Three balls of equal mass m are connected by light insulating inextensible threads of length $l$ each and kept on a level smooth non conducting ground. The balls A and B are given charge Q each. The strings are all taut. The string connecting A and B suddenly snaps. What is the maximum speed (in m/s) of C during the resulting motion ?
$Q = 1 μ C, l = 1.5 m, m a s s m = 1 g m .$

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answer is 2.

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Detailed Solution

Takes all three bodies as system electrostatic force
(1) Apply law of conservation of momentum in direction
(2) Apply law of conservation of energy,
So $m_{A} {\vec{v}}_{A} + m_{B} {\vec{v}}_{B} + m_{C} {\vec{v}}_{C} = 0 \Rightarrow v_{C} = - 2 v_{A} (a s V_{A} = V_{B})$
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Change in electrostatic P.E. = Increase in KE
$\frac{k Q^{2}}{l} - \frac{k Q^{2}}{2 l} = \frac{1}{2} m_{A} v_{A}^{2} + \frac{1}{2} m_{B} v_{B}^{2} + \frac{1}{2} m_{C} v_{C}^{2} [v_{A} = v_{B}, v_{c} = - 2 v_{A}]$
So $v_{C} = 2 m / s$

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