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[Matt]

I went to a science show with my family at the University of Minnesota today.  There was a demonstration on conservation of momentum.

It was for the kids, but there's something that's always been hard for me to get my head around with this topic.

Momentum is related to velocity and kinetic energy is velocity², which means your energy grows faster than momentum as you speed up.  If I were half the weight and twice as fast, I would have the same momentum but 2x the kinetic energy.

What does that mean in real-world terms?  When looking at collisions where things aren't perfectly elastic, like two football players, is kinetic energy or momentum more meaningful? 

If you get 10% faster for each 10% reduction in weight, is it a win because energy comes out higher?  Or is it a wash because momentum stays unchanged?

From the math angle, if two bodies collide, momentum is preserved and some amount of kinetic energy becomes heat, noise, etc.  Let's call that h.  In an elastic collision, h is 0 and in an inelastic collision it must be related to the mass and velocity of the inputs.  Is there some Newtonian equation that describes h, given m₁, v₁, m₂, v₂, and a coefficient of stickiness (maybe called coefficient of restitution or cr)?  Maybe solving that equation would help me understand the relationship between kinetic energy and momentum.

Thanks for any help.  Just imagine how twisted up I'll be if I ever go on the Quantum Physics tour at the Soudan Mine here in Minnesota!

[jmone]

For some reason I read "Soudan Mine here Biggest Ball of Twine in Minnesota"

[John Gateley]

Momentum is the derivative of Kinetic Energy with respect to velocity.

[glynor]

I think... And INAP (I'm Not A Physicist) but...

I think that the problem lies in trying to think about it in terms of Newtonian Physics.  I think that momentum describes something that occurs, while energy describes something that is.

In General Relativity, there is no difference between mass and energy.  They are two facets of the same coin.  As you move faster, you become more.  The energy is the description of what makes up that more, but the net effect is that you have more impact on the fabric of spacetime.  Momentum is like the wave of spacetime that builds up behind you.  It is caused by a combination of your mass-at-rest and your kinetic energy, and you "ride the wave", but the wave isn't the energy and the energy isn't the wave.

As far as your question about the football player... If what I think I understand about it holds true, then it is the force of the player, not the momentum.  Not just the kinetic energy, but the total energy "budget" (mass + energy) of the "thing" that hits you.  The wave is made of spacetime, and may be a force carrier, but is not the force itself.

[JimH]

Where is jgreen when we need him?

[mwillems]

Momentum is related to velocity and kinetic energy is velocity², which means your energy grows faster than momentum as you speed up.  If I were half the weight and twice as fast, I would have the same momentum but 2x the kinetic energy.

I am also not a physicist, but I think the classical mechanical explanation might be helpful (and that explanation is still more or less true at speeds much lower than the speed of light).  At least in newtonian mechanics (with idealized bodies) kinetic energy isn't velocity squared, it's 1/2*mv². Momentum (also in Newtonian mechanics) is mv.  

The distinction between the two, as I understand it, is that in a closed system momentum is always conserved, but kinetic energy is not necessarily conserved (which is the difference between an elastic and inelastic collision).  Imagine a perfectly inelastic collision between two ideal objects of equal mass, one of which is moving at velocity v and the other of which is stationary.  Pre-collision the momentum of the moving object is mv, the kinetic energy is 1/2*mv².  Post-collision, the moving object is slowed by the collision, while the stationary object is accelerated by the collision.  In our ideal example both objects are now moving at v/2 in the same direction.  The momentum of the first object is now m*1/2v, but the momentum of the second object is also m*1/2v, which when added together equal=mv (conservation of momentum).  The kinetic energy of each object is now 1/2*(m*1/2v²), which when added together is less than 1/2*mv².  In a perfectly elastic collision with the same parameters, the first object would stop entirely and the second would be accelerated to v, so both momentum and kinetic energy remain the same (albeit transferred entirely to the second object).  

All real world collisions will be somewhere between those two poles.  Speed has a greater effect on how hard it is to stop something than mass does in all but perfectly inelastic collisions (but it also has a greater effect on how hard it is to get something moving to begin with).  If you could trade speed for mass straight across you'd have greater kinetic energy (and thus would be harder to stop), but higher speed requires exponentially more work so the trade is rarely "straight across." Having greater kinetic energy at a given momentum would always be "preferable" for the football player unless the collision was perfectly inelastic (in which case it would be a wash), because in any even slightly elastic collision the greater kinetic energy provides some delta above the mere transfer of momentum.  

Again, not a physicist (and working from a now many-year old physics comprehension), but I'm pretty sure that's the way it works at "regular" speeds (as opposed to relativistic ones).  You can actually see the examples illustrated over at the wiki: http://en.wikipedia.org/wiki/Momentum#Application_to_collisions

[jgreen]

Here he is.

For a brief practicum on mass vs energy, google "zapruder frame 313".  Talk about science show!

[mojave]

There are discussions about momentum vs kinetic energy when comparing bullet or arrow weight vs velocity for hunting. Here are a couple of interesting ones:

Tuffhead Broadheads:  Kinetic Energy vs. Momentum

Momentum vs kinetic energy for dummies: How a traditional bow kills

[Listener]

Momentum is related to velocity and kinetic energy is velocity², which means your energy grows faster than momentum as you speed up.  If I were half the weight and twice as fast, I would have the same momentum but 2x the kinetic energy.

What does that mean in real-world terms?  When looking at collisions where things aren't perfectly elastic, like two football players, is kinetic energy or momentum more meaningful? 

If you get 10% faster for each 10% reduction in weight, is it a win because energy comes out higher?  Or is it a wash because momentum stays unchanged?

From the math angle, if two bodies collide, momentum is preserved and some amount of kinetic energy becomes heat, noise, etc.  Let's call that h.  In an elastic collision, h is 0 and in an inelastic collision it must be related to the mass and velocity of the inputs.  Is there some Newtonian equation that describes h, given m₁, v₁, m₂, v₂, and a coefficient of stickiness (maybe called coefficient of restitution or cr)?  Maybe solving that equation would help me understand the relationship between kinetic energy and momentum.

Thanks for any help.  Just imagine how twisted up I'll be if I ever go on the Quantum Physics tour at the Soudan Mine here in Minnesota!

It's always a bad sign when someone brings up quantum mechanics in an audio forum.

Your questions are a bit vague.  Matters for what?

If you are driving at 60mph, put the transmission in neutral and coast, the kinetic energy you had at that point will determine how far you get before the car comes to a halt.  If you are going uphill, some of the kinetic energy will be converted into potential energy.

Bill

[Matt]

For a brief practicum on mass vs energy, google "zapruder frame 313".  Talk about science show!

A gun is a good example of something that isn't totally intuitive to me.

Conservation of momentum means the shooter recoils the same amount as the person that gets shot (assuming the bullet doesn't go through them).

However, a huge percentage of the energy from the explosion goes into kinetic energy of the bullet (since the bullet moves more than the shooter, and work is force times distance). 

So the bullet is really fast, has a huge kinetic energy, totally destroys the person it hits, yet only makes them recoil about the same as the shooter.

[glynor]

It's always a bad sign when someone brings up quantum mechanics in an audio forum.

I thought it was a great question.  It is something I've often thought about myself, so I've been very curious about the answers.

[MrC]

Interesting topic.

I think Matt has seen too many action movies where a person shot gets blown backwards twenty feet and flies through a window.  Total Bunk.

The recoil of the gun is relative to the mass of the bullet and the cartridge's explosive force (gas propulsion).  These are small forces.

The damage done to human does not "totally destroy".  Rather, the overall damage is quite minimal compared to the entire body mass ( and that's not to say it is non-lethal).  The damage done is due to the high velocity piercing of flesh, tissue distortion, shattering of bone and fragmentary damage, and heat burn.  A shot human hardly moves - because the mass of the bullet is very small, and the human body is very massive comparatively, and most of the energy goes into the piercing of tissue and bullet distortion.

[jgreen]

Well, if the conversation's turning this way, ballistic tables really do shed light on this.  The issue is essentailly, "big, slow projectile, or small, fast projectile"? This seems to be the kind of "energy" that Matt is pondering.  

Unfortunately (for them), the more you investigate, the more you'll find that--either way--chickens take the fall.

[pcstockton]

I think... And INAP (I'm Not A Physicist) but...

I think that the problem lies in trying to think about it in terms of Newtonian Physics.  I think that momentum describes something that occurs, while energy describes something that is.

In General Relativity, there is no difference between mass and energy.  They are two facets of the same coin.  As you move faster, you become more.  The energy is the description of what makes up that more, but the net effect is that you have more impact on the fabric of spacetime.  Momentum is like the wave of spacetime that builds up behind you.  It is caused by a combination of your mass-at-rest and your kinetic energy, and you "ride the wave", but the wave isn't the energy and the energy isn't the wave.

As far as your question about the football player... If what I think I understand about it holds true, then it is the force of the player, not the momentum.  Not just the kinetic energy, but the total energy "budget" (mass + energy) of the "thing" that hits you.  The wave is made of spacetime, and may be a force carrier, but is not the force itself.

Great summation in layman's speak.  Well done.  

You cannot bring this to our "experience" of the world.  We often forget that these are universal principles that have little meaning in our earth bubble.  Feathers fall slowly here......

-p

[pcstockton]

"big, slow projectile, or small, fast projectile"

The slow projectile penetrates the shield.

[jgreen]

Umm, ballistic tables were compiled using goats, like it or not.  Okay, they were actually papier-mache goats pinatas.

[fitbrit]

Remember that energy cannot be destroyed, only converted. So let's take a look at that bullet hitting a 100 kg target (in simplified terms).
The bullet eventually comes to rest, so all of the bullet's kinetic energy is converted into some other form of energy. Only some of that energy is used to transfer the (conserved) momentum to the target if it's not 'fixed'. So immediately you have your answer as to which is more important. The rest of the kinetic energy causes a lot of damage due to changing to heat, some sound etc., PLUS creation of momentum in all directions of parts of the target.

Also, do not forget that momentum is a vector, and kinetic energy is not. The energy of the impact of the bullet will cause some of the target material to disperse in all directions. These particles/fragments will conserve momentum too, so that a bigger chunk going off the right, might be balanced with smaller, multiple chunks with the same speed flying off to the left, or a single smaller chunk going to the left at higher speed, or any combination that conserves momentum along all axes. The part of the target that stays intact, itself, may not move that far in the direction of the bullet's trajectory, because of its mass. But the faster the bullet, for any given bullet mass, the more damage it will do in transferring kinetic energy (and infinite pairs of equal and opposite momentum vectors) to parts of the target.
In other words, the size of the impact 'crater', and hence 'damage', is going to depend a lot more on the kinetic energy (scalar) than on the momentum (vector), in part because the kinetic energy is not constrained in direction as it transfers and converts into other forms of energy.

I used to be a physicist, even went to Cambridge to do physics. I changed to focus more on Chemistry, and the biology, however, when I discovered that many of the Cambridge physicists emitted foul odours from their bodies, and I would never get a girlfriend if I continued to hang out with them.

[Matt]

Remember that energy cannot be destroyed, only converted. So let's take a look at that bullet hitting a 100 kg target (in simplified terms).
The bullet eventually comes to rest, so all of the bullet's kinetic energy is converted into some other form of energy. Only some of that energy is used to transfer the (conserved) momentum to the target if it's not 'fixed'. So immediately you have your answer as to which is more important. The rest of the kinetic energy causes a lot of damage due to changing to heat, some sound etc., PLUS creation of momentum in all directions of parts of the target.

This is well written.

So to my original question, it seems there should be a simple equation to relate how much kinetic energy becomes non-kinetic energy (h from my first post) given the velocities, masses, and Cr of the objects.

In other words, an equation that says "in a collision, given these few input variables, this much kinetic energy must become another form of energy so that conservation of momentum is fulfilled."

Is that right, or are there other variables?  And what is the equation?  I tried to figure it out here but ran out of whiteboard space

[glynor]

This is well written.

+1

That was great.  Thanks!

[rjm]

Q: Why are craters on the moon round and not oblong when common sense says the majority of impacts would not have been perpendicular to the surface?

A: Because at very high speeds an impact becomes an explosion and the force radiates out in a perfect circle.

[fitbrit]

This is well written.

So to my original question, it seems there should be a simple equation to relate how much kinetic energy becomes non-kinetic energy (h from my first post) given the velocities, masses, and Cr of the objects.

In other words, an equation that says "in a collision, given these few input variables, this much kinetic energy must become another form of energy so that conservation of momentum is fulfilled."

Is that right, or are there other variables?  And what is the equation?  I tried to figure it out here but ran out of whiteboard space

You may want to look at this Wiki page, especially towards the bottom:

http://en.wikipedia.org/wiki/Coefficient_of_restitution

[Listener]

This is well written.

So to my original question, it seems there should be a simple equation to relate how much kinetic energy becomes non-kinetic energy (h from my first post) given the velocities, masses, and Cr of the objects.

In other words, an equation that says "in a collision, given these few input variables, this much kinetic energy must become another form of energy so that conservation of momentum is fulfilled."

Is that right, or are there other variables?  And what is the equation?  I tried to figure it out here but ran out of whiteboard space

There is no requirement to turn energy into some other form of energy.

My take is that there are three equations that must be satisfied for :

1. conservation of momentum as a vector sum (in absence of an external force such as gravity)


2. Conservation of energy ( summed over all masses before the event and after the event)


3. Conservation of mass (summed over masses present before and after the event)

A set of masses present after the event and velocities for those masses  must fit these equations.

Notes:
  no allowance for conversion to other forms of energy.
  no external forces present
  speeds too low to get Einstein  excited.

----
I'd answer rjm's question a bit differently.  I'd describe the process as two events:

The first event converts the kinetic energy of the meteor into heat and potential energy (deformation of the surface).
The second event converts those forms of energy back into kinetic energy of a lot of particles. (the explosion.)
----
So Matt, are you going to add a first person shooter game to MC?

Bill

[Yaobing]

A gun is a good example of something that isn't totally intuitive to me.

Conservation of momentum means the shooter recoils the same amount as the person that gets shot (assuming the bullet doesn't go through them).

However, a huge percentage of the energy from the explosion goes into kinetic energy of the bullet (since the bullet moves more than the shooter, and work is force times distance).  

So the bullet is really fast, has a huge kinetic energy, totally destroys the person it hits, yet only makes them recoil about the same as the shooter.

The bullet carries KE that is M/m times (M being the shooter's mass, m being the bullet's mass) that of the shooter.  The dead person, assuming he has the same mass as the shooter, and assuming the bullet stays inside the person, will gain approximately  the same KE as the shooter.  So the bullet's KE is not completely converted to the dead person's KE.  The difference is the heat generated inside the dead person's body and other types of energy.  The process of the bullet striking the person is definitely an inelastic collision (in the case when the bullet stays inside the person, it is called completely inelastic collision).

[Yaobing]

If you get 10% faster for each 10% reduction in weight, is it a win because energy comes out higher?  Or is it a wash because momentum stays unchanged?

What do you mean by "win"?  In the case of collision of football players, if one player decides to reduce his weight so he can gain more kinetic energy during collision, that is definitely not a win for him.

[Yaobing]

I used to be a physicist, even went to Cambridge to do physics. I changed to focus more on Chemistry, and the biology, however, when I discovered that many of the Cambridge physicists emitted foul odours from their bodies, and I would never get a girlfriend if I continued to hang out with them.

Really? Wow, I guess I am glad I am no longer a physicist.

Does a chemist do better?  Does your girl friend ever tell you that you smell like certain chemicals? 

[Matt]

So to my original question, it seems there should be a simple equation to relate how much kinetic energy becomes non-kinetic energy (h from my first post) given the velocities, masses, and Cr of the objects.

In other words, an equation that says "in a collision, given these few input variables, this much kinetic energy must become another form of energy so that conservation of momentum is fulfilled."

I'm going to try to solve the amount of kinetic energy that's lost (h) for a completely inelastic collision.  I'll probably goof up, so tell me where I go wrong:

Conservation of momentum in an inelastic collision:
m₁v₁ + m₂v₂ = (m₁ + m₂)v₃

Solve for v₃ (the velocity of the combined objects):
v₃ = (m₁v₁ + m₂v₂) / (m₁ + m₂)

Kinetic energy in original system:
ke = (m₁v₁²)/2 + (m₂v₂²)/2

Kinetic energy in final system:
ke^' = ((m₁ + m₂)v₃²)/2

Loss of kinetic energy (h):
h = ke - ke^'
h = (m₁v₁²)/2 + (m₂v₂²)/2 - ((m₁ + m₂)v₃²)/2

Use conservation of momentum to fill in v₃:
h = (m₁v₁²)/2 + (m₂v₂²)/2 - ((m₁ + m₂)((m₁v₁ + m₂v₂) / (m₁ + m₂))²)/2

In other words, the amount of kinetic energy lost, given an inelastic collision with m₁, v₁, m₂, v₂ is defined as (simplified version of the above):
h = m₁m₂(v₁ - v₂)² / 2(m₁ + m₂)

Does that make sense?

(* I used an online calculator to simplify the algebra.  I'm in over my head doing it long hand.  Actually, I'm just in over my head.)

[Yaobing]

That is correct.  For completely inelastic collision, there is a definite answer.  So is there one for elastic collision.  Anything in between, you have to have a variable parameter that specifies the degree of loss of KE (i.e. coefficient of restitution).