Physics - Circular Motion Question with Solution | TestHub

By energy conservation between A and B

$\Rightarrow \text{Mg}\frac{2\text{R}}{5}+0=\frac{\text{MgR}}{5}+\frac{1}{2}{\text{Mv}}^2$

$\text{v}=\sqrt{\frac{2\text{gR}}{5}}$



Now, radius of curvature after just leaving B is

$\text{r}=\frac{{\text{v}}_{\perp }^2}{{\text{a}}_{\text{r}}}=\frac{2\text{gR}/5}{\text{g cos 37}}=\frac{\text{R}}{2}$