A 1 185.0 kg car traveling initially with a speed of 25.000 m/s in an easterly d
ID: 1484070 • Letter: A
Question
A 1 185.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 300.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east. (a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) WebAssign will check your answer for the correct number of significant figures.20.89. m/s east (b) What is the change in mechanical energy of the car–truck system in the collision? (c) Account for this change in mechanical energy.
Explanation / Answer
How to find all of this:
m1 = 1185.0 kg
u1 = 25.000 m/s
v1 = 18 m/s
m2 = 9,300 kg
u2 = 20.0000 m/s
Here's the equation... so enter values and solve
a) Velocity of the truck right after the collision :
v2 = u1 * { [2 * m1 ] / [ m1 + m2 ] } - { [ m1 -m2 ] / [ m1 + m2 ] } * u2
v2 = (25.0 m/s) * { [ 2 * (1185 kg) ] / [ (1185 kg) + (9300 kg) ] } - { [ (1,185 kg) - (9300 kg) ] / [ (1,185 kg) + (9300 kg) ] } * (20.0 m/s)
v2 = (25.0 m/s) * { [ 2,370 kg ] / [ 10,485 kg ] } - { [ -8115 kg ] / [ 10,485 kg ] } * (20.0 m/s)
v2 = (25.0 m/s) * { 0.22 } - { -0.77 } * (20.0 m/s)
v2 = (5.5 m/s) - (-15.4 m/s)
v2 = 20.9 m/s
b) change in energy:
Calculate total kinetic energy pre-collision, KE = 0.5 * m * v^2
Car
KE = 0.5 * (1,185 kg) * (25.0 m/s)^2
KE = (550 kg) * (625 m^2/s^2)
KE = 370,3125 J
Truck
KE = 0.5 * (9300 kg) * (20.0 m/s)^2
KE = (4,650 kg) * (400 m^2/s^2)
KE = 1,860,000 J
Total pre-collision energy = 1,860,000 + 370,3125 = 5,563,125 J
Post-collision energy
Car
KE = 0.5 * (1185kg) * (18.0 m/s)^2
KE = (592.5 kg) * (324 m^2/s^2)
KE = 191,970 J
Truck
KE = 0.5 * (9300 kg) * (20.9 m/s)^2
KE = (4,450 kg) * (436.81 m^2/s^2)
KE = 19438045 J
Total post-impact energy = 2,135,7745 J
A loss of 1,579,4620 Joules.
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