Use the exact values you enter in previous answer(s) to make later calculation(s
ID: 2136784 • Letter: U
Question
Use the exact values you enter in previous answer(s) to make later calculation(s).
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 8,700-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.
Explanation / Answer
How to find all of this
m1 = 1,115 kg
u1 = 25.0 m/s
v1 = 18 m/s
m2 = 8,700 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 * (1,115 kg) ] / [ (1,114 kg) + (8,700 kg) ] } - { [ (1,115 kg) - (8,700 kg) ] / [ (1,115 kg) + (8,700 kg) ] } * (20.0 m/s)
v2 = (25.0 m/s) * { [ 2,230 kg ] / [ 9,815kg ] } - { [ -7,585 kg ] / [ 9,815 kg ] } * (20.0 m/s)
v2 = (25.0 m/s) * { 0.2272 } - { -0.7727 } * (20.0 m/s)
v2 = (5.68 m/s) - (-15.45 m/s)
v2 = 21.1 m/s
Calculate total kinetic energy pre-collision, KE = 0.5 * m * v^2
Car
KE = 0.5 * (1,115 kg) * (25.0 m/s)^2
KE = 348,437.5 J
Truck
KE = 0.5 * (8,700 kg) * (20.0 m/s)^2
KE = 1,740,000 J
Total pre-collision energy = 2,088,437.5 J
Post-collision energy
Car
KE = 0.5 * (1,115 kg) * (18.0 m/s)^2
KE = 180,630 J
Truck
KE = 0.5 * (8,700 kg) * (21.1 m/s)^2
KE = 1936663.5 J
Total post-impact energy = 2,117,293.5 J
A loss of 28,856 Joules.
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