geerky42
  • geerky42
Need help on #11, please. (Momentum)
Physics
schrodinger
  • schrodinger
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geerky42
  • geerky42
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geerky42
  • geerky42
I solved #10 and the answer is 109.217 m/s if this help you all.
geerky42
  • geerky42
I'm not sure exactly what #11 mean; I don't see how this system lost energy during collision.

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anonymous
  • anonymous
By the way, how did you find the velocity?
geerky42
  • geerky42
Are you telling me that the block lost energy to the spring?
anonymous
  • anonymous
\[\frac{ 1 }{ 2 }m _{b}v _{b1}^{2}=\frac{ 1 }{ 2 }m _{b}v _{b2}^{2}+\frac{ 1 }{ 2 }kx\] was the formula I got for 10. \[\sqrt{2[\frac{ 1 }{ 2 }(.003kg)(500\frac{ m }{ s })^{2}-\frac{ 1 }{ 2 }(859\frac{ N }{ m })(.02m)]/.003kg}=v _{b2}\]
anonymous
  • anonymous
my final answer for #10 was 494.24 m/s which makes more sense to me than 109.217 m/s
anonymous
  • anonymous
but I'm not sure. are you sure your answer for #10 is correct?
geerky42
  • geerky42
Well, I submitted my answer on online homework and it said it's correct.
anonymous
  • anonymous
ok, I'm going to retry it then.
geerky42
  • geerky42
Any idea for #11, @wio ?
anonymous
  • anonymous
oh sorry I was idle...
anonymous
  • anonymous
So you figured out how much energy was transferred to the block, and the remaining energy was used to calculate velocity, right?
anonymous
  • anonymous
Did you try putting the elastic energy as energy lost?
geerky42
  • geerky42
I did and the answer is incorrect...
anonymous
  • anonymous
Give me a second...
anonymous
  • anonymous
What about the conservation of momentum?
anonymous
  • anonymous
Was momentum give to the block on the spring?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
If that doesn't work then I'm really stumped here.
anonymous
  • anonymous
It's a really unfair question in my opinion.
anonymous
  • anonymous
@geerky42 Have you been looking at the momentum?
geerky42
  • geerky42
None, It doesn't work... I guess I need to ask my physics teacher a question. Thanks.
anonymous
  • anonymous
@geerky42 I'm not getting your 109 velocity through energy methods alone
geerky42
  • geerky42
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.
anonymous
  • anonymous
How did you find the speed of the block then?
geerky42
  • geerky42
"Sorry I just use conservation of energy to find the speed of block"
anonymous
  • anonymous
Yeah well the block started with no energy right?
geerky42
  • geerky42
Right.
anonymous
  • anonymous
It got elastic energy
anonymous
  • anonymous
you converted that into kinetic energy?
geerky42
  • geerky42
Well, it moves so it should have kinetic energy.
anonymous
  • anonymous
How did you find out how much bullet energy went into kinetic energy then?
geerky42
  • geerky42
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.
geerky42
  • geerky42
Kinetic energy instantly after collision.
geerky42
  • geerky42
I don't see how system lost the energy, spring absorb energy from block that has been hit by bullet.
geerky42
  • geerky42
The surface is frictionless. I don't know what else cause it to lost energy...
anonymous
  • anonymous
Oh, did you try entering the change in kinetic energy of the bullet?
geerky42
  • geerky42
No, how can this help me?
anonymous
  • anonymous
Because it would be the amount of energy the bullet lost
geerky42
  • geerky42
None, still don't work...
anonymous
  • anonymous
What are you getting for it though?
geerky42
  • geerky42
17.143 J
anonymous
  • anonymous
No, that's the new KE of the bullet. I said the change.
anonymous
  • anonymous
Meaning the first KE minus the new KE
geerky42
  • geerky42
I did this, I got this answer.
anonymous
  • anonymous
your original KE wasn't 375?
anonymous
  • anonymous
and new KE isn't 17?
geerky42
  • geerky42
Forgot to square one of the velocity, I got 357.107 J
anonymous
  • anonymous
Yes, that's what I'm getting as the change in bullet eneryg.
geerky42
  • geerky42
Submitted this answer; it's still wrong...
geerky42
  • geerky42
I guess I have to see my teacher and talk about it.
anonymous
  • anonymous
yeah sorry
geerky42
  • geerky42
Thanks for your time and patience, though.

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