How is momentum conserved if you lose kinetic energy?

[Minki]

The energy component of KE is the velocity. Momentum is mass x velocity so, in a collision containing in-elasticity, if KE is lost to heat then that heat energy must have been supplied by the velocity of the object. And since velocity has been lost to supply the heat then the overall momentum after the collision must be less than before the collision. The only way I can think that you could say that momentum is conserved is if you are including heat vibration in the overall momentum, including heat that has been dissipated to the environment. Thanks for any help on this.

 

[phinds]

Have you taken into account the fact that momentum is a vector not a scalar?

 

[Minki]

Yes. As far as I can see, the speed component of velocity is lost therefore the resultant vector must be smaller.

 

[Chestermiller]

Momentum goes as mv and kinetic energy goes as mv^2, so, mathematically, it is possible for momentum to be conserved while kinetic energy is lost. Take as an example the totally inelastic collision of a moving mass m with an initially stationary mass, also of mass m. Initially the momentum is mv and afterwards both masses are moving with half the original velocity of the initially moving mass, so the momentum is still mv. But, what do you calculate for the initial and final kinetic energy of the system?

 

[A.T.]

As far as I can see...

Do the math.

 

[russ_watters]

The energy component of KE is the velocity.

I can't make sense of this statement. Energy is energy. Velocity is not energy. When I look at the kinetic energy equation I see more than just velocity and when I look at the animation @A.T. provided I see that other thing isn't constant.

 

[Minki]

Thanks to both.
Chestermiller: Yes I've done the math and it does show that KE is lost and momentum is conserved. But from a practical viewpoint, I can't see how this can be the case - for the reason that I gave about the transfer of velocity being needed to dissipate energy.

russ_watters: Velocity is not energy? Well that's turned my whole understanding of physics on its head! Where is the energy then in half m v squared?

 

[A.T.]

Yes I've done the math and it does show that KE is lost and momentum is conserved. But from a practical viewpoint, I can't see how this can be the case

Your view is not really practical if it disagrees with basic math.

 

[Minki]

A.T: Not so much a view - more a need to understand what is happening in a demonstrable way, and to show how the math model works.

 

[russ_watters]

russ_watters: Velocity is not energy? Well that's turned my whole understanding of physics on its head! Where is the energy then in half m v squared?

The entire expression is energy. That's what the equals sign means!

Is a wheel a car or just part of a car?

To me it also seems like you are not paying attention to what the math says/means. Clearly if KE=V then [kinetic] energy is velocity. But you are clearly aware that that isn't the equation for kinetic energy. So I don't understand why you keep saying it.

 

[Buzz Bloom]

Hi @Minki:

I am surprised that you have not asked the question,

"What happened to the missing energy?"​

The total mechanical energy of the system before the collision is

E = (1/2) mv².​

After the collision, it is

E = (1/4) mv².​

Regards,
Buzz

​

 

[Minki]

Ok I think I get it. I see energy as mass under motion. If m or v is increased, then the energy is increased, but increasing the velocity is more significant than increasing the mass, hence the v squared. In a 100% inelastic condition, if a mass m with velocity v collides with an equal mass, then the mass doubles and the velocity halves. Halving the velocity is not fully compensated for by the doubling of the mass because the velocity is more significant than the mass when it comes to energy. But momentum is unaffected because that measurement gives mass and velocity equal significance.

Buzz: Does the missing energy becomes vibration in the masses?

 

[CWatters]

Thanks to both.
Chestermiller: Yes I've done the math and it does show that KE is lost and momentum is conserved. But from a practical viewpoint, I can't see how this can be the case - for the reason that I gave about the transfer of velocity being needed to dissipate energy.

russ_watters: Velocity is not energy? Well that's turned my whole understanding of physics on its head! Where is the energy then in half m v squared?

Its in the combination of mass and velocity.

You shouldn't assume mass is always a constant. With inelastic collisions two masses can become merged into one. Rockets loose mass as fuel is consumed or stages dropped. So both m and v can change.

 

[Buzz Bloom]

Does the missing energy becomes vibration in the masses?

Hi Minki:

Well, the missing energy has to go somewhere. One possibility is (internal mechanical) vibrations, as you suggest. Another is a different kind of vibration (at the molecular level) which is an increase in temperature. I can imagine that there are more alternative ways depending on the physical/chemical structure of the masses, but I do not have the knowledge to be more specific.

Regards,
Buzz

 

[Minki]

Yes, that's what I thought Buzz.

Well I'm glad I got that sorted out, and thanks to everyone for your help.

 

[jbriggs444]

if a mass m with velocity v collides with an equal mass, then the mass doubles and the velocity halves

Or if a mass m with velocity v/2 collides with an equal mass and velocity -v/2, the mass doubles and the velocity goes away.

By no coincidence, the reduction in kinetic energy is the same as when one mass had velocity v and the other had velocity 0.

 

[Dale]

Have you taken into account the fact that momentum is a vector not a scalar?

Yes. As far as I can see, the speed component of velocity is lost therefore the resultant vector must be smaller.

No, I don’t think that you have taken into account the vector nature of momentum.

Say you have two objects with identical masses and equal and opposite velocity. No matter what the magnitude velocity is, the momentum is zero. However, the KE can be annywhere between zero and infinity. The vector nature of momentum is the key. Once you truly do grasp that the conclusion is clear: two momenta can cancel each other, two KEs cannot

 

[russ_watters]

No, I don’t think that you have taken into account the vector nature of momentum.

Say you have two objects with identical masses and equal and opposite velocity. No matter what the magnitude velocity is, the momentum is zero. However, the KE can be annywhere between zero and infinity. The vector nature of momentum is the key. Once you truly do grasp that the conclusion is clear: two momenta can cancel each other, two KEs cannot

Hopefully this is not a derail, but I'm not sure I see how this distinction helps the OP. I recognize I myself tripped over this distinction in a recent thread (and it might have been you or @jbriggs who pointed it out), but in common everyday use, they don't seem to act much different in terms of vector vs scalar nature, particularly when looked at in 1D: 1+1=2 (or 1-1=0) regardless of if those numbers are vectors or scalars. What is the difference/value of one vs the other? Or maybe another way: is a purely 1D momentum problem necessarily a vector problem? The direction seems to me to be little more than a placeholder.

Or perhaps more concisely: what is the difference between a 1D vector and a scalar?

To me what seems like the relevant difference between energy and momentum for the OP is that energy is universally conserved whereas momentum is conserved for the objects. The objects have the same combined momentum before and after the collision, but the excess energy has been released into the universe. This explains to me why two different equations can produce two different answers yet both are "conserved" in their own way.

And as for why kinetic energy is a scalar and not a vector, I know we don't generally like sources from places like quora, but their answer makes a lot of sense: because we choose to define it that way - it could have been a vector if it proved useful. Being a scalar makes it easier to use conservation of energy because you don't have to worry about directions when converting between different forms of energy, particularly for forms of energy that don't have an inherrent directionality.

Am I off base on anything here?

 

[robphy]

The energy component of KE is the velocity.

I don't think that is a good way to think about it.
It's true that KE deals with the speed of a macroscopic object...
but it might be good to see some details of what is going on.

Here's a way to think about it,
which I learned from teaching mechanics from Tom Moore's Six Ideas That Shaped Physics Unit C.

Total Momentum conservation comes from Newton's Third Law.
So, that holds in any interaction of particles.

Now for kinetic energy...
If, in that interaction,
the transfer of momentum to an object (i.e. the impulse) changes the magnitude of the velocity vector of the object,
then nonzero work is done on the object. Essentially, look at the dot product of $d\vec p$ and $\vec v$.
The change in kinetic energy is then $\int\vec v \cdot d\vec p$

 

[brainpushups]

Not sure if this will offer additional insight to the OP, but thought I'd mention it:

For elastic collisions (consider 1D for simplicity) the relative velocity between two colliding objects is the same before and after the collision. If the collision is inelastic the relative velocity differs before and after the collision (the ratio of relative velocities is known as the coefficient of restitution).

If you know the how the relative velocities before and after the collision will differ then you can still solve the system of equations for an inelastic collision in which momentum is conserved and kinetic energy has changed.

 

[Dale]

To me what seems like the relevant difference between energy and momentum for the OP is that energy is universally conserved whereas momentum is conserved for the objects. The objects have the same combined momentum before and after the collision, but the excess energy has been released into the universe. This explains to me why two different equations can produce two different answers yet both are "conserved" in their own way.
...
Am I off base on anything here?

I don’t think that you are off base, but I don’t think that is the issue that bothered the OP. He seemed to recognize that KE was not conserved and that it could go out as you say into other forms of energy. My understanding of the issue is that he sees an inevitable coupling between KE and momentum since they are both functions of v.

Regarding the 1D approach. In 1D the distinction would be that energy is non negative while momentum is signed.

 

[Minki]

Ok, thanks for those extra insights. Something else I'd like to get cleared up: I've read that momentum is always conserved, for both elastic and inelastic collisions. But is it still said to be conserved if there is friction? Momentum will be transferred to the particles making up the material which is causing the friction, and I would have thought that the momentum would then be said to have been dissipated?

 

[jbriggs444]

Ok, thanks for those extra insights. Something else I'd like to get cleared up: I've read that momentum is always conserved, for both elastic and inelastic collisions. But is it still said to be conserved if there is friction? Momentum will be transferred to the particles making up the material which is causing the friction, and I would have thought that the momentum would then be said to have been dissipated?

If there is friction, there is something you are rubbing against. If you transfer momentum to that something, that something transfers equal and opposite momentum to you. Newton's third law in action.

If you skate across a rink and do a stop in front of the goalie, the momentum you lost is transferred to the rink. The rink gains exactly as much momentum as you lost.

 

[Minki]

So it's still conserved in the original mass because whatever you transfer it to will have transferred it straight back? Even if that transfer ends up as heat?

 

[jbriggs444]

So it's still conserved in the original mass because whatever you transfer it to will have transferred it straight back? Even if that transfer ends up as heat?

Momentum is transferred to something else. It does not come back. It goes to the something else. The total for the closed system consisting of you plus the something else remains unchanged.

Kinetic energy can end up as heat. Momentum is forever.

 

[Minki]

Ok, so it can only be said to be conserved if you consider the original mass and the environment which the momentum was transferred to.

 

[Minki]

Ok thanks.

 

[russ_watters]

Ok, so it can only be said to be conserved if you consider the original mass and the environment which the momentum was transferred to.

Specifically it is my understanding that every time you start or stop moving (or accelerate) on Earth, you change the Earth's rotation rate.

 

[Minki]

Yes, when you start moving the Earth's state is changed, but when you stop the Earth is returned to its original state.

 

[A.T.]

Ok, so it can only be said to be conserved if you consider the original mass and the environment which the momentum was transferred to.

The environment is not different than the second mass in a collision. The total momentum is conserved, not the individual momenta of each mass.

 

[CWatters]

Perhaps the OP should also read up about system boundaries. These are imaginary lines drawn around part or all of the things you are interested in analysing. Not really a physical line that has location or dimensions, its more a conceptual idea of what you are considering to be "part of your system". Generally everything inside your system boundary will feature in your conservation equations but there might be terms for energy or momentum that leave your system.

Momentum is only conserved by a "closed system" (no forces crossing the system boundary).
Energy is only conserved by a "closed system" (no energy crosses the system boundary).

Typically when someone askes "where does the energy go" or "doesn't this break conservation of momentum" it means they have forgotten about something that either should have been considered as part of the system (eg they drew their system boundary in the wrong place) or it crosses their system boundary (it's not "closed").

Common things people forget about are:

Energy that escapes the system as heat, perhaps due to friction, deformation of materials or an unaccounted for chemical reaction.
Momentum that escapes the system by transfer to/from planet Earth.

So when considering two objects that collide, if you want the make statements like "momentum is/isn't conserved" you first need to say what you are including in your system boundary... Are you talking about the momentum of one of the objects or both of the objects? Is it an elastic collision (no energy lost as heat) or inelastic (energy lost as heat).

PS: There are probably better more formal ways to describe system boundaries).

 

[CWatters]

Specifically it is my understanding that every time you start or stop moving (or accelerate) on Earth, you change the Earth's rotation rate.

Back of the envelope calculation...

10,000kg truck accelerates from rest to 50mph (22m/s) on the Earth which has a radius of 6,371,000 meters

Angular momentum L = Iω or rmv
where
r is the radius of the earth
m mass of truck
v velocity of truck

So the angular momentum of the truck increases to:

L_t = 6,371,000 * 10,000 * 22
= 1.4 * 10¹² kg.m².s⁻¹

Depending on which direction the truck goes this will increase of decrease the angular momentum of the planet by the same amount. Let's assume east or west..

Google suggests the moment of inertia of the planet is around 8 * 10³⁷ kg.m². It rotates at one revolution a day which is an angular velocity of 7.3 * 10^-5 R.s^-1. So the angular momentum of the Earth is about

L_e = 8 * 10³⁷ * 7.3 * 10^-5
= 5.8 * 10³³ kg.m².s⁻¹

So the truck accelerating causes the length of a day to change by about..

= 1.4 * 10¹² * 100 / 5.8 * 10³³
= 2.4*10^-20 %

If it kept going all day without stopping that would make his day longer or shorter by about 2*10^-17 seconds.

PS: I haven't checked my figures very carefully.

 

[robphy]

Possibly interesting reading:
from Edward Purcell's "Back of the Envelope" Problems in AJP
"If every car and truck in the US were driven 300 miles north and left there, by how much would the length of the day be changed?"
http://web.mit.edu/rhprice/www/Readers/Purcell/May1984-Problem1.pdf

...but, of course, this is a digression from the original question.

 

[russ_watters]

Possibly interesting reading:
from Edward Purcell's "Back of the Envelope" Problems in AJP
"If every car and truck in the US were driven 300 miles north and left there, by how much would the length of the day be changed?"
http://web.mit.edu/rhprice/www/Readers/Purcell/May1984-Problem1.pdf

...but, of course, this is a digression from the original question.

Not much of a digression. I considered using the figure skater example earlier, but we weren't dealing with rotation. The figure skater adds or subtracts kinetic energy by extending or retracting her arms (with w=fd against centrifugal force), changing the moment of inertia and rotation rate while keeping angular momentum constant.

Yours is another version of the same problem.

 

[Dr Dr news]

One of my favorite demonstrations in Mechanics is to take two carts of identical mass on a track with velcro affixed to their front bumpers and give them equal and opposite velocities. The system of two carts thus have zero momentum before colliding and stopping in the middle of the track. The momentum is conserved but all of the kinetic energy is converted to thermal energy. Another favorite is to take a mass of lead shot in a cardboard tube and repeatedly invert the tube doing work on the shot, converting the work to potential energy to kinetic energy ending in an inelastic collision, pour the shot into a calorimeter and measure the increased temperature. N(mgh) = mcΔT. The work is ultimately converted to an increase in the internal energy of the shot.