a
when block hits ridge at \(O\), it will start rotating about an axis passing through \(O\) and perpendicular to the plane of paper. No external torque acts on block hence angular momentum is conserved Angular momentum before hitting,

\(Li = M \cdot V \times AC\)

\(= MV \cdot( a / 2)=[( mva ) / 2]\)

after hitting, \(L_{f}=I_{0} \omega\)

\(I_{0}\) is MI of block about axis through \(O\) and perpendicular to plane of block.

if \(I_{c}\) is MI about \(C\) then \(I_{C}=\left[\left(M a^{2}\right) / 6\right]\)

From parallel axes theorem,

\(I_{0}=I_{c}+M r^{2}\)

and \(r^{2}=(a / 2)^{2}+(a / 2)^{2}=\left(a^{2} / 2\right)\)

\(I_{0}=(1 / 6) M a^{2}+(1 / 2) M a^{2}=(2 / 3) M a^{2}\)

\(L _{ f }=(2 / 3) Ma ^{2} \cdot \omega\)

also \(L _{ i }= L _{ f }\) hence

\([( mva ) / 2]=(2 / 3) Ma ^{2} \omega\) hence \(\omega=(3 v / 4 a )\)