Q172 / 200 · WBJEE 2014 · Physics
A uniform solid spherical ball is rolling down a smooth inclined plane from a height $h$. The velocity attained by the ball when it reaches the bottom of the inclined plane is v. If the ball is now thrown vertically upwards with the same velocity $v$, the maximum height to which the ball will rise is
- A$\frac{5 h}{8}$
- B$\frac{3 h}{5}$
- Correct answerC$\frac{5 h}{7}$
- D$\frac{7 h}{9}$
Solution
We know that total kinetic energy of a body rolling without slipping
$K_{\text {total }}=K_{\text {rot }}+K_{\text {trans }}$ For solid spherical ball, $I=\frac{2}{5} m R^{2}$ (along to diameter) and $v=R \omega$ where $R$ is radius of spherical ball
So.
$$
K_{\text {total }}=\frac{1}{2}\left(\frac{2}{5} m R^{2}\right) \omega^{2}+\frac{1}{2} m R^{2} \omega^{2}
$$
$$
\begin{array}{l}
=\frac{7}{10} m R^{2} \omega^{2} \\
K=\frac{7}{10} m v^{2}
\end{array}
$$
Potential energy = Kinetic energy
$$
\begin{aligned}
m g h &=\frac{7}{10} m v^{2} \\
v^{2} &=\frac{10}{7} g h
\end{aligned}
$$
For vertical projection,
$$
\begin{array}{c}
v^{2}=u^{2}+2 g h^{\prime} \\
\frac{10}{7} g h=0+2 g h^{\prime} \Rightarrow h^{\prime}=\frac{5}{7} h
\end{array}
$$
Hence, the correct answer is option: 'C'.