A door 1.0 m wide, of mass 15 kg, is hinged to one side so that it can rotate wi

Question

A door 1.0 m wide, of mass 15 kg, is hinged to one side so that it can rotate without friction about a vertical axis. It is unlatched. Police officer fires a bullet with a mass of 10g and a speed of 400 m/s into the exact center of the door, in a direction perendicular to the plane of the door (see figure). Note that the moment of inertia of a door of width d and mass M is Idoor=Md2/f.

1.find the angular speed of the door just agter the bullet imbeds itself in the door.

2.calculate the kinetic energy of the sys tem right before the bullet hits the door and right after.

3.I s energy conserved in the collision?

Explanation / Answer

1)By law of conservation of angular momentum,

Li = Lf

Libullet = Lf(door+bullet)

mvr = I(door+bullet)

mvd = (1/3*Md^2)+ [m(d)^2]

Plugging values,

0.010*400*0.5 = (1/3*15*0.5^2)+ (0.010*0.5^2)    => =1.6 rad/s

2)KEi = 1/2mv^2 = ½*0.010*400^2 = 800 J

KEf = 1/2I^2 = 1/2*[(1/3*15*0.5^2)+(0.010*0.5^2)]*1.6^2 = 1.6 J

3) Since KEi > KEf . energy not conserved.

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