roberts.spurs19
  • roberts.spurs19
Two identical air pucks are connected by an open-wound spring which can expand or compress. The pucks move without friction on a horizontal surface. Describe and explain the subsequent motion when the pucks are pulled apart and (a) are released together (b) puck A is held still when puck B is released. Puck A is then released at the instant the spring is reduced to its natural length Thank you! (Sorry it's a lot to read)
Physics
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SOLVED
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katieb
  • katieb
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theEric
  • theEric
Hi! (a) The pucks are identical, so they have the same mass! So you pull the spring apart. They both have a "restoring force," a force that will seek to restore the spring to its natural length. So the pucks will move inward. They move symmetrically. Once they hit the natural length, there is no more restoring force. But the pucks will keep going as described by inertia. They have momentum, and they'll keep going until some force changes that. Once the pucks compress the spring to less than its natural length, the restoring force will now push them apart. And they'll slow down, start moving the opposite direction, stretch the spring beyond its natural length, and the cycle will repeat. (b) Here, the one puck is shot towards the other, to reduce the spring to its natural length. Then you release the puck. Now the puck has momentum. It compresses the spring which in turns lessens its momentum and pushes the other puck. The spring will constantly try to get to its natural length while the momentum of the pucks prevent that. And so it will be like the first example, but it will be moving to the side because the system had momentum going to the side because of the first puck being released (to the side). Before it was symmetrical. This time, the system had net momentum.
roberts.spurs19
  • roberts.spurs19
Thank you so much for your help! I understand that now!
theEric
  • theEric
Awesome! Let me know if you have any questions! Also, just a learning tip that you might or might not know: Make sure you can repeat this without looking at it. In fact, try to think of it later! It's the only way to make sure you \(\sf know\) it. It takes a bit of ability to follow it, but that alone doesn't mean you know it. You know the entirety of it when you can repeat it on your own. I figure that's relevant because there was a lot of information above. I often notice with math problems that I can agree and justify each step in a solution, but that doesn't necessarily mean I can repeat all the steps on my own. :)

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theEric
  • theEric
Take care!
roberts.spurs19
  • roberts.spurs19
Thank you I will try and answer the question again later and see if I can remember it properly! You too!
theEric
  • theEric
Cool! Thanks!

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