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

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TestHub · Accepted Answer

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

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