I'll assume that 1.00 103 means 1x10^3
a)
Yes. Any object which has a velocity vector which does not pass directly through the chosen axis has angular momentum about that axis.
b)
No, mechanical energy is not conserved. Angular momentum is conserved, and any excess energy is dissipated within the door.
c)
W.before = W.after
Solve for both, then solve for final w. (note that W is angular momentum, w is angular velocity)
W.before = I.bullet * w.bullet + I.door * w.door
I.bullet = m.bullet * (.91 m)^2
w.bullet = V.bullet / (.91 m)
W.before = m.bullet * V.bullet * (.91 m) + 0
W.after = I.bullet * w.bullet + I.door + w.door
W.after = m.bullet * (.91 m)^2 * w.bullet + 1/3 * m.door * (1m)^2 * w.door
since w.bullet = w.door, we'll just call that w
W.after = w * (m.bullet * (.91 m)^2 + 1/3 * m.door * (1m)^2)
Set W.before = W.after and solve for w
m.bullet * V.bullet * (.91 m) + 0 = w * (m.bullet * (.91 m)^2 + 1/3 * m.door * (1m)^2)
.006 kg * 1000 m/s * .91 m = w * (.006 kg * (.91 m)^2 + 1/3 * 18.2 kg * (1 m)^2)
5.46 kgm^2/s = w * 6.0716 kgm^2
w = .8992 radians/second
d)
Before:
Ek = 1/2 * m.bullet * V.bullet^2
Ek = 1/2 * .006 kg * 1000000 m^2/s^2
Ek = 3000 joules
After:
Ek = I.bullet * w^2 + I.door * w^2
Ek = (.8992 rad/sec)^2 * (.006 kg * (.91 m)^2 + 1/3 * 18.2 kg * (1m)^2)
Ek = 4.909 joules
Note that kinetic energy after is significantly less than kinetic energy before.
A 0.006 00-kg bullet traveling horizontally with a speed of 1.00 x 10^3 m/s enters an 20.1-kg door, imbedding itself 10.9 cm from the side opposite the hinges as in the figure below. The 1.00-m wide door is free to swing on its frictionless hinges.
At what angular speed does the door swing open immediately after the collision?
Calculate the energy of the bullet-door system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision (what is the final and initial KE's)
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