Question 68 of 183

A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $3m{s}^{-1}$ (as shown in figure). Maximum height with respect to the initial position covered by it will be _____ $cm$
(take, $g=10m{s}^{-2}$)

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Answer: 75

The total kinetic energy of the ball at the bottom of the curve can be calculated as follows:
$\begin{matrix}K & = & \dfrac{1}{2}M{v}^{2}+\dfrac{1}{2}\cdot \dfrac{2}{3}M{R}^{2}{\omega }^{2} \\ & = & \dfrac{1}{2}M{v}^{2}+\dfrac{1}{3}M{\left(R\omega \right)}^{2} \\ & = & \dfrac{1}{2}M{v}^{2}+\dfrac{1}{3}M{v}^{2}\left[∵v=\omega R\right] \\ & = & \dfrac{5}{6}M{v}^{2}...\left(1\right)\end{matrix}$
The formula to calculate the potential energy acquired by the ball when it attains the maximum height is given by
$U=Mgh...\left(2\right)$
Equate equation (1) and (2) and simplify to obtain the required height.
$\begin{matrix}Mgh & = & \dfrac{5}{6}M{v}^{2} \\ & \Rightarrow & h=\dfrac{5{v}^{2}}{6g}...\left(3\right)\end{matrix}$
Substitute the values of the known parameters into equation (3) to calculate the required height attained by the ball.
$\begin{matrix}h & = & \dfrac{5\times {\left(3m{s}^{-1}\right)}^{2}}{6\times 10m{s}^{-2}} \\ & = & 0.75m\times \dfrac{100cm}{1m} \\ & = & 75cm\end{matrix}$