A 0 00400 Kg Bullet Traveling Horizontally With Speed 1 00 103 M S Strikes A 16 1 Kg Door Embedding Itself 10 2 Cm From 1 (42.1 KiB) Viewed 18 times
A 0.00400 kg bullet traveling horizontally with speed 1.00×103 m/s strikes a 16.1 kg door, embedding itself 10.2 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? (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.
(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? * cm
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