What If? Imagine now that the door is hanging vertically downward, hinged at the top, so that the figure is a side view of the door and bullet during the collision. What is the maximum height (in cm) that the bottom of the door will reach after the collision?
I've already answered parts a-e. I'm just stuck on part f.
Here are the (correct) answers I got for parts a-e
(a) Yes
(b) 5.31 kg * m^2 / s
(c) no
(d) 0.964 rad/s
(e) KEf = 2.56 J; KEi = 3000 J
A 0.00600 kg bullet traveling horizontally with speed 1.00 103 m/s strikes a 16.5 kg door, embedding itself 11.5 cm from the side opposite the hinges as shown in the figure below. The 1.00 m wide door is free to swing on its frictionless hinges.
(a) Before it hits the door, does the bullet have
(b) If so, evaluate this angular momentum (in kg · m2/s). (If not, enter zero.)
(c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation.
(d) At what angular speed (in rad/s) 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. (Enter your answers in J.)
(f) What If? Imagine now that the door is hanging vertically downward, hinged at the top, so that the figure is a side view of the door and bullet during the collision. What is the maximum height (in cm) that the bottom of the door will reach after the collision?
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