A bullet of mass m1=6.00 g traveling with a speed of v1= 1600 m/s impacts a door
ID: 1465236 • Letter: A
Question
A bullet of mass m1=6.00 g traveling with a speed of v1= 1600 m/s impacts a door with a witdth w=1.80 m and mass of m2=28.0 kg embedding itself at x=20.0 cm from the edge of the door oppisite the hinges. the door is free to swing on its hinges.
a) At what angular speed, w does the door swing open immediately after the collision? Hints: use the conservation of angular momentum. The door has the same moment of inertia as a rod with axis at one end.
b) Calculate the change of the kinetic energy of the door-bullet system during the collision, Delta K
Explanation / Answer
a) Apply cosnervation of angular momentum
let w is the angular speed of the door
I2*w = m1*v1*R
(m2*w^2/3 + m1*R^2)*w = m1*v1*R
(28*1.8^2/3 + 0.006*(1.8 - 0.2)^2)*w = 0.006*1600*(1.8 - 0.2)
30.25*w = 15.36
w = 15.36/30.25
= 0.508 rad/s <<<<<<-----------Answer
b) KI = 0.5*m1*v1^2
= 0.5*0.006*1600^2
= 7680 J
KF = 0.5*I2*w^2
= 0.5*(28*1.8^2/3 + 0.006*(1.8 - 0.2)^2)*0.508^2
= 3.9 J
delta_K = Ki - Kf
= 7680 - 3.9
= 7676 J <<<<<<-----------Answer
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