Question

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. $5h/8$
• B. $3h/5$
• C. $5h/7$
• D. $7h/9$

Answer 1

Answer: (C)

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, $K_{\text {total }}=\frac{1}{2}\left(\frac{2}{5} m R^{2}\right) \omega^{2}+\frac{1}{2} m R^{2} \omega^{2}$
$=\frac{7}{10} m R^{2} \,\omega^{2}$
$K=\frac{7}{10} \,m v^{2}$
Potential energy $=$ Kinetic energy
$m g h =\frac{7}{10} m v^{2} $
$v^{2} =\frac{10}{7} g h\,\,\,...(i)$
For vertical projection,
$v^{2}=u^{2}+2 g h^{'}$
$\frac{10}{7} g h=0+2 g h^{'}$
$ \Rightarrow h^{'}=\frac{5}{7} h$