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## Homework Statement

A proton and an alpha particle (q= +2e, m=4u) are fired directly at one another from far way, each with an initial velocity of 0.01c. What is their distance of closest approach, as measured between their centres?

## Homework Equations

m

_{i}v

_{i}= m

_{f }v

_{f}

and KE

_{i }+U

_{i}= KE

_{f}+ U

_{f}

U = (q1 x q2 ) / (4επr)

KE= 0.5mv^2

Mass of proton: 1.67E-27

Charge of Proton: 1.692E-19

c = 3E8

## The Attempt at a Solution

I'm sorry this question has been asked before but I didn't understand the explanations in those threads.

I tried using the conservation of momentum:

(1.67E-27 * 0.01c) - (4 * 1.67E-27 * 0.01c) = -(1.67E-27 * velocity of proton) + (4* 1.67E-27 * velocity of alpha particle)

∴ -9000000= -velocity of proton + 4*velocity of alpha particle

but then I don't understand how to find the velocities of the particles individually or how to get this equation into the KE equation.

When I tried using the conservation of energy I got:

½*1.67*10^-27*(0.01c)^2 + ½*4*1.67*10^-27*(0.01c)^2 = (3*1.602*10^-19) /4επr

∴ r = 115008.95

which is so far off, as the answer in the back of the book says: 1.93x10^-14

Which equation am I using wrong and how can I fix it?