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ggrree Group Title

http://gyazo.com/318bc0f81daf2cb7a082db69073c8a4e What's the best way to approach this question? (I will show you my approach in the next post)

  • 2 years ago
  • 2 years ago

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  1. ggrree Group Title
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    I thought that since both objects are on the same spring, they both experience the same force (as per Hooke's law), and since one of the objects has a smaller mass, it would experience a greater acceleration (as per newton's F =ma). therefore, the smaller object will go twice as high as the larger.

    • 2 years ago
  2. ggrree Group Title
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    what is wrong with my reasoning?

    • 2 years ago
  3. Shane_B Group Title
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    Was that the wrong answer? I don't see an issue with it. The force would be split among both objects and the one with the 1/2 mass would end up with twice the acceleration.

    • 2 years ago
  4. ggrree Group Title
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    the correct answer was C.

    • 2 years ago
  5. ggrree Group Title
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    Why should they both reach the same height?

    • 2 years ago
  6. goxenul Group Title
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    If you still care: they are both attached to the same spring, so there's no way they could rise to different heights.

    • 2 years ago
  7. CarlosGP Group Title
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    Both objects are launched at the same speed, then same height. Answer is C

    • 2 years ago
  8. ggrree Group Title
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    thanks, guys. but do you have any better explanation as to why they can't rise to different heights / why they are both launched at the same speed?

    • 2 years ago
  9. ggrree Group Title
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    The possible formuals I can work with are limited to F=ma, F =-k x, and Energy equations (kinetic, gravitational potential, elastic potenital.)

    • 2 years ago
  10. Shane_B Group Title
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    Since the answer is C, I suppose that the larger mass is limiting the acceleration of both objects. If they leave with the same acceleration (same dV/dT) they will reach the same height in the same amount of time.

    • 2 years ago
  11. goxenul Group Title
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    If they were seperately put on that spring then right, the smaller one would go higher, because, as you say, it would gain more velocity and so on. But in order to give masses a specific velocity, the spring needs to push them with that velocity. It can't push "faster" the smaller object, it pushes them both with the same speed. (That's not very formal, but rather intuitive.)

    • 2 years ago
  12. CarlosGP Group Title
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    The moment they are launched, the only acceleration that affects them is gravity, as they have been released, the spring will no longer make any force on them

    • 2 years ago
  13. Vincent-Lyon.Fr Group Title
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    You can reason with or without energy. Both objects are on the same board, hence they have the same velocity when launched. Either you 'know' an object's motion under gravity-only does not depend on its mass. Then it is pretty clear that both objects will reach the same height. Or you argue that KE is proportional to mass. Then A will have twice as much KE as B. PE is also proportional to mass, so, at the same height, A will have twice as much PE as B. When all their KE is converted in PE, A and B will then reach the same height.

    • 2 years ago
  14. ammubhave Group Title
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    Well let's reason that out. 1st fact to consider is that after both the blocks leave contact with the spring, only gravity affects both of them and we know that gravitational force will produce an accelaration independent of their masses. So the only differentiating factor would be their inital velocity with which they are released. If the initial velocities are the same, they are going to rise the same height. What does the inital velocity depend upon? Of course it depends ONLY upon how fast they were pushed up. AND NOT how hard they were pushed up. They rest on the same platform and hence they will move together therefore there is no way one can have a different inital velocity because that would mean the playform split up in two parts, one went faster up and the other slower and this is surely not the case. Hence, we concluded that they will have same inital velocity and hence will rise to the same height.

    • 2 years ago
  15. Shane_B Group Title
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    To be clear on my earlier post, the acceleration during the recoil of the spring is held constant for both loads since it's just one spring and a shared platform. Having the same acceleration over the same time interval implies their launch velocities will be equal. Once they have launched, -g is the obviously the only acceleration acting on the masses at that point.

    • 2 years ago
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