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 becomes 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=intN dt=m.(v)/(2)-(-mv)=(3)/(2)mv`........(i)
`mu JR=intmu(N dt)R=((2)/(5)mR^(2)omega-0)=(2)/(5)mR^(2)omega` ............(ii)

From Eqs. (i) and (ii), we get
`(3)/(2)mvR mu=(2)/(5)mR^(2)omega`......(iii)
Let V and `V_(1)` be the speeds of the plank and the sphere, respectively , in the horizontal direction.
`mu J=intmu N dt=MV=mV_(1)`
From Eqs. (i) and (iv), `mu((3)/(2))mv=MV`
`V=(3)/(2)(mumv)/(M)=(3)/(2)(4)/(15)(mRomega)/(M)=(2)/(5)(m)/(M)Romega`
And `V_(1)=(2)/(5)Romega=2m//s`.