# Walter Lewin demonstrates free momentum in an MIT lecture

Discussion in 'The Guru's Pub' started by UnrealGaming, Jun 13, 2021.

1. ### UnrealGamingAncient Guru

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The twice as heavy car should be going at half the speed for conservation of momentum to work. It goes faster, he got free momentum. If you keep adding twice as heavy cars at some point you'll end up with absurdly higher momentum than what u started with.

the setup is light car with a spring colliding into a heavy car with a spring

Can any physicist explain the mechanics of this? How is this even a thing?

timestamp is 21:03. ( linking to a timestamp doesn't work here )

Last edited: Jun 13, 2021
2. ### NoisivAncient Guru

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Only if the post-collision speed of the light car is zero. Which only happens if m1=m2.

No you won't. You'll end up with the light car going the same speed, opposite direction. And the heavy car not even moving. <- limit case of ball hitting the wall (infinite mass, representing your ultra heavy car composition) and simply bouncing back with the same speed You're overthinking it. Just follow the equation. You can rationalize later.

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3. ### UnrealGamingAncient Guru

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Yeah i know the math works out if you subtract the stuff that bounces back. But in reality if you add another car that's twice as heavy, and another one and another one... the last one will have a lot more momentum than what you started with. If you had a wall that the small car couldn't knock over, but you did that setup, the final car with twice or 3x more momentum will be able to knock it over. You kinda get this amplification effect.
And the stuff bouncing back is also free, you haven't lost any momentum in the lets call it "forward" direction. In fact you're gaining momentum, as you can see in the experiment.

Or am I overlooking something here?

Last edited: Jun 13, 2021
4. ### UnrealGamingAncient Guru

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I'm not referring to the original light car hitting heavier and heavier cars btw ( if that's what you meant here ). I'm talking about having a row of: a light car ---> twice as heavy car ---> twice as heavy car ... etc. ( the mass difference between two interacting cars is always just 2x )

A light car with a spring hits twice as heavy car with a spring, the twice as heavy car leaves with more momentum than the momentum it got hit by, as you can see in the video ( which is the weird part ). Then you just have another car in front of it, that's twice as heavy, so IT will leave with more momentum. And so on. And yeah half the mass cars always bounce back, which is also free. The car they hit leaves with extra momentum AND they bounce back with as much momentum as that extra was.

Last edited: Jun 13, 2021

5. ### NoisivAncient Guru

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yes, you would get more momentum but less speed.
with each consecutive heavy car goes getting 4/3 of previous car momentum but only 2/3 of its speed.

1 = -1/3 + 2x 2/3

heavy car number 1000 would barely move - having speed (2/3)^1000 of initial speed. its mass 2^1000 times bigger than initial mass.

If you let this to infinity you get a somewhat absurd situation which would look like an infinitely slow car with infinite momentum and infinite mass hitting twice as heavy car.

HOWEVER
you are the one who created this somewhat absurd and unintuitive situation of a car having infinite momentum when you decided to let this scenario slide into infinity.

introduce infinity (mass) - get rewarded with infinite momentum

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6. ### UnrealGamingAncient Guru

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So free momentum is possible.... i guess? That's interesting...

7. ### UnrealGamingAncient Guru

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@Noisiv btw, if you don't mind me asking, are you a physicist? If it's too personal or whatever, you don't have to answer ofc. Or do you know anyone else that could take a look at this? isn't this a "free energy" scenario kind of thing?

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8. ### AirbudMaster Guru

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Though I'm nowhere near as smart as you guys I've always wondered how a baseball player could hit a ball coming at him at 100 mph/km/h farther versus one that was sitting still?

I know for every action there's an equal and opposite but seems like it would be a negative reaction.

9. ### AsiJuAncient Guru

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No, that's a misconception.

Conservation of momentum states that the net momentum is the same before and after collision:

m1v1 + m2v2 = m1v1' + m2v2'

where ' denotes the velocities after the collision.
The SUM of momentum is equal before and after collision, no energy is lost in ideal conditions (and never created).
In practice there's always energy loss due to compression of objects, friction etc.

v2' ie. the speed of the heavier car after the collision is determined by the equation:

v2' = (m1v1 + m2v2 - m1v1') / m2

which when v2 = 0 - the heavier object is stationary initially - simplifies to:

v2' = (m1v1 - m1v1') / m2
= m1(v1 - v1') / m2
= (m1 / m2)(v1 - v1')

and if we denote m1 as 1/2m2 it becomes:

v2' = (1/2)(v1 - v1')

so actually the heavier object's velocity is half of the difference between lighter object's velocities before and after collision.
In the particular case that v2 = 0 and m2 = 2m1 -> m1 = 1/2 m2

(Note that in the video exercise he is solving for both v1' and v2' using conservation of momentum and conservation of kinetic energy, you need a pair of equations to solve for 2 unknowns.
The expression I wrote wouldn't help there as the final form still contains both unknowns v1' and v2'.)

PS: not a physicist but as a mechanical engineer I've studied basic Newtonean mechanics (and mathematics) and their applications in engineering a fair bit Calling this free momentum is misunderstanding the concept of conservation of momentum.
Having 2 objects with a mass ratio of 2 : 1 collide elastically does not imply the heavier object should travel at exactly half the initial velocity of the lighter object after collision.

Last edited: Jun 14, 2021
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10. ### AsiJuAncient Guru

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lecturer has solved that v1' = -1/3v1 from pair of equations, for mass m2 = 2m1

Substituting that to above expression gives:

v2' = (1/2)(v1 - (-1/3v1)) = (1/2)(v1 + 1/3v1) = (1/2)(4/3v1) = 2/3v1

matching the solution for v2' in the video and confirmed by air track results.

In other words:

- the heavier object will travel at two thirds of the initial speed of the lighter object
- the lighter object will bounce back at one third of its initial speed

after the collision.

Net momentum is conserved, let's assume initial velocity v1 = 1 for simplicity:

- initially m1 had momentum 1
total 1

- after the collision m1 has momentum -0.33 (=opposite direction, vector quantity)
- m2 has momentum 1.33 (= 2/3 velocity * 2 mass = 4/3)
total 1

Last edited: Jun 14, 2021
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11. ### UnrealGamingAncient Guru

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Thank you for taking the time to write all of that down, but as I've said, i get that the math works out. That's not in question.

The heavier car leaves with more momentum than the momentum it got hit by. That's actual, interactable, usable extra momentum ( seemingly ) out of no where.

And if you do an experiment I've described above, you can get multiple times more momentum on the other end, than what you started with. Without intervening at all.

Last edited: Jun 14, 2021
12. ### NoisivAncient Guru

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Indeed, a physicist.

No, you didn't brake anything.
Because cars are getting so slow, but they are getting even more heavier than they are getting slow - you get infinite momentum.
In math terms: sequence of car masses is converging to infinity faster than it's speed sequence is converging to zero. So momentum which is the multiplication of two is converging to infinity.

speed of n-th car: v = (2/3) ^n
mass of n-th car: m = 2^n

Momentum of n-th car: p = m* v = (2/3) ^n * 2^n = (4/3)^n = infinity when n goes to infinity

-------------------
Otoh energies are converging to zero. Because of speed coming as square term in kinetic energy, and therefore square of speed converging to zero faster than mass is converging to infinity.

Energy of n-th car: E ~ m*v^2 = (2/3) ^n^2 * 2^n = (4/3)^n * (2/3)^n = (8/9)^n = zero when n goes to infinity

So no free energy To get a useful work, energy is what matters, not momentum

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13. ### UnrealGamingAncient Guru

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Okay, now we are getting there. Let's ignore kinetic energy for a moment. Are you sure something with, lets say, 3x more momentum can't do more work?

And thank you for the replies btw.

14. ### NoisivAncient Guru

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The max you can get is (almost) two times momentum. Which would be a case of car hitting infinitely heavy car, and bouncing back with (almost) equal speed as if hitting a wall.

m1v1 = m1v2 + MV

1 = -1 + 2

You are free to ignore the energy, but it will still be conserved.

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15. ### AsiJuAncient Guru

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Work-energy principle (very shortly):

http://hyperphysics.phy-astr.gsu.edu/hbase/work.html

Energy is the ability to do work like said and in the infinite scenario kinetic energy tends to zero = work (able to be) performed by the object tends to zero.
So something with more momentum but less energy can do less work.

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16. ### UnrealGamingAncient Guru

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Let's say a car is going a 1000 m/s. Hits twice as heavy car. That car is now going 666, but has a 33% extra momentum. That car now hits twice as heavy car. Now the 3rd car is going 440, but it got another 33% momentum. Now we run it into another car. The 4th car is now going 290, but it got another 33% momentum. Now we run that one into another car. 5th car is now is now going 191, but it has another 33% extra momentum. We are already well over 2x the initial car's momentum. And we're nowhere near infinities of any kind.

Also, this is important, are you sure something with 2x or 3x ( or whatever ) more momentum can't do more work? ( Ignoring the theory of kinetic energy and it's relation to work for a moment. )

Are you sure a 191 m/s 16 KG car can't do more work than a 1000 m/s 1KG car can? If we engineer the experiment to not have bias for small fast things vs heavy slower things.

Last edited: Jun 14, 2021
17. ### NoisivAncient Guru

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No, you can't get more work, because no matter how many collisions you do, the total energy of the system will remain the same. And maximum amount of work that can be extracted from the system is equal to total energy of the system.

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18. ### UnrealGamingAncient Guru

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Yeah, I'm saying lets ignore the kinetic energy theory for a moment. Ultimately, Is there a well done, clear cut, experiment you can link me to showing 191 m/s 16 KG object ( aka something with 2-3 more momentum ) not being able to push a bigger ship in space vs a 1000 m/s 1KG object, get more rotations out of a turbine, compress a bigger spring etc. I'm genuinely interested. If you ( or anyone ) have some well done experiments showing more momentum not being able to do more work i'd like to take a look. And ofc thank you all for the replies so far.

19. ### NoisivAncient Guru

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Watch your video again, 4:40 timestamp.
Both conservation of momentum and conservation of energy has been used to determine v1 and v2.

1. p1= p2+p3
2. E1 = E2 +E3

You have two unknowns, v1 and v2, so at the very least you need two equations to solve the collision.
You are free to ignore the conservation of energy, but in that case you're not going anywhere - it's unsolvable.

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