Posted by Amy on .
A 0.005kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 20.3kg door, imbedding itself 10.6 cm from the side opposite the hinges as in the figure below. The 1.00mwide door is free to swing on its hinges.
(c) At what angular speed does the door swing open immediately after the collision? (The door has the same moment of inertia as a rod with axis at one end.
(d) Calculate the energy of the doorbullet system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.

physics 
bobpursley,
Use conservation of momentum (angular).
massbullet*velocitybullet*radius=massdoorbullet*w door
d) you do that, but energy was lost... 
physics 
Amy,
So (.005kg)(1000m/s)(1.06m)=(20.305kg)(w door)
w door=.26101? 
physics 
bobpursley,
I didn't check the math, but you need units ALWAYS. If your math is correct, w= .26 radians/sec