Two Protons Move From Location A and B to C

[burnst14]

Homework Statement

Two protons are released from rest, one from location 1 and another from location 2. When these two protons reach location 3, the first proton has a speed that is 4 times the speed of the second proton. If the electric potentials at locations 1 and 2 are 231 V and 115 V, respectively, what is the electric potential at location 3?

Homework Equations

V =kq/r
?

The Attempt at a Solution

I honestly don't know where to start. I feel like I have way too few pieces of information.

 

[ShayanJ]

Use the equation \Delta E_{kinetic}=q\Delta V.

 

[burnst14]

Oh that's a nice equation to have.

 

[burnst14]

Wait, with that equation, I can solve for ΔE, but what can I do with that?

 

[ShayanJ]

You know that the speed of one of the protons is 4 times the speed of the other,so you can find out that its kinetic energy is how many times the other one's.And from this and the equation I gave you,I think you can find your answer.

 

[burnst14]

Alright so I now have ΔE. This is equal to k\frac{Q}{r²} . Is there anything else it's equal to? Because with that equation, all I can do is calculate r and that has nothing to do with kinetic energy.

 

[burnst14]

P.S. - What's wrong with my LaTeX code?

 

[burnst14]

I seriously have no idea what to do. I've been just staring at this problem. I don't have any r values and I can't calculate field strength or voltage without those. I don't have anything I need except the charge of a proton and the mass of a proton. That's it. What am I missing?

 

[nasu]

You don't need any specific formula for electric field or voltage.
All you need is the work energy formula as given by Shyan.
Write it down for each proton. The initial KE is zero for both protons. And write ΔV as a difference, V3-V1 for example.

 

[burnst14]

Wait is ΔE kinetic energy in the formula given?

 

[burnst14]

Assuming that is the case, this problem and the help Shyan has been giving me makes much more sense. I now have:
4KE₂ = 1.6E-19(V₃ - 231)
and
KE₂ = 1.6E-19(V₃ - 115)

Now I can solve for V₃ no problem.

 

[burnst14]

Wrong answer. Calculated as such:
1.6E-19(V₃ - 231) = 4(1.6E-19(V₃ - 115))
V₃ = 76.33

That's wrong, where did I fall of the bus?

 

[ShayanJ]

KE_1=\frac{1}{2}mv_1^2 so if we have v_2=av_1 then,KE_2=\frac{1}{2}mv_2^2=\frac{a^2}{2}mv_1^2=a^2 KE_1 so if you multiply the speed by 4,kinetic energy will be multiplied by 16!

 

[burnst14]

I've never seen someone use an a in this equation. What does it represent?

 

[burnst14]

1.6e-19(V₃ - 231) = 16(1.6e-19(V₃ - 115))
V₃ = 12251.27 V

Also incorrect.

 

[ShayanJ]

a was just a representative for any coefficient,like your 4!

 

[burnst14]

Okay thanks. My WebAssign has a "Practice Another Version" and within it a "Show Solution". I always come to you guys because you let me figure it out, but this time it doesn't seem to be working. I'm just too incompetent on this topic. So anyway, here's what the "Show Solution" says:

"Let the electric potential at location 3 be V3. From conservation of energy we know that
ΔPE + ΔKE = 0
where KE is the kinetic energy of the proton and PE is the electric potential energy. We also know that
ΔPE = qΔV.We combine the above two equations to get
qΔV + ΔKE = 0
or
q(Vf − Vi) = −(KEf − KEi).
Since the protons are released from rest, we have
KEi = 0.
We are given that
v1 = 3v2
and therefore
KE1f = 9KE2f.We now apply the conservation of energy principle to each proton keeping in mind that proton 1 moves from location 1 to location 3 while proton 2 moves from location 2 to location 3.

Proton 1:
q(V3 − V1) = −KE1fProton 2:
q(V3 − V2) = −KE2fDividing one equation by the other we get
q(V3 − V1)
q(V3 − V2)
=
−KE1f
−KE2f
.
Substitute the values
V1 = 231 V, V2 = 115 V, and KE1f = 9KE2f
and solve for V3.
V3 = 101 V
This makes sense, as a positive charge will gain kinetic energy as it moves from a high potential point to a low potential point."

Why are they dividing the entire equations?

 

[nasu]

Dividing the equations is one way to proceed with solving them. But you don't need to do it.

How did you do it yourself?
Sow how you got that number in your previous post. The equation was OK. Probably you had some mistake in the algebra.

 

[burnst14]

All right. Excuse the faux scientific notation, it's just easier and avoids other sources of error.

1.6e-19(V₃ - 231) = 16(1.6e-19(V₃ - 115))

1.6E-19V₃ - 3.696E-17 = 25.6E-19V₃ - 2944E-17

2.94E-14 = 24E-19V₃

V₃ = 12251.27 V

 

[nasu]

I think that the error is in the term 2944e-17

But it is a really bad idea NOT to simplify the value of q right in the beginning. It will be less likely to make mistakes if the numbers are nicer.
But the method is OK.

 

[burnst14]

But it is a really bad idea NOT to simplify the value of q right in the beginning.

What do you mean "simplify"? It's already only two sig figs.

 

[burnst14]

Correct answer!

 

[nasu]

I mean that when you have
q*(V3-231)=16*q*(V3-115)
you divide both sides by q. No need to carry this through all your calculations.
Then you have
(V3-231)=16*(V3-115)

What do you mean by "Correct answer"?

 

[burnst14]

Oh okay. I've shied away from doing that because there were a couple situations in which that didn't work for some reason. I meant that by fixing the 2944E-17, I had gotten the correct answer.