A 1300-kg car runs into the rear of a 1700-kg minivan that was moving at 5 m/s.

ID: 2023498 • Letter: A

Question

A 1300-kg car runs into the rear of a 1700-kg minivan that was moving at 5 m/s. The two vehicles are locked together after the collision and move with a speed of 10 m/s.

a)What was the speed of the car before the collision? 16.54 m/s

b)How much kinetic energy is “lost” in the collision? 49,071J

c)How high above the ground would the locked-together vehicles have to be raised in order for their potential energy to equal the “lost” kinetic energy in the collision? 1.67m

d)How much work would it take to raise the vehicles to that height? 49,071J

Explanation / Answer

(a). Mass of car m = 1300 kg Mass og minivan M = 1700 kg Initial velocity of van U = 5 m / s Initial velocity of car u = ? Final velocity of car & van v = 10 m/ s From law of conservation of momentum , mu + MU = (m+M) v 1300 u + 8500 = 30000 u = 16.54 m / s (b).Initial kinetic energy K = ( 1/ 2) mu 2 + ( 1/ 2) M U 2 = 177821.54 + 21250 = 199071.54 J Final kinetic energy K ' = ( 1/ 2) (m+ M ) v 2 = 150000 J Kinetic energy lost = K - K ' = 49071.54 J (c). Potential energy P = 49071.54 J (m+M) gh = 49071.54 J From this height h = 1.67 m (d). work done = Work against gravitational work = ( m+ M) g * h where ( m+ M) g = gravitational force = 49071.54 J = 150000 J Kinetic energy lost = K - K ' = 49071.54 J (c). Potential energy P = 49071.54 J (m+M) gh = 49071.54 J From this height h = 1.67 m (d). work done = Work against gravitational work = ( m+ M) g * h where ( m+ M) g = gravitational force = 49071.54 J

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A 1300-kg car runs into the rear of a 1700-kg minivan that was moving at 5 m/s.

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