# Elastic Collision: how do the forces and work done make sense?

• alkaspeltzar
In summary: I know has to be there, seems redundant but yet I know that it is there.They are two ways of looking at the same interaction. From the perspective of ball 1 it does positive work on ball 2 and comes to a stop. From the perspective of ball 2 it does negative work on ball 1 until ball 1 comes to a stop. Another way of saying it is that they are an action/reaction pair according to Newton's 3rd law. These two distinct forces exist because of 1 interaction and cannot be separated.
alkaspeltzar
Okay, say we have two balls(equal mass and size), 1 and 2. #1 has kinetic energy and #2 is at a standstill, they collide. Ignore all friction, heat, sound losses etc..

Now I know that 1 exerts a force on 2, hence doing work, which in turn uses its kinetic energy up. Therefore, 2 speeds up, gaining the work done.

But ball 2 exerts a force on 1, doing negative work, hence taking energy. So I am confused. Doesn't ball 1 exerting a force, doing work, mean it will naturally have less kinetic energy and slow down. If so, then wouldn't ball 2's force in slowing it down be redundant?

What am I missing here? My confusion is that in ball 1 doing work, it is using the kinetic energy, as it speed up ball 2 which should mean it is slowing down. so the reaction force, which I know has to be there, seems redundant but yet I know that it is there.

They are two ways of looking at the same interaction. From the perspective of ball 1 it does positive work on ball 2 and comes to a stop. From the perspective of ball 2 it does negative work on ball 1 until ball 1 comes to a stop.

Another way of saying it is that they are an action/reaction pair according to Newton's 3rd law. These two distinct forces exist because of 1 interaction and cannot be separated. You cannot talk about one without the other because they arise from the same interaction.

Dale
Rayoub, I agree with that according to Newton's 3rd law. However, when looking at it from the energy/work side, given there is positive work and negative work, aren't they the same thing? So how can both be going on?

My mind is thinking that the energy was spent in positive work making ball 2 move and ball 1 come to a stop, so if there is a reaction for from ball 2, doing negative work, then there would be no energy for it to take, as it is already gone. Hence the redundancy.

Or maybe I am confusing it because as both happen simultaneously. Yes ball 1 is exerting force, doing work and spending its energy, but then ball 2 must be applying an opposite force, slowing ball 1 and absorbing energy

However, when looking at it from the energy/work side, given there is positive work and negative work, aren't they the same thing? So how can both be going on?

I think you answered your own question here. They are both going on at the same time because they are the same thing - part of the same interaction.

I remember being confused by this same thing. Hopefully, I hit upon the source of misunderstanding for you. This applies to all action/reaction pairs.

I wonder if what is confusing me too is that energy is doesn't come in equal and opposite pairs, but forces do!

So yes there is a force on ball1 equal and opposite to ball 2. However, do to this interaction, only energy flows really in one direction. ball 1 force, causes ball 2 to accelerate, doing work, thus decreasing ball 1 kinetic energy and if looking at the energy flow from ball 2, it is the same just opposite way of looking at it.

There is positive work being done on B and negative work being done on A simultaneously. The positive work on B increases the energy of B and the negative work on A decreases the energy of A. By conservation of energy the positive work on B is equal and opposite to the negative work on A.

So far there is no double counting or confusion because we were always talking about the work done on A or B. The confusion only arises if you shift to talking about work done by A or B. That is where you think things are being double counted but they are not. The work done on A is not just equal to the work done by B, it is the same thing. So when you say “work done on A” and then you say “work done by B”, you have simply repeated yourself. There is no additional energy nor any additional work involved, it is just a different name for the exact same thing.

To be precise we should never say the word “work” without explicitly stating two systems. So we should never say “the work done on A” nor “the work done by B”, but instead it should always be “the work done on A by B”. In a simple scenario like this the missing part of the statement can be inferred, but if you ever find yourself becoming confused again, then explicitly write your statements using the full “on A by B” terminology. Your confusion should disappear

hutchphd
I guess my confusion is this: doesn't the work done on A by B use it's kinetic energy up of B? If so, then the work done on B by A makes me think it's redundant as there would be no energy for it to take.

Isn't work done just the energy spent by a force thru a distance? So if one object has kinetic energy, it can therefore so work(spend energy) right? So the opposing object how is it taking energy(negative work) that is already gone?

alkaspeltzar said:
I guess my confusion is this: doesn't the work done on A by B use it's kinetic energy up of B? If so, then the work done on B by A makes me think it's redundant as there would be no energy for it to take.
The work done on A by B will change the energy of A. The work done on B by A will change the energy of B. By conservation of energy the work done on A by B is equal and opposite the work done on B by A.

But isn't work energy expendedor used? I've see that as definition. So isn't work using the kinetic energy up?

Maybe can someone help me understand work. Saying f times d doesn't help. Nor does just saying it's energy transfer. What does that mean?

alkaspeltzar said:
But isn't work energy expendedor used? I've see that as definition. So isn't work using the kinetic energy up?
alkaspeltzar said:
Maybe can someone help me understand work. Saying f times d doesn't help. Nor does just saying it's energy transfer. What does that mean?
There are two basic definitions of work.

One is that work is ##f\cdot d##. That is the one that I have been using here. In that case you can derive the fact that the work done on A increases the energy of A regardless of B. Then you can further derive that the work done on A by B is equal and opposite the work done on B by A. This leads to the conservation of energy, but note that in this formulation it is always the work done on the object that changes its energy.

The other basic definition of work is that work is a transfer of energy. In that definition of work there is no need to prove energy conservation, it is built into the definition. The work is the transfer of energy itself, so there is only one work in an interaction between two objects.

So in the first approach there are two works for one interaction, each separately increases the energy of the object that the work is on, but you can prove that they are equal and opposite. In the second approach there is only one work for one interaction and it increases one object’s energy by the amount it decreases the other object’s energy.

I think that you are getting confused because you are trying to mix both definitions. So you are trying to have two works in one interaction like the first approach but have each work change both energies like the second approach.

That would make sense as to why I'm getting confused. I have been reading that work is energy input/transfer/energy expended. So when I think about, as work being done by both interaction forces, according to newtsons third law, seems like it is redundant.

Also doesn't it take energy for work and visa versa? So doesn't work on an object assume then Energy is being used/outputted as force thru distance?

Thanks for help

Dale can you answer my last question? See I guess I learned that work was energy expended as a force thru distance. Given Newton's third law, then how can I have two works?

One work would be using energy to apply a force to accelerate the non moving object, but yet the energy would already be gone since the non moving object is doing work as well? That's where I'm stuck

alkaspeltzar said:
Dale can you answer my last question? See I guess I learned that work was energy expended as a force thru distance. Given Newton's third law, then how can I have two works?
Yes, with opposite signs.

alkaspeltzar said:
Dale can you answer my last question?
Sorry, I think I already have answered it, multiple times. I don’t have any more ways to explain this.

I have laid out very organized ways to consistently get the analysis right. If you have a disorganized way of thinking that is leaving you confused, why not use the organized thought patterns I have laid out above?

If the way you are thinking about it confuses you then why do you persist in that way of thinking when a non- confusing alternative is presented?

Last edited:
I guess i keep asking because when one is stuck and doesn't it get it, I keep trying to understand it.

Below is how I understand work. Energy expended as force thru a distance. My question arises from that.
https://ccrma.stanford.edu/~jos/pasp05/Work_Force_times.html

So when I think of work, i think the block A(which has kinetic energy) is expending energy, using it up to apply a force thru a distance on Block B, which then it moves and has gained kinetic energy.

When I apply Newton's 3rd law, there is a reaction force. It does work, but it is negative. Does this mean it is just applying a force against block A and in the end, accepting energy being used and thus slowing down? The energy provided is the same as that lost, thus there is really one work?

alkaspeltzar said:
So when I think of work, i think the block A(which has kinetic energy) is expending energy, using it up to apply a force thru a distance on Block B, which then it moves and has gained kinetic energy.
It doesn't matter if the KE changes. With other forces acting A & B can be both moving at constant velocity, while A is doing work on B.

alkaspeltzar said:
When I apply Newton's 3rd law, there is a reaction force. It does work, but it is negative. Does this mean it is just applying a force against block A and in the end, accepting energy being used and thus slowing down?
The slowing down part doesn't matter. A transfers energy to B. So A is doing positive work on B, and B is doing negative work on A.

alkaspeltzar said:
The energy provided is the same as that lost,
Not always. If there is slippage, the negative work on A can be greater than the positive work on B. The difference represents the energy dissipated by dynamic friction.

A.T. said:
If there is slippage, the negative work on A can be greater than the positive work on B. The difference represents the energy dissipated by dynamic friction.
When forces are acting at a distance this also occurs. Repulsion between objects moving closer together absorbs energy (a piston on a compression stroke as the gasses get hotter). Repulsion between objects moving apart releases energy (the same piston on the power stroke). Attraction between objects moving closer together releases energy (an object falling under the force of gravity as potential energy is released). Attraction between objects moving apart absorbs energy (a baseball having been hurled into the air and now slowing down as potential energy is stored).

alkaspeltzar said:
Below is how I understand work.
I recommend understanding work differently. If you change your understanding of work to either of the ones that I described above then the confusion disappears.

alkaspeltzar said:
When I apply Newton's 3rd law, there is a reaction force. It does work, but it is negative. Does this mean it is just applying a force against block A and in the end, accepting energy being used and thus slowing down? The energy provided is the same as that lost, thus there is really one work?
That would be the second of the two basic definitions of work that I described. Please re-read my post 12, especially the last paragraph. You need to pick one or the other definition for now and just use that until you feel more comfortable. Only add the second concept when you are solid with the first.

Dale, so I have thought about what you have said and I am trying to understand the interaction where there are two works;

"There is positive work being done on B and negative work being done on A simultaneously. The positive work on B increases the energy of B and the negative work on A decreases the energy of A. By conservation of energy the positive work on B is equal and opposite to the negative work on A."

But what stops me is how is energy being conserved? Wouldn't positive work being done on B(be creating energy) and negative work on A(be taking energy away). With that, yes, I see the system as a whole is not having an overall change of energy. But yet positive work and negative work are independently changing/creating/taking energy which I thought cannot be done. Energy cannot be created or destroyed right?

Thanks for the help

alkaspeltzar said:
Wouldn't positive work being done on B(be creating energy) and negative work on A(be taking energy away).
Yes.

alkaspeltzar said:
But yet positive work and negative work are independently changing/creating/taking energy which I thought cannot be done. Energy cannot be created or destroyed right?
Energy cannot be created or destroyed is a "one work" concept. In the "two work" approach, as long as a positive work must always correspond to an equal and opposite negative work then energy is conserved since energy created must equal energy destroyed. This is similar to momentum, a force in the positive direction makes positive momentum and a force in the negative direction independently makes a negative momentum, but as long as those two are equal and opposite then momentum is conserved.

The difference in the two approaches is only that in the "two work interaction" approach work is the fundamental definition and conservation of energy is derived whereas in the "one work interaction" approach the conservation of energy is the fundamental principle and work is defined with that already built in. They are completely equivalent.

Because these two approaches are completely equivalent, just use the approach that you prefer (sounds like the "one work interaction" for you). Use that one consistently and exclusively, you are only getting in trouble because you keep trying to mix the two.

Dale , I think I am getting it. I don't mind the two work approach. I do see the differences now.

And as for the Law of conservation of energy, in reality, energy as whole cannot be destroyed or created,,it in a system is conserved. So in the above, work can "in a sense" cause there to be energy put into one object(positive work) knowing that energy is being released and lost in the second object. It like energy that IS BEING LOST is really the cost for the object to put energy into the other. Does that make sense?

So based on this, what happens is like you said. When the two balls collide, there are equal and opposite forces, each doing work, increasing and decreasing the energy of the other simultaneously, such that the one ball would stop, other would be traveling away at the initial velocity(assuming perfect world).

Right?

Dale
Yes, that seems like you are getting it now!

## 1. What is an elastic collision?

An elastic collision is a type of collision between two objects in which both the momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy of the system before and after the collision are equal.

## 2. How do the forces in an elastic collision make sense?

In an elastic collision, the forces involved are equal and opposite. This is due to Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. This means that the force exerted by one object on the other is equal in magnitude but opposite in direction to the force exerted by the second object on the first.

## 3. How is work done in an elastic collision?

In an elastic collision, no work is done. This is because work is defined as the force applied over a certain distance, and in an elastic collision, the forces are equal and opposite, resulting in a net force of zero. Therefore, the objects do not undergo any displacement and no work is done.

## 4. How do the laws of conservation apply in an elastic collision?

The laws of conservation of momentum and kinetic energy apply in an elastic collision. This means that the total momentum and kinetic energy of the system before and after the collision are equal. This is due to the fact that in an elastic collision, no external forces are acting on the system, so the total momentum and energy must remain constant.

## 5. What factors affect the elasticity of a collision?

The elasticity of a collision is affected by the materials and surfaces of the objects involved. Objects made of more elastic materials, such as rubber, will have a more elastic collision compared to objects made of less elastic materials, such as clay. Additionally, the speed and angle of the collision can also affect the elasticity of the collision.

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