculate this answer.) kJ Account for this loss in energy.Explanation / Answer
a)
by the law of momentum conservation
m1u1 + m2u2 = m1v1 + m2v2
1140 x 25 + 9200 x 20 = 1140 x 18 + 9200 x v2
v2 = 20.8 m/s
b)
KE(initial) = 1/2 x m1 x u1^2 + 1/2 x m2 x u2^2
KE(i) = 1/2 x 1140 x 252 + 1/2 x 9200 x 202
KE(i) = 2196250 J
KE(final) = 1/2 x m1 x v1^2 + 1/2 x m2 x v2^2
KE(f) = 1/2 x 1140 x 182 + 1/2 x 9200 x 20.82
KE(f) = 2174824 J
Loss in KE = KE(i) - KE(f)
= 21426 J
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