Question

Three particles $\text{A, B}$ and $\text{C}$ each of mass $m$ are lying at the corners of an equilateral triangle of side $L$. If the particle $\text{A}$ is released, keeping the particles $\text{B}$ and $\text{C}$ fixed, the magnitude of instantaneous acceleration of $\text{A}$ is

• A. $\sqrt{5}\frac{Gm}{L^2}$
• B. $\sqrt{7}\frac{Gm}{L^2}$
• C. $\sqrt{2}\frac{Gm}{L^2}$
• D. $\sqrt{3}\frac{Gm}{L^2}$

Answer 1

The correct option is D $\sqrt{3}\frac{Gm}{L^2}$

For the shown configuration, the forces acting on the particle $\text{A}$ is, gravitational forces due to particle $\text{B}$ and $\text{C}$.

Since the $\Delta \text{ABC}$ is equilateral and masses are equal, the force $F$ will be the same.

From law of gravitation,

$\Rightarrow F=\frac{G{m}^2}{L^2}(\because F=\frac{G{m}_1{m}_2}{r^2})$

And we know that

$F_{net}=\sqrt{F^2+F^2+2F^2\cos {60}^o}$

$\Rightarrow F_{net}=\sqrt{3F^2}=\sqrt{3}F$

$\Rightarrow F_{net}=\frac{\sqrt{3}G{m}^2}{L^2}$

Thus, the acceleration of mass $m$ at point $\text{A}$,

$a=\frac{F_{net}}{m}=\frac{\sqrt{3}G{m}^2}{L^2\times m}$

$\therefore a=\frac{\sqrt{3}Gm}{L^2}$

Hence, option $\left(d\right)$ is the correct answer.


$$\begin{array}{c}\text{why this question ?}\\ \text{Tip: Net force experienced by the body}\\ \text{is the vector sum of force exerted by}\\ \text{adjacent masses as per law of gravitation.}\end{array}$$