A 1,115-kg car traveling initially with a speed of 25.0 m/s in an easterly direc

Question

A 1,115-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 8,200-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
Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (east)

(b) How much mechanical energy is lost in the collision?

_______________________ J

Describe this loss in energy.

Explanation / Answer

As the car has not deviated from the earlier direction, we assume that the collision is one-dimensional. It also makes calculations simple.
As losses are present, we can take only conservation of momentum for our calculations.
M1 * U1 + M2 * U2 = M1 * V1 + M2 * V2.
1 stands for car, 2 for truck. M is mass, U is initial velocity (before collision) and V is final velocity (after collision).
V2 = [1115*25 + 8200*20 - 1115*18]/8200 = 20.95 m/sec

To calculate losses, we will find the kinetic energies before & after collision. Any difference would give us the losses (in energy form).
(KE)1 = (KE)2 + Losses.
(KE)1 = 0.5 * (M1 * U1^2 + M2 * U2^2) = 0.5*(1115*25^2 + 8200*20^2) = 1988437.5 J
(KE)2 = 0.5 * (M1 * V1^2 + M2 * V2^2) = 0.5*(1115*18^2 + 8200*20.95^2)= 1980130.25 J

Losses = (KE)1 - (KE)2 = 1988437.5 - 1980130.25 = 8307.25 Joules

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