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geerky42

  • 2 years ago

Need help on #11, please. (Momentum)

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  1. geerky42
    • 2 years ago
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  2. geerky42
    • 2 years ago
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    I solved #10 and the answer is 109.217 m/s if this help you all.

  3. geerky42
    • 2 years ago
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    I'm not sure exactly what #11 mean; I don't see how this system lost energy during collision.

  4. wio
    • 2 years ago
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    By the way, how did you find the velocity?

  5. geerky42
    • 2 years ago
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    Are you telling me that the block lost energy to the spring?

  6. VeritasVosLiberabit
    • 2 years ago
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    \[\frac{ 1 }{ 2 }m _{b}v _{b1}^{2}=\frac{ 1 }{ 2 }m _{b}v _{b2}^{2}+\frac{ 1 }{ 2 }kx\] was the formula I got for 10. \[\sqrt{2[\frac{ 1 }{ 2 }(.003kg)(500\frac{ m }{ s })^{2}-\frac{ 1 }{ 2 }(859\frac{ N }{ m })(.02m)]/.003kg}=v _{b2}\]

  7. VeritasVosLiberabit
    • 2 years ago
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    my final answer for #10 was 494.24 m/s which makes more sense to me than 109.217 m/s

  8. VeritasVosLiberabit
    • 2 years ago
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    but I'm not sure. are you sure your answer for #10 is correct?

  9. geerky42
    • 2 years ago
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    Well, I submitted my answer on online homework and it said it's correct.

  10. VeritasVosLiberabit
    • 2 years ago
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    ok, I'm going to retry it then.

  11. geerky42
    • 2 years ago
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    Any idea for #11, @wio ?

  12. wio
    • 2 years ago
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    oh sorry I was idle...

  13. wio
    • 2 years ago
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    So you figured out how much energy was transferred to the block, and the remaining energy was used to calculate velocity, right?

  14. wio
    • 2 years ago
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    Did you try putting the elastic energy as energy lost?

  15. geerky42
    • 2 years ago
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    I did and the answer is incorrect...

  16. wio
    • 2 years ago
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    Give me a second...

  17. wio
    • 2 years ago
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    What about the conservation of momentum?

  18. wio
    • 2 years ago
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    Was momentum give to the block on the spring?

  19. wio
    • 2 years ago
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    Figure out the momentum before hand and figure out the momentum later. That would tell you how much momentum the block received. Compare that to the elastic energy.

  20. wio
    • 2 years ago
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    I mean, use the momentum of the block to find its velocity and figure out the kinetic energy of the block. Then compare that to the elastic energy.

  21. wio
    • 2 years ago
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    That might be what they're asking. It's true that I find it hard to tell. Obviously this is a closed system an no energy is being lost. So maybe what they mean is how much kinetic energy do we expect the block to have, which it won't have.

  22. wio
    • 2 years ago
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    If that doesn't work then I'm really stumped here.

  23. wio
    • 2 years ago
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    It's a really unfair question in my opinion.

  24. wio
    • 2 years ago
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    @geerky42 Have you been looking at the momentum?

  25. geerky42
    • 2 years ago
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    None, It doesn't work... I guess I need to ask my physics teacher a question. Thanks.

  26. wio
    • 2 years ago
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    @geerky42 I'm not getting your 109 velocity through energy methods alone

  27. geerky42
    • 2 years ago
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    Sorry I just use conservation of energy to find the speed of block after collision, I then plug the speed of block into conservation of momentum to find the speed of bullet after collision. My brain is dead, sorry about that.

  28. wio
    • 2 years ago
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    How did you find the speed of the block then?

  29. geerky42
    • 2 years ago
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    "Sorry I just use conservation of energy to find the speed of block"

  30. wio
    • 2 years ago
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    Yeah well the block started with no energy right?

  31. geerky42
    • 2 years ago
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    Right.

  32. wio
    • 2 years ago
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    It got elastic energy

  33. wio
    • 2 years ago
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    you converted that into kinetic energy?

  34. geerky42
    • 2 years ago
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    Well, it moves so it should have kinetic energy.

  35. wio
    • 2 years ago
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    How did you find out how much bullet energy went into kinetic energy then?

  36. geerky42
    • 2 years ago
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    I don't find it from bullet, I did work "backward." We know that the spring has been compressed by 2 cm and from that we can find elastic energy and then from that, we can convert it into kinetic energy. That's how I find the speed of block.

  37. geerky42
    • 2 years ago
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    Kinetic energy instantly after collision.

  38. geerky42
    • 2 years ago
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    I don't see how system lost the energy, spring absorb energy from block that has been hit by bullet.

  39. geerky42
    • 2 years ago
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    The surface is frictionless. I don't know what else cause it to lost energy...

  40. wio
    • 2 years ago
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    Oh, did you try entering the change in kinetic energy of the bullet?

  41. geerky42
    • 2 years ago
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    No, how can this help me?

  42. wio
    • 2 years ago
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    Because it would be the amount of energy the bullet lost

  43. geerky42
    • 2 years ago
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    None, still don't work...

  44. wio
    • 2 years ago
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    What are you getting for it though?

  45. geerky42
    • 2 years ago
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    17.143 J

  46. wio
    • 2 years ago
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    No, that's the new KE of the bullet. I said the change.

  47. wio
    • 2 years ago
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    Meaning the first KE minus the new KE

  48. geerky42
    • 2 years ago
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    I did this, I got this answer.

  49. wio
    • 2 years ago
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    your original KE wasn't 375?

  50. wio
    • 2 years ago
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    and new KE isn't 17?

  51. geerky42
    • 2 years ago
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    Forgot to square one of the velocity, I got 357.107 J

  52. wio
    • 2 years ago
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    Yes, that's what I'm getting as the change in bullet eneryg.

  53. geerky42
    • 2 years ago
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    Submitted this answer; it's still wrong...

  54. geerky42
    • 2 years ago
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    I guess I have to see my teacher and talk about it.

  55. wio
    • 2 years ago
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    yeah sorry

  56. geerky42
    • 2 years ago
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    Thanks for your time and patience, though.

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