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Question

A uniform solid cylinder rolls without slipping on a rough horizontal floor, its centre of mass moving with a speed v. It makes an elastic collision with a smooth vertical wall. After impact:

A
its centre of mass will move with a speed v initially
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B
its motion will be rolling without slipping immediately
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C
its motion will be rolling with slipping initially and its rotational motion will stop momentarily at some instant
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D
its motion will be rolling without slipping only after some time.
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Solution

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^iv^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=ωR (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.

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