Physics - Gravitation Question with Solution | TestHub

We choose reference point, infinity, where total energy of the system is zero. So, initial energy of the system $=0$ Final energy $$=\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}_1^2+\frac{1}{2}{\mathrm{m}}_2{\mathrm{v}}_2^2-\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{\mathrm{d}}$$ From conservation of energy, Initial energy $=$ Final energy $$\therefore 0=\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}_1{}^2+\frac{1}{2}{\mathrm{m}}_2{\mathrm{v}}_2{}^2-\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{\mathrm{d}}$$ or $\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}_1^2+\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}_2^2=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{\mathrm{d}}\dots (1)$ By conservation of linear momentum $${\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2=0$$ or $\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=-\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}\Rightarrow {\mathrm{v}}_2=-\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}{\mathrm{v}}_1$ Putting value of ${\mathrm{v}}_2$ in equation (1), we get $${\mathrm{m}}_1{\mathrm{v}}_1^2+{\mathrm{m}}_2{\left(-\frac{{\mathrm{m}}_1{\mathrm{v}}_1}{{\mathrm{m}}_2}\right)}^2=\frac{2{\mathrm{Gm}}_1{\mathrm{m}}_2}{\mathrm{d}}$$ $$\begin{array}{rl}& \frac{m_1{m}_2{v}_1^2+m_1^2{v}_1^2}{m_2}=\frac{2{\mathrm{Gm}}_1{m}_2}{\mathrm{d}}\\ & {\mathrm{v}}_1=\sqrt{\frac{2{\mathrm{Gm}}_2^2}{\mathrm{d}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}}={\mathrm{m}}_2\sqrt{\frac{2\mathrm{G}}{\mathrm{d}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}}\\ & \text{ Similarly }{\mathrm{v}}_2=-{\mathrm{m}}_1\sqrt{\frac{2\mathrm{G}}{\mathrm{d}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}}\end{array}$$