Question

A small block slides with velocity $0.5\sqrt{gr}$ on the horizontal frictionless surface as shown in the figure. The block leaves the surface at a point $C$ . The value of $\cos \theta =\frac{x}{4}$ , find $x$ .

Answer 1

Answer: 3

$mg\cos \theta -N=\frac{m{v}^2}{r};N=mg\cos \theta -\frac{m{v}^2}{r}$
$N=0\Rightarrow mg\cos \theta =\frac{m{v}^2}{r}$
$v=\sqrt{gr\cos \theta }$ or $\cos \theta =\frac{v^2}{gr}$....(i)
Now, from the conservation of energy, we have,
$\frac{m{v}_0^2}{2}+mgr(1-\cos \theta )=\frac{m{v}^2}{2}\dots 2$
Substituting the value of $m{v}^2$ from equation 1 in equation 2 ,
$\frac{m{v}_0^2}{2}+mgr(1-\cos \theta )=\frac{mgr\cos \theta }{2}$
$\Rightarrow mgr(1-\cos \theta )=\frac{mgr\cos \theta }{2}-\frac{m0.5{\sqrt{gr}}^2}{2}$
$\Rightarrow 1-\cos \theta =\frac{\cos \theta }{2}-\frac{1}{8}$
$\Rightarrow \cos \theta =\frac{3}{4}$
Thus, $x=3$.