# Momentum & Friction

OME9A
Conservation of momentum only applies when there are no external forces, so how would I add friction into the equation?

For a sticky collision,

$$p_{A1}+p_{B1}=p_{A2}+p_{B2}$$

But I've seen that the momentum after is usually less than the momentum before (for a sticky, linear collision of 2 carts on a track)

Would I add the force of friction of each cart to the left side of the equation? So

$$p_{A1}+p_{B1}+f_{A1}dt+f_{B1}dt=p_{A2}+p_{B2}$$

But the collision itself seems like .0001 seconds so it doesn't really help... How could I form an equation to explain the loss of momentum?

And finally, if two carts of the same weight are traveling at each other, their forces of friction would cancel out so the conservation of momentum would apply?

Thank you!

Momentum is always conserved. The effect of friction is to change some of the kinetic energy into heat.

OME9A
Hmmm... but in all my experiments it is never conserved! So it's just "human error?" That sucks... I was hoping there was a cool explanation!

Homework Helper
Hmmm... but in all my experiments it is never conserved! So it's just "human error?" That sucks... I was hoping there was a cool explanation!

It's not really human error. There will always be friction due to the motion of the masses in relation to the surface as well as the air. That is why in your experiments you'll find that the values differ.

OME9A
It's not really human error. There will always be friction due to the motion of the masses in relation to the surface as well as the air. That is why in your experiments you'll find that the values differ.

Right, but I thought that explained the loss in Kinetic Energy, not momentum...

Homework Helper
Right, but I thought that explained the loss in Kinetic Energy, not momentum...

The presence of friction means that the system is not closed anymore.

OME9A
The presence of friction means that the system is not closed anymore.

so if it isn't closed, we have no way of calculating the correct change in momentum? Even if we know the external forces?

You still seem to be confused between the loss of kinetic energy (true) and loss of momentum (false). If you are doing experiments, you may be neglecting some effect, such as momentum transfer to the track.

OME9A
You still seem to be confused between the loss of kinetic energy (true) and loss of momentum (false). If you are doing experiments, you may be neglecting some effect, such as momentum transfer to the track.

I understand that momentum is always conserved when there are no external forces, but I've read that if there are external forces, the conservation of momentum does not hold... So how can we create an equation that factors in the external forces since the loss of momentum is false?

Is there anyway to estimate this momentum transfer to the track?

OME9A
The presence of friction means that the system is not closed anymore.

I read that even with friction it is still closed because matter is not exchanged... Only heat/energy is exchanged. So then the conservation of momentum still applies, and the only reason for the difference is human error...right?

Homework Helper
I read that even with friction it is still closed because matter is not exchanged... Only heat/energy is exchanged. So then the conservation of momentum still applies, and the only reason for the difference is human error...right?

Well it depends on what you consider the system as. But most likely it is due to error in measurement.

If possible, you could do an energy balance and include friction and see if that will get you what the final velocity is.