Question

A point $\text{P}$ lies on the axis of a fixed ring of mass $M$ and radius $R$, at a distance $2R$ from its centre $\text{O}$. A small particle which is at rest starts from $\text{P}$ and reaches $\text{O}$ under the influence of gravitational attraction only. Its speed at $\text{O}$ will be

• A. $\sqrt{\frac{2GM}{R}(1-\frac{1}{\sqrt{5}})}$
• B. $\sqrt{\frac{2GM}{R}}$
• C. $\sqrt{\frac{2GM}{R}(\sqrt{5}-1)}$
• D. Zero

Answer 1

The correct option is A $\sqrt{\frac{2GM}{R}(1-\frac{1}{\sqrt{5}})}$

Since no work is done by the external forces on the system (ring + mass), mechanical energy of the system will be conserved.

The potential at a point along the axis of the ring at a distance $r$ from the center is given as
$$V=-\frac{GM}{\sqrt{(R^2+r^2)}}$$ Potential at point $\text{P}$,

$V_P=-\frac{GM}{\sqrt{(R^2+(2R{)}^2)}}=-\frac{GM}{R\sqrt{5}}$

Potential at point $\text{O}$,

$V_O=-\frac{GM}{\sqrt{(R^2+(0{)}^2)}}=-\frac{GM}{R}$

Let, $m$ be the mass of the particle placed at point $\text{P}$ and $v$ be its velocity at point $\text{O}$.

Now, conserving mechanical energy between the initial and final positions:

$P.E_P+K.E_P=P.E_O+K.E_O$

$\Rightarrow m{V}_P+0=m{V}_O+\frac{1}{2}m{v}^2$

$\Rightarrow -\frac{GMm}{R\sqrt{5}}+0=-\frac{GMm}{R}+\frac{1}{2}m{v}^2$ ,

$\Rightarrow v^2=\frac{2GM}{R}(1-\frac{1}{\sqrt{5}})$

$\Rightarrow v=\sqrt{\frac{2GM}{R}(1-\frac{1}{\sqrt{5}})}$

Hence, Option (a) is the correct answer.

Key Concept: The potential of a point on the axis of the ring at a distance $r$ from the center is given as $$V=-\frac{GM}{\sqrt{(R^2+r^2)}}$$ Why this question: To make students apply energy conservation for extended objects. The crucial aspect is to remember the result for gravitational potential of various mass distributions.