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geerky42

  • one year ago

Need help on #11, please. (Momentum)

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

  3. geerky42
    • one year 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
    • one year ago
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    By the way, how did you find the velocity?

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

  6. VeritasVosLiberabit
    • one year 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
    • one year 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
    • one year ago
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    but I'm not sure. are you sure your answer for #10 is correct?

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

  10. VeritasVosLiberabit
    • one year ago
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    ok, I'm going to retry it then.

  11. geerky42
    • one year ago
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    Any idea for #11, @wio ?

  12. wio
    • one year ago
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    oh sorry I was idle...

  13. wio
    • one year 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
    • one year ago
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    Did you try putting the elastic energy as energy lost?

  15. geerky42
    • one year ago
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    I did and the answer is incorrect...

  16. wio
    • one year ago
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    Give me a second...

  17. wio
    • one year ago
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    What about the conservation of momentum?

  18. wio
    • one year ago
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    Was momentum give to the block on the spring?

  19. wio
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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    If that doesn't work then I'm really stumped here.

  23. wio
    • one year ago
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    It's a really unfair question in my opinion.

  24. wio
    • one year ago
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    @geerky42 Have you been looking at the momentum?

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

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

  27. geerky42
    • one year 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
    • one year ago
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    How did you find the speed of the block then?

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

  30. wio
    • one year ago
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    Yeah well the block started with no energy right?

  31. geerky42
    • one year ago
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    Right.

  32. wio
    • one year ago
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    It got elastic energy

  33. wio
    • one year ago
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    you converted that into kinetic energy?

  34. geerky42
    • one year ago
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    Well, it moves so it should have kinetic energy.

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

  36. geerky42
    • one year 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
    • one year ago
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    Kinetic energy instantly after collision.

  38. geerky42
    • one year 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
    • one year ago
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    The surface is frictionless. I don't know what else cause it to lost energy...

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

  41. geerky42
    • one year ago
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    No, how can this help me?

  42. wio
    • one year ago
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    Because it would be the amount of energy the bullet lost

  43. geerky42
    • one year ago
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    None, still don't work...

  44. wio
    • one year ago
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    What are you getting for it though?

  45. geerky42
    • one year ago
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    17.143 J

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

  47. wio
    • one year ago
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    Meaning the first KE minus the new KE

  48. geerky42
    • one year ago
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    I did this, I got this answer.

  49. wio
    • one year ago
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    your original KE wasn't 375?

  50. wio
    • one year ago
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    and new KE isn't 17?

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

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

  53. geerky42
    • one year ago
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    Submitted this answer; it's still wrong...

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

  55. wio
    • one year ago
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    yeah sorry

  56. geerky42
    • one year ago
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    Thanks for your time and patience, though.

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