# Distance of Closest Approach

## 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

mivi = mf vf
and KEi +Ui = KEf + Uf

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?

## Answers and Replies

rude man
Homework Helper
Gold Member
Using energy arguments, equate the initial k.e. of both particles to the potential energy of the particles when all their k.e. is gone.

rude man
Homework Helper
Gold Member

## 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

mivi = mf vf
and KEi +Ui = KEf + Uf

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

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
Use the correct units. What is 0.01c in SI units? What is the formula for potential energy of two charges a distance r apart? Nothing like (3*1.602*10^-19) /4επr for charges e and 2e.

gneill
Mentor
I think you'll want to do the calculation in the center of momentum frame of reference. Because the masses are not equal, for observers in any other frame the closest approach will happen when the system still has kinetic energy.

Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
The absolutely easiest way of solving this problem is to ask the following questions:
What is the velocity of the particles at the time of closest approach?
What is the kinetic energy at this point?
What must the potential energy be?
At what distance is this fulfilled?

You can go to the CoM frame but it is not necessary. I also suggest you keep the symbolic representations of your quantities and only insert them at the very end.