Broken String and Speeding Charges

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It is clear that the velocity of C will be maximum at the state given above.

The velocity of the outer masses will be half of the velocity of the middle mass due to Momentum conservation.

Given that $\displaystyle U_o = 1J$ ,
$\Rightarrow\displaystyle 3\frac{kq^2}{l} = 1$

$\displaystyle\Rightarrow \frac{kq^2}{l} = \frac{1}{3}$

Conserving Energy between the initial and final state,

$PE_i = PE_f+KE_f$

$\displaystyle\Rightarrow 3\frac{kq^2}{l} = (\frac{kq^2}{l}+\frac{kq^2}{l}+\frac{kq^2}{2l}) + (\frac{1}{2}mv^2 + 2(\frac{1}{2}m(\frac{v}{2})^2))$

$\displaystyle\Rightarrow \frac{kq^2}{l} = \frac{3}{2}mv^2$

$\displaystyle\Rightarrow \frac{1}{3} = \frac{3}{2}\times 10^{-3}v^2$

$\displaystyle\Rightarrow v = \sqrt{\frac{2000}{9}}\approx\boxed{14.9\frac{m}{sec}}$

Momentum is conserved. So, things will happen similarly in the spheres (finally at the edges) between whom the string had been cut and it would be balanced by the sphere in the middle of the two. The velocities are as follows: image

Let the length of each side of the triangle be $a$ and the charges be $q$ .

Given, $3k \frac{q^2}{r}=1$

So, by conservation of energy we get:

$Energy_{initial}=Energy_{final} \implies 3k \frac{q^2}{r}=\frac{5}{2} k \frac{q^2}{r} + \frac{1}{2} m (v^2+4v^2+v^2) \implies v^2= \frac{10^3}{18}$

So, $2v=14.9$

as the forces after breaking of sting are only internal so com is unchanged from there we get Vc=2Va=2Vc so now by using conservation of energy this vc max can be found where the charges ar alligned in a straight line