A solid ball of mass `m` and radius `r` spinning with angular velocity `omega` falls on a horizontal slab of mass `M` with rough upper surface (coefficient of friction `mu`) and smooth lower surface. Immediately after collision the normal component of velocity of the ball remains half of its value just before collision and it stops spinning. Find the velocity of the sphere in horizontal direction immediately after the impact (given: `Romega= 5`).

Answer

2

Explanation

`J=int N dt =mv/2 -(-mv)=3/2,mv`………..i
`muJR=intmu(N dt)R=(2/5mR^(2)omega-0)=2/5mR^(2)omega`……..ii

From eqn i and ii we get
`3/2mv rmu=2/5mR^(2)omega`……….iii
let `V` and `V_(1)` be the speeds of the plank and the sphere, respectively in the horizontal direction.
`muJ=intmj Ndt=Mv=mV_1`........iv
Form eqn i and iv `mu(3/2)mv=MV`
`V=3/2(mumv)/M=3/2 4/15 (mRomega)/M=2/5Romega`
and `V_(1)=2/5Romega=2m//s`