# momentum

Discussion in 'MCAT Study Question Q&A' started by chiddler, 04.24.12.

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1. ### chiddler 5+ Year Member

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With a massless spring + block attached to end of spring, momentum is not conserved, is it? Because as the spring compresses, the massless spring does not absorb the momentum since it is massless. But it gives it back to the mass as it expands.

Is this right?

3. ### kasho11

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I think you're mixing up energy and momentum. As long as there are no external forces, momentum is conserved. If this is a vertical spring where gravity affects it, momentum is not conserved. If it is a frictionless horizontal spring with no external forces, momentum is conserved.

4. ### chiddler 5+ Year Member

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it's a TBR passage and here's the basic setup. maybe i'm missing the point:

Frictionless surface, horizontal spring. A block slides into a spring with a mass on its tip and attaches to it. The spring begins to oscillate.

Considering momentum and potential energy of the two masses, after the oscillation begins:
Answer: momentum is not conserved, but potential energy is.

I understand PE; their height isn't changing. I'm trying to understand momentum, though.

OH IS IT BECAUSE WE ARE ONLY CONSIDERING THE MASSES AS THE SYSTEM? and therefore the spring is not part of the system!

RIGHT?

5. ### kasho11

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Yes, it depends on what you define as the system.

6. ### milski1K member 5+ Year Member

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Momentum is always preserved. If it seems that it is not, your system is not big enough and something is not included as part of it.

In your example, if you account for the momentum of whatever the spring is attached to at its other end (Earth?), you'll still find that momentum is preserved. Of course, that's not very useful for any practical calculations but that's a whole different story.

7. ### chiddler 5+ Year Member

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ok. i misunderstood why momentum is not conserved in this question. TBR didn't provide any explanation.

yes i still remember your explanation a while back with medpr and myself. but in this specific case, i was trying to rationalize it given the system consisted of spring + masses. the realization is that the question is only considering masses.

thanks for responses.

8. ### milski1K member 5+ Year Member

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Not sure what you're getting at about the masses. A real spring with a mass won't change much on its own - eventually, it will compress and the momentum of the spring and the object will become zero.

I don't want to confuse you if you already understand what's going on but I don't exactly get your point about masses.

9. ### chiddler 5+ Year Member

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i was thinking p=mv. if m = 0, then there is no momentum in an ideal spring as it compresses.

10. ### milski1K member 5+ Year Member

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True. But there's no momentum in a real compressed spring either - v=0 in that case. Just spring+object is not the whole system (from momentum conservation point of view), be it ideal (massless) or not.

11. ### chiddler 5+ Year Member

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12. ### pfaction 5+ Year Member

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Wait what, how is momentum not conserved here? Momentum is always conserved....

13. ### milski1K member 5+ Year Member

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It is, you need to consider the object attached on the other end of the spring as part of the system. If you're looking only at the spring itself and the object hitting it, you'll have a point where the total momentum is 0.

14. ### RetailBoy 2+ Year Member

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I think you and BerkTech are talking about same problem:

"As Rabolisk has pointed out, momentum is not conserved when there is a net force acting on the system. As he also stated, this concept is built into Newton's second law. Basically, if a force is acting on an object (or system), then the object (or system) is experiencing an acceleration, which means it's velocity is changing. If the linear velocity changes for an object of constant mass, then the linear momentum changes. (The linear part is not the key part here, but for the system in question it will help).

Once the moving block collides with the stationary block resting against the spring, the two blocks will move together against the resistive force of the spring. That means there is a force acting on the two-block system (namely the restoring force of the spring). That force is against the motion of the blocks, so the blocks will slow down, and thereby lose velocity (and momentum). Momentum was not conserved after collision, because of the spring force acting on the system (F = -kx). "