Physics problem! (momentum and energy)

[Jittles]

Thank you ergeorge you made the connection I couldn't think of.



A 5.4g bullet (m1) is fired at 345 m/s at a stationary 2.93kg block (m2) attached to a spring with k = 2000 N/m. The spring compresses 1.4 cm and the bullet comes through the block, what is the velocity of the bullet after it leaves the block?

Here is a really really sad illustration of it.


What I did:
m1v1 = (m1 + m2)v2 <---- solve for v2
1/2 (m1 + m2) v2^2 = .59 <-- plug in v2
1/2 k x^2 = .39 <--- spring energy
.59 - .39 = .2 <----- kinetic E - spring E
.2 = 1/2 m1 vf^2 <----- energy difference (loss) = kinetic energy of bullet after it exits the block.
vf = 12.1 m/s

Can anybody confirm this or show me where I went wrong? I think I did it correctly but I confused myself thoroughly while doing it and I'm not sure anymore.

 

[Jittles]

Ok I don't know what just happened with my other post of a blank message and all. This is where the physics fun is at.

 

[GroundZero]

n

 

[Jittles]

Originally posted by: GroundZero
n

alright maybe somebody else can offer a little bit more insight?

 

[Moonbeam]

I think you have to calculate the energy it would require to compress a spring to that extent. Once you know that you can figure out how fast a spring compressed with that force would accelerate m2 over the distance it was compressed. You would then know it's effective momentum and that that was supplied by the bullet. You can then calculate the original momentum of the bullet and subtract the momentum of M2 and then calculate the velosity of m1 required to have that momentum. But this is just a guesss since I'm a liberal.

PS what you did means nothing to me since I don't know the forumlas you are using or what you ae solving for.

 

[Eli]

Originally posted by: Moonbeam
I think you have to calculate the energy it would require to compress a spring to that extent. Once you know that you can figure out how fast a spring compressed with that force would accelerate m2 over the distance it was compressed. You would then know it's effective momentum and that that was supplied by the bullet. You can then calculate the original momentum of the bullet and subtract the momentum of M2 and then calculate the velosity of m1 required to have that momentum. But this is just a guesss since I'm a liberal.

PS what you did means nothing to me since I don't know the forumlas you are using or what you ae solving for.

:Q

 

[Jittles]

Oh yeah I'm using p = mv (momentum) and kinetic energy K = 1/2 m v^2 and spring energy 1/2 k x^2

 

[freakflag]

I think you have to calculate the energy it would require to compress a spring to that extent. Once you know that you can figure out how fast a spring compressed with that force would accelerate m2 over the distance it was compressed. You would then know it's effective momentum and that that was supplied by the bullet. You can then calculate the original momentum of the bullet and subtract the momentum of M2 and then calculate the velosity of m1 required to have that momentum. But this is just a guesss since I'm a liberal.

You forgot to take into account friction on the bullet from passing through the block.
But then, you liberals are always forgetting something.

 

[Jittles]

Originally posted by: freakflag

I think you have to calculate the energy it would require to compress a spring to that extent. Once you know that you can figure out how fast a spring compressed with that force would accelerate m2 over the distance it was compressed. You would then know it's effective momentum and that that was supplied by the bullet. You can then calculate the original momentum of the bullet and subtract the momentum of M2 and then calculate the velosity of m1 required to have that momentum. But this is just a guesss since I'm a liberal.

You forgot to take into account friction on the bullet from passing through the block.
But then, you liberals are always forgetting something.

Please help solve the physics problem above ignore the argument nothing to see there just ignore them yes move along. And friction doesn't matter. That is essentially what caused the block to compress the spring, and that is where some energy might be lost but I can't find a way to account for it.

 

[freakflag]

Please help solve the physics problem above ignore the argument nothing to see there just ignore them yes move along. And friction doesn't matter. That is essentially what caused the block to compress the spring, and that is where some energy might be lost but I can't find a way to account for it

You need to know the dimensions of the block and of the bullet. With the dimensions and mass you can calculate the density of the block. With the dimensions of the bullet, you can calculate the surface area that comes into contact. With all of that, you can calculate the friction. There's a little more to it to be 100% accurate, but, that's close enough for a class.

 

[Yossarian]

I first have to ask whether you have a permit for the firearm in question??!!

 

[Moonbeam]

Jittles, you don't need the density of the block or any of that crap. It's an idealized physics problem. What did you think of my solution. Was it understandable or what you did? You need to calculate the momentum of M2 at it's original location it it were accellerated from the compression point by the force supplied by the spring. That will give you the momentum given up by the bullet regardless of the size or density of the mass. Once you have the momentum given up, you can calculate the new velocity since you know the mass of the bullet. NO?

 

[Moonbeam]

You could probably do the same thing by determining the kenetic energy of the bullet and then determine how far that much energy would depress the spring. The actual distance of derpession over the total possible depression should equal the new velocity over the old.

 

[freakflag]

Jittles, you don't need the density of the block or any of that crap. It's an idealized physics problem. What did you think of my solution. Was it understandable or what you did? You need to calculate the momentum of M2 at it's original location it it were accellerated from the compression point by the force supplied by the spring. That will give you the momentum given up by the bullet regardless of the size or density of the mass. Once you have the momentum given up, you can calculate the new velocity since you know the mass of the bullet. NO?

See...that's so typical of liberals. Always willing to shortcut the facts just to get the desired answer.

What you really need to remember Jittles, if you really want to succeed, is that are no true short cuts in anything. Short cuts are all about what you're willing to give up to get what you want.
That's why there aren't really many liberal physicists.
In physics, you deal with immutable laws of nature. There are only right answers and answers that lead to more questions. There are no shortcuts.

 

[Moonbeam]

Speaking of immutable laws, mass is mass and density is freakflag.

If you shot a steel bullet into a magnetic field in a vacuum and the magnet field acted on the bullet to slow it to the extent it subtracted from its momentum that momentum imparted to a 2.93kg block accellerated frictionlessly by a spring of 2000N/m compressed 1.4cm, you would get the same answer. A spring doesn't care a whit about the density it accellerates, only the mass. It can be ten girl scouts or the equivalent uranium, they are all going to have the same effect on a spring scale and all come flying off at the same velocity if gravity suddenly disappeared.

 

[freakflag]

frictionlessly

That's the key word. There IS friction when passing through a "block". So, how can you calculate friction without the density of the material the bullet is passing through?

?

And how can you calculate the true exit velocity without factoring the friction?

But then, you'd only need that to get an ACCURATE, real world solution.
In an idealized problem, I guess it really wouldn't matter.

 

[freakflag]

A spring doesn't care a whit about the density it accellerates, only the mass.

Oh, and, we're talking about decelleration here. Just like a liberal to change the facts, though.

 

[Moonbeam]

And how can you calculate the true exit velocity without factoring the friction?
------------------
Easy, frictionless bullet. And besides the density still doesn't matter. Less dense longer trip through, more time for friction. AHAHAHAHAHA

So did I get a cigar. I'm meeting Monica tonight.

 

[Armitage]

Originally posted by: Jittles

Originally posted by: freakflag

I think you have to calculate the energy it would require to compress a spring to that extent. Once you know that you can figure out how fast a spring compressed with that force would accelerate m2 over the distance it was compressed. You would then know it's effective momentum and that that was supplied by the bullet. You can then calculate the original momentum of the bullet and subtract the momentum of M2 and then calculate the velosity of m1 required to have that momentum. But this is just a guesss since I'm a liberal.

You forgot to take into account friction on the bullet from passing through the block.
But then, you liberals are always forgetting something.

Please help solve the physics problem above ignore the argument nothing to see there just ignore them yes move along. And friction doesn't matter. That is essentially what caused the block to compress the spring, and that is where some energy might be lost but I can't find a way to account for it.

Friction does matter because come of that energy will go into heat instead of compressing the spring.
But it's really more then friction ... you are physically damaging the block ... bending & tearing metal/plastic whatever. That takes energy.

But I suspect this is an idealized problem where that sort of thing is neglected.

For what it's worth, we did an experiment similar to this in highschool ... suspended a telephone book as a pendulum and shot it with a crossbow & .22 short. Data was captured was a polaroid camera with the lens held open, and an led taped to the phone book. From that, you were able to figure out roughly the velocity of both projectiles (again, neglecting friction/heat, and the energy expended in tearing through the phone book).

My physics teacher rocked!
Could never do that kind of experiment in today's environment, and our schools are much poorer for it.

 

[freakflag]

Easy, frictionless bullet. And besides the density still doesn't matter. Less dense longer trip through, more time for friction. AHAHAHAHAHA

What??? Liberal gibberish. Sheesh.

Could never do that kind of experiment in today's environment, and our schools are much poorer for it.

Totally agreed. It's sad, really.

 

[Moonbeam]

^ Can we get this man an answer?

 

[jaydee]

Lol, poor kid...

 

[Spyro]

Originally posted by: jaydee
Lol, poor kid...

 

[jaydee]

As a Libertarian, my official stance on the issue, is that everyone in your school should aquire a gun and try it out for themselves and record their own data. Make sure the kindergarteners aim carefully...

BTW, I have to agree, that is a really, really sad illustration.

I'm working on it, give me a few minutes.

 

[jaydee]

What I did:
m1v1 = (m1 + m2)v2 <---- solve for v2
1/2 (m1 + m2) v2^2 = .59 <-- plug in v2
1/2 k x^2 = .39 <--- spring energy
.59 - .39 = .2 <----- kinetic E - spring E
.2 = 1/2 m1 vf^2 <----- energy difference (loss) = kinetic energy of bullet after it exits the block.
vf = 12.1 m/s

Can anybody confirm this or show me where I went wrong? I think I did it correctly but I confused myself thoroughly while doing it and I'm not sure anymore.

Ok, either I'm a moron, or 1/2k*x^2 = 0.196 J (k=2000 N/m, x=0.014m). I think you forgot to divide by 2 there.

I think overall you're wrong, it's not that simple. For conservation of energy, you have the bullet originally, the energy "absorbed" by the spring, the kinetic energy of the moving block (m2), and the kinetic energy of the bullet as it exits the block.

Your 2 unknowns are going to be the velocity of the block after striking, and the bullet after exiting, you can solve for that, because you also know that momentum is conserved (back up to the previous sentence with "momentum" in place of "energy").

I'm on vacation, I don't feel like solving for the 2 unknowns, but you should be able to take it from there, or anyone else interested can continue.