Given,
Mass of the rod, M=0.9kg
Mass of particle, m=0.1kg
Length of the rod, l=1m
Velocity of particle, v=80ms−1
No external torque acting on the system of rod and particle, so angular momentum of system about the suspension is conserved, i.e.; Li=Lf
⇒mvl=Iω...(1)
Where, I is the momentum of inertia of rod and particle system, ω is the angular velocity of system just after the collision.
⇒mvl=(3Ml2+ml2)ω...(2)
Substitute the values in equation (2)
⇒0.1×80×1=(30.9×(1)2+0.1×(1)2)×ω
ω=20rads−1
