# 1-D elastic collision between proton and nucleus

## Homework Statement

A proton of mass m is moving with initial speed v0 directly toward the center of a nucleus of mass 31m, which is initially at rest. Because both carry positive electrical charge, they repel each other. Find the speed v' of the nucleus for the following conditions:

a) the distance between the two is at it's smallest value.

b) the distance between the two is very large.

p0=m*v0

KE=1/2*m*v0^2

v.cm=(v0*m)/32m

## The Attempt at a Solution

At first I guessed 0 for both, just because.

So, since it's an elastic collision, the p and the KE remain the same before and after the collision.

p=m*v0

p=m*vf+31m*v'

KE=1/2*m*vo^2

KE=1/2*m*vf^2+1/2*31m*v'^2
~~~~~~~~~~~~~~~~~~
m*v0=m*vf+31m*v'
divide both sides by m
v0=vf+31v'
or vf=v0-31v'

1/2*m*vo^2=1/2*m*vf^2+1/2*31m*v'^2
multiply by 2/m
v0^2=vf^2+31v'^2

exchange vf

v0^2=v0^2-62v0*v'+962v'^2+31v'^2

some algebra

62v0*v'=993v'^2

v'=62/993v0
or 0.0624*v0

checking this, this is the case during (b) but not during (a)

I feel like I've missed something, probably not considering the fact that the two are pushing against each other at close distances but I'm not 100% sure how to account for that.

TSny
Homework Helper
Gold Member
For part (a), what type of energy besides kinetic energy do you need to take into account?

TSny
Homework Helper
Gold Member
To answer question (a) you actually don't need to use energy at all. Just think about how the velocity of one particle compares to the velocity of the other particle at the instant of closest approach.

My guess would be to include electrostatic potential energy. So, if Q=q, then U=kQq/r becomes U=k/r, and if r is very small, and some arbitrary value 1, then U=1.

So, very close the total energy would be KE+U, but then the U would just cancel out.

Well, if you look at it as a regular system, the instant before the collision, the nucleus' velocity is 0 and the proton's velocity is v0. The instant after, the nucleus' velocity is v' and the proton's velocity is vf. I know that the v.cm is 1/32*v0.

HA! I'm not sure I understand, but this is about what I've got: since the center of mass is moving at 1/32*v0 to the right, after the collision the nucleus moves at 1/32*v0. I'm not sure I understand why, but I got it.

TSny
Homework Helper
Gold Member
The key concept for part (a) is conservation of momentum. As the proton approaches the nucleus, the nucleus starts to recoil and pick up speed as the proton slows down due to the mutual electric repulsion. Eventually there is an instant of time when the speed of the nucleus matches the speed of the proton. If you think about it, that's the instant at which the distance between them has reached a minimum. The total momentum of the system at that instant must equal the initial momentum of the system.