A 0.005-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 20.3-kg door, imbedding itself 10.6 cm from the side opposite the hinges as in the figure below. The 1.00-m-wide door is free to swing on its hinges.

Question

(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 door-bullet system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.

Answer 1

Use conservation of momentum (angular).

massbullet*velocitybullet*radius=massdoorbullet*w door
d) you do that, but energy was lost...

Answer 2

So (.005kg)(1000m/s)(1.06m)=(20.305kg)(w door)
w door=.26101?

Answer 3

I didn't check the math, but you need units ALWAYS. If your math is correct, w= .26 radians/sec