A thin uniform square plate with side l and mass M can rotate freely about a stationary vertical axis coinciding with one of its sides. A small ball of mass m flying with velocity v at right angles to the plate strikes elastically the centre of it. Find:
(a) the velocity of the ball `v^'` after the impact,
(b) the horizontal component of the resultant force which the axis will exert on the plate after the impact.

Explanation

(a) In the process of motion of the given system the kinetic energy and the angular momentum relative to rotation axis do not vary. Hence it follows that
`1/2mv^2=1/2mv^('^2)+1/2((Ml^2)/(3))omega^2`
and `mvl/2=mv^'l/2+(Ml^2)/(3)omega`
From these equations we obtain
`v^'=((3m-4M)/(3m+4M))v`. and `omega=(4v)/(l(1+4m//3M))`
As `overset(rarr')vuarruarrvecv`, so in vector form `overset(rarr')v=((3m-4M)/(3m+4M))vecv`
(b) Obviously the sought force provides the centripetal acceleration to the C.M. of the rod and is
`F_n=mw_(cn)`
`=Momega^2l/2=(8Mv^2)/(l(1+4M//3m)^2)`