In the figure shown in the text, if the block is a cube of side `a`
find
(a). `omega` just after impact
(b). Loss of mechanical energy during impact
(c) minimum value of v so as the block overcomes the obstacle and does not turn back.

Explanation

(a). From conservation of angular momentum about `O`
`L_(i)=L_(f)`
`mv_(c)r_(bot)=I_(o)omega=(I_(c)+mr^(2))omega`
`impliesmv((a)/(2))=[(ma^(2))/(6)+m((a)/(sqrt(2)))^(2)]omega`
`impliesomega=(3)/(4)((v)/(a))`
(b). loss of mechanical energy
`=E_(i)-E_(f)`
`=(1)/(2)mv^(2)-(1)/(2)I_(o)omega^(2)`
`=(1)/(2)mv^(2)-(1)/(2)[(ma^(2))/(6)+m((a)/(sqrt(2)))^(2)]((3)/(4)(v)/(a))^(2)`
`=(5)/(16)mv^(2) `,Brgt (c)
Block overcomes the obstacle at `O` if centre of mass rises upto a height `(a)/(sqrt(2))` as shown in figure (from the initial height `(a)/(2))`
Because after that torque of `mg` about `O` will itself rotate the block on other side as shown in figure
`therefore` decrease in rotational kinetic energy `=` increase in gravitational potential energy
`therefore(1)/(2)I_(o)omega^(2)=mg((1)/(sqrt(2))-(a)/(2))`
or `(1)/(2)[(ma^(2))/(6)+m((a)/(sqrt(2)))^(2)][(3)/(4)(v)/(a)]^(2)=mg` `((a)/(sqrt(2))-(a)/(2))`
`thereforev=sqrt(1.1ga)`