A 1200-kg car is backing out of a parking space at 5.0 m/s . The unobservant dri

Question

A 1200-kg car is backing out of a parking space at 5.0 m/s . The unobservant driver of a 1900-kg pickup truck is coasting through the parking lot at a speed of 3.2 m/s and runs straight into the rear bumper of the car.What is the change in internal energy of the two-vehicle system if the velocity of the pickup is 1.5 m/s backward after they collide? Express your answer with the appropriate units. Enter positive value if the energy increases and negative value if the energy decreases. B-What is the coefficient of restitution for this collision? A-

Explanation / Answer

This is an inelastic collision, linear momentum is conserved but KE is not conserved

let us take the direction of the car from parking lot as +ve x-axis and we will put signs for all other velocities appropriateely. It is one dimensional collision.

let v1 be the velocity of the car after collision, conserving the momentum we will have

1200*5.0 -1900*3.2 = 1200*v1 +1900*1.5

from the above we will have v1 = -2.44 m/s, -ve sign indicates the car is moving in opposite direction after collision.

KE before collision = 0.5(1200*5.0^2 +1900*3.2^2) = 24728 J

KE after collision = 0.5(1200*2.44^2 +1900*1.5^2) = 5709 J

change in internal energy = 5709-24728 = -19018

co-efficient of restitution e = vel.seperation/vel.approach

                                          = (5.0-(-3.2))/(1.5-(-2.44)) = 0.48

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