- #1
kingsmaug
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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.
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.