Question

A small block slides with velocity $0.5$$\sqrt{gr}$ on the horizontal frictionless surface as shown in figure. The block leaves the surface at L. The angle a is

• A. ${\cos }^{-1}\frac{3}{4}$
• B. ${\cos }^{-1}\frac{4}{3}$
• C. ${\sin }^{-1}\frac{3}{4}$
• D. ${\sin }^{-1}\frac{4}{3}$

Answer 1

The correct option is A ${\cos }^{-1}\frac{3}{4}$

Here $\cos a=\frac{r-h}{r}$
or $-h=r\cos a-r$
or $h=r(1-\cos a)$
At L, $mg\cos a=\frac{m}{r}{v}^2$
$=\frac{m}{r}[v_0^2+(\sqrt{2gh}{)}^2]$
or $mg\cos a=\frac{m}{r}(\frac{1}{4}gr+2gh)$
$=\frac{m}{r}(\frac{1}{4}gr+2gr1-\cos \alpha )$
or $\cos a=\frac{1}{4}+2-2\cos a$
or $3\cos a=\frac{9}{4}$ or $\cos \alpha =\frac{3}{4}$
or $a={\cos }^{-1}3/4$