Figure shows a cubical wooden block of side length a and mass `M` which is resting over a horizontal surface and constrained to rotate about its right lower edge `D`. A bullet of mass m moving horizontally, directly towards its centre with velocity `v` strikes and gets embedded in it. During subsequent motion the block just topples, calculate `v`. Assume `m lt lt M`.

Answer

`(sqrt(2)-1)`

Explanation

Moment of inertia of block about its edge `D` is equal to `2/3 Ma^(2)`
When bullet strikes the blocks and gets embedded into it even then moment of inertia remains practically equla to `I=2Ma^(2)//3` because `m lt lt M`
During the collision, a reaction is offered by `D`. Hence momentum of the system dies not remain constant but the angular momentum of the sytem remains constant about `D` because angular impulse produced by the reaction is zero about `D` itself.
If angular velocity of the block, just after the collision is `w` then according to law of coservation of angular momentum,
`I.omega=mv.a/2implies omega=(3mv)/(4Ma)`
The block will jus topple, if the block reaches to the positioin shown in figure
Here `/_\PE=-/_\KE` of the system
`Mg(a/sqrt(2)-a/2)=1/2Iomega^(2)`

`implies v=(2M)/Msqrt(2/3 ga(sqrt(2)-1))`