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(B) Let the mass of the proton be $m_1 = m$ and its charge be $q_1 = e$. Let the mass of the $\alpha$-particle be $m_2 = 4m$ and its charge be $q_2 =

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$5 e^2 / 4 \pi \varepsilon_0 m v^2$

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(B) Let the mass of the proton be $m_1 = m$ and its charge be $q_1 = e$. Let the mass of the $\alpha$-particle be $m_2 = 4m$ and its charge be $q_2 = 2e$.Since the $\alpha$-particle is free to move,we analyze the system in the center-of-mass frame. The reduced mass $\mu$ of the system is given by $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m \cdot 4m}{m + 4m} = \frac{4m}{5}$.The initial kinetic energy of the system in the center-of-mass frame is $K_i = \frac{1}{2} \mu v^2 = \frac{1}{2} (\frac{4m}{5}) v^2 = \frac{2}{5} mv^2$.At the point of minimum separation $r$,the relative velocity of the particles is zero. By the law of conservation of energy,the initial kinetic energy is equal to the electrostatic potential energy at distance $r$:$\frac{1}{2} \mu v^2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r}$Substituting the values: $\frac{2}{5} mv^2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{e \cdot 2e}{r}$$\frac{2}{5} mv^2 = \frac{2e^2}{4 \pi \varepsilon_0 r}$Solving for $r$: $r = \frac{5 e^2}{4 \pi \varepsilon_0 m v^2}$.

$A$ proton of mass $m$ and charge $e$ is projected from a very large distance towards an $\alpha$-particle with velocity $v$. Initially,the $\alpha$-particle is at rest,but it is free to move. If gravity is neglected,then the minimum separation along the straight line of their motion will be:

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