A 0.004 00-kg bullet traveling horizontally with a speed of 1.00 times 103 m/s e

Question

A 0.004 00-kg bullet traveling horizontally with a speed of 1.00 times 103 m/s enters an 16.7-kg door, imbedding itself 11.5 cm from the side opposite the hinges as in the figure below. The 1.00-m wide door is free to swing on its frictionless hinges. Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? Yes No If so, evaluate this angular momentum. (If not, enter zero.) kg middot m2/s Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation. Yes No At what angular speed does the door swing open immediately after the collision? rad/s Calculate the 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. KEf = KEj =

Explanation / Answer

a) YES b) 0.004*1000*0.115 = 0.46 kg-m/s c)NO, energy is lost in collision d) conserving ANGULAR MOMENTUM 0.46 = 16.704*w w = 0.46/16.704 = 0.027 rad/sec

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