Speed of charged balls after collision

[Bling Fizikst]

Homework Statement

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Relevant Equations

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I tried to apply energy conservation . $$\frac{-kQq}{l}=\frac{m}{2}(v_1^2+v_2^2)-\frac{kqQ}{2r}$$ Now conserving momentum : $$0=mv_1-mv_2$$ Solving for $v_1=v_2=v'$ we get : $$v'=\sqrt{\frac{kQq}{m}\left(\frac{l-2r}{2r}\right)}$$

Since the balls are elastic , so they should collide elastically?After collision , i am assuming they are having the same speeds $v$ at infinity .
Again by momentum conservation : $$2mv=-mv_1+mv_2$$
From this , i am getting $$v=0$$ which is obviously wrong.
I believe i am attempting it in a too oversimplified manner as i am unable to get into the depths of this problem .

 

[kuruman]

And what is your answer that is incorrect? How did you get it? How can we help you if we don't know what you did?

 

[Bling Fizikst]

And what is your answer that is incorrect? How did you get it? How can we help you if we don't know what you did?

Extremely sorry for my careless post . I have edited it now,

 

[kuruman]

Extremely sorry for my careless post . I have edited it now,

Thank you.

i am assuming they are having the same speeds v at infinity .

Is that a good assumption? The equation $$\frac{-kQq}{l}=\frac{m}{2}(v_1^2+v_2^2)-\frac{kqQ}{2r}$$expresses energy conservation before the collision. What is the corresponding expression after the balls separate? Remember that the spheres are conducting and have different charges.

 

[jbriggs444]

Since the balls are elastic , so they should collide elastically?After collision , i am assuming they are having the same speeds $v$ at infinity .

So, in the scenario as you envision it, the two balls collide and neutralize each other. They then separate at constant velocity thereafter?

What if $Q \ne q$?

Edit, I see that @kuruman made the same point. Before I did. He tends to do that.

 

[Bling Fizikst]

What is the corresponding expression after the balls separate? Remember that the spheres are conducting and have different charges.

I am imagining that the balls after collision travel a certain distance , stop and returns back again as they have unlike charges?
When the balls collide , they will have redistributed their charges by induction to have a total charge of $$\frac{Q+(-q)}{2}$$ each .

By energy conservation , $$\frac{m}{2}(v_1^2+v_2^2)-\frac{kQq}{2r}=\frac{m}{2}v^2\cdot 2 + \frac{k(Q-q)^2}{4\cdot 2r}$$
From the first enegry equation from my earlier post , $$\frac{m}{2}(v_1^2+v_2^2)-\frac{kQq}{2r}=-\frac{kQq}{l}$$
Plugging this and simplifying gives : $$v=\sqrt{-\frac{k}{m}\left(\frac{(Q-q)^2}{8r}+\frac{Qq}{l}\right)}$$ which is contradictory . I am not sure where i went wrong .

 

[kuruman]

Plugging this and simplifying gives :

Plugging what in what?
I understand that the first energy equation from the earlier post is the mechanical energy of the two-particle system just before the collision. Where is the corresponding equation immediately after the collision? Is mechanical energy conserved through the collision?

which is contradictory .

What is contradictory. What did you expect?

 

[Bling Fizikst]

Where is the corresponding equation immediately after the collision? Is mechanical energy conserved through the collision?

I believe it is : $$\boxed{\frac{m}{2}(v_1^2+v_2^2)-\frac{kQq}{2r}}=\frac{m}{2}v^2\cdot 2 + \frac{k(Q-q)^2}{4\cdot 2r}$$

Plugging what in what?

$$\boxed{\frac{m}{2}(v_1^2+v_2^2)-\frac{kQq}{2r}}=-\frac{kQq}{l}$$ in the above equation . It happens to be the LHS of the above equation . Or rather equating the boxed expressions .

 

[Bling Fizikst]

What is contradictory. What did you expect?

The expression inside the square root is negative .

 

[kuruman]

I believe it is : $$\boxed{\frac{m}{2}(v_1^2+v_2^2)-\frac{kQq}{2r}}=\frac{m}{2}v^2\cdot 2 + \frac{k(Q-q)^2}{4\cdot 2r}$$

In this equation $v_1$ and $v_2$ on the LHS are the speeds before the collision. I agree that they the same so use the same symbol, say $v_i$ like you did for the speeds after the collision. After the collision, the speeds are also the same. Let's call them $v_{\!f}$.

Key question to answer: Is $v_{\!f}$ equal to $v_i$? Why or why not? Remember that the collision is elastic and the masses are equal.

 

[haruspex]

Again by momentum conservation : $$2mv=-mv_1+mv_2$$

Wouldn’t it be $mv-mv=-mv_1+mv_2$?

as they have unlike charges?

They are conductors and have just made contact.

 

[Bling Fizikst]

Key question to answer: Is $v_{\!f}$ equal to $v_i$? Why or why not? Remember that the collision is elastic and the masses are equal.

By definition of elastic collision , the kinetic energy remains conserved . So , $$\frac{m}{2}v_i^2\cdot 2=\frac{m}{2}v_f^2\cdot 2\implies v_i=v_f ?$$

 

[Bling Fizikst]

Wouldn’t it be $mv-mv=-mv_1+mv_2$?

Yeah my bad . This means we again end up at $$v_1=v_2$$ as in the first momentum equation .

 

[kuruman]

By definition of elastic collision , the kinetic energy remains conserved . So , $$\frac{m}{2}v_i^2\cdot 2=\frac{m}{2}v_f^2\cdot 2\implies v_i=v_f ?$$

Yes. So is mechanical energy conserved through the collision?