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
$mgh=\frac{7}{10}m{v}^2$
$v^2=\frac{10}{7}gh...(i)$
For vertical projection,
$v^2=u^2+2g{h}^{{}^{\prime }}$
$\frac{10}{7}gh=0+2g{h}^{{}^{\prime }}$
$\Rightarrow h^{{}^{\prime }}=\frac{5}{7}h$