Where did I go Wrong? Solving for Particle Speed with Mass & Charge

[hidemi]

Homework Statement

Two particles, each having a mass of 3.0 μg and having equal but opposite charges of magnitude 5.0 nC, are released simultaneously from rest when the two are 5.0 cm apart. What is the speed of either particle at the instant when the two are separated by 2.0 cm?

The answer is 1.5 m/s.

Relevant Equations

F = kQ^2/r^2

F= 9*10^9 (5*10^-9)^2 / (0.05)^2

a=F/m = F/(3*10^-6)

v^2 = u^2 + 2aS = 0 + 2a*0.03
v = 60

Where did I do wrong? Thanks!

 

[Steve4Physics]

Homework Statement:: Two particles, each having a mass of 3.0 μg and having equal but opposite charges of magnitude 5.0 nC, are released simultaneously from rest when the two are 5.0 cm apart. What is the speed of either particle at the instant when the two are separated by 2.0 cm?

The answer is 1.5 m/s.
Relevant Equations:: F = kQ^2/r^2

F= 9*10^9 (5*10^-9)^2 / (0.05)^2

a=F/m = F/(3*10^-6)

v^2 = u^2 + 2aS = 0 + 2a*0.03
v = 60

Where did I do wrong? Thanks!

What happens to the force as the particles get closer? So what happens to the acceleration?
Is there an easier approach which doesn't require the force and acceleraton?

 

[haruspex]

Homework Statement:: Two particles, each having a mass of 3.0 μg and having equal but opposite charges of magnitude 5.0 nC, are released simultaneously from rest when the two are 5.0 cm apart. What is the speed of either particle at the instant when the two are separated by 2.0 cm?

The answer is 1.5 m/s.
Relevant Equations:: F = kQ^2/r^2

F= 9*10^9 (5*10^-9)^2 / (0.05)^2

a=F/m = F/(3*10^-6)

v^2 = u^2 + 2aS = 0 + 2a*0.03
v = 60

Where did I do wrong? Thanks!

The force is not constant.

 

[PeroK]

Where did I do wrong?

First, your attempted solution is very poor: it's just a number salad with no units.

Second, you made a fundamental, basic mistake.

 

[hidemi]

I'm thinking about using this equation: U = 1/2mv^2 + kQq/r
E1 = 1/2 * 3*10^-9 * 0^2 + 9*(-25)*10^-9/0.05^2
E1 = 1/2 * 3*10^-9 * V^2 + 9*(-25)*10^-9/0.02^2
V = 67 (m/s)
I cannot find a matching answer.

 

[haruspex]

kQq/r
9*(-25)*10^-9/0.05^2

?

 

[hidemi]

?

E1 = 1/2 * 3*10^-9 * 0^2 + 9*(-25)*10^-9/0.05
E2 = 1/2 * 3*10^-9 * V^2 + 9*(-25)*10^-9/0.02
E1 = E2, so V = 67 (m/s)

Sorry it was a typo

 

[PeroK]

I'm thinking about using this equation: U = 1/2mv^2 + kQq/r
E1 = 1/2 * 3*10^-9 * 0^2 + 9*(-25)*10^-9/0.05^2
E1 = 1/2 * 3*10^-9 * V^2 + 9*(-25)*10^-9/0.02^2
V = 67 (m/s)
I cannot find a matching answer.

It's no surprise that you keep getting the wrong answer, because you are lacking the basic techniques needed for a more advanced topic like EM:

You generally don't use units; you plug numbers into equations without thinking; you don't do any algebra; all you have is a number salad.

May I ask where you are studying physics? Do none of your professors comment on this? This is the very basics of EM. You have some serious mathematics coming up if you continue your studies. Has no one suggested you need to learn how to do algebra? You cannot make further progress in physics without developing mathematically.

 

[haruspex]

E1 = 1/2 * 3*10^-9 * 0^2 + 9*(-25)*10^-9/0.05
E2 = 1/2 * 3*10^-9 * V^2 + 9*(-25)*10^-9/0.02
E1 = E2, so V = 67 (m/s)

Sorry it was a typo

Looks like a tie, one error by you and one by the problem setter.
You have counted the PE of the system but the KE of only one particle.
At a guess, the problem setter turned a μg into 10^-6kg.

Btw, I agree with @PeroK. It is far better to work entirely symbolically until the final step.

 

[Delta2]

Though this seems to be an electrodynamics problem, you have to use the two well known laws of classical mechanics: Conservation of energy and conservation of momentum. Be sure when you apply conservation of energy to correctly count for the potential and kinetic energy of both particles.

 

[PeroK]

@hidemi to be constructive, here's what I would have expected you to do.

First, state that you have a system of two particles and that the total potential energy for such a system is: $$U = \frac{kq_1q_2}{r} = -\frac{kq^2}{r}$$ Where $q = 5.0nC$ in this particular case.

Second, you should at least have thought about whether this is the potential energy of the system or of each particle. And, you could have asked here for confirmation - that it is indeed the total PE of both particles combined.

Third, you should have identified the total KE of the system as: $$KE = \frac 1 2 mv_1^2 + \frac 1 2 mv_2^2$$
Fourth, you should argued (as above) that the particles must have the same speed - as they have the same mass and the same magnitude of Coulomb force acting on them.

That would allow you to rewrite the KE of the system as: $$KE = mv^2$$
Fifth, you use conservation of energy: $$mv^2 -\frac{kq^2}{r_1} = -\frac{kq^2}{r_0}$$ which gives you: $$v^2 = \frac{kq^2}{m}(\frac 1 {r_1} - \frac 1 {r_0})$$
Finally, you can plug the numbers into that.

This is what you should be training yourself to do. To think through a problem like this.

 

[hidemi]

Thank you all for pointing out my mistakes! I finally got the correct answer 1.5m/s

 

[haruspex]

Thank you all for pointing out my mistakes! I finally got the correct answer 1.5m/s

But 1.5m/s is wrong. How did you get that? Should the 3μg be 3mg?

 

[hidemi]

But 1.5m/s is wrong. How did you get that? Should the 3μg be 3mg?

Yes, the 3μg is 3mg. And my calculation is as attached. Thanks!