A cube of mass m and height `H` slides with a speed `v`. It strikes the obstacle of height `h = (H/4)`. Find the speed of the `CM` of the cube just after the collision.

Answer

(sqrt(5))/6`

Explanation

`mv(H/2-h)=m[((H^(2)+H^(2)))/12=(H/2)^(2)+(H/2-h)^(2)]omega`

`omega=((H/2-h)v)/((H^(2))/6+(H^(2))/4+(H/2-h)^(2))`
`v=romega=((H/2-h)v.sqrt((H^(2))/4+(H/2-h)^(2)))/(5/12H^(2)+(H/2-h)^(2))`
If `h=H/4` we have
`v'=(H/4v.sqrt((H^(2))/4+(H^(2))/16))/((5/12+1/16)H^(2))`
`(v/4.sqrt(5)/4)/((15+3)/48)=(3sqrt(5))/18vimpliesv' (sqrt(5))/6v`