MCAT: Physics Question

[TieuBachHo]

Two protons initially are at rest at a distance 10 nm from each other. They are released and accelerate away from each other. How fast are they both going after they are very far apart?
(mass= 1.66 x 10^-27kg, e = 1.6 x 10^-19 C, k = 9 x 10^9 Nm^2/C^2)

The answer choice is 3.8 x 10^3 m/s.

This seems to be a relatively easy question but I can't seem to come up with a correct answer Thanks?

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[TRN1983]

Okie dokie. Here is my attempt to answer this (caveat: my math didn't agree with the answer you have). I'm running on a rusty memory here, but I'm pretty sure this is energy conservation.

In short: The initial electric potential energy that each proton has at its initial point (10nm separation) will be transferred entirely into kinetic energy when the separation of the two protons approaches infinity (i.e. there is no more repulsive force between them). In other words: (delta)U = (delta)KE

Again, I'm going off of a rusty memory, but I believe that the following equations apply:

U=kQq/r (where r=d) and you can set that equal to KE=1/2mv^2 and solve for v.

v=sqrt(2kQq/mr)

HOWEVER: When I do this I get 5.3*10^3 m/s

Note: You could also use work energy theorem here. I.e. W=Fd=(delta)KE; where d = r and F=kqQ/r^2 you get F*d=(kqQ/r^2) * r = kqQ/r, which you can then set equal to (delta)KE.

Oh, and what book is this from? This would be a very poor MCAT problem b/c it has far too much calculation in it.

 

[TieuBachHo]

Thanks, I got the same thing too but I doubted it because it didn't match the given answer

When I tried to use the reference frame, it didn't came out right either ... I have no idea how they got that answer

It's from a KAPLAN book

 

[TRN1983]

I'd imagine its a typo.

 

[fileserver]

Maybe both protons are moving and the potential energy split up. Then you would get the same answer as the book.

The above post only show that 1 of the protons is moving while the other is stationary. (If this question is on the MCAT, I would have missed it too)

 

[TRN1983]

Maybe both protons are moving and the potential energy split up. Then you would get the same answer as the book.

The above post only show that 1 of the protons is moving while the other is stationary. (If this question is on the MCAT, I would have missed it too)

No, that wouldn't change the answer. The initial potential energy must eventually be turned entirely into kinetic energy b/c of the following.

U=kqq/r --> When the distance approaches infinity, then U approaches ZERO. Assuming that the transfer of energy goes only to potential energy then we know that U=kqq/r where r=10nm HAS to equal the KE when r=infinity in order to satisfy the first law of thermodynamics. So, the only thing you need to know is the U of each proton (both are the same and both are described by U=kqq/r) at the point, where r=10nm, and the mass of the protons.

I'm pretty sure that the book must have a typo in it. U=kQq/r describes the potential energy that charge Q has as a result of q's electric field. U=kqQ/r describes the potential energy that charge q has as a result of Q's electric field. Both of those answer's yields the same result b/c q=Q and they both have the same r (10nm). Each of their individual U's will turn into their individual KE's.

 

[fileserver]

So, the question is how much work put in initially so the the protons are where they are.

 

[TRN1983]

So, the question is how much work put in initially so the the protons are where they are.

That would depend on their starting positions W=F*d

 

[fileserver]

Touche, haha.

Let's say that two protons are infinite apart and you move one to 10nm to the other proton. So amount of work you put in is the same as th potential energy, U. Therfore, the total initial energy of the system is U.
The final total energy must be the same which is in the form of KE of the 2 protons. So each proton has 1/2U in KE. V would equal to the answer the book gave.

So it comes down to that same question, is it U or 2U initially.

 

[TieuBachHo]

Thanks all,

Is this what you mean fileserver?
1/2U(i) = KE(f)

I had already tried that one out also but it wouldn't come out right either V final would be rounded up to 3.73 x 10^3 (m/s) and not 3.8 x 10^3 (m/s).

I hope it's a typo or else I wouldn't know what else is there that I am missing.