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A 0.00600-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters a 17.5-kg door,...

A 0.00600-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters a 17.5-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. A) 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.) B) 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. (Give the kinetic energy of the door-bullet system just after collision.)
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From the law of conservation of the angular momentum, the initial angular momentum of the bullet about rotational axis is equal to the final angular momentum of the system. (0.006000 kg)(1.00x103 m/s) (1.00 m-0.20 )17.5 kg+0.006000 kg) (1.00 m) o Therefore, ( 0.006000 kg)(1.00 x 103 ms)(1.00 m-0.20 m) 17.5 kg+0.006000 kg) (1.00 m a = 0.8225 rad s The rotational kinetic energy of the system of door and bulet is, 17.5 kg+0.006000 kg) (1.00 m ) (0.8225 rad/s) 213) As the collision is inelastic, the final kinetic energy of the system is less than that of the initial kinetic energy of the bullet.

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