Thread: What percentage of the system's original kinetic energy was lost?

A 0.390 kg piece of wood is at rest on a frictionless table. A 8.30 g bullet, moving with a speed of 495 m/s, strikes the piece of wood, and is embedded in it. After the collision, the piece of wood and the bullet move slowly down the table. What percentage of the system's original kinetic energy was lost?

The table is frictionless, so no energy is lost.
KE = 1/2 * m * (v^2)

the KE of the wood is initially zero, because of the zero (v) velocity.
The KE of the Bullet is found by substituting the values.
Add them together. Notice that if the bullet combines with the wood, it will slow down. It will also have more mass. The KE would end up being the same, but since there is no final velocity, there is no way to calculate the change in KE...
So, like I said, No friction, no energy loss.
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energy lost = 0%
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initial momentum = (0.0083)(495) = 4.1085 kg-m/s
final momentum = 4.1085 = (0.390 + 0.0083)V = 0.3983V
V = 4.1085/0.3983 = 10.32 m/s

initial KE of bullet = 1/2(0.0083)(495)² = 1017 J
final KE of bullet & wood = 1/2 (0.3983)(10.32)² = 21. J
lost energy = (1017 - 21)/1017 = 996/1017 = 98% of KE lost ANS