Question

Three equal masses each of mass M are placed at the three corners of an equilateral triangle of side 'a'. If two particles are kept fixed and the third particle is released. Then speed of the particle when it reaches the mid-point of the side connecting other two masses:

• A. $\sqrt{\frac{2Gm}{a}}$
• B. $2\sqrt{\frac{Gm}{a}}$
• C. $\sqrt{\frac{Gm}{a}}$
• D. $2\sqrt{\frac{Gm}{2a}}$

Answer 1

Answer: (B)

$2\sqrt{\frac{Gm}{a}}$

$v_a=\frac{-G{m}^2\times 2}{a}{u}_a=0$
$v_b=\frac{-G{m}^2\times 2}{\frac{a}{2}}=\frac{-4G{m}^2}{a}=u_b$
$\frac{-G{m}^2\times 2}{a}+\frac{1}{2}m{u_a}^2=\frac{-4G{m}^2}{a}+\frac{1}{2}m{u_b}^2$
$\Rightarrow \frac{2G{m}^2}{a}=\frac{1}{2}m{u_b}^2$
$\Rightarrow 2\sqrt{\frac{Gm}{a}=u_b}$