Question

A 0.005kg bullet travelling horizontally with a speed of 1000m/s enters an 18kg door, embedding itself 10cm from the side opposite the hinges as in figure. The 1m wide door is free to swing on the hinges.

a) Before it hits the door, does the bullet have angular momentum relative the door’s axis of rotation? Explain.

b) Is mechanical energy conserved in this collision?

c) If the door opens immediately after collision, calculate the angular speed the door swing. 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 or equal to the kinetic energy of the bullet before collision.

Answer 1

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

1. Of course m * R x V = .005 * .9 * 1000
2. No reason it should be, heat generated when bullet strikes. However Angular momentum does not change.
3. I am going to ignore the I of the bullet after collision which is .005*.81
so
.005 * .9 * 1000 = I w = (1/3)18(1)^2 w^2
So w = 0.866 radians/second

Answer 2

I think you can do (1/2) m v^2 and (1/2) I w^2