a bullet of mass of 12 grams is fired into a large block of mass 1.5 kg suspended from light vertical wires. the bullet imbeds in the block and the whole system rises 10 cm. find the velocity of the bullet just before the collision.

Answer 1

First find the momentum of the block and embedded bullet right after the collision from conservation of energy:
(1/2) m v^2 = m g h
note m cancels
h = 10 cm = .1 meter
solve above for v
m = 1.5 + .012 kg
now momentum before = momentum after
.012 Vbullet + 0 Vmass = 1.512 v
so
Vbullet = (1.512 / .012) v