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geerky42
 3 years ago
Need help on #11, please. (Momentum)
geerky42
 3 years ago
Need help on #11, please. (Momentum)

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0By the way, how did you find the velocity?

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1Are you telling me that the block lost energy to the spring?

anonymous
 3 years ago
Best 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. \[\sqrt{2[\frac{ 1 }{ 2 }(.003kg)(500\frac{ m }{ s })^{2}\frac{ 1 }{ 2 }(859\frac{ N }{ m })(.02m)]/.003kg}=v _{b2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my final answer for #10 was 494.24 m/s which makes more sense to me than 109.217 m/s

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but I'm not sure. are you sure your answer for #10 is correct?

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1Well, I submitted my answer on online homework and it said it's correct.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, I'm going to retry it then.

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1Any idea for #11, @wio ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh sorry I was idle...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So you figured out how much energy was transferred to the block, and the remaining energy was used to calculate velocity, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did you try putting the elastic energy as energy lost?

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1I did and the answer is incorrect...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What about the conservation of momentum?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Was momentum give to the block on the spring?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Figure 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
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0That 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
 3 years ago
Best ResponseYou've already chosen the best response.0If that doesn't work then I'm really stumped here.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's a really unfair question in my opinion.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@geerky42 Have you been looking at the momentum?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@geerky42 I'm not getting your 109 velocity through energy methods alone

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry 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
 3 years ago
Best ResponseYou've already chosen the best response.0How did you find the speed of the block then?

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1"Sorry I just use conservation of energy to find the speed of block"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah well the block started with no energy right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It got elastic energy

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you converted that into kinetic energy?

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1Well, it moves so it should have kinetic energy.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How did you find out how much bullet energy went into kinetic energy then?

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1I 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
 3 years ago
Best ResponseYou've already chosen the best response.1Kinetic energy instantly after collision.

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

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1The surface is frictionless. I don't know what else cause it to lost energy...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, did you try entering the change in kinetic energy of the bullet?

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1No, how can this help me?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because it would be the amount of energy the bullet lost

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1None, still don't work...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What are you getting for it though?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, that's the new KE of the bullet. I said the change.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Meaning the first KE minus the new KE

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1I did this, I got this answer.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your original KE wasn't 375?

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1Forgot to square one of the velocity, I got 357.107 J

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, that's what I'm getting as the change in bullet eneryg.

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1Submitted this answer; it's still wrong...

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1I guess I have to see my teacher and talk about it.

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1Thanks for your time and patience, though.
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