Question

# A 0.00400-kg bullet traveling horizontally with speed 1.00 103 m/s strikes a 17.7-kg door, embedding itself...

A 0.00400-kg bullet traveling horizontally with speed 1.00 103 m/s strikes a 17.7-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? Yes No (b) If so, evaluate this angular momentum. (If not, enter zero.) kg · m2/s If 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. Yes No (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 kinetic energy of the bullet before the collision. KEf = J KEi = J

a)

yes , bullet had angular momentum relative the door's axis of rotation

b)

r = distance of point where bullet hits from axis of rotation = 10.2 cm = 0.102 m

Lbullet = angular momentum of bullet = mbullet vbullet r = (0.004) (1000) (0.102) = 0.41 kgm/s

c)

since the bullet gets embedded in the door, hence it is an inelastic collision , and energy is not conserved in inelastic collision

d)

Idoor = moment of inertia of door = M w2 /3= 17.7 (1)2/3 = 5.9

Ibullet = m r2 = (0.004) (0.102)2 = 0.000042

W = angular velocity

using conservation of angular momentum

Lbullet = (Idoor + Ibullet) W

0.41 = (5.9 + 0.000042) W

W = 0.07 rad/s