Conservation of Momentum for Magnets

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Homework Statement



Given 2 magnets of equal weight (w0) and each with a magnetic strength (t0) placed inside a straight hollow cylindrical tube. One of the magnets is attached to one end of the tube and the other magnet is free to move inside the tube. The other end of the tube is open to space/vacuum.

The tube is travelling in space with a velocity (v0) such that it is oriented vertically towards direction of motion. The end with the magnet attached is the leading end (towards direction of motion).

Consider 2 cases:

Case 1: The free flowing magnet and the tube both are travelling at the same velocity (v0) and the free flowing magnet is at the initial distance (d0) such that both magnets exert can pull on each other.

Case 2: The tube and free magnet are moving at velocity v0. Their distance of initial separation is (d1) such that the magnets exert no pull at each other (so d1 remains same unless an external force is applied). Assume, at some time (t0) an external push to the free magnet (in direction of motion) such that it is now moving with a velocity (v0+v1) and the tube at the same velocity (v0).

What does the conservation of momentum predict when both magnets finally come in contact.

Homework Equations




The Attempt at a Solution



My understanding is as follows:

Case 1: Both magnets when they physically come in contact and stick together, the tube would continue travelling in the direction of motion with the same initial velocity (v0).

Case 2: When both magnets come in contact and stick together, due to conservation of momentum the new velocity of tube would increase and would be exactly (v0+v1/2).

P.S. The tube has negligible weight so we assume it as 0.

Is my interpretation correct? Or does the resultant velocity depend on other factors like the magnetic strength too?
 

Answers and Replies

  • #2
Is my interpretation correct? Or does the resultant velocity depend on other factors like the magnetic strength too?
Yes, what you have done looks correct. You might want to explain your reasoning in case 1 though.
 
  • #3
rude man
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In case 2 the only external force was a small push to get the movable magnet moving towards the fixed one. The rest of the velocity gain of the moving magnet is due to the attractive force with the fixed magnet, meaning no change in system momentum for that part of momentum gain of the moving magnet. So the new final velocity of the tube + magnets will be only slightly greater than v0. The system momentum gain will be the time integral of the applied force over the duration of the push.
 
  • #4
haruspex
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Their distance of initial separation is (d1) such that the magnets exert no pull at each other
I struggle to understand how this can be, which makes me think I do not understand the set-up.
 
  • #5
rude man
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I struggle to understand how this can be, which makes me think I do not understand the set-up.
Think static friction ... then for case 2 it's overcome.
 
  • #6
haruspex
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Think static friction ... then for case 2 it's overcome.
It says the magnets exert no pull. What has that to do with friction?
 
  • #7
rude man
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It says the magnets exert no pull. What has that to do with friction?
There is stickiness between the magnets and the containing tube. If said stickiness exerts a greater static friction force on the magnet than the pull from the other magnet, the movable magnet will not move. Then when you give it a push, it does. Exactly same idea as static frfiction force μsmg, mutatis mutandis.
 
  • #8
haruspex
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There is stickiness between the magnets and the containing tube. If said stickiness exerts a greater static friction force on the magnet than the pull from the other magnet, the movable magnet will not move. Then when you give it a push, it does. Exactly same idea as static frfiction force μsmg, mutatis mutandis.
Yes, I understand that, but it says the magnets exert no pull. That is quite different from the net force being zero.
 
  • #9
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Yes, I understand that, but it says the magnets exert no pull. That is quite different from the net force being zero.

If the magnets are far enough the magnetic pull would be almost 0. The static friction would be the dominant force. I suppose we would have used the term 'for all practical purposes' the magnetic pull is 0. Or, the net force is 0. It should have been more clearer. Apologies.
 
  • #10
haruspex
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If the magnets are far enough the magnetic pull would be almost 0. The static friction would be the dominant force. I suppose we would have used the term 'for all practical purposes' the magnetic pull is 0. Or, the net force is 0. It should have been more clearer. Apologies.
No need to apologise if the wording you posted is exactly as it was given to you. Is it?
 
  • #11
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No need to apologise if the wording you posted is exactly as it was given to you. Is it?

Yes. The original problem was what I was given.
 
  • #12
haruspex
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Yes. The original problem was what I was given.
Then if rude man's interpretation is correct I can only say the problem is very badly worded indeed.
 
  • #13
rude man
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OP, I suggest you proceed with the solution. The wording is a bit addling but I find the scenario and intent intuitively obvious. As I said, I don't agree with your finding of case 2 and I think you should show us your argument as to why the new system speed should be v0 + v2/2.

Remember that, for any system, momentum doesn't change except for any impulse applied from outside the system: F(t)dt = Δp.
 
  • #14
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OP, I suggest you proceed with the solution. The wording is a bit addling but I find the scenario and intent intuitively obvious. As I said, I don't agree with your finding of case 2 and I think you should show us your argument as to why the new system speed should be v0 + v2/2.

Remember that, for any system, momentum doesn't change except for any impulse applied from outside the system: F(t)dt = Δp.

my intuitive reasoning was as follows: In Case 1 the velocity after the magnets stick together is (v0) i.e. no change in initial velocity. Now in Case 2, half of the system has an additional velocity of (v1) when it is pushed towards the pipe. Thus, that half has an additional momentum of (w0*v1) towards the direction of motion.

Now when the two magnets stick the weight becomes (2*w0). But the momentum gained due to the push will have to be conserved. Thus, the stuck magnets would gain an additional velocity of (v1/2). The velocity is halved as the weight doubles so the momentum due to push is conserved.

Is there a flaw in this argument (I am unable to see it)?
 
  • #15
rude man
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my intuitive reasoning was as follows: In Case 1 the velocity after the magnets stick together is (v0) i.e. no change in initial velocity. Now in Case 2, half of the system has an additional velocity of (v1) when it is pushed towards the pipe. Thus, that half has an additional momentum of (w0*v1) towards the direction of motion.
Realize that most of the extra momentum gained by the moving magnet is at the expense of the stationary magnet. The stationary magnet, along with the whole tube, have slowed down compared to before the push until contact is made.

Think again about what I said before: "Remember that, for any system, momentum doesn't change except for any impulse applied from outside the system". Assume it takes just a small nudge to get the moving magnet moving. Then, next to no external impulse (force x time) has been applied to the system (consisting of the two magnets plus the tube). So, little change in momentum to the system, and the new v is only slightly greater than the old.
 
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  • #16
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Realize that most of the extra momentum gained by the moving magnet is at the expense of the stationary magnet. The stationary magnet, along with the whole tube, have slowed down compared to before the push until contact is made.

Think again about what I said before: "Remember that, for any system, momentum doesn't change except for any impulse applied from outside the system". Assume it takes just a small nudge to get the moving magnet moving. Then, next to no external impulse (force x time) has been applied to the system (consisting of the two magnets plus the tube). So, little change in momentum to the system, and the new v is only slightly greater than the old.
[/QUOTE]

I am trying to understand but struggling a bit. The increase in velocity (v1) (as I understand) in the second magnet is due to an external force. The magnet attached to the other end of the tube would be unaffected. Thus, after the external force/impulse/nudge is applied, the tube should be travelling at the velocity v0 and the other magnet at v0+v1.

I am unclear where this line/approach is wrong.
 
  • #17
rude man
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I am trying to understand but struggling a bit. The increase in velocity (v1) (as I understand) in the second magnet is due to an external force.
. No , the increase in velocity of the movable magnet is mostly due to the attractive force of the other magnet. Now, admittedly, if it took a lot of 'push' to get the movable magnet moving, then the new v would be closer to v0 + v1/2, might be more even. But, no way would it be exactly v0 + v1/2.
The magnet attached to the other end of the tube would be unaffected.
. No, it would be affected, along with the tube. Newton's law: "for every action a reaction"!

Try to understand my last paragraph in post 15.
 
  • #18
rude man
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Here's maybe a helpful picture: at first, the free magnet is stuck and the external force F is pushing on it to get it loose. During that time the tube accelerates from v0 to some higher v = v2 given by Fdt = M(v2-v0) where M is the total system mass (both magnets plus the tube).

Then, the free magnet starts to move and the external force is removed. Pretend you're situated alongside the tube, halfway between the magnets, at the system c.g. (assuming they have equal mass and the tube itself is massless), and moving at constant speed v2, and you can see inside the tube. What you see is the two magnets moving toward each other until they collide. But you're always situated halfway between the magnets. So during the time the magnets are moving towards each other you're actually moving faster than the tube since the tube itself is actually slowing down from your view since it moves with the fixed magnet. The c.g. of the system does not move with respect to you. That's the important point. It always moves at v = v2, even after the magnets have collided.

Finally, the magnets collide and the tube moves at v = v2 again. The tube has traveled backwards a total distance d1/2 with respect to your viewpoint.
 

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