A 0.00600-kg bullet traveling horizontally with speed 1.00 x 103 m/s strikes a 20.5-kg door, embedding itself 10.5 cm fromthe side opposite the hinges as shown in the figure below. The 1.00-m wide door is free to swing on its frictionless hinges.Hinge(a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation?O YesNo(b) If so, evaluate this angular momentum. (If not, enter zero.)|kg · m²/sIf not, explain why there is no angular momentum.This answer has not been graded yet.(c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation.YesNo(d) At what angular speed 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 kineticenergy of the bullet before the collision.KE: =KE, =

Given,Mass of the bullet travelling, mb =0.00600 kgHorizontally with speed of bullet, v = 1×103…

Answered: A 0.00600-kg bullet traveling…

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