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