Figure An overhead view of a bullet striking a door. Hinge 18.0 kg - 0.005 00 kg

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Chapter12: Static Equilibrium And Elasticity
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A 0.005 00-kg bullet traveling horizontally with a speed of 1.00 × 10
3 m/s strikes an 18.0-kg door, embedding itself 10.0 cm from the side opposite the hinges as shown in Figure. The 1.00-m wide door is free to swing on its frictionless hinges. (a) Before it hits the door, does
the bullet have angular momentum relative to the door’s axis of rotation? (b) If so, evaluate this angular momentum. If not, explain why there is no angular momentum. (c) Is the mechanical energy of the bullet–door system constant during this collision? Answer without doing a calculation. (d) At what angular speed does the door swing open immediately after the
collision? (e) Calculate the total energy of the bullet–door system and determine whether it is less than or equal to the kinetic energy’ of the bullet before the collision. (f) What If? Imagine now that the door is hanging vertically downward, hinged at the top, so that Figure  is a
side view of the door and bullet during the collision. What is the maximum height that the bottom of the door will reach after the collision?

Figure
An overhead view of a bullet striking a door.
Hinge
18.0 kg -
0.005 00 kg
Transcribed Image Text:Figure An overhead view of a bullet striking a door. Hinge 18.0 kg - 0.005 00 kg
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