Asked by Anonymous

A 0.018 kg bullet traveling 222 m/s penetrates a 3.9 kg block of wood, and exits the block with a velocity of 72 m/s. If the block was stationary on a horizontal frictionless surface before impact, how fast does it move after the bullet exits?
Answered by Damon
momentum before = momentum after

.018*222 = .018*72 + 3.9 * v
Answered by Anonymous
thank you
Answered by Damon
You are welcome :)
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