JEE Main 2014 — Physics Mechanics

We choose reference point, infinity, where total energy of the system is zero. So, initial energy of the system $=0$=0 Final energy $$=\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1^2+\frac{1}{2} \mathrm{~m}_2 \mathrm{v}_2^2-\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}$$=21​ m1​v12​+21​ m2​v22​− dGm1​ m2​​ From conservation of energy, Initial energy $=$= Final energy $$\therefore 0=\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1{ }^2+\frac{1}{2} \mathrm{~m}_2 \mathrm{v}_2{ }^2-\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}$$∴0=21​ m1​v1​2+21​ m2​v2​2− dGm1​ m2​​ 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}} \ldots(1)$21​ m1​v12​+21​ m1​v22​= dGm1​ m2​​…(1) By conservation of linear momentum $$\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2=0$$m1​v1​+m2​v2​=0 or $\frac{\mathrm{v}_1}{\mathrm{v}_2}=-\frac{\mathrm{m}_2}{\mathrm{~m}_1} \Rightarrow \mathrm{v}_2=-\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{v}_1$v2​v1​​=− m1​m2​​⇒v2​=− m2​m1​​v1​ Putting value of $\mathrm{v}_2$v2​ in equation (1), we get $$\mathrm{m}_1 \mathrm{v}_1^2+\mathrm{m}_2\left(-\frac{\mathrm{m}_1 \mathrm{v}_1}{\mathrm{~m}_2}\right)^2=\frac{2 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}$$m1​v12​+m2​(− m2​m1​v1​​)2= d2Gm1​ m2​​ $$\begin{aligned} &\frac{m_1 m_2 v_1^2+m_1^2 v_1^2}{m_2}=\frac{2 \mathrm{Gm}_1 m_2}{\mathrm{~d}} \\ &\mathrm{v}_1=\sqrt{\frac{2 \mathrm{Gm}_2^2}{\mathrm{~d}\left(\mathrm{~m}_1+\mathrm{m}_2\right)}}=\mathrm{m}_2 \sqrt{\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right)}} \\ &\text { Similarly } \mathrm{v}_2=-\mathrm{m}_1 \sqrt{\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right)}} \end{aligned}$$​m2​m1​m2​v12​+m12​v12​​= d2Gm1​m2​​v1​= d( m1​+m2​)2Gm22​​​=m2​d(m1​+m2​)2G​​ Similarly v2​=−m1​d(m1​+m2​)2G​​​