A 0.006-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 21.1-kg door, imbedding itself 10.4 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.
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C) Angular momentum of the system is the same before and after. Assume rotation about the door hinge.
d) before (1/2)m v^2
after =(1/2) I w^2
it better be much less after.
Angular momentum of bullet alone about door hinge L = m v r = .006 * 10^3 * (1.000-.104) = 5.376 kg m^2/s
Angular momentum of system after collision (ignore mass of bullet)
L = I w where I = ( 1/3) m b^2 and w is angular velocity and b is door width.
L = (1/3)(21.1)(1)^2 w = 7.033 w
so
7.033 w = 5.376
w = .7644 rad/s
=
A 0.006-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 21.1-kg door, imbedding itself 10.4 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.
(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.)
Answer: 0.749rad/s
(d) Calculate the energy of the door-bullet system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.
Answer:_____J
I can't figure out C
5 answers
I'm sorry I meant I cant figure out part D.
energy before = (1/2) m v^2
energy after = (1/2) I w^2
energy after better be much smaller than energy before
energy after = (1/2) I w^2
energy after better be much smaller than energy before
before (1/2) (.006)(10^3)^2 = 3000 Joules
after (1/2)(7.003)(.7644)^2 = 2.046 Joules
the bullet entering the wood turned most of the energy into heat
after (1/2)(7.003)(.7644)^2 = 2.046 Joules
the bullet entering the wood turned most of the energy into heat
By the way, we did not get the same answer for part c
3 answers