A10g bullet traveling at 450m/s strikes a 7.0kg , 1.1m--wide door at the edge opposite the hinge. the bullet embeds itself in the door, causing the door to swing open. what is the angular velocity of the door just after impact?

Using momentum balance, we can say that the momentum before the strike and after the strike are equal. Therefore:

m1 v1 + m2 v2 = m’ v’

However the door is not moving before the strike therefore v2 = 0:

0.010 kg * 450 m/s + 0 = (7.010 kg) v’

v’ = 0.64 m/s 

6/12=1/2

step-by-step explanation:

pe = weightxheight = 2000x1000=2,000,000joules