Question

A 1100 kg car traveling initially with a speed of 26.00 m/s in an easterly direction crashes into the back of a 9 450 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 and direction of the truck right after the collision? (Give your answer to at least four significant figures.)

(b) What is the change in mechanical energy of the car-truck system in the collision?

Answer 1

As the car has not deviated from the earlier direction, we assume that the collision is one-dimensional. It also makes calculations simple.

a) 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).

M2 * V2 = 1100 * 26 + 9450 * 20 - 1100 * 18 = 197800

This gives V2 = 197800/ 9450 = 20.9312m/s

b) 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 = 1/2 * (M1 * U1^2 + M2 * U2^2) = (1100 * (26)^2 + 9450 * (20)^2) = 4523600J

(KE)2 = 1/2* (M1 * V1^2 + M2 * V2^2) =(1100 * (18)^2 + 9450 * (20.9312)^2 ) = 4496588J

Losses = (KE)1 - (KE)2 = 4523600 - 4496588 Joules = 27011.98899 Joules

Hope this helped!