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geerky42 Group TitleBest ResponseYou've already chosen the best response.1
I solved #10 and the answer is 109.217 m/s if this help you all.
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
I'm not sure exactly what #11 mean; I don't see how this system lost energy during collision.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
By the way, how did you find the velocity?
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
Are you telling me that the block lost energy to the spring?
 one year ago

VeritasVosLiberabit Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ 1 }{ 2 }m _{b}v _{b1}^{2}=\frac{ 1 }{ 2 }m _{b}v _{b2}^{2}+\frac{ 1 }{ 2 }kx\] was the formula I got for 10. \[\sqrt{2[\frac{ 1 }{ 2 }(.003kg)(500\frac{ m }{ s })^{2}\frac{ 1 }{ 2 }(859\frac{ N }{ m })(.02m)]/.003kg}=v _{b2}\]
 one year ago

VeritasVosLiberabit Group TitleBest ResponseYou've already chosen the best response.0
my final answer for #10 was 494.24 m/s which makes more sense to me than 109.217 m/s
 one year ago

VeritasVosLiberabit Group TitleBest ResponseYou've already chosen the best response.0
but I'm not sure. are you sure your answer for #10 is correct?
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
Well, I submitted my answer on online homework and it said it's correct.
 one year ago

VeritasVosLiberabit Group TitleBest ResponseYou've already chosen the best response.0
ok, I'm going to retry it then.
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
Any idea for #11, @wio ?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
oh sorry I was idle...
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
So you figured out how much energy was transferred to the block, and the remaining energy was used to calculate velocity, right?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
Did you try putting the elastic energy as energy lost?
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
I did and the answer is incorrect...
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
Give me a second...
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
What about the conservation of momentum?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
Was momentum give to the block on the spring?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
If that doesn't work then I'm really stumped here.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
It's a really unfair question in my opinion.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
@geerky42 Have you been looking at the momentum?
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
None, It doesn't work... I guess I need to ask my physics teacher a question. Thanks.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
@geerky42 I'm not getting your 109 velocity through energy methods alone
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
How did you find the speed of the block then?
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
"Sorry I just use conservation of energy to find the speed of block"
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
Yeah well the block started with no energy right?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
It got elastic energy
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
you converted that into kinetic energy?
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
Well, it moves so it should have kinetic energy.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
How did you find out how much bullet energy went into kinetic energy then?
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
Kinetic energy instantly after collision.
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
I don't see how system lost the energy, spring absorb energy from block that has been hit by bullet.
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
The surface is frictionless. I don't know what else cause it to lost energy...
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
Oh, did you try entering the change in kinetic energy of the bullet?
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
No, how can this help me?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
Because it would be the amount of energy the bullet lost
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
None, still don't work...
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
What are you getting for it though?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
No, that's the new KE of the bullet. I said the change.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
Meaning the first KE minus the new KE
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
I did this, I got this answer.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
your original KE wasn't 375?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
and new KE isn't 17?
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
Forgot to square one of the velocity, I got 357.107 J
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.2
Yes, that's what I'm getting as the change in bullet eneryg.
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
Submitted this answer; it's still wrong...
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
I guess I have to see my teacher and talk about it.
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
Thanks for your time and patience, though.
 one year ago
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