A 1.36 103-kg car traveling initially with a speed of 20.0 m/s in an easterly di
ID: 2175074 • Letter: A
Question
A 1.36 103-kg car traveling initially with a speed of 20.0 m/s in an easterly direction crashes into the rear end of a 9.00 103-kg truck moving in the same direction at 15.0 m/s. The velocity of the car right after the collision is 13.0 m/s to the east. What is the velocity of the truck right after the collision? (Assume east is the positive direction. Indicate the direction with the sign of your answer.)How much mechanical energy is lost in the collision? (Use the exact value you entered the first part to make this calculation.)
Explanation / Answer
(a) Velocity of the truck = 20.5 m/s (b) Change in energy = 75,437 Joules lost How to find all of this m1 = 1,100 kg u1 = 25.0 m/s v1 = 18 m/s m2 = 8,900 kg u2 = 20.0 m/s Here's the equation... so enter values and solve v2 = u1 * { [2 * m1 ] / [ m1 + m2 ] } - { [ m1 -m2 ] / [ m1 + m2 ] } * u2 v2 = (25.0 m/s) * { [ 2 * (1,100 kg) ] / [ (1,100 kg) + (8,900 kg) ] } - { [ (1,100 kg) - (8,900 kg) ] / [ (1,100 kg) + (8,900 kg) ] } * (20.0 m/s) v2 = (25.0 m/s) * { [ 2,200 kg ] / [ 10,000 kg ] } - { [ -7,800 kg ] / [ 10,000 kg ] } * (20.0 m/s) v2 = (25.0 m/s) * { 0.22 } - { -0.75 } * (20.0 m/s) v2 = (5.5 m/s) - (-15 m/s) v2 = 20.5 m/s Calculate total kinetic energy pre-collision, KE = 0.5 * m * v^2 Car KE = 0.5 * (1,100 kg) * (25.0 m/s)^2 KE = (550 kg) * (625 m^2/s^2) KE = 343,750 J Truck KE = 0.5 * (8,900 kg) * (20.0 m/s)^2 KE = (4,450 kg) * (400 m^2/s^2) KE = 1,780,000 J Total pre-collision energy = 2,123,750 J Post-collision energy Car KE = 0.5 * (1,100 kg) * (18.0 m/s)^2 KE = (550 kg) * (324 m^2/s^2) KE = 178,200 J Truck KE = 0.5 * (8,900 kg) * (20.5 m/s)^2 KE = (4,450 kg) * (420.25 m^2/s^2) KE = 1870113 J Total post-impact energy = 2,048,313 J A loss of 75,437 Joules.
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