Question

Three solid spheres each of mass m and radius R are placed at three corners of an equilateral triangle of side ‘d’ released. The speed of any one sphere at the time of collision would be:

• A. $\sqrt{Gm[\frac{1}{d}-\frac{3}{R}]}$
• B. $\sqrt{Gm[\frac{3}{d}-\frac{1}{R}]}$
• C. $\sqrt{Gm[\frac{2}{R}-\frac{1}{d}]}$
• D. $\sqrt{Gm[\frac{1}{R}-\frac{2}{d}]}$

Answer 1

The correct option is D $\sqrt{Gm[\frac{1}{R}-\frac{2}{d}]}$


They all collide at the centroid. During collision, the distance between the centers of any two spheres will be 2R. Applying WET,
$k_i+U_i+w_{NC}=k_f+U_f$
$0–\frac{3G{m}^2}{d}+0=3\times \frac{1}{2}m{v}^2–\frac{3G{m}^2}{2R}$
$\Rightarrow \frac{v^2}{2}=Gm[\frac{1}{2R}-\frac{1}{d}]$
$v=\sqrt{Gm[\frac{1}{R}-\frac{2}{d}]}$