Consider 3 cases where momentum is not conserved

[SaintJude]

So the total momentum of a system is conserved as long as no external forces act on the system. Easy enough to remember...but I don't understand why certain scenarios are not conserved momentum situations (my fortune cookie yesterday said "I need understanding" )

The following situations are cases where momentum is NOT conserved. Why?

A block of a certain mass m collides with another block off M that was attached to a spring, and then they stuck together and oscillated after the collision.

2 pendulums, one held at 90o to the left and one held at 90o to the right. They are let go and collide at the bottom (zero degree). is momentum conserved? the answer is no because gravity is acting on both pendulums.

How can friction and gravitation force prevent conservation of momentum?

Two guns, separated by 5m and pointed at each other are tilted so that each gun points at an angle theta above the horizontal. Both guns are fired simultaneously, with each bullet leaving its gun at a speed of 250m/s. The bullets collide inelastically in midair, falling together to the ground.

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[milski]

It's not a matter of what forces are acting, it's just a matter of including all object exerting forces in the system. For example, the two pendulums have gravity acting on them. That's exerted by the Earth itself. Either you need to include the Earth in the system and account for the transfer of momentum from the pendulum to the Earth or you have to consider the Earth as an external force. The former has no external forces and has a constant momentum, the later does not since gravity is an external force. Same type of thinking can be applied for the other two examples.

Think about a canary in a cage, then let the canary out of the cage but in your room. If you treat the cage as the system, the canary is not preserved. But if you expand your system to include more objects (your room), the canary is preserved.

 

[MedPR]

To simplify (maybe) what milski said, momentum is always conserved if you choose the right system.

 

[milski]

To simplify (maybe) what milski said, momentum is always conserved if you choose the right system.

Yes, if it's not conserved, you need a bigger system.

 

[SaintJude]

But on the MCAT/practice exam they won't tell you "consider the ball and the Earth as part of the system"....Then what?

The answer in each case were

B. Neither the momentum nor the kinetic energy was conserved.

 

[chiddler]

"2 pendulums, one held at 90o to the left and one held at 90o to the right. They are let go and collide at the bottom (zero degree). is momentum conserved? the answer is no because gravity is acting on both pendulums."

this is conserved, isn't it? if they collide and then bounce back up in an elastic collision.

it is not conserved if they just bump together as they drop and just stop = inelastic collision.

 

[MedPR]

"2 pendulums, one held at 90o to the left and one held at 90o to the right. They are let go and collide at the bottom (zero degree). is momentum conserved? the answer is no because gravity is acting on both pendulums."

this is conserved, isn't it? if they collide and then bounce back up in an elastic collision.

it is not conserved if they just bump together as they drop and just stop = inelastic collision.

Yup, momentum would be conserved even if they stuck together and kept moving in one direction.

I don't understand how momentum isn't conserved though. Even if they were to hit each other and stop at the bottom, momentum might still be conserved. Say they both had mass 10kg and speed 10m/s. The initial momentum of the system would be 0, since the two masses are traveling at the same speed but opposite direction and if the final velocity was 0, then momentum is conserved.. right?

 

[SaintJude]

Ah, some guy on another website has answered my question:

If gravity causes objects to fall, but the Earth is not mentioned in the problem, then AAMC would claim, confusingly, that momentum is not conserved.

So, by including only objects in the system, they are implicitly defining the system. I guess that's common sense.

 

[MedPR]

Ah, some guy on another website has answered my question:



So, by including only objects in the system, they are implicitly defining the system. I guess that's common sense.

That's a lame answer. That's like if the only value they give you in a free fall problem is the delta h value and don't tell you "it is on earth" and expect you to calculate the time. Any of the given answers could be true since they don't tell you what the acceleration is.

Idk, I guess it makes sense, but I think it would be worded better on the mcat.

 

[chiddler]

Yup, momentum would be conserved even if they stuck together and kept moving in one direction.

I don't understand how momentum isn't conserved though. Even if they were to hit each other and stop at the bottom, momentum might still be conserved. Say they both had mass 10kg and speed 10m/s. The initial momentum of the system would be 0, since the two masses are traveling at the same speed but opposite direction and if the final velocity was 0, then momentum is conserved.. right?

no because their momentum has dissipated into heat. if it was conserved then the final momentum must be the sum of their initial momentum. what you wrote, initial is not equal to final.

It's not a matter of -10 + 10 = 0. It's a matter of 10 + 10 = 20 total momentum initial.

 

[MedPR]

no because their momentum has dissipated into heat. if it was conserved then the final momentum must be the sum of their initial momentum. what you wrote, initial is not equal to final.

It's not a matter of -10 + 10 = 0. It's a matter of 10 + 10 = 20 total momentum initial.

Huh? If momentum is conserved, mvi=mvf. If two objects are traveling in opposite directions, one has a positive velocity and the other has a negative velocity.

If two objects with mass 50kg and speed 10m/s collide head on, they will either collide elastically or inelastically. If elastic, they will both be going 10m/s after the collision, just in the opposite direction. If completely inelastic, they will stick together and not move because momentum is conserved.

Initially m1v1+m2v2 = 50(10)+50(-10)=0.

In Elastic:
final m1v1+m2v2= 50(-10)+50(10)=0

In Inelastic:

final m1v1+m2v2 = 100(0)=0

 

[milski]

That's a lame answer. That's like if the only value they give you in a free fall problem is the delta h value and don't tell you "it is on earth" and expect you to calculate the time. Any of the given answers could be true since they don't tell you what the acceleration is.

Idk, I guess it makes sense, but I think it would be worded better on the mcat.

It's not really lame. Consider that when you have the pendulums just released at the top, their total momentum is 2mv, pointed downwards. Just before they hit each other, the momentum of the system is close to 0. Any change of something significant to zero is generally non-trivial and cannot be ignore. The only explanation for the 'loss' of momentum in this case is to consider the gravity force as an external force.

 

[milski]

no because their momentum has dissipated into heat. if it was conserved then the final momentum must be the sum of their initial momentum. what you wrote, initial is not equal to final.

It's not a matter of -10 + 10 = 0. It's a matter of 10 + 10 = 20 total momentum initial.

Momentum cannot dissipate into heat. Only energy can convert back and forth into heat. Momentum is always preserved. If the total moment is changed, you have not included all the objects that participate in the system.

 

[MedPR]

It's not really lame. Consider that when you have the pendulums just released at the top, their total momentum is 2mv, pointed downwards. Just before they hit each other, the momentum of the system is close to 0. Any change of something significant to zero is generally non-trivial and cannot be ignore. The only explanation for the 'loss' of momentum in this case is to consider the gravity force as an external force.

Why is the momentum close to 0 just before they collide?

Before you drop the pendulum bobs, all of their energy is potential right, mgh? So when they hit the bottom, all of the energy is kinetic, 1/2mv^2. So they went from 0 velocity to maximal velocity.

Oh.. but since they both have maximal velocities and in opposite directions, they are about to cancel each other out in order to conserve momentum, which initially was 0.


But the problem says momentum is NOT conserved.. So I'm still confused.

 

[milski]

Why is the momentum close to 0 just before they collide?

Before you drop the pendulum bobs, all of their energy is potential right, mgh? So when they hit the bottom, all of the energy is kinetic, 1/2mv^2. So they went from 0 velocity to maximal velocity.

Oh.. but since they both have maximal velocities and in opposite directions, they are about to cancel each other out in order to conserve momentum, which initially was 0.


But the problem says momentum is NOT conserved.. So I'm still confused.

You seem to be confusing energy and momentum. These are two different things. Energy is a scalar, KE+PE and will stay constant in that scenario.

Momentum is a vector, m*v. When you are releasing them at 90 degrees, the momentum of each is mv, pointed downward. The total momentum of the system is 2mv, since they're parallel. Just before they hit the momentum of each pendulum is pointed directly towards the other pendulum and the total moment of the system is 0 since they cancel each other.

When you consider the collision you normally talk about the moment right before and right after. Most of the time you can ignore any other changes in momentum since that's a very short period and treat the momentum as a constant. If you extend that 'moment' further as the pendulums get further away from each other this assumption becomes more and more incorrect.

 

[MedPR]

You seem to be confusing energy and momentum. These are two different things. Energy is a scalar, KE+PE and will stay constant in that scenario.

Momentum is a vector, m*v. When you are releasing them at 90 degrees, the momentum of each is mv, pointed downward. The total momentum of the system is 2mv, since they're parallel. Just before they hit the momentum of each pendulum is pointed directly towards the other pendulum and the total moment of the system is 0 since they cancel each other.

When you consider the collision you normally talk about the moment right before and right after. Most of the time you can ignore any other changes in momentum since that's a very short period and treat the momentum as a constant. If you extend that 'moment' further as the pendulums get further away from each other this assumption becomes more and more incorrect.

I wasn't confusing them, I was just trying to use them to explain it to myself. I get what you are saying, but the problem says that they are dropped from 90deg, which to me implies they are dropped from rest. So initial momentum = 0.

Here is why I was using energy conservation. If we assume no air resistance, and consider only one of the pendulums at a time, don't we see that just before it hits 0 it is at maximal velocity? If it is at maximal velocity, how can its momentum be approaching 0? I understand that when they hit their individual momentums will cancel out because they have the same mass and are moving in opposite directions, but just before they hit, don't they have maximal momentum?

 

[milski]

I wasn't confusing them, I was just trying to use them to explain it to myself. I get what you are saying, but the problem says that they are dropped from 90deg, which to me implies they are dropped from rest. So initial momentum = 0.

Here is why I was using energy conservation. If we assume no air resistance, and consider only one of the pendulums at a time, don't we see that just before it hits 0 it is at maximal velocity? If it is at maximal velocity, how can its momentum be approaching 0? I understand that when they hit their individual momentums will cancel out because they have the same mass and are moving in opposite directions, but just before they hit, don't they have maximal momentum?

You're right about the moment being 0 at top but that's more of a coincidence than anything else. Consider the momentum when they are at 45 degrees down. Some of the moment will cancel but m*v*sqrt(2)/2 of it will be pointed down and will add to a total of mv*sqrt(2).

We are discussing the moment of the system - that means the moment of everything in it, in other words the sum of the moments of the two pendulums. Even when their velocity is at max, they are virtually in opposite direction, making the total momentum of the system mv + (-mv) = 0.

So you have a change of momentum from 0 to sqrt(2)mv to 0 again. You cannot explain that without involving an external force or including Earth as part of the system.

 

[chiddler]

Momentum cannot dissipate into heat. Only energy can convert back and forth into heat. Momentum is always preserved. If the total moment is changed, you have not included all the objects that participate in the system.

it can when it is an inelastic collision which is what medpr described, isn't it:

"stuck together and kept moving in one direction."

A completely elastic collision does not make this possible, so kinetic energy can be lost to heat.

 

[milski]

it can when it is an inelastic collision which is what medpr described, isn't it:

"stuck together and kept moving in one direction."

A completely elastic collision does not make this possible, so kinetic energy can be lost to heat.

Momentum will be preserved, regardless of the type of collision. You may have to include the spring but if they stick together, they will continue in the initial direction at a speed somewhere between the speed of the two.

The only way to transfer momentum is impulse, which is f*t. From Newton's 3rd law whenever you have a force, you have the opposite force on the other body. Whatever impulse one of the bodies is gaining, the other is losing, making the total sum a constant.

Energy may be preserved or lost.

 

[MedPR]

Momentum will be preserved, regardless of the type of collision. You may have to include the spring but if they stick together, they will continue in the initial direction at a speed somewhere between the speed of the two.

The only way to transfer momentum is impulse, which is f*t. From Newton's 3rd law whenever you have a force, you have the opposite force on the other body. Whatever impulse one of the bodies is gaining, the other is losing, making the total sum a constant.

Energy may be preserved or lost.

Where does the energy go when momentum is not conserved?

 

[milski]

Where does the energy go when momentum is not conserved?

Momentum is always preserved.

Not sure what you mean with the question?

Here is something of a big picture, maybe it will help. There is a universe and objects in it. As they interact, they transfer momentum between themselves. Whenever momentum is transferred, it's never lost. So while some subsets may gain or lose some momentum, if you look at a high enough level, all of the momentum still stays there and is accounted for.

Mechanical energy (PE+KE) is different - as objects transfer it around between themselves, they lose some of it as heat. Even if you account for the whole universe, at some point you convert mechanical energy to heat and in that sense you lose mechanical energy.

 

[chiddler]

oh.

that makes sense. which is why as you wrote it depends on the system you're looking at when you say it is or is not conserved.

 

[MedPR]

Momentum is always preserved.

Not sure what you mean with the question?

Here is something of a big picture, maybe it will help. There is a universe and objects in it. As they interact, they transfer momentum between themselves. Whenever momentum is transferred, it's never lost. So while some subsets may gain or lose some momentum, if you look at a high enough level, all of the momentum still stays there and is accounted for.

Mechanical energy (PE+KE) is different - as objects transfer it around between themselves, they lose some of it as heat. Even if you account for the whole universe, at some point you convert mechanical energy to heat and in that sense you lose mechanical energy.

Ah I see. Energy is conserved if you look at a high enough level, but mechanical energy is not. Like total energy, momentum is conserved if you go high enough as well.

Thanks!

Again, if you're not getting a 15 on PS, I'm probably going to get a 5.

 

[chiddler]

Ah I see. Energy is conserved if you look at a high enough level, but mechanical energy is not. Like total energy, momentum is conserved if you go high enough as well.

Thanks!

Again, if you're not getting a 15 on PS, I'm probably going to get a 5.

yeah. that's a threat milski!

 

[MedPR]

yeah. that's a threat milski!

milski is going to find a bunch of errors on his (or her) mcat and the aamc is going to give him/her a 20

 

[milski]

Ah I see. Energy is conserved if you look at a high enough level, but mechanical energy is not. Like total energy, momentum is conserved if you go high enough as well.

Thanks!

Again, if you're not getting a 15 on PS, I'm probably going to get a 5.

Exactly. Energy has different forms and can 'mutate' between them, while momentum is stuck as it is and cannot hide.

 

[milski]

milski is going to find a bunch of errors on his (or her) mcat and the aamc is going to give him/her a 20

Only Chuck Noris can do better.

 

[SaintJude]

Just quick recap question for the MCAT pendulum scenario:

In this problem, the Earth is not part of the defined system. So the answer would be "The momentum is not conserved and neither is the kinetic energy." right?

 

[milski]

Just quick recap question for the MCAT pendulum scenario:

In this problem, the Earth is not part of the defined system. So the answer would be "The momentum is not conserved and neither is the kinetic energy." right?

That's technically correct. The kinetic energy loss is really minimal so in most problems you will not account for that and will use that the total mechanical energy is constant. (preserved).