A 0.00600 kg bullet traveling horizontally with speed 1.00×103 m/s strikes a 21.1 kg door, embedding itself 11.1 cm from

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A 0 00600 Kg Bullet Traveling Horizontally With Speed 1 00 103 M S Strikes A 21 1 Kg Door Embedding Itself 11 1 Cm From 1 (79.73 KiB) Viewed 96 times

A 0.00600 kg bullet traveling horizontally with speed 1.00×103 m/s strikes a 21.1 kg door, embedding itself 11.1 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 angular momentum relative the door's axis of rotation? \begin{tabular}{l} Yes \\ No \\ \hline \end{tabular} (b) If so, evaluate this angular momentum (in kg⋅m2/s ). (If not, enter zero.) kg⋅m2/s If not, explain why there is no angular momentum. (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? rad/s (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.) KEf​=1 KE1​=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 door and bullet during the collision. What is the maximum height (in cm ) that the bottom of the door will reach after the collision? x cm