Bicycle Questions: Static Friction & Gear Ratios

[Orodruin]

Basically what i had in mind was that the ground "steals" rotational kinetic energy from the wheel simultaneously giving it back as translational kinetic energy.

Thus doing a grand total of zero work, just as the centripetal force does zero work to rearrange kinetic energy from being due to movement in the x-direction to being due to movement in the y-direction.

The x and y directions are separate degrees of freedom in the orbital example just as much as the wheel angle and bike travel distance are in the bike example. There then happens to be reaction forces constraining the x and y coordinates to a one-dimensional subspace, but this is also in complete analogy to the friction constraining the bike's travel distance to the wheel's rotation angle. In both cases, the reaction forces do not do any work but affect the movement of the system.

 

[Nugatory]

Basically that's the whole point. Why it is up to friction whether the work of chain will go solely as rotational KE (case of no friction) or partially to rotational and partially to translation KE (case with friction).

They're different physical systems, with different forces acting on them (static friction in one case but not the other). - it would be more surprising if they didn't behave differently.

This bit about a static friction and work has been discussed extensively before. Try this thread and the thread it in turn references: https://www.physicsforums.com/showthread.php?t=745038

 

[nasu]

Basically that's the whole point. Why it is up to friction whether the work of chain will go solely as rotational KE (case of no friction) or partially to rotational and partially to translation KE (case with friction).

The internal forces can change KE but they cannot change the total momentum or angular momentum.

In absence of any external forces not only the wheels will spin but the whole bicycle and bicyclist will rotate in opposite direction so that total angular momentum is zero.

You need an external force not to change KE but to change the momentum. Or the momentum distribution, if you wish. The external force will "produce" a linear momentum and its torque around the CM will prevent the bicycle to rotate as a whole. The normal force contribute too, not only the friction.

 

[A.T.]

What work is that of the wheel to the axis...

Think of the wheel as a lever.
- pivot : ground contact
- input (work done on the wheel) : the pull of the chain at the sprocket
- output (work done by the wheel) : the push of the axle at the bike frame

If you mean the work from some sort of internal force between the wheel and the axis, those are opposite and equal (if the wheel increases the translational KE of the axis then the axis reduces the translational KE of the wheel by the same amount) and cannot change the total translational KE of the system wheel+axis that we know its happening when a friction force is present.

The work done by the chain on the wheel is greater than the work done by the wheel on the bike frame. The reminder goes into accelerating the wheel.

This is not the case i wonder why you even mention this.

It is a valid way to model the system where the ground contact force is doing work. But it should not be confused with a model that includes the wheel as a separate body.

 

[mattt]

It is the work of internal forces of the system. Force due to biker's feet does work on them, the pedals do work on the axis, the axis on the chain and the chain on the wheel.

Kinetic energy of the bike increases due to work of internal forces, and the energy necessary for that comes from energy in biker's muscles.

Presence of friction just allows redirection of this energy into translational kinetic energy of the bike, but is not supplying any additional energy.

This.

It is not that difficult to understand.

 

[Delta2]

The work done by the chain on the wheel is greater than the work done by the wheel on the bike frame. The reminder goes into accelerating the wheel.

Thats unclear to me, it seems to me that the so called reminder goes into rotational acceleration.

 

[Delta2]

The internal forces can change KE...

I don't seem to get this sorry. Can u explain it abit more. Ok at first glance they seem they can change total rotational KE but i am not sure about the TOTAL translational KE.

You need an external force not to change KE but to change the momentum. Or the momentum distribution, if you wish. The external force will "produce" a linear momentum and its torque around the CM will prevent the bicycle to rotate as a whole. The normal force contribute too, not only the friction.

Well as you say we need external force to change linear momentum, thus the magnitude of velocity thus the translational KE. I ve to say this is the point where it gets hard for me to grasp, how is it possible for the external force to change the linear momentum but not the translational KE.

 

[Delta2]

This.

It is not that difficult to understand.

Well as i said in other posts for me personally the hard point to understand is how this external friction force does the redirection of energy without doing any work.

 

[mattt]

Well as you say we need external force to change linear momentum, thus the magnitude of velocity thus the translational KE. I ve to say this is the point where it gets hard for me to grasp, how is it possible for the external force to change the linear momentum but not the translational KE.

It is exactly the same when you start to walk. The exterior force (static friction) changes the total linear momentum (of you), but the work of this static friction force is zero.

The total work (which in this case is equal to the work done by internal forces) is not zero, and it is just equal to the change in total kinetic energy (from 0 when you are at rest, to some KE when you start to walk).

They are just two different theorems:

\vec{F}_{ext}(t) = \frac{d \vec{P}(t)}{dt}

(that is, the total exterior force is equal to the rate of change of the total linear momentum vector of the system of particles, assuming internal forces are "Newtonian").


\int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt = \Delta KE

(that is, the total work, sum of all works of all forces on all particles, is equal to increment in total kinetic energy of the system of particles).


They are two different mathematical theorems and they state different things.

 

[A.T.]

Thats unclear to me, it seems to me that the so called reminder goes into rotational acceleration.

Linear and rotational. Or just assume a massless wheel, which doesn't require energy to accelerate.

 

[A.T.]

Well as i said in other posts for me personally the hard point to understand is how this external friction force does the redirection of energy without doing any work.

Consider a stick falling over under gravity, with the lower end fixed on the ground.

 

[Delta2]

It is exactly the same when you start to walk. The exterior force (static friction) changes the total linear momentum (of you), but the work of this static friction force is zero.

The total work (which in this case is equal to the work done by internal forces) is not zero, and it is just equal to the change in total kinetic energy (from 0 when you are at rest, to some KE when you start to walk).

They are just two different theorems:

\vec{F}_{ext}(t) = \frac{d \vec{P}(t)}{dt}

(that is, the total exterior force is equal to the rate of change of the total linear momentum vector of the system of particles, assuming internal forces are "Newtonian").

Νοw that you mentioned this you made me think in terms of impulse. So from that relation follows \vec{p(t)}-\vec{p(0)}=\int_0^t{\vec{T(t)}}dt. So its the impulse of friction T(t) that changes the linear momentum of the bike (since there are no other external forces on the x-axis). So if friction is zero the impulse is zero hence p(t)=p(0)=constant, hence translational KE also constant.

However if the friction is not zero then it follows that the change in the linear momentum (hence the change in translational KE=p^2(t)/2m) is due to the impulse of the friction. We need not provide work from internal forces to increase translational KE.

 

[A.T.]

... changes the linear momentum of the bike...

If you have a body called "bike" in your model, which combines the entire bike and rider into one translating block, then the horizontal ground force will do work on this body, because the interface between "bike" and ground is not static anymore in this coarse model. It doesn't matter that it could be treated as static friction in a more detailed model with wheels, because in physics you do calculations based on your actual model, and not on some extra info that you didnt include in the model. And this coarse model discards the information about how wheels work, and the details of their interaction with the ground, so this interaction is not modeled as static friction anymore.

 

[Delta2]

If you have a body called "bike" in your model, which combines the entire bike and rider into one translating block, then the horizontal ground force will do work on this body, because the interface between "bike" and ground is not static anymore in this coarse model. It doesn't matter that it could be treated as static friction in a more detailed model with wheels, because in physics you do calculations based on your actual model, and not on some extra info that you didnt include in the model. And this coarse model discards the information about how wheels work, and the details of their interaction with the ground, so this interaction is not modeled as static friction anymore.

Τhat equation holds regardless of the modeling of the bike or whether you apply it only to wheel or the whole bike, since there arent any other external forces. To make you see it better, notice that the impulses due to internal forces will cancel each otther and only the impulse of external force will remain. To make u see it even better i ll apply it to the wheel and the rest of bike separately and i assume only the internal force between the wheel and the axis (you can also include the internal force of the chain but the treatment will be similar because one end of chain is on the wheel while the other end of chain is on the pedal which is on the rest of bike). So for the linear momentum of the wheel it will be

P_w(t)-P_w(0)=impulse(T,t)+impulse(F_internal,t)

and for the linear momentum of the rest of bike will be

P_b(t)-P_b(0)=impulse(-F_internal,t). Adding those two together the two impulses of F_internal will cancel out and what will remain will be the equation we want. ( I am really busy now and i avoid typing in tex, sorry anyway).

 

[mattt]

Νοw that you mentioned this you made me think in terms of impulse. So from that relation follows \vec{p(t)}-\vec{p(0)}=\int_0^t{\vec{T(t)}}dt. So its the impulse of friction T(t) that changes the linear momentum of the bike (since there are no other external forces on the x-axis). So if friction is zero the impulse is zero hence p(t)=p(0)=constant, hence translational KE also constant.

However if the friction is not zero then it follows that the change in the linear momentum (hence the change in translational KE=p^2(t)/2m) is due to the impulse of the friction. We need not provide work from internal forces to increase translational KE.

That statement is misleading (to say the least). If there were no internal forces (in the man + bicycle system) or just if the work of internal forces were zero in this system, then the bicycle and man would remain at rest and total KE would remain as zero.Exactly the same in the "man or robot starting to walk" case. If there were no internal forces or just the work of these internal forces (in the man or in the walking robot) were zero, then the man (also the robot) would not move. Total kinetic energy would remain as zero.The problem, more than anything, is "the way you state it". I think you already know that in this system (man + bicycle, or equivalently "man or robot starting to walk") the work of the static friction force is zero, that the static friction force is equal to the rate of change of total linear momentum vector, and that the work of all internal forces is equal to the increment of total kinectic energy. I think you understand all these facts, but you "decide" to state it in a very misleading way.

 

[Delta2]

To tell you the truth i just don't know. Maybe i am wrong on how i am thinking and how i am stating things. This thread has tired me. Deep down what i understand is that regardless if it is exactly one or more contact points, friction "steals" rotational kinetic energy from the back wheel and gives it as translational (and probably the opposite is happening in the front wheel where we don't have chain to do work, there friction counteracts the force of the axis and gives it back as rotational kinetic energy). Oh well i am just tired, maybe when i come back in this thread after some weeks and i read again i understand better.

 

[A.T.]

Τhat equation holds regardless of the modeling of the bike or whether you apply it only to wheel or the whole bike, since there arent any other external forces.

You equation is about momentum, not work.

 

[nearlynothing]

For Delta²:

Internal forces can change the total kinetic energy of your system. Whereas they can never change the total momentum of it, only external forces can do that. There's no contradiction here they are just two different theorems, as someone stated above. I think what's confusing you is the differentiation you make between "translational" and "rotational" kinetic energy, ultimately they are the same thing, namely energy due to motion of material particles.

Consider a system composed of a spring attatched to a block, with the spring against a wall, when you compress the spring and release it, the "translational" kinetic energy of the block + spring system increases, the only external force here is the normal from the wall, would you say that force does work?

 

[Delta2]

For Delta²:

Internal forces can change the total kinetic energy of your system. Whereas they can never change the total momentum of it, only external forces can do that. There's no contradiction here they are just two different theorems, as someone stated above. I think what's confusing you is the differentiation you make between "translational" and "rotational" kinetic energy, ultimately they are the same thing, namely energy due to motion of material particles.

Consider a system composed of a spring attatched to a block, with the spring against a wall, when you compress the spring and release it, the "translational" kinetic energy of the block + spring system increases, the only external force here is the normal from the wall, would you say that force does work?

Oh well i am tempted again to write in this thread...
Nice example. I could say again that the impulse of the normal changes the momentum of the system and since at least in this case the total translation kinetic energy is p^2/2M the impulse is what changes the kinetic energy of the system.

There is some sort of other thing that puzzles me also: What is happening to the work of the force from the block to the spring? It appears to me this work is negative yet it increases the kinetic energy of the spring (work from the normal is zero anyway). We have a violation of the work-energy theorem (applied to the spring only) or the theorem doesn't hold in the case of spring?

Also back to the bike . I wonder what is happening to the front wheel where there is no chain attached. How this weel is gaining rotational kinetic energy? Does the friction via its torque do work in the front wheel? I really don't understand the work of the torque of friction is different from the work of friction in this case?

 

[mattt]

Oh well i am tempted again to write in this thread...
Nice example. I could say again that the impulse of the normal changes the momentum of the system

Yes, that was the first of the two mathematical theorems I wrote in an earlier post.

and since at least in this case the total translation kinetic energy is p^2/2M the impulse is what changes the kinetic energy of the system.

After the interaction (when the spring + block leave the wall) the kinetic energy of the system will be \sum_{i}\left(\frac{1}{2}m_i v^2_i\right)

if all the particles move with the same velocity vector (which will be equal to the velocity vector of the center of mass) then that is the same as p^2/2M, but in this example it does not hold.

(the spring, if ideal, will for ever be compressing and stretching after it leaves the wall).

But you still are confused about the following:

You seem to think that the change in total kinetic energy of a system of particles is a function of the total linear momentum vector change. That is not correct in general. That is true only when (before the interaction, and also after the interaction) all the particles of the system have exactly the same velocity vector, which will be equal to the velocity vector of the center of mass.

In a given system of Newtonian particles, the same change in total linear momentum vector can result in totally different changes in total kinetic energy of the system.

Imagine this very simple example:

A system of two point particles at rest with different masses:

Particle 1 is at rest at P1(0,1) (for example), and particle 2 is also at rest at P2(0,0). Imagine that mass of particle 2 is = 2* mass of particle 1.

case 1: A constant force \vec{F} (horizontal, to the right) is exerted on particle 1 during a time interval [t_1, t_2].

case 2: That "same" force (only difference now will be the point of application) is exerted on particle 2 during the same time interval.

(for simplicity, imagine there are no internal forces in this example).


The change in total linear momentum vector of the system will be exactly the same in case 1 and case 2.

But the change in total kinetic energy of the system will be different in case 1 and case 2 ( in case 1 the total kinetic energy change will be = 2* total kinetic energy change in case 2 ).

So the "same" force (only difference being the point of application) acting during the same time interval, produces exactly the same change in total linear momentum vector of the system, but it produces (in this simple example) double of change in total kinetic energy of the system in case 1 wrt case 2, depending on the point of application of this "same" force.

I hope this helps you to understand the subtleties of these concepts.


In the "spring + block system", the exterior force (the normal force the wall exerts on one end of the spring) is equal to the rate of change of total linear momentum vector of the system (ignoring gravity for simplicity). But the work of this normal force is zero.

The total change in kinetic energy of the system is equal to the total work of all internal forces in this case.

There is some sort of other thing that puzzles me also: What is happening to the work of the force from the block to the spring? It appears to me this work is negative yet it increases the kinetic energy of the spring (work from the normal is zero anyway).

Try to think about the whole system made up of thousands of point particles (this always helps). Put your attention to one concrete given particle (it does not matter if it is a particle of the spring or a particle of the block). During the time interval when the spring is stretching and about to leave the wall (i.e., during the interaction wall-system), this particle you are putting your attention on, is accelerating to the right (imagine the wall is vertical and the system is at the right side). That is because the total force on this concrete particle during this time interval, is non-zero and pointing to the right (and this total force on this concrete particle is equal to the total INTERNAL force on this concrete particle, ignoring the weight for simplicity).

Obviously the work of this total force ( = total internal force ) on this concrete particle (of the spring or of the block, it is the same) is not zero, it is positive.

If you think exactly the same with ANY other particle of the spring or of the block, you realize the total work of internal forces is obviously non-zero, it is positive, and it is equal to the total change in kinetic energy of the system.

We have a violation of the work-energy theorem (applied to the spring only) or the theorem doesn't hold in the case of spring?

It is a mathematical theorem. It can not be violated :-)

It could not apply depending on the mathematical-physical hypothesis of our model, but in this example-model, it does apply.

Also back to the bike . I wonder what is happening to the front wheel where there is no chain attached. How this weel is gaining rotational kinetic energy? Does the friction via its torque do work in the front wheel? I really don't understand the work of the torque of friction is different from the work of friction in this case?

I will explain this later, I got to go now.

 

[jbriggs444]

There is some sort of other thing that puzzles me also: What is happening to the work of the force from the block to the spring? It appears to me this work is negative yet it increases the kinetic energy of the spring (work from the normal is zero anyway). We have a violation of the work-energy theorem (applied to the spring only) or the theorem doesn't hold in the case of spring?

If you look at center-of-mass work (force times displacement of the center of mass), the work-energy theorem applies. The net force on the spring from the wall and from the block combined is away from the wall. The center of mass moves away from the wall. Positive center-of-mass work is done and linear kinetic energy increases.

If you look at total work (force times displacement of the material at the point of contact), conservation of energy applies. The wall does zero work on the spring and the block does negative work on the spring. It must be so to balance the energy books. The block carried away kinetic energy and that energy had to have come from somewhere. The potential energy of the spring makes up for the resulting energy deficit.

You can easily see that the work done by the block is negative -- were it not for that interaction, the spring would *ZING* across the room with essentially all of its potential energy having been converted to kinetic and vibrational energy in the spring and none having been converted to kinetic energy in the block.

Always keep straight which kind of work you are talking about.

 

[A.T.]

There is some sort of other thing that puzzles me also: What is happening to the work of the force from the block to the spring? It appears to me this work is negative yet it increases the kinetic energy of the spring

The kinetic energy of the spring comes from internal work of the spring (stored potential energy), not from the work of the block on the spring.

I wonder what is happening to the front wheel where there is no chain attached. How this weel is gaining rotational kinetic energy?

How do you gain rotational kinetic energy, when somone pushes you over? In the rest frame of the ground, the bike frame is doing work on the front wheel. It applies a force at the axle, which creates a torque around the static ground contact point.

 

[Jano L.]

Oh well i am tempted again to write in this thread...
Nice example. I could say again that the impulse of the normal changes the momentum of the system and since at least in this case the total translation kinetic energy is p^2/2M the impulse is what changes the kinetic energy of the system.

You're right in what you mean, but you're saying it incorrectly. In general, kinetic energy changes as a result of work of some force. The interesting thing about this is that the force can be any force, even internal force. In the case of block with a spring, the kinetic energy of the block increases as a result of the force due to spring. The kinetic energy of the spring increases due to work of internal stress forces in the spring on the spring itself.

There is some sort of other thing that puzzles me also: What is happening to the work of the force from the block to the spring? It appears to me this work is negative yet it increases the kinetic energy of the spring (work from the normal is zero anyway).

This work is negative, so it cannot be the sole work done on the spring! There is additional work done on the spring due to internal elastic forces. The work-energy theorem is valid, we just have to include all mass points and all forces that act on them.

Also back to the bike . I wonder what is happening to the front wheel where there is no chain attached. How this weel is gaining rotational kinetic energy? Does the friction via its torque do work in the front wheel? I really don't understand the work of the torque of friction is different from the work of friction in this case?

In the frame of the road, the fork pushes the axis forward and since the axis moves in the direction of this force, the fork does work on the axis. Friction force enables transfer of part of this work to rotational energy, but it does not contribute any work of its own.

 

[mattt]

To Delta^2:

As you can see in the last answers, you can "translate" from a "microscopic view" (system of point particles, the way I tried to explain it to you) to "macroscopic view" (talking about "the spring", "the block", the work this subsystem "does" on this other subsystem,...the way the last few posters tried to explain it to you) and vice versa.

They are equivalent ways of stating the same things (the same results), if done correctly, obviously.

Depending on the person, some find it easier to "look at it" from the "microscopic=system of point particles" point of view, some others find it easier to "think about it" from the "macroscopic=subsystems doing work to each other" point of view.

As I said, if done correctly, it is totally equivalent.

Personally, when I start to have some doubts about a given problem (in Newtonian mechanics), I usually see it more clearly when I go down to the "microscopic"="system of point particles" point of view. But probably other people feel the other way around :-)

 

[Delta2]

Good answers regarding the kinetic energy of the spring, however i am not satisfied about the other two questions.

As even matt stated, changes in the linear momentum vector result in changes in translational kinetic energy. The change in translational KE is always a function of the momentum, which exact function depends in the case.
So still can say impulse of external forces changes momentum p which changes translational KE(p).

Jano.L sorry you have to explain better how the friction which is applied at the contact point in the ground, transforms the work of the force from the fork to the axle into rotational kinetic energy. What i see is that the torque of friction gives rotational acceleration and that the force from the axle gives no torque. Its kinda the same situation, we have the friction which changes the angular momentum of the wheel which changes the rotational kinetic energy of the wheel which is function of its angular momentum.

 

[mattt]

Good answers regarding the kinetic energy of the spring, however i am not satisfied about the other two questions.

As even matt stated, changes in the linear momentum vector result in changes in translational kinetic energy. The change in translational KE is always a function of the momentum, which exact function depends in the case.
So still can say impulse of external forces changes momentum p which changes translational KE(p).

Did you really read my post? :-)

I said change in total kinetic energy of a system of particles IS NOT a function of change of total linear momentum vector of this system of particles.

I even wrote an example of a system in which THE SAME change in total linear momentum vector give rise to two totally different changes in total kinetic energy of the system.

I even stated when it is and when it is not. :-(

Jano.L sorry you have to explain better how the friction which is applied at the contact point in the ground, transforms the work of the force from the fork to the axle into rotational kinetic energy. What i see is that the torque of friction gives rotational acceleration and that the force from the axle gives no torque. Its kinda the same situation, we have the friction which changes the angular momentum of the wheel which changes the rotational kinetic energy of the wheel which is function of its angular momentum.

More on this later on.

 

[Delta2]

Did you really read my post? :-)

I said change in total kinetic energy of a system of particles IS NOT a function of change of total linear momentum vector of this system of particles.

I even wrote an example of a system in which THE SAME change in total linear momentum vector give rise to two totally different changes in total kinetic energy of the system.

I even stated when it is and when it is not. :-(



More on this later on.

yes i read your post and jano.l's which both i found most usefull to me. I read also the other two posts but were rather confusing or i didnt find em satisfying.
Matt, can you find me a case where we have a change in the magnitude of the linear momentum, yet we have no change in the translational KE?

What i am saying is that

external impulse cause change in linear momentum which cause change in translational KE.

It doesn't matter if the same change of momentum in two different cases cause two different changes in KE , all it matters is that it causes the change.

 

[nearlynothing]

external impulse cause change in linear momentum which cause change in translational KE.

It doesn't matter if the same change of momentum in two different cases cause two different changes in KE , all it matters is that it causes the change.

If I understand correctly, you're saying that without this external force which changes the momentum, and so changes the velocity, there wouldn't be a change in traslational KE, so that the external force must be the factor that indeed changes the traslational KE, and so we can say that it performs work. Tell me if I'm wrong.

The problem here is that then you have to define work done by a force as "the amount of change in the kinetic energy that wouldn't happen without this particular force". That's not the definition we use in physics. It would be a pretty hard to apply definition also.

 

[Delta2]

If I understand correctly, you're saying that without this external force which changes the momentum, and so changes the velocity, there wouldn't be a change in traslational KE, so that the external force must be the factor that indeed changes the traslational KE, and so we can say that it performs work. Tell me if I'm wrong.

The problem here is that then you have to define work done by a force as "the amount of change in the kinetic energy that wouldn't happen without this particular force". That's not the definition we use in physics. It would be a pretty hard to apply definition also.

Hm no, i am saying an external force changes momentum which in the general case changes the KE. So i require some explanation how this happens whithout the external force doing work. Some answers are confusing some are satisfying, overall i can't say i have a clear picture. Maybe i can ask it more clear , is it possible for an external force to change the magnitude of the total linear momentum without doing work?

I guess it is all we need is impulse and not work such as in the case of the normal of wall in your example.

 

[mattt]

yes i read your post and jano.l's which both i found most usefull to me. I read also the other two posts but were rather confusing or i didnt find em satisfying.
Matt, can you find me a case where we have a change in the magnitude of the linear momentum, yet we have no change in the translational KE?

Yes.

Imagine two point masses:

Particle 1 of mass = 1 Kg is at rest at point (0,1) and particle 2 of mass = 2 Kg is moving horizontally to the left at constant velocity \vec{v}_2 = -\frac{1}{\sqrt{2}} \vec{i} m/s.

Suppose there are no internal forces in this example.

An exterior force \vec{F}_a = 1 \vec{i} N acts on particle 1 during 1 second of time.

Another exterior force \vec{F}_b = \sqrt{2}\vec{i} N acts on particle 2 also during 1 second of time.

Before the exterior forces are applied, we have:

\vec{p} = \vec{p}_1 + \vec{p}_2 = \vec{0} + m_2\vec{v}_2 = 2 \frac{-1}{\sqrt{2}}\vec{i} = -\frac{2}{\sqrt{2}}\vec{i} = -\sqrt{2}\vec{i} Kg m/s

After the exterior forces have been applied, we have (do yourself the computations) :

\vec{p} = \vec{p}_1 + \vec{p}_2 = 1 \vec{i} + \vec{0} = 1 \vec{i} Kg m/s

So the total linear momentum vector has changed in magnitude (and also in sense).

The total kinetic energy before the exterior forces have been applied was:

\frac{1}{2}m_1 v^2_1 + \frac{1}{2}m_2 v^2_2 = 0 + \frac{1}{2}2\frac{1}{2} = \frac{1}{2} J.

The total kinetic energy after the exterior forces have been applied is:

\frac{1}{2}m_1 v^2_1 + \frac{1}{2}m_2 v^2_2 = \frac{1}{2}.1. 1^2 + 0 = \frac{1}{2} J.


So the total linear momentum vector has changed in magnitude, but the total kinetic energy of the system remains the same.

What i am saying is that

external impulse cause change in linear momentum which cause change in translational KE.

It doesn't matter if the same change of momentum in two different cases cause two different changes in KE , all it matters is that it causes the change.

As you can see in this last example, that is not true. :-)