Two small particles, each of mass m carrying positive charge q each are attached to the ends of a non-conducting light thread of length 2l. A third particle of mass $2 m$ is attached at mid-point of the thread. The whole system is placed on a smooth horizontal floor and the particle of mass $2 m$ is given a velocity v as shown in Fig. minimum distance between the two charged particles during the process of motion is expressed as $(\frac{a . q^{b} l}{c d \in_{0} m V^{2} l + q^{e}})$ find $a + b + c + d + e$ to nearest integer

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answer is 13.

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Detailed Solution

At the instant of minimum separation all three move with same velocity $= V_{C H} = \frac{V}{2}$
$\frac{1}{4 π ε_{0}} \frac{q^{2}}{2 l} + \frac{1}{2} .2 m V^{2} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{d} + \frac{1}{2} . 4 m \frac{V^{2}}{4} \frac{q^{2}}{4 π ε_{0} 2 l} + \frac{m V^{2}}{2} = \frac{1}{4 π ε_{0}} \frac{q^{2}}{d}$
Solving we get