A 0.006 00-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters

Question

A 0.006 00-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 15.6-kg door, imbedding itself 8.60 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. evaluate this angular momentum. At what angular speed does the door swing open immediately after the collision? 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.

Explanation / Answer

Li = m r v Lf = (m r^2 + 1/3 M L^2 ) w 0.006*(1-.086)*1E3 = ( .006*(1-.086)^2 + 1/3*15.6*1^2)w w=1.0536 Ei = 1/2 mv^2=0.5*0.006*(1E3)^2=3000 Ef = 1/2 ((m r^2 + 1/3 M L^2 ) w^2=0.5*( .006*(1-.086)^2 + 1/3*15.6*1^2)1.0536^2 =2.89 so it is less than just bullet

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