Kinetic energy object vs opposing KE object?

AI Thread Summary
The discussion centers around a scenario involving a ball and a cart colliding, analyzing the kinetic energy (KE) and momentum of both objects. The ball, with a mass of 0.150 kg moving at 90 m/s, has a significant amount of kinetic energy compared to the 35 kg cart moving at 4.5 m/s. Participants debate whether the remaining kinetic energy after the collision can do work on the cart, with some arguing that energy itself does not do work; rather, it is the force exerted during the collision that changes momentum and kinetic energy. The conversation highlights the importance of understanding the conservation of momentum and the role of external forces in collision events. Ultimately, the consensus emphasizes that while kinetic energy can be transformed, the interaction forces during a collision are what facilitate work being done.
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Homework Statement


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?

Homework Equations


TMEi+Wnc=TMEf

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.
 
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normal_force said:
however, in this scenario some energy is not taken into other forms but is left for work.
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.
 
normal_force said:
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?
it appears your scenario is not portraying the physical event which is taking place- ask what is the flaw?
 
drvrm said:
it appears your scenario is not portraying the physical event which is taking place- ask what is the flaw?
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
 
normal_force said:
What is the "flaw"?
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.
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.
 
normal_force said:
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.
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.
 
normal_force said:
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.
You seem to think that KE one way cancels KE the other way in equal measure. That's how momentum works, but not energy.
normal_force said:
the only time MOMENTUM is NOT conserved is when a net external non-conservative force is working on it...which is my scenario..
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.
 
ehild said:
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.
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...
 
haruspex said:
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.
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.
 
  • #10
drvrm said:
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.
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 into work...non-conservative external work.
 
  • #11
normal_force said:
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...

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.
 
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  • #12
ehild said:
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.
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
 
  • #13
normal_force said:
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.
But that is not an argument for the two bodies doing equal work in the collision.

normal_force said:
SINCE this is happening, there is a external force,
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.
 
  • #14
haruspex said:
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.

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.
 
  • #15
haruspex said:
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.
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...
 
  • #16
normal_force said:
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...
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.
 
  • #17
haruspex said:
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.

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 I am saying?
 
  • #18
normal_force said:
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 I am saying?
No.
 
  • #19
haruspex said:
No.
Sorry, I made a MISTAKE...I meant 200 Joules, sorry
 
  • #20
normal_force said:
Sorry, I made a MISTAKE...I meant 200 Joules, sorry
You seem to have changed the scenario... where does this upswing come from?
 
  • #21
haruspex said:
You seem to have changed the scenario... where does this upswing come from?[/QUOTE
No change, look my scenario is that energy is lost but some remains to DO work, thus force times distance...
 
  • #22
normal_force said:
No change...
In post #17, you asked me to imagine that the bullet only lost 200J of work. In the scenario I set up that is not the case, therefore you have changed the scenario.
 
  • #23
because it was off topic and I am trying to get a point off...the point being is that my scenario is correct but hard to find in nature.
 
  • #24
normal_force said:
because it was off topic and I am trying to get a point off...the point being is that my scenario is correct but hard to find in nature.
Which scenario, the ball and cart in your original post?
 
  • #25
haruspex said:
Which scenario, the ball and cart in your original post?

Yes, the energy one...
 
  • #26
normal_force said:
Yes, the energy one...
In what way is that scenario hard to find in nature?
I assume you mean this one:
normal_force said:
Ball=405 Joules
Cart=201 Joules
The ball ... has 204 remaining Joules which it will use to push back the cart doing work.
Let the force the ball exerts on the cart at some moment be a vector F (to the right). The force the cart exerts on the ball is -F. Suppose the point of contact at some instant is a distance x to the right of its initial position, in the inertial frame. The work the ball has done is ##\int_0^x F.dx##, and the cart has done minus that. But that does not mean the ball has done 201J of work on the cart. Much of the energy has gone into internal deformation of each. Indeed, if we take two equal masses with the same initial speed, the point of contact does not move, so neither does any work on the other. Each loses its KE to work done internally.

You are misleading yourself by thinking in terms of KE doing work. As others on this thread have pointed out, forces do work. When you kick a ball, the force your boit exerts on the ball does work on the ball, while the reaction force from the ball does (negative) work on your boot, detracting from its KE.
 
  • #27
haruspex said:
In what way is that scenario hard to find in nature?
I assume you mean this one:

Let the force the ball exerts on the cart at some moment be a vector F (to the right). The force the cart exerts on the ball is -F. Suppose the point of contact at some instant is a distance x to the right of its initial position, in the inertial frame. The work the ball has done is ##\int_0^x F.dx##, and the cart has done minus that. But that does not mean the ball has done 201J of work on the cart. Much of the energy has gone into internal deformation of each. Indeed, if we take two equal masses with the same initial speed, the point of contact does not move, so neither does any work on the other. Each loses its KE to work done internally.

You are misleading yourself by thinking in terms of KE doing work. As others on this thread have pointed out, forces do work. When you kick a ball, the force your boit exerts on the ball does work on the ball, while the reaction force from the ball does (negative) work on your boot, detracting from its KE.

Yes, But KE is mechanical energy...You the ball hits the goalies HEAD let's say, it hits his head doing work on him...as it slows down it exerts a force...

Another example? Think friction..breaking car...

seriously...momentum is not everything, this is a work-energy question
 
  • #28
haruspex said:
In what way is that scenario hard to find in nature?
I assume you mean this one:

Let the force the ball exerts on the cart at some moment be a vector F (to the right). The force the cart exerts on the ball is -F. Suppose the point of contact at some instant is a distance x to the right of its initial position, in the inertial frame. The work the ball has done is ##\int_0^x F.dx##, and the cart has done minus that. But that does not mean the ball has done 201J of work on the cart. Much of the energy has gone into internal deformation of each. Indeed, if we take two equal masses with the same initial speed, the point of contact does not move, so neither does any work on the other. Each loses its KE to work done internally.

You are misleading yourself by thinking in terms of KE doing work. As others on this thread have pointed out, forces do work. When you kick a ball, the force your boit exerts on the ball does work on the ball, while the reaction force from the ball does (negative) work on your boot, detracting from its KE.

Yes, but KE is mechanical energy, it can do work...that soccer ball has Kinetic energy, it can do work...when it slows down after it hits something, it DOES WORK...force TIMES distance...

It does work when it comes to rest...KE=TME====>Wnc

I don't think you guys get it or understand work-energy
 
  • #29
normal_force said:
Yes, But KE is mechanical energy...You the ball hits the goalies HEAD let's say, it hits his head doing work on him...as it slows down it exerts a force...

Another example? Think friction..breaking car...

seriously...momentum is not everything, this is a work-energy question
If I understand your arguments correctly, if a mass m1, KE E1, has a head-on collision with a mass m2, KE E2<E1, each does E2 of work, leaving E1-E2 for their subsequent combined KE. Is that your thinking?
If so, consider m2 initially stationary, so E2=0. It would follow that no work is done in the collision, so it must be completely elastic (!)
 
  • #30
normal_force said:
WORK...force TIMES distance...
Yes, but the distance there is the distance through which the point of application moves in the direction of the force. If two equal masses collide head on at the same speed the point of application never moves, so neither does any work on the other. All the work done is internal.
 
  • #31
haruspex said:
If I understand your arguments correctly, if a mass m1, KE E1, has a head-on collision with a mass m2, KE E2<E1, each does E2 of work, leaving E1-E2 for their subsequent combined KE. Is that your thinking?
If so, consider m2 initially stationary, so E2=0. It would follow that no work is done in the collision, so it must be completely elastic (!)

Not combining...but subtracting KE...KE is used up...not added in this scenario...KE goes into work...
 
  • #32
normal_force said:
Not combining...but subtracting KE...KE is used up...not added in this scenario...KE goes into work...
Not sure what you mean. Please express that in terms of equations. What I wrote seems to match the logic you applied in post #1.
 
  • #33
haruspex said:
Not sure what you mean. Please express that in terms of equations. What I wrote seems to match the logic you applied in post #1.

Okay, a moving object posses Kinetic energy...thus mechanical energy, thus the ability to do work...even without a constant accelerating force. Thus if an object has Kinetic energy, it has the ability to do work because of its energy by a virtue of motion.

Here is my equation

(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.
 
  • #34
normal_force said:
Okay, a moving object posses Kinetic energy...thus mechanical energy, thus the ability to do work...even without a constant accelerating force. Thus if an object has Kinetic energy, it has the ability to do work because of its energy by a virtue of motion.

Here is my equation

(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.
How are you calculating those lost energies?
 
  • #35
haruspex said:
How are you calculating those lost energies?

For the sake of getting the question, I just made up a value...thats all
 
  • #36
normal_force said:
For the sake of getting the question, I just made up a value...thats all
That's a strange thing to do. How will you ever find the final velocities with any certainty that way? If we are given that the two objects coalesce you can use conservation of momentum to find the final velocity and the energy lost.
 
  • #37
normal_force said:
For the sake of getting the question, I just made up a value...thats all
... Moreover, consider the equal initial KEs case, with no internal losses of energy. This should be an elastic collision, but your formula says there is no KE left.
 
  • #38
well, this isn't a conservation of momentum problem, its better to use Work-energy, as this is what it is...its an example of mechanical energy to do work, that a small object with superior KE, inferior mass and momentum can stop a cart and do work on it.
 
  • #39
normal_force said:
well, this isn't a conservation of momentum problem, its better to use Work-energy, as this is what it is...its an example of mechanical energy to do work, that a small object with superior KE, inferior mass and momentum can stop a cart and do work on it.
That is your claim, but everyone else on the thread disagrees with you. Please try to answer my post #37.
 
  • #40
haruspex said:
That is your claim, but everyone else on the thread disagrees with you. Please try to answer my post #37.
Oops, sorry, didn't see it...

Okay, well that is because the KE is taken up to do work...

Look, Imma try to explain this a lot better...but you are not making sense.
 
  • #41
normal_force said:
Okay, well that is because the KE is taken up to do work...
You cannot have it both ways. According to the equation you posted at #33 there should be no energy left for KE after the collision.
 
  • #42
I showed in my equation the energy lost which was not 100% of the energy, a great deal of energy dissipated...but there is some energy left to do work...as it clearly shows.
 
  • #43
normal_force said:
I showed in my equation the energy lost which was not 100% of the energy, a great deal of energy dissipated...but there is some energy left to do work...as it clearly shows.
In my post #37, I applied your formula from your post #33 to the case of two masses of equal initial KE colliding head on elastically. This is the case I am challenging your logic on. According to your formula there is no KE left.
 
  • #44
haruspex said:
In my post #37, I applied your formula from your post #33 to the case of two masses of equal initial KE colliding head on elastically. This is the case I am challenging your logic on. According to your formula there is no KE left.
Well, in short

(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.

(405 JOULES)+ (-201 Joules cart) = 204 Joules remaining of ball...so it stopped the cart...but as long as its in contact with the cart...and is not elastic like a spring and compresses too much...it will use up the remaining Kinetic energy in non-conservative work...so the ball doesn't rebound, all energy is used up...no elastic potential...so in the end...

The ball stopped the cart, and pushed back the cart.
 
  • #45
normal_force said:
Well, in short

(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.

(405 JOULES)+ (-201 Joules cart) = 204 Joules remaining of ball...so it stopped the cart...but as long as its in contact with the cart...and is not elastic like a spring and compresses too much...it will use up the remaining Kinetic energy in non-conservative work...so the ball doesn't rebound, all energy is used up...no elastic potential...so in the end...

The ball stopped the cart, and pushed back the cart.
Stop dodging the issue. As I already explained twice, I'm asking you to apply the same logic to a slightly different set-up: one in which the two masses have the same initial KE and collide elastically.
 
  • #46
haruspex said:
Stop dodging the issue. As I already explained twice, I'm asking you to apply the same logic to a slightly different set-up: one in which the two masses have the same initial KE and collide elastically.

Okay...well, the final velocity: -80.2 m/s for ball and cart is 5.22 M/s...

I get that, what is the point?
 
  • #47
normal_force said:
Okay...well, the final velocity: -80.2 m/s for ball and cart is 5.22 M/s...

I get that, what is the point?
No, not your ball and cart; the situation I posed in post #37: two arbitrary masses with equal KE. According to your reasoning in post #33, the KEs should completely cancel, leaving none.
 
Last edited:
  • #48
haruspex said:
No, not your ball and cart; the situation I posed in post #37: two arbitrary masses with equal KE. According to your reasoning in post #33, the KEs should completely cancel, leaving none.

Okay, but I think you are miss reading me...

thinking of this, compression occurs, maybe a little...like 0.000098 meters...KE lost, thus it is doing work, Wnc to be correct. So, KE turns into work, that what stops it...
 
  • #49
normal_force said:
Okay, but I think you are miss reading me...

thinking of this, compression occurs, maybe a little...like 0.000098 meters...KE lost, thus it is doing work, Wnc to be correct. So, KE turns into work, that what stops it...
We are only going to resolve this if you make clear what your reasoning is in a form that can be applied to all circumstances. In your post #33 (and in post #1 I think) your reasoning appears to be that if two masses collide head-on with KEs E1 and E2 < E1 then E2 of the one cancels E2 of the other, leaving E1-E2 as remaining total KE. If that is not your reasoning please state carefully what it is. (You have a habit of writing incomplete sentences, trailing off with ..., which leaves them rather vague. You also keep answering in terms of specific examples, which though useful leaves the general principle unclear.)
 
  • #50
haruspex said:
We are only going to resolve this if you make clear what your reasoning is in a form that can be applied to all circumstances. In your post #33 (and in post #1 I think) your reasoning appears to be that if two masses collide head-on with KEs E1 and E2 < E1 then E2 of the one cancels E2 of the other, leaving E1-E2 as remaining total KE. If that is not your reasoning please state carefully what it is. (You have a habit of writing incomplete sentences, trailing off with ..., which leaves them rather vague. You also keep answering in terms of specific examples, which though useful leaves the general principle unclear.)
Okay, no, what I am saying is totally different.
 

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