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## Homework Statement

A proton of mass

*m*is moving with initial speed

*v*0 directly toward the center of a nucleus of mass 31

*m*, 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.

## Homework Equations

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.