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?
mivi = mf vf
and KEi +Ui = KEf + Uf
U = (q1 x q2 ) / (4επr)
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?