A disc of mass `M` and radius `R` is rolling purely with centre's velcity `v_(0)` on a flat horizontal floor when it hits a step in the floor of height `R//4` The corner of the step is sufficiently rough to prevent any slippoing of the disc against itself. What is the velocity of the centre of the disc just after impact?

Options

A. `4v_(0)//5`
B. `4v_(0)//7`
C. `5v_(0)//6`
D. none of these

Explanation

Angular momentum will remain conserved at the point of impact P, and just after impact it start rotating about point P.
`thereforeMv_(0)(R-(R)/(4))+(1)/(2)MR^(2)((v_(0))/(R))=(3)/(2)MR^(2).omega`
`thereforeomega=(5v_(0))/(6R)`
`thereforev_(COM)=omegaR=(5v_(0))/(6)`