A 0.00525–kg bullet traveling horizontally with a speed of 1.00 103 m/s enters a
ID: 1471918 • Letter: A
Question
A 0.00525–kg bullet traveling horizontally with a speed of 1.00 103 m/s enters a 21.5–kg door, embedding itself 20.0 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.
(a) Before it hits the door, does the bullet have angular momentum relative to the door's axis of rotation? Yes No Correct: Your answer is correct. Explain. This answer has not been graded yet.
(b) Is mechanical energy conserved in this collision? Answer without doing a calculation. Yes No Correct: Your answer is correct.
(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.) rad/s
(d) Calculate the energy of the door–bullet system. (Enter the kinetic energy of the door–bullet system just after collision.) J
Explanation / Answer
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.
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 * (.8 m)^2
w.bullet = V.bullet / (.8 m)
W.before = m.bullet * V.bullet * (.8 m) + 0
W.after = I.bullet * w.bullet + I.door + w.door
W.after = m.bullet * (.8m)^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 * (.8 m)^2 + 1/3 * m.door * (1m)^2)
Set W.before = W.after and solve for w
m.bullet * V.bullet * (.8 m) + 0 = w * (m.bullet * (.8 m)^2 + 1/3 * m.door * (1m)^2)
.00525 kg * 1000 m/s * .8 m = w * (.00525 kg * (.8 m)^2 + 1/3 * 21.5 kg * (1 m)^2)
4.2kgm^2/s = w * 7.1668 kgm^2
w = 0.5857 radians/second
d)
After:
Ek = I.bullet * w^2 + I.door * w^2
Ek = (.5857 rad/sec)^2 * (.00525 kg * (.8 m)^2 + 1/3 * 21.5 kg * (1m)^2)
Ek = 2.459 joules
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