Posted by Ana on Friday, March 19, 2010 at 7:36pm.
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) 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
7.033 w = 5.376
w = .7644 rad/s
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
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
By the way, we did not get the same answer for part c
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