A solid ball rolls down a parabolic path ABC from a height h as shown in the figure. The portion AB of the path is rough while BC is smooth. How high will the ball climb in BC ?

Question

A solid ball rolls down a parabolic path ABC from a height h as shown in the figure. The portion AB of the path is rough while BC is smooth. How high will the ball climb in BC ? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-ex8htodzlnnt.png" /> (A)  H = (5/7) h â¡ (B) H=(5/2) h (C) H=(7/5) h (D) H=(3/7) h

Tardigrade Admin · Accepted Answer

Using the conservation of mechanical energy between

A

and

B

At

B

, total kinetic energy

= m g h

Here,

m =

mass of the ball
The ratio of rotational to translational kinetic energy would be,

\frac{K _{R}}{K _{T}} = \frac{2}{5}

∴ K_{R} = \frac{2}{7} m g h

and

K_{T} = \frac{5}{7} m g h

In portion

BC

, friction is absent.
Therefore, rotational kinetic energy will remain constant and translational kinetic energy will convert into potential energy.
Hence, if

H

be the height to which ball climbs in

BC

, then

m g H = K_{T}

or

m g H = \frac{5}{7} m g h

or

H = \frac{5}{7} h

Q. A solid ball rolls down a parabolic path $A BC$ from a height $h$ as shown in the figure. The portion $A B$ of the path is rough while $BC$ is smooth. How high will the ball climb in $BC$ ?

$H = \frac{5}{7} h$

$H = \frac{5}{2} h$

$H = \frac{7}{5} h$

$H = \frac{3}{7} h$

Q. A solid ball rolls down a parabolic path ABC from a height h as shown in the figure. The portion AB of the path is rough while BC is smooth. How high will the ball climb in BC ?

H=75​h⁡

H=25​h

H=57​h

H=73​h

Q. A solid ball rolls down a parabolic path $A BC$ from a height $h$ as shown in the figure. The portion $A B$ of the path is rough while $BC$ is smooth. How high will the ball climb in $BC$ ?

$H = \frac{5}{7} h$

$H = \frac{5}{2} h$

$H = \frac{7}{5} h$

$H = \frac{3}{7} h$