Electric potential and kinetic energy equations with particles

[fisixC]

Homework Statement

Particles A (of mass m and charge Q) and
B (of m and charge 5 Q) are released from
rest with the distance between them equal to
1.411 m.
If Q = 25 µC, what is the kinetic energy
of particle B at the instant when the particles
are 3.411 m apart?
Answer in units of J.

Homework Equations

U = (k(q1)(q2))/(r)
K = .5((m)(v^2))

The Attempt at a Solution

I know that it wants me to set up an energy equation on particle B so I have:
Ui + Ki = Uf + Kf

It tells me that the particles start at rest so that means Ki cancels so we have:
Ui = Kf + Uf

Now I isolate Kf to find the kinetic energy final at that distance away:
Kf = Ui - Uf

Then substitute in the equations noting that substituting the kinetic energy equation is not necessary because it wants to know how much kinetic energy there is at that point not anything which is in the kinetic energy equation:
Kf = (((k)(q1)(q2))/(r1)) - (((k)(q1)(q2))/(r2))

And after the numbers are in I come up with 11.6711218613 J, which is incorrect.

So where am I going wrong if I am at all, or is it just my plugging in algebra skills? I'm totally stumped I'm pretty sure this is how you do that problem.
Thank you in advance for the help!

 

[fisixC]

Also I did take into consideration that particle B has five times the amount of charge that particle A has.

additionally:
ki = initial kinetic energy
kf = final kinetic energy
ui = initial potential energy
uf = final potential energy

k = Coulomb's constant = 8.98755E9

 

[fisixC]

Could my answer be correct just that it is negative instead of positive? But why?

*EDIT*
Never mind it wasn't negative...then I'm just wrong, but why.

 

[fisixC]

Anyone could help me at all? I'm so lost right now..

 

[fisixC]

Desperate, anyone..

 

[gneill]

Hi fisixC. I've had a look at the problem.

Both charges are positive, so the force between them is repulsive. By Newton's third law, the forces will be equal and opposite. Further, since their masses are the same, they should both experience the same accelerations and have equal kinetic energy. So it looks like the center of mass of the system is the place to set up our frame of reference.

The first thing to determine is how much energy we're playing with (always find out how big the bang will be before turning the switch!) So, can you tell me what the electrical potential energy is for the two particle system when they're at rest at their starting positions?

 

[fisixC]

Both particle's potential energy at the starting position would be the same:
((k)(25e-6)5*(25e-6))/(1.411) = 19.9050983345 J

 

[gneill]

Both particle's potential energy at the starting position would be the same:
((k)(25e-6)5*(25e-6))/(1.411) = 19.9050983345 J

Okay. That's the potential energy of the system (the two particles comprise the system in question).

Now, when they're released, they are exerting equal and opposite forces on each other. Since their masses are equal, they both accelerate at the same rate. Now, thanks to conservation of energy we know that the original potential energy is being traded for kinetic energy (when the particles eventually "reach infinite separation", all of the potential energy will have been converted to kinetic energy).

But we're only interest in what the situation is at 3.411 meters separation. At that point there's still plenty of potential energy left in the system. What is the potential energy for the two particles when they're separated by a distance of 3.411 meters?

 

[fisixC]

Potential Energy at 3.411m :
((k)(25e-6)(5*25e-6))/(3.411) = 8.23397647318 J

 

[gneill]

Excellent. So, look at the potential energy you just computed and compare it to the original potential energy. Where has the difference gone, and what do you think you might do with it?

 

[fisixC]

The difference has been converted to kinetic energy so when you subtract the two numbers you should get the kinetic energy at 3.411 m.

 

[gneill]

The difference has been converted to kinetic energy so when you subtract the two numbers you should get the kinetic energy at 3.411 m.

Bingo! Now, given that you now have the kinetic energy of the system in hand, how are you going to assign it to the particles?

 

[fisixC]

Divide by two because the kinetic energy for each particle is equal?

 

[gneill]

Divide by two because the kinetic energy for each particle is equal?

Yes. That will be your answer. It is the kinetic energy of each of the particles as "seen" from the original frame of reference which was tied to the center of mass of the system (this is just incidental information that you might like to be aware of).

You can fancy-up the answer by supplying an equation for the total energy (the sum of the potential and kinetic energies) and stating that the sum is a constant.

 

[fisixC]

But because the particle is going away from the origin does that do anything to the sign of the energy? Negative/positive?

 

[gneill]

Kinetic energy is always positive (remember: KE = (1/2)mv²). The velocity is squared, so it's always positive. Only differences in KE can be negative.

 

[fisixC]

Thank you so much, you have no idea how much you've helped me. Now I can get some sleep for the test tomorrow.

Sorry for double posting I didn't realize it went to the next page.

 

[fisixC]

Thank you so much, now I can get some sleep for class.

 

[gneill]

Glad to help. Good luck!