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

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Ronak Agarwal by Ronak Agarwal
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