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. $\frac{5h}{8}$
• B. $\frac{3h}{5}$
• C. $\frac{5h}{7}$
• D. $\frac{7h}{9}$

Answer 1

Answer: (C)

$\frac{5h}{7}$

Let the mass and radius of the spherical ball be $m$ and $r$ respectively.

Moment of inertia of solid spherical ball $I=\frac{2}{5}m{r}^2$

For inclined plane motion :

Let the angular velocity of the ball be $w$.
$\therefore$ $v=r\omega$ (pure rolling) ...............(1)

Work-energy theorem :

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{r}^2\times \frac{v^2}{r^2}$ (using (1))

$mgh=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2$ $⟹v^2=\frac{10}{7}gh$

For vertically upward motion :

Let the maximum height attained by the ball be $h^{\prime }$. Velocity the ball at this point is zero i.e $v^{\prime }=0$

Using $(v^{\prime }{)}^2-v^2=2a{h}^{\prime }$

$0-v^2=2(-g)h^{\prime }$ $⟹$ $v^2=2g{h}^{\prime }$

$2g{h}^{\prime }=\frac{10}{7}gh$ $⟹$ $h^{\prime }=\frac{5}{7}h$