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A bullet of mass 10 g and speed 500 m/s ...

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

Text Solution

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Mass of the bullet `m=10g=0.01kg`
Velocity of the bullet `v=500m//s`
Width of the door `l=1.0m`
Radius of the door `r=(1)/(2)m`
Mass of the door `M=12kg`
Bullet is in the middle of door, so its distance from fixed axis `r=(l)/(2)`
`therefore r=(l)/(2)m`
angular momentum imparted by the bullet on the door
`therefore L=mvr`
`=0.01xx500xx(1)/(2)`
`=2.5JS`
Moment of inertia of door about vertical axis from one end of door
`I=(Ml^(2))/(3)`
`=(12xx(1)^(2))/(3)`
`=4kgm^(2)`
but angular momentum `L=Iomega`
`omega=(L)/(I)`
`=(2.5)/(4)`
`=0.625" rad/s"`
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