Question

A small block slides with velocity $0.5 \sqrt{g r}$ on the horizontal frictionless surface as shown in the figure- $3.85$. The block leaves the surface at point $C$. The angle $\theta$ in the figure is :

• A. $\cos ^{-1}(4 / 9)$
• B. $\cos ^{-1}(3 / 4)$
• C. $\cos ^{-1}(1 / 2)$
• D. None of the above

Answer 1

Answer: (B)

From conservation of energy,

Let velocity of block at $C$ is $v$
$\frac{1}{2} m v_{0}^{2}+m g r(1-\cos \theta) =\frac{1}{2} m v^{2} $
$v =\sqrt{v_{0}^{2}+2 g r(1-\cos \theta)}$
when block leaves the surface,
$ N =0 $
$m g \cos \theta =\frac{m v^{2}}{r} $
$ r g \cos \theta =v_{0}^{2}+2 g r(1-\cos \theta) $
$ 3 r g \cos \theta =\frac{r g}{4}+2 r g $
$3 \cos \theta =\frac{9}{4} $
$\cos \theta =\frac{3}{4} $
$ \theta =\cos { }^{-1}\left(\frac{3}{4}\right) $