Conservation of momentum, collisions - what about friction?

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SUMMARY

The discussion centers on the conservation of momentum and energy during elastic collisions, specifically involving two pucks of masses 3m and 2m on a frictionless surface. The participants confirm that momentum is conserved during the collision, expressed through the equation m1v1 + m2v2 = m1u1 + m2u2. They clarify that while friction can affect momentum, its impact is negligible during the brief duration of a collision. The conversation emphasizes the importance of understanding elasticity, as the absence of elasticity information complicates the ability to solve for post-collision velocities.

PREREQUISITES
  • Understanding of elastic collisions and momentum conservation
  • Familiarity with kinetic energy equations
  • Basic knowledge of friction and its effects on motion
  • Ability to solve simultaneous equations
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  • Learn how to apply conservation of momentum in various collision scenarios
  • Explore the effects of friction on momentum and energy during collisions
  • Investigate the role of elasticity in determining collision outcomes
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gelfand
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I would like to check my understanding for this problem :

A puck with mass ##3m## is stationary on a horizontal friction-less surface. It is
being impacted in an elastic head-on collision by another puck with the mass
##2m## traveling with speed ##u## to the right. Find the speed and direction of
motion of each puck after collision in terms of ##u##

**********

So for this I would note that because there's no friction there's going to be no
loss of momentum, and energy is always conserved.

So for the equality of momentum before and after the collision I will have:

$$
m_1 v_1 + m_2 v_2 =
m_1 u_1 + m_2 u_2
$$

And for the equality of energy I have:

$$
W + PE_0 + KE_0 =
PE_f + KE_f + \text{Energy(Lost)}
$$

Here there's no work being done during the movement of the puck, and there's no
energy lost due to friction. The pucks are on the ground, so there's no
potential energy either, which gives$$
KE_0 = KE_f
$$

The kinetic energy for the system is

$$
\frac{1}{2}m_1u_1^2 +
\frac{1}{2}m_2u_2^2 =
\frac{1}{2}m_1v_1^2 +
\frac{1}{2}m_2v_2^2
$$

I can multiply all by ##2##, set the second term of the LHS to zero and input the
given quantities here.

Once this is done I have a set of simultaneous equations that can be solved.

But what I'm not sure about -

What if there's friction? How do I consider this?

It seems that If I have friction then I'll no longer have

$$
m_1 v_1 + m_2 v_2 =
m_1 u_1 + m_2 u_2
$$

Though I will still have

$$
W + PE_0 + KE_0 =
PE_f + KE_f + \text{Energy(Lost)}
$$But this one equation (energy conservation) wouldn't allow me to solve for two
unknowns?

I see there are other questions appearing that have similar meanings - I have tried to be explicit in my understanding here though, so if there's anything that "doesn't look right" please say.

Thanks
 
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The principle of conservation of momentum still holds if friction is considered, but you would have to add in the momentum that the Earth gains from the friction with the puck.
 
gelfand said:
What if there's friction? How do I consider this?
You can almost always ignore the effects of friction that take place during a collision. That is because a collision takes place quickly. There is little time for any frictional force to have an effect on momentum. And there is little displacement over which the frictional force can do any work.

This holds when the frictional force is small relative to the force of the collision. For instance, consider hockey pucks sliding across the surface of a table and colliding. The force of friction between pucks and table is non-zero. But it is small compared to the very high and very brief force of the collision.

In some cases, the force of friction is part and parcel of the collision. Consider, for instance, a rotating rubber ball bounced onto a floor at an angle. The normal force of the ball on the floor during the collision is very brief and very high. Similarly the force of friction during the collision can be very brief and very high. Friction cannot be ignored in such a case.

[If you've ever bounced a "Super Ball" (a highly elastic rubber ball with pretty high friction) onto the floor with some spin, you may have experienced this first hand]
 
FactChecker said:
The principle of conservation of momentum still holds if friction is considered, but you would have to add in the momentum that the Earth gains from the friction with the puck.
thanks - could you elaborate a little? I'm not too sure what you're meaning is - or whether you're suggesting considerations such as Earth's mass etc?
 
jbriggs444 said:
You can almost always ignore the effects of friction that take place during a collision. That is because a collision takes place quickly. There is little time for any frictional force to have an effect on momentum. And there is little displacement over which the frictional force can do any work.

This holds when the frictional force is small relative to the force of the collision. For instance, consider hockey pucks sliding across the surface of a table and colliding. The force of friction between pucks and table is non-zero. But it is small compared to the very high and very brief force of the collision.

In some cases, the force of friction is part and parcel of the collision. Consider, for instance, a rotating rubber ball bounced onto a floor at an angle. The normal force of the ball on the floor during the collision is very brief and very high. Similarly the force of friction during the collision can be very brief and very high. Friction cannot be ignored in such a case.

[If you've ever bounced a "Super Ball" (a highly elastic rubber ball with pretty high friction) onto the floor with some spin, you may have experienced this first hand]

Cheers - I'm not familiar with the super ball etc.

So we should consider the materials for this type of problem?

If the question was posed **without** the information on the elasticity, would I have had to change my approach ? I mean - I guess that it would still be elastic, but this would be something that I couldn't take for granted. I'm not sure how I would factor that.

I wish I had found this forum at the start of the module rather than two weeks before the exam! :')
 
gelfand said:
If the question was posed **without** the information on the elasticity, would I have had to change my approach ?
Without information on elasticity, there are many possible outcomes from the collision. You have two unknowns and only one equation (conservation of momentum). There would have been no way to solve the problem.
 
jbriggs444 said:
Without information on elasticity, there are many possible outcomes from the collision. You have two unknowns and only one equation (conservation of momentum). There would have been no way to solve the problem.
ah right ok - thanks.

I wasn't sure if there was some kind of way to test whether it's elastic.
 
gelfand said:
thanks - could you elaborate a little? I'm not too sure what you're meaning is - or whether you're suggesting considerations such as Earth's mass etc?
Yes, I'm talking about the mass of the Earth. So that is usually ruled out by saying there is no friction. If you are talking about the heat generated by internal friction, that does not change the principle of conservation of momentum. Heat is random (directionless average) motion with 0 momentum. Kinetic energy can be converted to heat but momentum will not be.
 
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FactChecker said:
Yes, I'm talking about the mass of the Earth. So that is usually ruled out by saying there is no friction. If you are talking about the heat generated by internal friction, that does not change the principle of conservation of momentum. Heat is random (directionless average) motion with 0 momentum. Kinetic energy can be converted to heat but momentum will not be.
Ok thanks - I'm not sure that I've seen anything that takes these into consideration. So I think that I'll chalk it up to general knowledge rather than something that (for now) I should apply.
 

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