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Conservation of momentum vs conservation of kinetic energy

  1. Nov 1, 2005 #1
    if i hit a still 1 kg billiard ball with another 1kg ball at 10 m/s the second ball will stop and the first ball will acquire a speed of 10 m/s, both conservation of momentum and energy acomplish

    but if i hit a still 10 kg ball with a 1 kg ball at 10 m/s the 10 kg ball will acquire a speed of 1 m/s and the 1 kg ball will stop, conservation of momentum is accomplished but conservation of energy seems not to because the kinetic energy of the 1 kg ball is 0.5*100=50 while the kinetic energy transferred to the 10 kg ball will be 5*1=5

    so the kinetic energy has reduced 10 times, how is this possible if this is an ideal situation in which kinetic energy is not transformed into heat nor anything alike so where has this kinetic energy gone to?
  2. jcsd
  3. Nov 1, 2005 #2
    The answer, of course, is that this wouldn't happen because of conservation of energy. In a real situation, the 1 kg ball would rebound with a speed close to its original speed but in the opposite direction, while the 10 kg ball would move off with low speed.

    The exact values (in such 1D examples) are given by:

    v1 = (u1(m1 - m2) + 2m2u2)/(m1 + m2)

    where m is mass, u is original velocity and v is final velocity. 1 and 2 are the two balls. Which is 1 and which is 2 is arbitrary.
  4. Nov 1, 2005 #3


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    The final speed of the 10kg ball would be 1.818m/s, the final speed of the 1kg ball would be -8.18m/s.
    Last edited: Nov 1, 2005
  5. Nov 1, 2005 #4
    gracias hombre invisible and thanks daniel i could have figured it out myself testing it by hitting a big coin with a small one as i have just done but im not still satisfied so lets change the scenerio

    i tense a spring with a 1kg mass at 10 m/s transforming the kinetic energy into potential energy stored in the spring ending the 1kg mass at 0 speed,
    now i substitute the 1 kg mass with the 10 kg mass and let the spring go loose

    what will be the speed of the 10 kg mass?

    1 m/s as conservation of linear momentum states or 1/2*1*100=50 so 1/2*10*10=50 so the speed would be 3.3 m/s in the case of the 10 kg mass to keep truth the conservation of energy but 1 m/s to keep truth the conservation of momentum

    can anyone explain this apparent contradiction between conservation of momentum and conservation of kinetic energy?
  6. Nov 1, 2005 #5
    Okay, first a couple of things on conservation laws.

    Kinetic energy is not a conserved quantity. In fact, you demonstrated that yourself refering to "transforming the kinetic energy into potential energy stored in the spring" (in this case mechanical energy is conserved). Energy is a conserved quantity, but this energy does not have to stay in the same form.

    Secondly, when thinking about conservation of any quantity, you have to consider every single thing wherein that quantity is transfered to or from, in part or in whole, as being part of that system.

    The mechanics referred to in your question involve ideal systems, not real-life ones. In the above example, we tend to ignore the conservation of momentum for simplicity, and with good reason: we can get along without thinking about it.

    In truth, the initial momentum of the 1 kg mass (10 kg m/s) is being transferred to the spring, as well as anything the spring is attached to and anything causing friction. If you place your hand on the mass when the spring is at its maximum extension, the movement of the spring will decay quite quickly to nothing because of this. Momentum is conserved on the fundemental level: various atoms will be moving more vigorously due to friction, etc, but we cannot account for every momentum transference in such a complex system.

    When you replace the 1 kg mass with a 10 kg mass, YOU yourself are changing the momentum of the masses, and yourself come to that.

    There are times when conservation of momentum can be analysed: in collisions for instance. There are times when it cannot due to complexity. Your scenario falls into the latter.

    As for kinetic energy, this will be equal to that of the 1 kg mass for any given extention x of the spring. Since the 10 kg mass is 10 times heavier, it will have 1/10th the speed: a maximum of 1 m/s.
  7. Nov 1, 2005 #6
    if the 10 kg reach a speed of 1 m/s the system would be ideal because the momentum is completly conserved 10kg*1m/s=1kg*10m/s

    but being the system ideal where have the energy gone to because energy has gone from 1/2*1*10*10=50 to 1/2*10*1*1=5

    and what if i start with the 10 kg at 1 m/s to tense the spring and then put the 1 kg when releasing the spring energy so it acquires a speed of 10 m/s

    the energy would have gone from 5 to 50

    as for the recoil the earth would have after tranforming the momentum into stored energy it would be the same in both cases so it doesnt really matter, but anyhow when the spring stops extending the earth will also stop recoling

    i understand you are saying that the final speed of the 10 kg will be 1 m/s in an ideal situation and close to it in a real one

    by the way i hope you dont hate me like my teachers used to because of asking too many questions
  8. Nov 1, 2005 #7

    Doc Al

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    The amount of energy stored in the spring will equal the KE of the initial mass: [itex]1/2 m v^2 = 1/2 (1) (10)^2 = 50 \ {J}[/itex].

    If you stop the spring at full compression, then replace the 1 kg mass with a 10 kg mass, that 50 J of energy will now go into the KE of the 10 kg mass. Its speed (assuming an ideal spring) will end up being about 3.2 m/s.

    What makes you think momentum would be conserved in this situation?
  9. Nov 1, 2005 #8
    i start with 1 kg at 10 m/s and then have 10kg at 3.2 m/s,so the inertial momentum has increased 3 times without applying energy to the system

    now i repeat the spring process and transfer the kinetic energy now from the 10 kg to 100kg and triple the inertia again,then i transfer it again from 100 kg to 1000 kg and triple the inertia again

    lets say i triple the inertia until i reach a inertia of a million, no energy has been added so far and i started with an inertia of 10, now i have a giantic mass at a very low speed

    so you must agree that from an initial inertia of 10 i get an inertia of a million without having used any energy by repeating the spring process

    now seems to me that an object with a momentum of inertia of a million like a million tons ship going at 1 m/s will be harder to brake than a motorbike of 100 kg at a speed of 100 m/s that has an inertia of 10000,their kinetic energy will be the same but the inertia will be much bigger in the case of the ship

    seems obvious to me that it will be much more difficult to brake the ship than the motorbike because the inertia in the ship is much bigger

    but then the harder the braking is the more energy is released into heat

    in the case of the ship that has 100 times more inertia than the motorbike though the same kinetic energy it will be released 100 times more heat

    this seems absolutly contradictory i refuse to believe that an object with an inertia of 10 will release the same energy when braking than an object with an inertia of 1 million however their kinetic energy is the same

    i study nautics and ive had a motorbike and i know how both a ship or bike brake depending on the inertia
  10. Nov 1, 2005 #9

    Doc Al

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    I assume by "inertial momentum" you mean momentum. OK. So?

    Why take so many steps? Just swap the 1 kg mass with the 1 million kg mass in one step. :smile:

    The gigantic mass at low speed will have the same KE as the small mass at high speed. (But its momentum will be greater.) So?

    Nonetheless, if you apply the same stopping force to each, they will come to rest in the same distance and require the same work to stop. (But the bigger object will take more time to stop.)

    The same amount of "heat" will be released.

    Believe whatever you like. But if you want to really understand things, study some physics.
  11. Nov 1, 2005 #10
    You are quite right. That was a miscalculation on my part.

    EDIT: since 0.5 * 1 kg * (10 m/s)^2 = 0.5 * 10 kg * v^2
    v^2 = (10 m/s)^2 / 10 m/s
    v = ROOT(10)

    I went mad (EDIT: twice - it's the wine!). It happens.

    Exactly the same in reverse. The key is that the maximum kinetic energies of the two masses will be the same. The reason is that the potential energy of the masses is mass-independant.

    Yes, by which time all of the momentum of the mass has been dissipated, that is, transferred to its surroundings such as the Earth.

    Why would you say that? That's what the forum is for. You are most welcome and your questions are interesting and no doubt helpful to others too. If your teachers didn't like you asking questions, perhaps they weren't confident about giving the answers. You don't have to be afraid of asking too many questions here. There are plenty of people to answer them and all (well, most) people are all too happy to do so.
    Last edited: Nov 1, 2005
  12. Nov 1, 2005 #11
    thanks a lot for your patient doc

    maybe is my teachers fault, he taught to me that big ships are very difficult to brake even at very low speeds because of their very big inertia but well according to physics seems that is kinetic energy what counts

    then at least you must admit that if you doubt what you are being taught sometimes you are right
  13. Nov 1, 2005 #12
    If I understand you correctly, you are saying that the momentum of the mass as it is increased through subsequent replacements of higher masses should be increasingly difficult for the compressed spring to bring to a stop. What you are perhaps not considering is that by the same token the greater mass will be much more difficult for the extended spring to accelerate. Since the same amount of work is done in accelerating the mass as is done in stopping it, there is no problem.

    To put it in context, if the first mass of 1 kg has a maximum speed of 10 m/s, a mass of 1 million kg will have a maximum speed of about 0.0003 m/s. So the force of the spring acting against it will have 33,333 times longer to bring it to a stop.

    Irrespective of which, Hooke's law only really holds under simple set-ups of ideal springs. In a realistic scenario, the spring wouldn't budge the million kg weight a single millimetre.
  14. Nov 1, 2005 #13
    a 0.01kg bullet at a speed of 333 m/s has the same kinetic energy than a 10000 kg train at 0.33 m/s

    heres my challenge i put myself against a wall with the hand cover with kevlar
    and ill brake the bullet feeling the same force than recoil felt the guy who shot me

    now if you trust mainstream physics so much you can put yourself against a wall and try to brake a 10 ton train at 0.33 m/s with your hand

    would you trust what you have studied so much as to do this experiment?

    do you really think that you wouldnt be smashed into the wall?

    physics may say whatever but my intuition wouldnt let me take the train chalenge and i wouldnt let you take it either
  15. Nov 1, 2005 #14
    by the way hombre invisible forgive for not answering but ive been thinking 14 hours in a row and i cant take no more
  16. Nov 1, 2005 #15
    I'm glad you wouldn't take it. The bullet would impart a mere 3.33 kg m/s of momentum upon you, while the train would impart 3300 kg m/s of momentum upon you. That would definitely squash your hand somewhat, but let's be sure of what you're talking about here: momentum, not kinetic energy. Not the same thing.
  17. Nov 2, 2005 #16

    Doc Al

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    Bullet versus Train challenge!

    Both bullet and train have the same kinetic energy (about 550 J) and will both require the exact same work (force times distance) to be brought to a stop. But, as El Hombre points out, this does not mean that the two situations are exactly the same. Momentum makes a difference!

    If you could somehow apply the same force to both bullet and train, then both would be stopped in the same distance. But that's a big if! The high speed (but low momentum) bullet allows a situation where you can generate a high impact force (the kevlar helps you absorb it without permanent tissue damage) and absorb the energy (and momentum) without being knocked off your feet.

    The low speed (but high momentum) of the train does not allow you to comfortably generate such a force without being run over. However, it's not that hard to stop the train, if you've got the room (and the friction) to do it with. If you can generate a measly 55 N of force (about 12 lbs), you can stop that train in 10 meters. (Just don't lose your footing and get run over.)

    I'm sure you've seen those strongman competitions where men move huge train cars by pulling a harness. It's the same thing, only in reverse. (And they've got to overcome friction!)

    (And your instructor is of course correct: It's certainly difficult to brake huge ships even at low speeds, for the same reasons we're discussing.)
  18. Mar 18, 2009 #17
    I happened to be confused in the same way.

    The thing is as said in the first reply the new velocities of the colliding objects
    are such that the conservation of energy is satisfied.

    So i tried to solve the following system:

    m1u1^2+m2u2^2 = m1v1^2+m2v2^2 (conserv energy)
    m1u1+m2u2 = m1v1+m2v2 (conserv momentum)

    u1,2 - velocities before collision
    v1,2 - celocities after the collision

    The resulting quadratic equation yielded two possible solutions:
    a) v1 = u1 - yes, that certainly satisfies the above stated laws - no collision happened :-)
    b) and v2 = (u2(m2-m1) +2m2m1)/(m1+m2) (as in the first reply)

    so the conservation of energy is an additional constraint for the conservation of momentum

  19. Oct 30, 2010 #18
    We know we are in trouble when someone starts talking about "trust[ing] mainstream physics." Classical Newtonian physics has been solved. With very simple equations. For centuries now. There's nothing controversial about it.

    Seriously, it really is a serious issue when someone who just doesn't understand something starts trying to make it sound controversial by using such descriptors as "mainstream."

    It's particularly disturbing when politicians talk that way, but at least they usually are talking about issues that aren't nearly as cut-and-dried as simple kinetic energy / momentum problems. At least for now.
  20. Mar 23, 2012 #19
    I'm wondering too why must we apply conservation of Momentum rather only conservation of energy.
    In a 5g bullet,250m/s, 2.495kg block and k=40N/M using COM and then COE,
    KE at start =PE in spring at end
    [itex]\frac{1}{2}(2.5)(0.5)^2=\frac{1}{2}(40)x^2[/itex] ....from worked example

    But if we apply only COE then
    Energy added to the system =PE in spring at end
    [itex]\frac{1}{2}(0.005)(250)^2=\frac{1}{2}(40)x^2[/itex] totally disagreed with above.

    Does by applying COM we can find energy loss in collision?
    What about if different property of block used, eg. a putty?
    What if no energy loss in collision, do we have to resort to COM too?

    Maybe i have an analogy.
    A bank started business with no money.
    First customer deposited x dollars but his book only stated he had y dollars.
    Why his money less, due to COM.
    Last edited: Mar 23, 2012
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