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I have this problem to solve:

Two protons are on the x axis at x=-1 and x=0. An alpha particle is placed at x=2, and the proton on the left is released, find its speed at infinity.

I know that this involves the change in kinetic and potential energies of the particles, but I dont understand which particles are included in that calculation. Is the proton on the left neglected because it is the one released? An explanation in general would be appreciated.

For the purposes of following the rules in asking a homework question properly, the specific physics topics I was asking about were: which particles are included in the kinetic and potential energy calculation for determining the speed at infinity of a proton, and why. My effort in solving this question was finding out that work is the change in kinetic energy + change in potential energy, but not knowing which particles to include in the calculation. This should clear up any confusion that this question is off topic.

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closed as off-topic by Brandon Enright, Jim, John Rennie, Qmechanic Apr 16 '14 at 12:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Brandon Enright, Jim, John Rennie, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ This question has already been solved by the answer given. I updated the question because it was marked as off topic. $\endgroup$ – ez4nick Apr 16 '14 at 19:09
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I'm assuming the proton at x=0 and the alpha particle at x=2 are held in place. I'll call these the "held" particles. There will be two forces which act on the "free" proton (at x=-1). I'm also assuming that this is just a classical mechanics problem, and will solve it that way.

Before we go into the calculations, let's think about what is going to happen. We expect the proton to feel a force from the two other particles. Due to the fact that every particle is positive, we expect the free proton to be repulsed by the other two particles and shoot off in the -x direction. Since the free proton was at rest when we released it (right?), it's going to try to convert all of its energy from potential energy to kinetic energy. At some infinite point, the proton will get rid of its potential energy, and get to its maximum speed. Sweet.

This proton got all of its energy from the held particles, right? So we need to calculate the potential energy from each of them. Since the free proton doesn't care about the interactions between the two held particles, we don't need to calculate that. We do need:

Energy #1: The force of the held proton on the free proton.

Energy #2: The force of the held alpha particle on the free proton.

Total Potential Energy= Energy #1 + Energy #2

The Total Potential Energy is all the energy that proton can use in its speedy retreat. Since it will convert all of its potential energy into kinetic energy, we just use a formula for kinetic energy (which has velocity and mass in it, not the other ones!), and set that equal to our total potential energy. Do a little math, and find the free proton's velocity! That's your answer!

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  • $\begingroup$ Thanks. I was able to get the correct solution and understand the problem better now. $\endgroup$ – ez4nick Apr 15 '14 at 23:34

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