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Kinetic energy object vs opposing KE object?

  1. Apr 3, 2016 #1
    1. The problem statement, all variables and given/known data
    There is a Ball that weights .150 kg and is moving at 90m/s =607.5 Joules and it is moving to the right, so its coming from the left. on the opposing side is a 35kg cart moving at 4.5m/s = 354.37 Joules
    The Cart has superior momentum and mass....
    however, in this scenario some energy is not taken into other forms but is left for work.
    Ball=405 Joules
    Cart=201 Joules
    The ball hits the cart, the ball should stop the cart in a VERY short distance, it has 204 remaining Joules which it will use to push back the cart doing work.
    This must be true with mechanical energy, correct?
    My question is, is this scenario true?

    2. Relevant equations
    TMEi+Wnc=TMEf

    3. The attempt at a solution
    Im sure this is correct, the math works out and it still shows that ability of mechanical energy in the form of KE to do work.
     
  2. jcsd
  3. Apr 3, 2016 #2
    i think you should take into consideration the m.v =momentum of the bodies into consideration as energy states may change and can be related to work done by the body or on the body by environment.
    but the collision has a feature of contact forces or field effects.
    the above perspective can lead you to understand the physical event which is happening.
     
  4. Apr 3, 2016 #3
    it appears your scenario is not portraying the physical event which is taking place- ask what is the flaw?
     
  5. Apr 3, 2016 #4
    What is the "flaw"?

    but this is done with objects and material that are able to do this. It shouldn't be that hard really. I mean, if we have energy that is taken into heat, sound, deformation, etc.....if we have Kinetic energy left, it can exert a non-conservative force, doing work, from its Kinetic energy. A good real life example is with wind and car, kinetic energy of wind can stop a car.

    In this case, think....bowling ball hits pin, doing work, using force....

    Most collisions have ELASTIC and INELASTIC results, but like with a car crash, you see work as both cars crumple, the force from the KE transfer to each vehicle...both vehicles deform and get smashed, they are doing work on each other...energy does into THAT, plus sound and heat.

    so, imagine if the objects didn't deform as much and were able to do non-conservative work....which by work-energy and momentum...the only time MOMENTUM is NOT conserved is when a net external non-conservative force is working on it....which is my scenario...Kinetic energy DOING work on the WHOLE body, not deformation , compression, etc.

    Since NET external forces are occurring, momentum is irrelevant as it is not conserved due to a external force from the objects mechanical energy....

    The cart in its stopping distance is compressing, not the whole body moving but in a way compressing, then stopped. The force changes KE and momentum, it does work....Wnc=∆KE, like I've shown...I made it realistic by adding a significant loss of energy to other forms....

    This should work like this.....

    (TMEi ball - TMEi ball energy lost) + (- TMEi cart - (-TME cart energy lost)) = TMEf System


    (TMEi ball - TMEi ball energy lost) + (- TMEi cart - (-TME cart energy lost)) = TMEf System
    (607.5J ball - 202.5J ball energy lost) + (- 354.37J cart - (-153.37 J cart energy lost)) = Ball 204 Joules Cart 0 Joules ---> Ball 204 Joules KE-->work-->cart displaced -->KE--> Work.
    | IMPACT POINT
    V
    O ball -->{IMPACT}<--[_____]CART ====> ________|______OBall[____]CART ---> ===> Ball stops, transfers energy to cart.

    I don't mean to be rude, but how is there a flaw? I should work out perfectly
     
  6. Apr 3, 2016 #5

    ehild

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    The kinetic energy does not do work. It is the force, applied to a body, that does work on that body, changing its kinetic energy.
    During the collision of two bodies, they interact with each other. The force of interaction changes the momentum of the bodies, and also their individual kinetic energies. The net momentum is conserved if the time of collision is so short that the effect of external forces can be ignored. The net energy might be conserved if the force of interaction is conservative. But it usually is not the same as the sum of the initial kinetic energies.
    If you want to resolve a collision event you need to use conservation of momentum and the coefficient of restitution.
     
  7. Apr 3, 2016 #6
    you are saying 'energy does the work' whereas our normal picture of world is that 'forces do work by changing the state of a physical system' and this approach leads to deeper analysis of 'interactive' forces in nature and the role of inertia of rest or of motion.
    in physical analysis we usually do model dependent calculations and the picture is the rate of change of momentum which leads to forces either 'external' or 'internal' to the system. it leads to open up the processes and leads to 'knowledge' ;sometimes new knowledge.
    'energy' is like your bank balance but 'interactions' are the processes of its state of motion/investments in doing production- an analysis based on 'envelope' of energy may give you info but one has to open the envelope.
     
  8. Apr 3, 2016 #7

    haruspex

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    You seem to think that KE one way cancels KE the other way in equal measure. That's how momentum works, but not energy.
    Momentum doesn't care about conservative/non-conservative. Momentum is conserved if there is no net external force on the system. In your ball and cart system, there is no external force mentioned.
     
  9. Apr 3, 2016 #8

    If it has Kinetic energy, it has mechanical energy, it has the ability to do work, thus exert a FORCE, as it slows down....it exerts a force....Like I said, Momentum is IRRELEVANT at this point because of the net external force....Fs=1/2m*v^2.....
     
  10. Apr 3, 2016 #9

    How so, AS you know, Kinetic energy is the ability to do work, it supplies force as it slows down....I've said it there is a NON-CONSERVATIVE FORCE because the balls KE is doing work., non-conservative external work.
     
  11. Apr 3, 2016 #10

    What you said violates the Work-energy theorem, that is outside the picture of the world....The SCENARIO in question is where Kinetic energy is left to do work on the whole body, where as it is able to do work, its lost some to other forms but it is doing work with the remaining KINETIC ENERGY which is turned in to work....non-conservative external work.
     
  12. Apr 3, 2016 #11

    ehild

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    Yes, it is usually said in school, that energy is the ability to do work. But it does not mean that energy does the work.
    The body that has energy can do work on an other body, trough the force it exerts on the other body.
    What do you mean with the formula Fs=1/2 m*v^2? 1/2 m*v^2=KE is kinetic energy, but what is Fs? And what external force is present during a collision????
    Work is defined as force F multiplied by the displacement s, W=F*s. The Work-Energy theorem states that the net work done on a point-like mass is equal to the change of the kinetic energy of that mass . W=ΔKE. If a body has velocity and collides with an other one, it can slow down and loose KE, as the other body exerts force on it, a force, opposite to its velocity and does work on it.
    In case of collision between two bodies, they make a system. During the collision, the two bodies interact, they exert force on each other, and these forces are opposite and of equal magnitude according to Newton's third law.
    The momentum p of a body, of mass m and moving with velocity v, is p=mv. The time derivative of the momentum is equal to the net force exerted on the body. dp/dt=F. In a two-body system, the momenta add up. In case there is no other force but their interaction, F12 is the force body "2" exerts on body "1" and F21=-F12, is the force the body "1" exerts on body "2", dp1/dt=F12 and dp2/dt=F21=-F12. Adding up the time derivatives of momenta, you get that the time derivative of the total momentum is d(p1+p2)/dt = F12+F21, but the two force cancel each other so p1+p2= constant. The total momentum is conserved in a collision.
    The total mechanical energy of the system usually is not the same after the collision as it was before it. Some is lost, transformed to other energies. It is conserved only when the force of interaction between the colliding bodies is conservative.
     
    Last edited: Apr 3, 2016
  13. Apr 3, 2016 #12

    The body of KINETIC energy has the ability to do work, it the body has kinetic energy it has the ability to do work, THUS IN THIS SCENARIO.....we have energy left to do work, it ti has KINETIC ENERGY, then it has MECHANICAL ENERGY then it MUST do work.

    The Work-energy theorem, Force * Distance = ∆ KE Wnc=TMEf - TMEi

    SINCE this is happening, there is a external force, momentum is IRRELEVANT
     
  14. Apr 3, 2016 #13

    haruspex

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    But that is not an argument for the two bodies doing equal work in the collision.

    You don't seem to understand what is meant by an external force. There is no force acting on the ball+cart system from some other body.
     
  15. Apr 3, 2016 #14
    They are NOT doing equal work, they are doing unequal work.

    Second, if BOTH bodies have KINETIC ENERGY, and in this scenario....some energy is left to do work on the WHOLE body....thus when its slowing down, its EXERTING a non-conservative EXTERNAL FORCE=Force applied.....

    No energy=NO WORK Kinetic energy is a form of mechanical energy, mechanical energy is the ability to do work. Since is does work, it applies a FORCE over a distance, thus it does by energy from motion. =KINETIC ENERGY.

    If a non-conservative force is applied, then momentum is not conserved.....

    Think of it THIS WAY. If I had a hammer that weighed .05 kg's and It was moving 90m/s, then it has 202.5 Joules of KINETIC ENERGY, I swung the hammer to get that Kinetic energy through a distance of 5 meters, thus I applied 40.5 newtons of force to get it to 90m/s and 202.5 Joules....THUS Wnc=∆KE. Now I use my hammer and I hit a nail, the hammer moving at a constant velocity of 90m/s is NOT accelerating so it has finite Kinetic energy....it has 202.5 Joules of work it can do....

    I hit a nail on a wall, I drive the nail 0.025 meters in so Wnc=∆KE ---> F * 0.025 = 202.5 J's KE 202.5/0.025=8100 NEWTONS OF FORCE......

    It was KINETIC ENERGY TRANSFORMED INTO WORK=non-conservative force applied * distance = change in Kinetic energy....

    that is my scenario.
     
  16. Apr 3, 2016 #15

    Okay, let me explain this to you.....

    Kinetic energy is the ability to do work, it is a form of mechanical energy, Wnc=∆KE FS=1/2m*v^2
    NO ENERGY=NO WORK objects with constant velocity have the ability to apply a force due to their KINETIC ENERGY, example bowling ball and hammer
    Imma make this quick and show you......I have a hammer, it weights 0.05kg's, I push on it with 40.5 Newtons of FORCE over a distance of 5 meters....it has 90m/s velocity and 202.5 Joules of KINETIC ENERGY from 202.5 Joules of WORK.....My hammer with finite KINETIC ENERGY as it is no acceleration hits a nail on the wall, doing 202.5 JOULES of work, it drives the nail in by 0.025 meters......it has the ability to do this because it has energy by a virtue of motion, energy needed to do WORK, you need ENERGY TO DO WORK......F*d=∆KE or F*s=KE so F* 0.025 meters = 202.5 Joules KE......202.5/0.025 meters = 8100 NEWTONS OF FORCE......

    with non-conservative forces....MOMENTUM IS NOT CONSERVED......

    In this problem, We are saying that in this case...not all energy goes into deformation, sound, heat....some is left over....to do work, so a lesser mass, lesser momentum object with superior Kinetic energy...does work to stop a cart and then with its remaining KE, turns that into work pushing the cart back.....
     
  17. Apr 3, 2016 #16

    haruspex

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    I have no objection to any of that, except at the end, where you appear to argue that the object with the greater initial KE 'wins' in some sense.
    A bullet mass .005kg at 400m/s to the right hits a block mass 1kg moving at 3m/s to the left.
    No external forces here, so the net momentum is 1kgm/s to the left. On embedding in the block, the bullet plus block will move at 1m/s (almost) to the left.
    Yet the bullet started with 400J of energy and the block with a mere 4.5J. If you want to think in terms work done by each, the bullet expended almost all its 400J (only .0025J left) while the block expended 4J.
     
  18. Apr 3, 2016 #17
    Okay, but imagine of that ENERGY hasn't gone into deformation or sound....as the bullet is passing into material it does work on material, internally...not on whole body....Imagine....if that bullet lost only 200 joules....and did the remaining 200 Joules of work on the block, the block would have 400 J's of KE...then as it swings up...400 Joules of Potential energy....you get what Im saying?
     
  19. Apr 3, 2016 #18

    haruspex

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    No.
     
  20. Apr 3, 2016 #19
    Sorry, I made a MISTAKE....I meant 200 Joules, sorry
     
  21. Apr 3, 2016 #20

    haruspex

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    You seem to have changed the scenario... where does this upswing come from?
     
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