Distance of closest approach

Classical Mechanics Level 5

A proton and an $\alpha$ -particle are projected with velocity $v =\sqrt { \frac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }mL } }$ each from infinity as shown.The perpendicular distance between their initial velocities is $L$ .Let the distance of their closest approach be $D$ .

Given $x={ (\frac { 8D }{ L } -5) }^{ 2 }$ .Find $x$

Details and assumptions

1)mass of proton= $m$ ,charge= $+e$

2)mass of $\alpha$ -particle= $4m$ ,charge= $+2e$

3) ${ \varepsilon }_{ 0 }$ =Vacuum Permittivity