A 0.00525–kg bullet traveling horizontally with a speed of 1.00 10 3 m/s enters
ID: 1481374 • Letter: A
Question
A 0.00525–kg bullet traveling horizontally with a speed of 1.00 103 m/s enters a 16.0–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.
At what angular speed does the door swing open immediately after the collision in rad/s? (The door has the same moment of inertia as a rod with axis at one end.)
(d) Calculate the energy of the door–bullet system in J. (Enter the kinetic energy of the door–bullet system just after collision.)
Explanation / Answer
let
m = 0.00525 kg
M = 16 kg
L = 1 m
R = L - 0.2
= 1 - 0.2
= 0.8 m
w = ?
a)
Apply conservation of angular momentum
m*v*R = I*w
m*v*R = (M*L^2/3 + m*r^2)*w
==> w = m*v*R/(M*L^2/3 + m*r^2)
= 0.00525*1000*0.8/(16*1^2/3 + 0.00525*0.8^2)
= 0.787 rad/s
b) KEf = 0.5*I*w^2
= 0.5*(M*L^2/3 + m*r^2)*w^2
= 0.5*(16*1^2/3 + 0.00525*0.8^2)*0.787^2
= 1.65 J
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