A proton and an positron (identical to an electron, except positivel

1. Feb 22, 2014

TwinCamGTS

A proton and an "positron" (identical to an electron, except positivel

1. The problem statement, all variables and given/known data
A proton and an "positron" (identical to an electron, except positively charged) are brought 6 µm apart and released from rest.

a.) What is the initial potential energy stored by this system? 3.84e-23 J

b.) In this problem, we'll let BOTH charges move. The collective kinetic energy of the charges (Ksys) will be drawn from Usys. However, since both charges are moving, they will need to "share" this energy.

Once in motion, what percentage of Ksys does each charge have at any given moment?
proton: K = . % of Ksys
positron: K = . % of Ksys
(In this scenario, how valid is our usual approximation to assign the entire Usys to one particle? Which particle should it be assigned to?)

What is the speed of each charge after they have repelled a long distance apart?
proton: vf = . m/s
positron: vf = m/s

2. Relevant equations

electric potential = kqq/r
energy conservation equations =
KEi+PEi=KEf+PEf
((1/2)mvi^2)+(kqq/r-initial)=((1/2)mvf^2)+(kqq/r-final)

momentum conservation
M1V1=m2v2

proton: mass (1.67E-27kg) positron: mass (9.11E-31kg)

3. The attempt at a solution

for part a)
electric potential = kqq/r
((9E9)(1.67E-27)(1.67E-27))/(6E-6) = 3.84E-23

i have problem on part b where i have to find the speed of positron and proton, also the percentage of KE of proton and positron have after being released and both moving away from each other

i set up the equations using energy conservation method to solve the velocity first because i thought if i can find the velocity then i know the KE ratio of proton and positron have

the formula to find the velocity is in the picture.
that is how i set it up but after i find the velocity and use it to find another velocity i got it wrong. can someone help me with this? i really appreciate it if you can show me the right way to solve for each velocity of proton and positron.
thank you so much

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2. Feb 22, 2014

voko

Do you really need the velocity? You have already found that $$v_2 = -{m_1 \over m_2} v_1$$ so $$v_2^2 = {m_1^2 \over m_2^2} v_1^2$$ and then $$K_2 = {m_2 v_2^2 \over 2} = ?$$

3. Feb 22, 2014

TwinCamGTS

In energy conservation equation. The initial potential energy later will be distributed to the final kinetic energy.

The initial potential energy is already found. While the kinetic energy is the speed of both proton and positron moving away from each other. and the momentum between proton and positron is conserved as well. Thats why i set uo my equation like that.

So, after i find out the velocity, i use it find the other velocity. By energy conservation law, there will be no energy loss in te process. Thats mean the kinetic energy (.5*m*v^2) of both proton and positron add up will be equal to the amount of potential energy the system have in the beginning. But my value is way more.

I just wondering if i set up the equation wrong of there is other way to solve it?

4. Feb 22, 2014

vela

Staff Emeritus
You might find it easier to work with momentum initially and write the kinetic energy of a particle as $K = \frac{p^2}{2m}$.