A solid cube of wood of side `2a` and mass `M` is resting on a horizontal surface. The cube is constrained to rotate about an axis passing through `D` and perpendicular to face `ABCD`. A bullet of mass `m` and speed `v` is shot at a height of `4a//3` as shown in the figure. The bullet becomes embedded in the cube. Find the minimum value of `v` required to topple the cube. Assume `mgt gtM`.

Explanation

Let `omega` be the angular velocity of cube just after collison about an axis passing through `D`. From conservation of the angular mometum about an axis passing through `D`.

`mv((4a)/3)=I_(D).omega`
Here `I_(D)=I_(com)+mr^(2)`
where `r=` perpedicular distance of axis of rotation passing through `D` from centre of mass (com) of the cube `=sqrt(2)a`
or `(4mva)/3=[M((4^(2)+4a^(2))/12)+M(2a^(2))]omega`
or `(4mva)/3=8/3Ma^(2).omega`
`omega=1/2 (mv)/(Ma)`........i
The cube will topple if its `COM` is just able to reach in a vertical height `h_(2)` as shown in figure.

Hence applying conservation of mechanical energy
`1/2I_(D)omega^(2)=Mg(h_(2)-h_(1))` (`:'I_(D)=8/3Ma^(2))`
or `1/2 (8/3Ma^(2)).{1/2(mv)/(Ma)}^(2)=Mg(sqrt(2-1)a`
or `1/2(m/M)^(2)v^(2)=ag(sqrt(2)-1)`
or `v=M/m[3ag(sqrt(2)-1)]^(1/2)`