The acceleration of two colliding particles

[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

 

[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.

 

[yrjosmiel]

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

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?

 

[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.

 

[yrjosmiel]

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

oh shoot i forgot electric repulsion is a thing dammit

Also, I just realized 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?

 

[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?

 

[Ibix]

How do we determine the average momentum change per particle?

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.

 

[Ibix]

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

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.

 

[yrjosmiel]

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.

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

 

[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.

 

[yrjosmiel]

Remember that there is a hemisphere of possible approaches to the wall, not just a semi-circle.

Does the third dimension really matter?

 

[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.

 

[Eleni_]

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
View attachment 208872
velocity-time graph of particle 2
View attachment 208873

Do the particles have the same mass?

 

[yrjosmiel]

Do the particles have the same mass?

Let's say I have 2 particles with equal mass:

Yes.

 

[rcgldr]

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

 

[Eleni_]

Yes.

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 don't know if this is the correct answer.

 

[Eleni_]

Yes.

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 don't know if this is the correct answer.

 

[yrjosmiel]

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 don't know if this is the correct answer.

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 realized that it really isn't since collisions are basically interactions of electric fields of electrons repelling each other).

 

[Eleni_]

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 realized that it really isn't since collisions are basically interactions of electric fields of electrons repelling each other).

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

 

[yrjosmiel]

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

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.

 

[Eleni_]

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.

the acceleration will be 0 but the velocity will be constant. Since no other force will be applied to our particulate, then it will be moving forever (Newton 1 law). There is a gadget, the Newton's cradle which describes (not entirely) our case.

 

[yrjosmiel]

the acceleration will be 0 but the velocity will be constant. Since no other force will be applied to our particulate, then it will be moving forever (Newton 1 law). There is a gadget, the Newton's cradle which describes (not entirely) our case.

Actually, there should be a force.
As you can see on the velocity-time graph, the velocity of both particles change at the time of collision. Since there is a change in velocity, there must be acceleration.

 

[Eleni_]

Actually, there should be a force.
As you can see on the velocity-time graph, the velocity of both particles change at the time of collision. Since there is a change in velocity, there must be acceleration.

Accelaration is a derivative of speed.dv/dt. The speed will change exactly as you mentioned but at a point it will remain constant and thus accelaration we decreasing and decreasing with lim=0

 

[Ibix]

In the model @yrjosmiel is using, the force and acceleration are Dirac delta functions - that is they are infinite, but are applied for zero time. The integral with respect to time, however, is well defined, so the momentum change is finite. As I noted earlier this model is unrealistic, but good enough for many circumstances. If you are playing billiards, for example, you can model the balls as rigid and having instantaneous collisions with infinite accelerations because ignoring this is a much smaller error than others - e.g. in your accuracy with the cue (certainly with my skill level at billiards).

In practice, all objects deform slightly on a very short timescale, so the force is always finite and applied over a finite time. For example, if you are playing tennis you can impart spin to the ball by striking it with a racquet moving slightly sideways, and you'll get different effects depending on the tension in the strings. You can't neglect the deformation of the strings if you want to analyse that case.

Basically, the graph in post #1 is an idealisation. In practice, if you stretch the graph horizontally so you can see short enough times, you'll find that there is a very steeply sloped line joining the two apparently discontinuous segments.