A bullet of mass 10g and speed `500ms^(-1)` is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0m wide and weighs 12kg. 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

Explanation

M.I of door about an axis at one end`=(ML^2)/(3)`
Given `m_b=10^(-2)kg" "v_b=500ms^(-1)`
`md_d=12kg,L=1m`
`:.I=(12xx1^2)/3=4kgm^2`
`K.E_b=1/2xxm_bv_b^2=1/2xx10^(-2)xx(500)^2=1250J`
but K.E`=1/2Iomega^2` i.e `omega^2=(2K.E)/I=(2xx1250)/4`
`omega^2=625` or `omega=25rads^(-1)`