A 1,290-kg car traveling initially with a speed of 25.0 m/s in an easterly direc

Question

A 1,290-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. +25.0 m/s +20.0 m/s +18.0 m/s Before After (a) What is the velocity of the truck right after the collision? 6.24 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? Account for this loss in energy

Explanation / Answer

a) Velicity of the truck after the collision is given by Vt

By conservation of momentum

m(c)*v(c) + m(t)*v(t) = m(c)*V(c) + m(t)*V(t)

1290*25 + 8200*20 = 1290*18 + 8200*V(t)

V(t) = (1290*25 + 8200*20 - 1290*18)/8200

V(t) = 21.10 m/s

b) KE lost = KE before – KE after

KE lost = (0.5*1290*25^2 + 0.5*8200*20^2) - (0.5*1290*18^2 + 0.5*8200*21.10^2)

= 8784 J

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