A uniform rod of mass `M` and length `l` is placed on a smooth- horizontal surface with its one end pivoted to the surface. A small ball of mass `m` moving along the surface with a velocity `v_(0)`, perpendicular to the rod, collides elastically with the free end of the rod. Find:
a. the angular velocity of the rod after collision.
b. the impulse applied by the pivot on the rod during collision. (Take `M//m = 2`)

Answer

`2/5mv_(0)`

Explanation

a. Since net torque about the pivot for (ball `+`rod) system is zero using conservation of agular momentum
`mv_(0)l=mvl+(ml^(2))/3omega`
`v_(0)=v+2/3omegal [:'M=2m]` ………i
Also, for elastic collision `e=1`
`omegal-v=v_(0)`…ii

Solving eqn i and ii we get
`2v_(0)=5/3omegal`
`:. omega=(6v_(0))/(5l)`
b.Now for rod, angular impulse about the pivot`=`change in angular momentum about the pivot
`Jl=(Ml^(2))/omega`

`:. J=2/3momegal=2/3mxx(6v_(0))/5=(4mv_(0))/5`
Now using imple momentum equation for the rod
`J+J'=(Momegal)/2`
`implies4/5mv_(0)+J'=2m(omegal)/2implies4/5mv_(0)+J'=6/5mv_(0)`
`:. J'=2/5mv_(0)`