Need help on #11, please. (Momentum)

- geerky42

Need help on #11, please. (Momentum)

- schrodinger

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- geerky42

##### 1 Attachment

- geerky42

I solved #10 and the answer is 109.217 m/s if this help you all.

- geerky42

I'm not sure exactly what #11 mean; I don't see how this system lost energy during collision.

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## More answers

- anonymous

By the way, how did you find the velocity?

- geerky42

Are you telling me that the block lost energy to the spring?

- 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

my final answer for #10 was 494.24 m/s which makes more sense to me than 109.217 m/s

- anonymous

but I'm not sure. are you sure your answer for #10 is correct?

- geerky42

Well, I submitted my answer on online homework and it said it's correct.

- anonymous

ok, I'm going to retry it then.

- geerky42

Any idea for #11, @wio ?

- anonymous

oh sorry I was idle...

- anonymous

So you figured out how much energy was transferred to the block, and the remaining energy was used to calculate velocity, right?

- anonymous

Did you try putting the elastic energy as energy lost?

- geerky42

I did and the answer is incorrect...

- anonymous

Give me a second...

- anonymous

What about the conservation of momentum?

- anonymous

Was momentum give to the block on the spring?

- 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

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

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

If that doesn't work then I'm really stumped here.

- anonymous

It's a really unfair question in my opinion.

- anonymous

@geerky42 Have you been looking at the momentum?

- geerky42

None, It doesn't work... I guess I need to ask my physics teacher a question. Thanks.

- anonymous

@geerky42 I'm not getting your 109 velocity through energy methods alone

- 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

How did you find the speed of the block then?

- geerky42

"Sorry I just use conservation of energy to find the speed of block"

- anonymous

Yeah well the block started with no energy right?

- geerky42

Right.

- anonymous

It got elastic energy

- anonymous

you converted that into kinetic energy?

- geerky42

Well, it moves so it should have kinetic energy.

- anonymous

How did you find out how much bullet energy went into kinetic energy then?

- 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

Kinetic energy instantly after collision.

- geerky42

I don't see how system lost the energy, spring absorb energy from block that has been hit by bullet.

- geerky42

The surface is frictionless. I don't know what else cause it to lost energy...

- anonymous

Oh, did you try entering the change in kinetic energy of the bullet?

- geerky42

No, how can this help me?

- anonymous

Because it would be the amount of energy the bullet lost

- geerky42

None, still don't work...

- anonymous

What are you getting for it though?

- geerky42

17.143 J

- anonymous

No, that's the new KE of the bullet. I said the change.

- anonymous

Meaning the first KE minus the new KE

- geerky42

I did this, I got this answer.

- anonymous

your original KE wasn't 375?

- anonymous

and new KE isn't 17?

- geerky42

Forgot to square one of the velocity, I got 357.107 J

- anonymous

Yes, that's what I'm getting as the change in bullet eneryg.

- geerky42

Submitted this answer; it's still wrong...

- geerky42

I guess I have to see my teacher and talk about it.

- anonymous

yeah sorry

- geerky42

Thanks for your time and patience, though.

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