Question

A bullet of mass 10g and speed 500m/s is fired into a door get embedded exactly at the center of the door. The door is 1.0m wide and weighs 12kg. It is hinged at one end and rotates about vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.

Answer 1

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Answer 2

Answer:

ω=0.625rad/s

Explanation:

mass of bullet=m=10×10  ⁻³kg

velocity of bullet = v= 500m/s

distance=r= 0.5m

Angular momentum imparted by bullet on the door=mvr

=(10×10  ⁻³ )×500×0.5 kgm² /s

Mass of door = M = 12kg

Length = L = 1m

Moment of inertia of the door, I=ML²/3=  

​  (1/3)×12×1²       =4kgm²

 Angular momentum of the system after the bullet gets embedded≈Iω

From conservation of angular momentum about the rotation axis,

mvr=Iω

⟹ω=0.625rad/s