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(A) Let the initial separation be $r_i = 4R$ and the final separation when they just touch be $r_f = 2R$. By the principle of conservation of energy,

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$\sqrt{\frac{k Q^2}{4 m R}\left(1-\frac{G m^2}{k Q^2}\right)}$

Vedclass · Answer

(A) Let the initial separation be $r_i = 4R$ and the final separation when they just touch be $r_f = 2R$. By the principle of conservation of energy, the total initial energy equals the total final energy. Initial Energy: $E_i = 2 	imes (\frac{1}{2} m u^2) - \frac{G m^2}{4R} + \frac{k Q^2}{4R} = m u^2 - \frac{G m^2}{4R} + \frac{k Q^2}{4R}$. Final Energy (at the moment of touching, speed is zero): $E_f = 0 - \frac{G m^2}{2R} + \frac{k Q^2}{2R}$. Equating $E_i = E_f$: $m u^2 - \frac{G m^2}{4R} + \frac{k Q^2}{4R} = - \frac{G m^2}{2R} + \frac{k Q^2}{2R}$. $m u^2 = \frac{k Q^2}{2R} - \frac{k Q^2}{4R} - \frac{G m^2}{2R} + \frac{G m^2}{4R}$. $m u^2 = \frac{k Q^2}{4R} - \frac{G m^2}{4R} = \frac{1}{4R} (k Q^2 - G m^2)$. $u^2 = \frac{1}{4 m R} (k Q^2 - G m^2) = \frac{k Q^2}{4 m R} (1 - \frac{G m^2}{k Q^2})$. $u = \sqrt{\frac{k Q^2}{4 m R} (1 - \frac{G m^2}{k Q^2})}$.

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