A ball rolls down an inclined plane and acquires a velocity $v_{r}$ when it reaches the bottom of the plane. If the same ball slides from the same height on smooth inclined plane and acquires a velocity $v_{B}$ at bottom. Which of the following statements is/are correct?

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$v_{r} < v_{s},$ because the rolling ball acquires rotational as well as translational kinetic energy

$v_{r} < v_{s},$ because a work is done by the rolling ball against the frictional force

$v_{r} = v_{s},$ because kinetic energy of two balls is same at bottom of planes

$v_{r} > v_{s},$ because the angular velocity acquired makes the rolling ball to travel faster

answer is D.

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Detailed Solution

When ball moves along smooth plane, the accelerating force on it is $m g \sin θ$ .
Hence, its acceleration is equal to $g \sin θ$

When the ball rolls down the rough inclined plane then $m g \sin θ$ acts down the plane but a friction comes into existence which acts up the plane. That friction produces angular acceleration on the ball & the net accelerating force on the ball becomes equal to $(m g \sin θ - f r i c t i o n)$ . Therefore, it has a smaller acceleration. Therefore, its velocity at the bottom of the plane is less than that of ball moving down a smooth plane or $v_{s} > v_{r}$ .
In fact at bottom of the planes, both the balls have same K.E. because loss of P.E. of both the ball is same. But the ball rolling down rough plane has translational as well as rotational K.E. at bottom of the plane while the ball sliding down smooth plane has translational K.E only. Therefore, translational kinetic energy of rolling ball will be less than the translational K.E. of ball sliding down the smooth plane. Hence (4) is correct.