(a) Before it hits the door, does the bullet have angular momentum relative the
ID: 1350663 • Letter: #
Question
(a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation?
YesNo
(b) If so, evaluate this angular momentum. (If not, enter zero.)
kg · m2/s
If not, explain why there is no angular momentum.
This answer has not been graded yet.
(c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation.
YesNo
(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.
Explanation / Answer
a and b.
Yes
distance of impact from hinge r = 1 m -10.5 cm = 0.895 m
So, the a ngular momentum of bullet is,
L = r p = 0.895 * (0.00600 * 1.00 * 103) = 5.37 kg-m2/s
c.
No. the collsion is an inelastic, because the bullet is embedded in the door. In this collision,
mechanical energy is not conserved.
d.
Apply the law of conservation of angular momentum:
L = I *
Moment of inertia of bullet about hinge:
I1 = m * r2 = 0.00600 * 0.895 * 0.895 = 4.8 * 10-3 kg-m2
Moment of inertia of the door:
I2 = (1/3) * M * d2 = 20.7 * 1*1 / 3 = 6.90 kg-m2
Total moment of inertia is,
I = I1 + I2= 4.8 * 10-3 + 6.90 = 6.9048 kg-m2
Now L = I *
The angular velocity of door is,
= L/ I = 5.37 / 6.9048 = 0.7777 rad/s
e.
Initial kinetic energy of bullet:
KEi = (1/2) * m * v2 = 0.5 * 0.00600 * (1.00 * 103)2 = 3.00 * 103 J
Final kinetic energy:
KEf = (1/2) * I * 2 = 0.5 * 6.9048 * 0.7777 = 2.685 J
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