conceptual physics

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pizza1994

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Momentum is conserved in which of the following cases?

A. Two automobiles breaking as they collide
B. A falling object colliding with another falling object
C. Two ships bumping into one another
D. Two disks colliding while moving across an air track

how do you do this...?

DrknoSDN

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D seems like the one they are aiming for but B would work if the objects were at terminal velocity.

Conservation of momentum just implies that initial total momentum = final total momentum.

For this question just look for the most elastic collision. Movable rigid objects usually have conservation of momentum because of how short a time they are in contact.
Objects that don't touch but can still repel each other (magnets or electric charges) have perfectly elastic collisions.

1bls2g

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Uh...I thought momentum was always conserved...? KE is only conserved is elastic collisions, but all collisions have conservation of momentum.

DrknoSDN

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Uh...I thought momentum was always conserved...? KE is only conserved is elastic collisions, but all collisions have conservation of momentum.
These scenarios are not isolated systems. You can lose momentum to work, or heat, friction, etc.

If you assume a large enough system then yes total momentum is always conserved. That's like saying when you jump, you're actually pushing the world down slightly, but given the context you just pick the best of the options given.

1bls2g

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You're describing loss of kinetic energy, not loss of momentum. Momentum can't be lost to heat.

Edit: to clarify, the only time it seems you do not conserve momentum is when some force acting on the objects alters their velocities. For example, in B, like DrknoSDN stated, they would have to be at terminal velocity because otherwise gravity would continue increasing their velocities and overall kinetic energies even right up until their collision and afterwards. If they are falling at different but constant terminal velocities, the one coming from the top has a faster velocity but when it collides it imparts an impulse that transfers the same amount of momentum out of itself that it does into the second, slower object. Therefore, the net change in momentum is zero. Essentially, there is no external force, but the internal forces are equal and opposite. If there were still an external force imparting an impulse, like gravity, then momentum would be changing and not conserved.

I think the most important thing is that the two objects travel at a constant velocity right before and right after the collision. In other words, no external force is providing acceleration in one direction or another. Then you can determine the comparison between momentum right before and right after collision.

In this case, two automobiles breaking as they collide may have conservation of momentum. The automobiles have momentum vectors in the opposite directions, so if their momentum magnitudes are equal (even if they have different masses they can have different initial velocities to make their overall magnitude of momentum equal) and then they collide and dissipate all kinetic energy, their momentum is conserved. However, if they had different magnitudes of momentum and were still traveling head on, their initial momentums would not cancel each other out and the initial momentum would not equal zero, so if all their kinetic energy were lost and there was no velocity at all after the collision, there would be no conservation of momentum. However, if by "breaking" it means that the cars break into tons of little pieces, like an exploding bullet, then the pieces' momentum vectors right after collision would add up to their initial car's momentum vector right before collision.

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DrknoSDN

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Yes, you can never really lose momentum. It is a fundamental law. But if a question asks "which scenario is momentum conserved?" like OP, then you much choose systems in a way where 3 of the 4 scenarios don't have conserved momentum. It's all about how you set up the system for each answer choice.

The easiest way to find conservation of momentum is to look at conservation of kinetic energy. If total kinetic energy is the same before and after the impact and no mass was lost or gained, then that's probably the best choice.

You're describing loss of kinetic energy, not loss of momentum. Momentum can't be lost to heat.
Momentum is predominantly lost due to friction. Just google "conservation of momentum friction".

http://teacher.nsrl.rochester.edu/phy_labs/Conservation_Laws/Conservation_Laws.html
"Despite their fundamental nature, the conservation laws are often difficult to observe in ordinary experiences, primarily because of the presence of friction. Friction between moving bodies and their surroundings means there are external forces acting on the system, therefore, the conservation laws do not apply. So, to observe the conservation laws, friction must be eliminated as much as possible."

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section3.rhtml
"The conservation of momentum only applies to systems that have no external forces acting upon them. We call such a system a closed or isolated system: objects within the system may exert forces on other objects within the system (e.g., the cue ball can exert a force on the eight ball and vice versa), but no force can be exerted between an object outside the system and an object within the system. As a result, conservation of momentum does not apply to systems where friction is a factor."

How would you explain rolling a ball on a flat surface and it coming to a stop? No impact, and yet momentum went from positive to zero.
The kinetic energy of the ball was transferred to vibrational energy of the molecules of the surface as heat.
Answer D has the lowest loss of momentum on impact because the object is rigid, the time the disks are in contact is extremely small, and the friction lost to the table is very small because it's an air track.

BerkReviewTeach

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Momentum is conserved in which of the following cases?

A. Two automobiles breaking as they collide
B. A falling object colliding with another falling object
C. Two ships bumping into one another
D. Two disks colliding while moving across an air track

how do you do this...?

Just a bit of a pet peeve in terms of etiquette here, but you should reference the question for two reasons: (1) so anyone who has not done that question does not get it spoiled by the explanation here and (2) it's intellectual property and should be referenced. Assuming you are not taking the lecture course from which this came, the point of this question was to emphasize that the typical physics question assumes that essentially frictionless environments are (a) disks on air tracks, (b) ice skating, and (c) billiard balls. That was a significant part of the class discussion of this question and the point the lecturer is trying to make.

D seems like the one they are aiming for but B would work if the objects were at terminal velocity.

You have a great understanding of so many things that it's downright impressive. I always appreciate your answers. The caveat you presented here is an aside that is actually mentioned in the course lecture when discussing this question. It is done so to make a point about exception hunting to make an answer choice work. It involves comparing an answer that is generally or universally true (in this case D) with an answer that is true only under a special-case condition (in this case B). Choice B, as you've presented it in terms of two objects at terminal velocity colliding, is a special case designed to make an answer correct that is otherwise not correct. The better answer choice is the one that is generally true rather than true only under a specific condition. That is one of the take-home messages of this question, as well as a few others, in the lecture.

To go a step further, your special case can be invalidated by a special case of that special case where if they are falling at the same terminal velocity they will never collide. Or, if the higher object has a lesser terminal velocity than the lower object, then they will never collide. They can only collide if the terminal velocity of the lower object is less than the terminal velocity of the higher object, in which case at the point of impact the drag force up would increase and in all likelihood there would in fact be a net force present until the terminal velocity of the combined pair (falling together after collision) reaches a new terminal velocity. In addition, there is a drafting factor that will come into play as the upper object enters the wake of the lower object, which in all likelihood will remove it from a state of terminal velocity (because of the reduced drag force resulting from less collisions with the air). So your caveat can only hold true for a femtosecond when two objects are at their respective terminal velocities in a uniform gaseous environment where buoyancy factors are equal and the object at higher elevation has a greater velocity at the point of impact than the lower object and the upper object cannot experience a change in its drag force due to drafting off of the lower object just prior to the collision. As long as all of that is true, then choice B can be valid.

Note to 1bis2g: The OP made a slight error when copying the answer choices from the handout. It is 'braking' and not 'breaking' in choice A.

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1bls2g

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Yes, you can never really lose momentum. It is a fundamental law. But if a question asks "which scenario is momentum conserved?" like OP, then you much choose systems in a way where 3 of the 4 scenarios don't have conserved momentum. It's all about how you set up the system for each answer choice.

The easiest way to find conservation of momentum is to look at conservation of kinetic energy. If total kinetic energy is the same before and after the impact and no mass was lost or gained, then that's probably the best choice.

Momentum is predominantly lost due to friction. Just google "conservation of momentum friction".

http://teacher.nsrl.rochester.edu/phy_labs/Conservation_Laws/Conservation_Laws.html
"Despite their fundamental nature, the conservation laws are often difficult to observe in ordinary experiences, primarily because of the presence of friction. Friction between moving bodies and their surroundings means there are external forces acting on the system, therefore, the conservation laws do not apply. So, to observe the conservation laws, friction must be eliminated as much as possible."

http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section3.rhtml
"The conservation of momentum only applies to systems that have no external forces acting upon them. We call such a system a closed or isolated system: objects within the system may exert forces on other objects within the system (e.g., the cue ball can exert a force on the eight ball and vice versa), but no force can be exerted between an object outside the system and an object within the system. As a result, conservation of momentum does not apply to systems where friction is a factor."

How would you explain rolling a ball on a flat surface and it coming to a stop? No impact, and yet momentum went from positive to zero.
The kinetic energy of the ball was transferred to vibrational energy of the molecules of the surface as heat.
Answer D has the lowest loss of momentum on impact because the object is rigid, the time the disks are in contact is extremely small, and the friction lost to the table is very small because it's an air track.

I think we're more or less on the same page - I agree with you and stated in my other reply that there should be no net external forces because otherwise momentum is not conserved. That includes friction. However, when you were saying momentum is lost to heat, that didn't make sense. It's the kinetic energy that is lost to heat. Momentum can be lost due to the net external force, but it's the energy that is converted from one form to another. You cannot make momentum into heat.

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DrknoSDN

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To go a step further, your special case can be invalidated by a special case of that special case where if they are falling at the same terminal velocity they will never collide. Or, if the higher object has a lesser terminal velocity than the lower object, then they will never collide. They can only collide if the terminal velocity of the lower object is less than the terminal velocity of the higher object, in which case at the point of impact the drag force up would increase and in all likelihood there would in fact be a net force present until the terminal velocity of the combined pair (falling together after collision) reaches a new terminal velocity. In addition, there is a drafting factor that will come into play as the upper object enters the wake of the lower object, which in all likelihood will remove it from a state of terminal velocity (because of the reduced drag force resulting from less collisions with the air). So your caveat can only hold true for a femtosecond when two objects are at their respective terminal velocities in a uniform gaseous environment where buoyancy factors are equal and the object at higher elevation has a greater velocity at the point of impact than the lower object and the upper object cannot experience a change in its drag force due to drafting off of the lower object just prior to the collision. As long as all of that is true, then choice B can be valid.
Thank you for the reply and compliment. =)

To simplify B I was looking at that scenario similarly to the air track in choice D. If you have 2 billiard balls falling with equal vertical positions (neither one is above or below the other), and identical terminal velocities, one ball could translate into the other ball, strike it, and conserve momentum. The terminal velocity of two identical balls would be the same. The velocity in the vertical direction would be constant (no acceleration), but one ball could hit the other horizontally. The upward force of air resistance emulates an air track that is frictionless and if you pushed one ball into the other total momentum would be conserved after the impact.

It may be easier to think of the scenario as one of those "skydiving simulators" where you are not falling, but large fans (air track) push enough air upward where the force of air resistance perfectly balances the gravitational force. If you had a perfectly non turbulent air-track (actual freefall at terminal velocity) then you could slide balls around and they would be free to collide with highly conserved momentum as if they were levitating, very similar to choice D.

I mostly mentioned B, because freefall at terminal velocity is similar to choice D. You just need a lot more air.

1bls2g

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Interesting that the non-isolated systems are tested on the MCAT. My actual physics course only covered isolated systems :/ So it's a bit harder to remember to think if there's any frictional/net external forces on the system.