# B The acceleration of two colliding particles

1. Aug 12, 2017

### yrjosmiel

Something about collisions has been bothering me.
Let's say I have 2 particles with equal mass: one is moving at velocity v towards the other particle. Kinetic energy and momentum is conserved during this collision. However, I have a question: what is the acceleration of the particles during the collision? Is it infinite (since the particles suddenly change velocity in an instant)? I know it shouldn't be, or should it?

velocity-time graph of particle 1

velocity-time graph of particle 2

2. Aug 12, 2017

### Ibix

It depends on your model. As you've drawn it the velocities are discontinuous, so you are modelling their acceleration as instantaneous. This is not strictly realistic, but is a good enough idealisation for many circumstances.

In practice, objects deform during a collision and there is a smooth velocity change - albeit frequently on a timescale of milliseconds.

3. Aug 12, 2017

### yrjosmiel

Question: do molecules deform?
For example: I have a sealed box with gas in it. The molecules of the gas are whizzing lighting fast. Of course, it will hit the side of the box. This is where the pressure comes from. How do I know how much force each molecule exerts on the wall per collision?

4. Aug 12, 2017

### Ibix

Molecules do deform, but interactions between molecules aren't really about two things colliding; they're about their electrons' electromagnetic fields interacting, which gets complicated (beyond me).

Regarding the force from a single interaction, it'll vary through the interaction. I'm not sure that it's the force you really want to know, though. The impulse (the momentum change) is more interesting in terms of thinking about pressure. You can easily calculate the momentum change in the turnaround, and multiply by the expected number of collisions per second to get an average force over that second.

5. Aug 12, 2017

### yrjosmiel

oh shoot i forgot electric repulsion is a thing dammit

Also, I just realised something: the force exerted to the wall by the particle can vary depending on the angle of collision.
How do we determine the average momentum change per particle?

6. Aug 12, 2017

### bahamagreen

Aren't collision "contacts" between fields... the particles don't touch?

PV=nRT ... does it look like gas pressure does not take the particle mass into account?

7. Aug 12, 2017

### Ibix

If you know the average speed of the gas molecules you can use the equipartition theorem to work out the average speed perpendicular to the wall.

8. Aug 12, 2017

### Ibix

It depends what your constraints are. What matters is the average energy of each gas particle. If everything is in thermal equilibrium then the only thing that matters is the temperature, which measures the average energy of a particle (kT). If you've got a mix of particles of different masses in the box, they'll different types will have different distributions of speeds, different average velocities, but the same average energy. But you might, for some reason, know something about your particle masses and velocities but not the temperature. Then you'd figure out the temperature from the masses and velocities - which seems to be where the OP is coming from.

9. Aug 12, 2017

### yrjosmiel

Is it possible to instead get the average perpendicular speed of molecules to the wall by getting the momentum of one particle (mv), multiplying it by the sum of cosθ from 0 to 90 with a step interval approaching 0, then dividing it all by 90*step interval?

θ = the angle of the trajectory of the particle relative to the normal
pardon me if that bunch of words made no sense

10. Aug 12, 2017

### Ibix

I think you need to do the integral formally if you want to go about it that way, rather than some discrete approximation. Remember that there is a hemisphere of possible approaches to the wall, not just a semi-circle. You need to do a two-dimensional integral.

11. Aug 13, 2017

### yrjosmiel

Does the third dimension really matter?

12. Aug 13, 2017

### Ibix

Yes. There is only one way to approach the wall perpendicular to it, but many ways to approach at an angle, and more and more ways the further off perpendicular you are.

The equipartition theorem will let you write down the answer in a couple of lines, which is why I recommended that.

13. Aug 13, 2017

### Eleni_

Do the particles have the same mass?

14. Aug 13, 2017

### yrjosmiel

Yes.

15. Aug 13, 2017

### rcgldr

To make the situation simpler, imagine that the two objects can share the same space (they intersect), and have some non-zero radius.

16. Aug 14, 2017

### Eleni_

I am a beginner so most likely what i am saying is not the right answer but i will give a try. Since the force F particle 1 has on particle two and vice versa then F1=F2 (Newton 3 law). Now according Newton second law F=ma since they have equal force and equal mass then they should have the same a, which at the time of the collision will be -a. Again i dont know if this is the correct answer.

17. Aug 14, 2017

### Eleni_

I am a beginner so most likely what i am saying is not the right answer but i will give a try. Since the force F particle 1 has on particle two and vice versa then F1=F2 (Newton 3 law). Now according Newton second law F=ma since they have equal force and equal mass then they should have the same a, which at the time of the collision will be -a. Again i dont know if this is the correct answer.

18. Aug 14, 2017

### yrjosmiel

I know they will have the same magnitude of acceleration. I am more concerned whether acceleration is infinite or not because the change in velocity is instantaneous (although it has already been realised that it really isn't since collisions are basically interactions of electric fields of electrons repelling each other).

19. Aug 14, 2017

### Eleni_

I guess that if there is no friction then the acceleration will be indeed infinite

20. Aug 14, 2017

### yrjosmiel

However, there is no such thing as infinite acceleration in this world we are in right now. It would require an infinite amount of energy to do that.