Physics - Gravitation Question with Solution | TestHub

In arrangements$\left(1\right)$and$\left(2\right),$

Momentum conservation,

$0=m{v}_1-2m{v}_2$...(i)

Energy conservation

$-\frac{Gm\cdot 2m}{a}=\frac{1}{2}m{v}_1^2+\frac{1}{2}2m{v}_2^2-\frac{Gm\cdot 2m}{a/2}$...(ii)

From Equation (i), we get

$v_1=2v_2$

Putting this value$\left(v_1=2v_2\right)$in Equation (ii), we get

$-\frac{2G{m}^2}{a}=\frac{1}{2}m{\left(2v_2\right)}^2+\frac{1}{2}2m{v}_2^2-\frac{4G{m}^2}{a}$

$3m{v}_2^2=\frac{2G{m}^2}{a}$

$v_2=\sqrt{\frac{2Gm}{3a}}$

Relative velocity$v_{1,2}=v_1+v_2$

$=3\sqrt{\frac{2Gm}{3a}}$

$=\sqrt{\frac{6Gm}{a}}$