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

Options

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

Answer

A

Explanation

Angular momentum of block w.r.t. O before collision with `O =Mv(a/2)`
On collision, the block will rotate about the side passing through O. Now its angualr momentum = `I Omega`
By law of conservation of angular momentum
`Mv(a/2) = I Omega rArr Mv(a/2) = ((Ma^(2))/6 + (Ma^(2))/2)`
`= Omega rArr Omega = (3v)/(4a)`
Where I is the moment of inertia of the block about the axis perpendicular to the plane passing through O.