A 0.00400-kg bullet traveling horizontally with speed 1.00 103 m/s strikes a 19.

Question

A 0.00400-kg bullet traveling horizontally with speed 1.00 103 m/s strikes a 19.8-kg door, embedding itself 11.7 cm from the side opposite the hinges as shown in the figure below. The 1.00-m wide door is free to swing on its frictionless hinges.

(a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? Yes No Correct: Your answer is correct.

(b) If so, evaluate this angular momentum. (If not, enter zero.) 4.027 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. kg · m2/s If not, explain why there is no angular momentum. there will be angular momentum Score: 1 out of 1 Comment:

(c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation. Yes No Correct: Your answer is correct.

(d) At what angular speed does the door swing open immediately after the collision? rad/s

(e) Calculate the total 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. KEf = J KEi = J

i need answer for( d) and (e)

Explanation / Answer

d)
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



e)
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.

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