Q. A point $P(R\sqrt{3},0,0)$ lies on the axis of a ring of mass $M$ and radius $R$. The ring is located in $y - \mathcal{z}$ plane with its centre at origin $O$. A small particle of mass $m$ starts from $P$ and reaches $O$ under gravitational attraction only. Its speed at $O$ will be
A) $\sqrt{\frac{GM}{R}}$
B) $\sqrt{\frac{Gm}{R}}$
C) $\sqrt{\frac{GM}{2R}}$
D) $\sqrt{\frac{Gm}{2R}}$
Explanation
(a) $\left( \begin{array}{r}
Total \\
mechanical \\
energy
\end{array} \right)_{P} = \left( \begin{array}{r}
Total\ final \\
mechanical \\
energy
\end{array} \right)_{O}$
$\Rightarrow \ \frac{1}{2}m(0)^{2} - \frac{GMM}{\sqrt{\left( \sqrt{3}R \right)^{2}R^{2}}} = \frac{1}{2}mv^{2} - \frac{GMm}{R}\ $
$\Rightarrow \ - \frac{GMm}{2R} - \frac{1}{2}mv^{2} - \frac{GMm}{R}$
$\Rightarrow \ v = \sqrt{\frac{GM}{R}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.