Choosing what consists of a "system" in Newton's laws of motion

[sachin]

Homework Statement

I am posting another query on my previous post in the below link

https://www.physicsforums.com/threads/tt.ps://www.physicsforums.com/threads/n-semi-circular-track.1044825-laws-block-sliding-up-a-frictionless-semi-circular-track.1044825/

The question says find the common horizontal velocity of the blocks when the smaller block of mass m reaches the point A of the bigger block of mass M, the smaller block was given an initial velocity v to the left as in the figure, all surfaces are frictionless.

Relevant Equations

m x v + M x 0 = (m + M)v'

The question is solved in a single step by taking the blocks as a system and using conservation of linear momentum in the horizontal direction as there is no net force acting in the horizontal direction.

Conserving the momentum we get,
m x v + M x 0 = (m+M)v',
so,,v' = mv/(m +M).where v' is the common horizontal velocities of the blocks at the point A.

My concern is quite trivial,
while choosing the system why did we consider only the two blocks,why did not we consider the Earth into the system ?

what is the necessary condition of choosing a system while using conservation of linear momentum ?

 

[PeroK]

I assume the large block moves without friction across the surface? In which case, there is no external horizontal force on the system.

 

[kuruman]

To add to the above, if you chose as system one of the masses, you would not be able to conserve momentum because the chosen mass experiences an external force from the other mass. To conserve momentum, you must choose a system that is isolated from all external forces.

 

[sachin]

I assume the large block moves without friction across the surface? In which case, there is no external horizontal force on the system.

yes,both the blocks move to the left,there is no friction between the larger block and ground.

 

[sachin]

To add to the above, if you chose as system one of the masses, you would not be able to conserve momentum because the chosen mass experiences an external force from the other mass. To conserve momentum, you must choose a system that is isolated from all external forces.

my concern was why do not we add the Earth into the system,in the picture the bigger block is placed on the ground i.e earrh.

 

[kuruman]

my concern was why do not we add the Earth into the system,in the picture the bigger block is placed on the ground i.e earrh.

Because the Earth does not exert a force in the horizontal direction along which momentum is conserved.

 

[sachin]

Because the Earth does not exert a force in the horizontal direction along which momentum is conserved.

even there was no force betwen the blocks as all surfaces are frictionless,still we chose both of them as a system.

 

[erobz]

even there was no force betwen the blocks as all surfaces are frictionless,still we chose both of them as a system.

But there is also the assumption of a locally flat Earth. So, while the small mass is sliding on the flat part of the sled, no momentum is being transferred to the sled. It is only when the small mass reaches the curvature on the sled that it begins to transfer its momentum to the sled. The Earth is also taken to have effectively infinite mass here, so it doesn't gain any momentum in the $y$ direction.

 

[PeroK]

Or, more simply, you have an approximately flat surface and an approximately constant external force of gravity. There is no need for any elaborate or unphysical assumptions.

 

[erobz]

Uh oh @PeroK is skeptical. What is wrong?

If the Earth were curved and not "infinetly massive" then the little mass would transfer momentum to the sled and the sled would transfer momentum to the Earth (at least possibly for some finite initial small mass of velocity).

 

[sachin]

But there is also the assumption of a locally flat Earth. So, while the small mass is sliding on the flat part of the sled, no momentum is being transferred to the sled. It is only when the small mass reaches the curvature on the sled that it begins to transfer its momentum to the sled. The Earth is also taken to have effectively infinite mass here, so it doesn't gain any momentum in the $y$ direction.

that means as long as m is sliding on the horizontal part,only m has to be taken as a system and only M as another system,both m and M can not be taken as a system to find their velocities.

 

[erobz]

that means as long as m is sliding on the horizontal part,only m has to be taken as a system and only M as another system,both m and M can not be taken as a system to find their velocities.

While the little mass is sliding on the flat part of the sled ,the velocity of the sled is $0$. So if you include $M$ in the "system" it has no momentum before or after any duration of time before the little mass reaches the curved part of the $M$(in the absence of friction).

 

[sachin]

While the little mass is sliding on the flat part of the sled ,the velocity of the sled is $0$. So if you include $M$ in the "system" it has no momentum before or after any duration of time before the little mass reaches the curved part of the $M$(in the absence of friction).

my concern was if the question is find the velocities of the blocks when m reaches the end of the horizontal part of M,we can not get the velocities by taking both as a system,but if we consider individually as systems,we can get the final velocities.

 

[sachin]

can we conclude that if two bodies are interacting i.e applying forces on each other without any other forces acting from outside,we can take them as a system ?

 

[erobz]

my concern was if the question is find the velocities of the blocks when m reaches the end of the horizontal part of M,we can not get the velocities by taking both as a system,but if we consider individually as systems,we can get the final velocities.

Do you mean " if there is friction between $m$ and $M$" we can't determine the final velocities? Because we can determine the final velocities if there is no friction between $m$ and $M$.

 

[sachin]

no,I meant no net external force on the system,took friction as an internal force,moreover if there is friction,I hope we can get the final velocities of the blocks without conserving momentum to the system but calculating individually also.

 

[sachin]

can we conclude "momentum conservation is a tool when forces are not clearly understood" like in the curved parts but when in horizontal parts its not required or can not be used even if there is friction.

 

[erobz]

If there is friction between $m$ and $M$, then there is a dissipative force that would be concealed by considering both masses together. I don't think impulse/momentum has anything to say about internal forces, but is about external forces acting on the system. So if you are going to capture that information, when there is friction (an internal force when both masses are considered) you are going to need to consider the masses as separate systems that are coupled through the shared force of friction between them as an external force acting on each mass.

 

[erobz]

can we conclude "momentum conservation is a tool when forces are not clearly understood" like in the curved parts but when in horizontal parts its not required or can not be used even if there is friction.

Beyond me, but others will be able to explain that here. There was a problem not too long ago that will be a good one to examine:

https://www.physicsforums.com/threads/a-disk-sliding-up-a-wedge.1045076/

 

[PeroK]

Uh oh @PeroK is skeptical. What is wrong?

You're wilfully complicating a simple scenario. And, in the process causing confusion and leading the OP astray. As can be seen in the posts above.

If the Earth were curved and not "infinetly massive"

The Earth is not infinitely massive. To require it to be so for a simple mechanics problem is nonsensical.

 

[PeroK]

no,I meant no net external force on the system,took friction as an internal force,moreover if there is friction,I hope we can get the final velocities of the blocks without conserving momentum to the system but calculating individually also.

Even if there is friction between the blocks momentum is conserved - this is a direct consequence of Newton's third law.

 

[PeroK]

can we conclude "momentum conservation is a tool when forces are not clearly understood" like in the curved parts but when in horizontal parts its not required or can not be used even if there is friction.

No. Momentum is conserved whenever there is no external force. In this case momentum is conserved in the horizontal direction, but not the vertical direction.

 

[PeroK]

If there is friction between $m$ and $M$, then there is a dissipative force that would be concealed by considering both masses together. I don't think impulse/momentum has anything to say about internal forces, but is about external forces acting on the system. So if you are going to capture that information, when there is friction (an internal force when both masses are considered) you are going to need to consider the masses as separate systems that are coupled through the shared force of friction between them as an external force acting on each mass.

This is wrong. You still have conserved momentum, but you need to calculate the energy lost to friction between the masses.

 

[sachin]

if all surfaces are frictionless,considering the motion only in the horizontal part,can we
take m and M as a system even though they themselves are isolated systems.

 

[PeroK]

if all surfaces are frictionless,considering the motion only in the horizontal part,can we
take m and M as a system even though they themselves are isolated systems.

They are not isolated because of their shape.

I really can't see the problem in considering $M$ and $m$ as a system whose centre of mass cannot move horizontally? What is the problem?

 

[sachin]

can we take two isolated systems as a single system and use momentum conservation.

 

[erobz]

You're wilfully complicating a simple scenario. And, in the process causing confusion and leading the OP astray. As can be seen in the posts above.

I would never willfully complicate the scenario. And the OP is at worst no closer to understanding (that includes up to this point - after your replies as well).

The Earth is not infinitely massive. To require it to be so for a simple mechanics problem is nonsensical.

Ok put the two bodies in deep space outside of external gravitational field of the Earth. They are gravitationally attracted to each other. For the sake of my argument $M = 10 \, \rm{kg}$ and $m = 1 \, \rm{kg}$, for some large enough initial velocity of $m$ the masses $M$ and $m$ will (at least temporarily) separate from each other?

EDIT: For the record, I wasn't trying to be sarcastic when I said "Uh oh @ PeroK is skeptical"...just humorous about potentially being incorrect. It seems like that might have been misinterpreted as "shots fired" given the nature of your reply.

 

[Lnewqban]

even there was no force betwen the blocks as all surfaces are frictionless,still we chose both of them as a system.

The link shown in the OP is not working.
The transfer of kinetic energy between both blocks does not happen because of friction, but because the vertical loop that the small block is forced to make.

That change of trajectory is the only thing that induces a force on the big block, which then moves horizontally only because the surface of the Earth restricts any downwards (vertical) movement.

In this case, Earth is a limit to the natural movement of the studied or observed M+m system, rather than a part of it.

 

[kuruman]

can we take two isolated systems as a single system and use momentum conservation.

You can take any number of masses as a "system" as long as you specify precisely what this system consists of. Let's make the Earth part of the system which consists of small mass A, large mass B on which A is sliding and Earth E. To a very good approximation, the Earth exerts only a vertical force on masses A and B. By "vertical" I mean perpendicular to the flat surface on which B is sliding and which is firmly attached to the Earth. To a very good approximation, our three-mass system is isolated from all external forces, i.e. the Moon, Sun, Jupiter, galaxies far far away etc. exert negligible forces.

Momentum of this three-mass system is conserved in both horizontal, $x$, and vertical, $y$, directions, $$\begin{align} &\left( \Delta P_{\text{x,A}}+ \Delta P_{\text{x,B}}\right)+\Delta P_{\text{x,E}}=0 \nonumber \\ & \left( \Delta P_{\text{y,A}}+ \Delta P_{\text{y,B}}\right)+\Delta P_{\text{y,E}}=0 \nonumber \end{align}$$The Earth exerts a vertical force on A and B but no horizontal force. This means that the Earth and either mass do not exchange momentum in the horizontal direction, i.e. $\Delta P_{\text{x,E}}=0$. The two equations become,$$\begin{align} &\left( \Delta P_{\text{x,A}}+ \Delta P_{\text{x,B}}\right)=0 \nonumber \\ & \left( \Delta P_{\text{y,A}}+ \Delta P_{\text{y,B}}\right)+\Delta P_{\text{y,E}}=0 \nonumber \end{align}$$In this particular problem only the top equation is relevant to the solution.