The correct option is
D its motion will be rolling without slipping only after some time.
During collision, all forces passes through the COM. Hence,
τcom=0,
⇒ΔL=0
⇒Iω=constant,
∴ω will remain same.
Collision is elastic, thus after the collision, linear velocity will be same in magnitude and opposite in direction (So option A is correct).
Initially, cylinder rolls without slipping.
Hence,
v=ωR (pure rolling)
After collision
→vP,g→ velocity of
P w.r.t. ground.
→vP,g=→vP,com+→vcom,g
where
→vP,com→ velocity of
P w.r.t COM
=−ωR^i
→vcom.g→ velocity of com with respect to ground
=−v^i
⇒→vP,g=−ωR^i−v^i
⇒→vP,g=−2ωR^i [From Eq. (i)]
Hence just after impact, motion of cylinder will be rolling with slipping. The friction will act rightward, which oppose the motion and reduce angular velocity.
Due to
α (angular acceleration due to friction),
ω decreases and becomes
0 . Due to
α after some time, angular velocity will be in anticlockwise direction and it will be equal to
v′=ω′R (it starts rotating without slipping) as shown. Then, there will be no slipping between ground and cylinder and no friction will act.
So options A, C, D are correct.