A solid sphere is rolling purely on a rough horizontal surface (coefficient of kinetic friction `=mu`) with speed of centre `=u`. It collides in-elastically with a smooth vertical wall at a certain moment, the coefficient of restituting being `(1)/(2)`. The sphere will begin pure rolling after a time.

Options

A. `(3u)/(7mug)`
B. `(2u)/(7mug)`
C. `(3u)/(5mug)`
D. `(2u)/(5mug)`

Answer

A

Explanation

`a=(mumg)/(m)=mug`

`omega=(u)/(R)impliesalpha=((mumg)R)/(2//5mR^(2))=(5mug)/(2R)`
Pure rolling will start when
`((u)/(2)-at)=R(alphat-omega)`
or `((u)/(2)-mugt)=R((5mug)/(2R)t-(u)/(R))`
solving we get, `t=(3u)/(7mug)`