Question

A uniform door initially at rest has a width W = 1.5m and mass M= 20 kg. The door is hinged on one side and can freely rotate about the hinges without friction. A policeman fires a bullet, m, = 15 g into the door at a speed v. = 500m/s. The bullet takes a horizontal path and strikes the door in a perpendicular direction at the exact center. The moment of inertia of the door about an axis passing through the hinges is given by I. = 1/3MW2. a) Find the angular velocity, o of the door just after bullet is embedded. b) What is the initial kinetic energy of the bullet door system?

Answer 1

I= mw 3 hing mb=15g Mezorg V6=500 m/s Initial kinetic energy = IC E of bullet Aqui 2 - f xoxlox500² = 1875 get et embeded- mbro bullet hit centre of door C. and It = lot In I mw+ mu Şx20x1.5 + 18x103 final energy = 2 I 그 2 - 15 m Iu w im kg angular velocity Ef 1 xlsx w

from conservation of energy Ei =. Ef 2 7.5w 1875 (@) W =15.81 r rad/see ab Initial kinetic energy 1875 joule.