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(B) Due to the repulsive electrostatic force, the particles will interact. At the point of closest approach, both particles must move with the same ve

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$\frac{1}{4\pi \varepsilon_0} \frac{4Q^2}{mv^2}$

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(B) Due to the repulsive electrostatic force, the particles will interact. At the point of closest approach, both particles must move with the same velocity $u$ along the line of motion.Using the law of conservation of linear momentum:$mv + m(0) = (m + m)u$$mv = 2mu \implies u = \frac{v}{2}$Now, applying the law of conservation of energy:Initial Energy = Final Energy (at closest distance $R$)$\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + \frac{1}{2}mu^2 + \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{R}$$\frac{1}{2}mv^2 = mu^2 + \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{R}$Substituting $u = \frac{v}{2}$:$\frac{1}{2}mv^2 = m(\frac{v}{2})^2 + \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{R}$$\frac{1}{2}mv^2 = \frac{1}{4}mv^2 + \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{R}$$\frac{1}{4}mv^2 = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{R}$Solving for $R$:$R = \frac{1}{4\pi \varepsilon_0} \frac{4Q^2}{mv^2}$

Two identical particles of mass $m$ carry a charge $Q$ each. Initially, one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards the first particle from a large distance with speed $v$. The closest distance of approach is:

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