GravitationHard
Question
Consider two identical metallic spheres of radius $R$ each having charge $Q$ and mass $m$. Their centers have an initial separation of $4R$. Both the spheres are given an initial speed of $u$ towards each other. The minimum value of $u$, so that they can just touch each other is :
(Take $k = \frac{1}{4\pi\epsilon_{0}}$ and assume $kQ^{2} > Gm^{2}$ where G is the Gravitational constant)
Options
A.$\sqrt{\frac{kQ^{2}}{4mR}\left( 1 - \frac{Gm^{2}}{kQ^{2}} \right)}$
B.$\sqrt{\frac{kQ^{2}}{4mR}\left( 1 + \frac{Gm^{2}}{kQ^{2}} \right)}$
C.$\sqrt{\frac{kQ^{2}}{2mR}\left( 1 - \frac{Gm^{2}}{kQ^{2}} \right)}$
D.$\sqrt{\frac{kQ^{2}}{2mR}\left( 1 - \frac{Gm^{2}}{2kQ^{2}} \right)}$
Solution
Using energy conservation
(2) $\left( \frac{1}{2}mu^{2} \right) - \frac{Gm^{2}}{4r} + \frac{KQ^{2}}{4r} = - \frac{Gm^{2}}{2r} + \frac{KQ^{2}}{2r}$
$$u = \sqrt{\frac{1}{4mr}\left( KQ^{2} - Gm^{2} \right)}$$
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