Question

A small frictionless block slides with velocity $0.6\sqrt{gr}$ on the horizontal surface as shown in the Figure. The block leaves the surface at point $C$. The angle $\theta$ in the figure is _____.

Answer 1

Forces acting on the block is normal force and gravity. The equation of motion is:

$N=\frac{m{V}^2}{r}+mg\cos \theta$

When the block leaves the surface, $N=0$

$\therefore \frac{m{V}^2}{r}=-mg\cos \theta$

$⟹\frac{V^2}{r}=-g\cos \theta$

$⟹\cos \theta =-\frac{V^2}{rg}$

$⟹\cos \theta =\frac{-0.6\times 0.6\times rg}{rg}$

$⟹\cos \theta =-0.36$

$⟹\theta ={\cos }^{-1}\left(0.36\right)$

$⟹\theta ={111.1}^o$