# Oxford PAT 2014, Question 17

## 12 thoughts on “Oxford PAT 2014, Question 17”

1. Hi,
Isn’t the final result v=sqrt(8kQ^2/3md)?
So, if you work out the potential energy you get: 2kQ^2/d
Let m=mass of particle 1, and v= velocity of particle 1, and m’=mass of particle 2, and v’= velocity of particle 2
Then the Kinetic Energy is (0.5mv^2)+(0.5m’v’^2)
Since m’=2m, then if they exert the same force on each other, a’=a/2, therefore v’=v/2
Therefore Kinetic Energy= (0.5mv^2)+(0.52m(v/2)^2)= 3/4mv^2
So, 3/4mv^2 = 2kQ^2/d
Therefore v=sqrt(8kQ^2/3md)

1. Your reasoning and calculations are both absolutely right, but you’ve calculated the speed of the particle with mass = m, and the question asked for the speed of the particle with mass = 2m.

(You’ve written “Let m=mass of particle 1, and v= velocity of particle 1, and m’=mass of particle 2, and v’= velocity of particle 2 … Since m’=2m, … Therefore v= …”).

1. Thanks for pointing that out-However, I’ve now rearranged it and I’ve got an answer of 2Qsqrt(k/3dm), which seems to be the same answer as Retnemmoc.

1. I’m pretty sure that your calculation for v was correct. (From your first comment): v=sqrt(8kQ^2/3md). As you also correctly state above, v’ = v/2.

So plugging your v’ = v/2 into your original calculation for v gives v’ = sqrt(8kQ^2/3md)/2 = sqrt(8kQ^2/6md) =sqrt(2kQ^2/3md) = Q√(2k/3md), which is the same as what I got.

2. Gabriel Maheson says:

Hey, quick question, Initially, why wouldn’t we say that the 2m charge has PE=(2kQ^2)/d J and so does the other, and so we can calculate the work done in losing PE and gaining Ek solely for the charge with mass 2m? What I’m trying to say is… would we say the initial PE of the combined charges is 2kQ^2, or is that for the individual charges? Hopefully you understand what I’m getting at? I can’t see how the combined PE can be equal to the PE of only one charge! Thanks for any clarification !

1. Funnily enough, I thought the same thing. Why is the electrostatic potential split between the two charges? Each particle experiences a force of 2kQ^2/d^2, so surely each particle will have electric potential energy of 2kQ^2/d.

1. Thanks both for asking this question because it is a subtle point. You don’t actually need to understand any of these details in order to get the right answer, and I think the people who set the test assumed that you would just blindly follow the instructions and treat the total potential energy as the formula given, in which case it is all easy. I’m afraid it’s the curse of physics that the more you think about it the harder it gets.

You are right about the force that the particles both experience when they are at distance d: it is 2kQ^2/d^2.

But let’s consider what happens when we start with both particles an infinite distance apart and bring one of them to a distance d from the other. It’s easiest to consider the case where we hold one still and move the other one closer, while exerting a force -2kQ^2/r^2 on it. We are exerting a force on both particles but only one of the forces is doing any work, because the other particle doesn’t move. In this case the total work done is the integral of -2kQ^2/r^2 as r goes from infinity to d, which is 2kQ^2/d. This is the total work done, so it is the total potential energy gained by the system; it is not the energy gained by each particle.

3. Retnemmoc says:

Shouldn’t 2.k.2Q.Q in the second to last term simplify to 4kQ^2, which would make the final term 2Qsqrt(k/3md) instead?

1. But the bottom is 2m (1 + 2m/m) d so the 4 k Q^2 is divided by 2. So I think my answer’s right. It wouldn’t be the first daft mistake I’ve made though: can you check your working again, and confirm if you’re sure and I can check again if necessary.

4. Lee says:

Should we take Gravitational Potential Energy into the equation too? since G.P.E is gained when the 2 masses move further from each other?

1. HI Lee, thanks for your comment.

The question describes the masses as ‘charged particles’, so I’m pretty sure we are supposed to ignore gravitation here, because the gravitational forces would be so low in comparison with the electrostatic forces. In fact for a normal particle (a proton, say) the gravitational force would be so low that you would have to be able to measure the distance d with impractically-high accuracy before the gravitational force would be comparable to the error in the electrostatic force resulting from tiny errors in measurement of d.