Question

In figure a solid ball rolls smoothly from rest $($starting at height $H=6.0\text{m})$ until it leaves the horizontal section at the end of the track, at height $h=2.0\text{m}$. The ball hit the ground at distance $x\text{m}$ horizontally from point $A$. Find the value of $x$.

$[$ Take $g=10{\text{m/s}}^2$$]$

Answer 1

On applying "Law of conservation of mechanical energy" between points $\text{B}$ and $\text{C}$.

$mgH=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2+mgh$ $($taking ground level as reference$)$

$\Rightarrow mgH=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{r}^2\times \frac{v^2}{r^2}+mgh$

$\Rightarrow gH=\frac{7}{10}{v}^2+gh$

Substituting the given values we get,

$10\times 6=\frac{7}{10}{v}^2+10\times 2$

$\Rightarrow v^2=\frac{400}{7}\Rightarrow v=\sqrt{\frac{400}{7}}=7.6\text{m/s}........\left(1\right)$

Motion along $\text{y-direction}$,

$-h=0\times t-\frac{1}{2}\times g{t}^2$

Substituting the given data we get,

$2=\frac{1}{2}\times 10\times t^2\Rightarrow t=\sqrt{\frac{4}{10}}=0.6.........\left(2\right)$

Motion along $\text{x-direction}$,

$d=vt\Rightarrow d=7.6\times 0.6=4.6\text{m}$ $[$from $\left(1\right)$ and $\left(2\right)]$