A cubical block of side a is moving with velocity v on a horizontal smooth palne as shown . It hits a ridge at point O. The angular sped of the block after it hits O is

Options

A. 3v/2a
B. `sqrt(3)v//sqrt(2)a`
C. 3v/4a
D. Zero

Answer

C

Explanation

Angular momentum with respect O before colliding with `O=Mv""(a)/(2)`
On collision , the block will rotate about the side passing through O.
By law of conservation of angular momentum

`Mv""(a)/(2)=l omega `
`implies Mv""(a)/(2)=((Ma^(2))/(6)+(Ma^(2))/(2))omega implies omega=(3v)/(4a)`
Where I is the axis perpendicular to the plane passing through O.