Physics problem! (momentum and energy)

[Armitage]

Originally posted by: 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.

The block is motionless (ie. has no kinetic energy) at the point when the spring is at its maximum compression. So your conservation of energy equation is simply:

KE0 = PE + KE1

KE0: initial kinetic energy of the bullet = 0.5 M1 V0^2 (V0: initial velocity, M1: mass of the bullet)
PE: Potential energy of the spring at its max compression = 0.5 K x^2
KE1: Kinetic energy of the bullet after leaving the block = 0.5 M1 V1^2

0.5 M1 V0^2 = 0.5 K x^2 + 0.5 M1 V1^2

M V0^2 - K x^2 = M V1^2

V1 = sqrt((M1 V0^2 - K x^2)/M1)

V1 = sqrt((0.0054*345^2 - 2000*0.014^2)/0.0054)

V1 = 344.89 M/s

Note that the mass of the block is irrelevent because we are working the problem at a point when it is at rest.

As I stated in a previous post, this is neglecting some important bits like friction, and the energy spent tearing a hole through the block and deforming the bullet.

 

[jaydee]

The block is motionless (ie. has no kinetic energy) at the point when the spring is at its maximum compression.

No where does he state either of those conditions have to be true. The block does not have to be motionless, because the spring isn't necesarily at it's maximum compression. Right? Why else we would we be given the mass of the block?

I don't know. If this is a high school problem, you're probably right egeorge, they do have a lot stupid questions like this. In college calc-physics though, we are NEVER given 'too-much' information, and the problems are typically of this caliber.

 

[Armitage]

Originally posted by: jaydee

The block is motionless (ie. has no kinetic energy) at the point when the spring is at its maximum compression.

No where does he state either of those conditions have to be true. The block does not have to be motionless, because the spring isn't necesarily at it's maximum compression. Right? Why else we would we be given the mass of the block?

I don't know. If this is a high school problem, you're probably right egeorge, they do have a lot stupid questions like this. In college calc-physics though, we are NEVER given 'too-much' information, and the problems are typically of this caliber.

He states that the "spring compresses to 1.4 cm"
Unless we have alot more information about the problem. At least the velocity of the block at that point. I think we have to take that as the maximum compression.

That point of maximum compression probably occurs at some time after the bullet has left the block, ie. the block still has some forward velocity at the time the bullet exits it. If we neglect air drag on the bullet, that's OK, because the block on the spring has its own conservation of energy going on. Say that, at the time the bullet leaves the block, the spring is only compressed 1cm. If the max compression is 1.4 cm, we can solve for the velocity of the block at the time the bullet exits:

0.5 M2 Ve^2 + 0.5 K Xe^2 = 0.5 K Xf^2

Ve: Velocity of the block at the time the bullet exits.
Xe: Compression of the spring at the time the bullet exits
Xf: Max compression of the spring

So, as long as the bullet has left the block at the time of max compression (which I think is almost certain), it doesn't matter when max compression occurs because the block on the spring is now an isolated system with its own conservation of energy going on.

If that's not the maximum compression of the spring, I don't think you have enough information about the problem.

 

[Moonbeam]

The spring is motionless because it's at it maximum extension. We need the mass of the target because the momentum it acquires to be compressed is exactly the momnetum it would acquire at the rest point when it recoils. The momentum necessary to supply that compression is supplied by the bullet and its exit velocity can be found by subtracting that amount of momentum from it's original momentum and reverse engineering the velocity from that new momentum at that mass.

First we have to find what velocity a spring of k=2000N/m compressed 1.4cm will accelerate a mass of 2.93 kg

Once we know the velocity we can find it's momentum at that velocity.

Once we have that momentum we can find the momentum of the bullet before hitting the target.

Once we have the original momentum we can subtract the momentum imparted to the target to get the momentum the bullet retained after leaving the target.

Once we have the new momentum of the bullet, since we know its mass we can find its new velocity.

I don't know or understand understand the formulas you gave. See, if you can, whether this doesn't give the answer.

 

[Moonbeam]

It doesn't matter when the bullet left the block. It imparts that energy required to compress a spring a distance. From that you can compute the momentum it imparted, exactly that the compressed spring will reaccelerate the block on rebound. Take that momentum from the bullet (subtract), which is where it came from and you have its new momentum. Find new velocity from that. NO?

 

[Armitage]

Originally posted by: Moonbeam
The spring is motionless because it's at it maximum extension. We need the mass of the target because the momentum it acquires to be compressed is exactly the momnetum it would acquire at the rest point when it recoils. The momentum necessary to supply that compression is supplied by the bullet and its exit velocity can be found by subtracting that amount of momentum from it's original momentum and reverse engineering the velocity from that new momentum at that mass.

First we have to find what velocity a spring of k=2000N/m compressed 1.4cm will accelerate a mass of 2.93 kg

Once we know the velocity we can find it's momentum at that velocity.

Once we have that momentum we can find the momentum of the bullet before hitting the target.

Once we have the original momentum we can subtract the momentum imparted to the target to get the momentum the bullet retained after leaving the target.

Once we have the new momentum of the bullet, since we know its mass we can find its new velocity.

I don't know or understand understand the formulas you gave. See, if you can, whether this doesn't give the answer.

If you do it with energy, you don't need to know the velocity or mass of the block, as long as you know the maximum compression of the spring.

 

[Armitage]

Originally posted by: Moonbeam
It doesn't matter when the bullet left the block.

Correct, so long as the bullet has left the block, and you know the maximum compression of the spring..
That's what I demonstrated above.

 

[Moonbeam]

Why would you not need the mass of the block? A light block would have compressed much more. The bullet has to overcome the inertia of the block and compress the spring. Both of those are taken into account by knowing the momentum of the block on recoil to staarting point. It's like a bow. It will shoot a light arrow fast and a heavy ons more slowly. The momentum of each, however, will be the same at the same draw length.

 

[Armitage]

Originally posted by: Moonbeam
Why would you not need the mass of the block? A light block would have compressed much more. The bullet has to overcome the inertia of the block and compress the spring. Both of those are taken into account by knowing the momentum of the block on recoil to staarting point. It's like a bow. It will shoot a light arrow fast and a heavy ons more slowly. The momentum of each, however, will be the same at the same draw length.

Yes, exactly, but instead of firing the arrow, we look at the energy in the bow.

The equation for the energy of a spring-mass system is:

E = 1/2 M V^2 + 1/2 K X^2

M: Mass
V: Velocity
K: Spring constant
X: Spring displacement

Energy is conserved, E is a constant.

So, you need to know both x and v to solve for E
But, for one special case, we already know v. At the point of maximum compression v=0
In your example, this is the point at which you've drawn your bow, but not released it.
The energy in that system is the same regardless of whether you have the light or heavy arrow notched.

Back to the energy equation...
If v=0, the term containing mass drops out, and you know the energy of the system simply by the displacement without regard to the mass of the system.

 

[jaydee]

A light block would have compressed much more.

That is false. We are given the spring constant, so mass is independant.

Now, if we were asked to find k by reverse Hooke's law, then mass would matter, but as it stands now, k is given.

The bullet has to overcome the inertia of the block and compress the spring. Both of those are taken into account by knowing the momentum of the block on recoil to starting point. It's like a bow. It will shoot a light arrow fast and a heavy ons more slowly. The momentum of each, however, will be the same at the same draw length.

In order to calculate the momentum of the block on recoil, you need the velocity. So what you're saying, is we need Hooke's law F=-kx, find the force, get the acceleration of the block, but we'd still need the time it took to recoil to compute the velocity using that method. Am I right?

There's also the SHM displacement equation: X(t)=Xm*sin(wt+phi), but we'd need the angular frequency (phase constant 'phi' would be 0), which we can find thanks to w=[k/m]^(1/2), so maybe he can work something along that route. Then taking the derivative of X(t) is velocity. You can shortcut that by saying V(t)=-w*Xm. So that perhaps would work.

If this is a HS problem, the above is all gibberish, because you shouldn't ever need to use SHM displacement, velocity, acceleration equations in friggin HS.

 

[jaydee]

eeerg. egeorge is right. My last post is kinda moot because

1/2k*x^2 = recoil = 1/2m(block)*v(block)^2

I think I'm making this vastly more complicated than it needs to be.

 

[Moonbeam]

I don't think you need time, though. A spring of known springiness should accelerate the same mass to the same velocity in the same time every time. That's got to be in the force of the spring somewhere I'd think. I don't know any of these laws. I was trying to suggest that the way to understand the problem was to assume that the momentum imparted to m2 could be subtracted away form m1 to find the answer and that that momentum would be the equivalent of what the spring would reimpart to m2 as it recoiled or decompressed. I don't know if that's right, but it's what I thought.

I see the point about the energy of the bow, by the way, but I still don't exactly see why the mass is irelevant. The energy of a bow is usually thought of, I guess, without regard to the weight of the string since it's negligible. But if the string woere neutronium you would have to be Samson to pull it and if you shot an arrow it might take all day to leave the bow. The bullet didn't just compress the spring as its energy contribution. It had to accelerate a mass. In other words there is potential energy in the compressed spring, but also potential energy in the displaced mass. The sum of each can be known by the kenetic energy of m2 at the point where it returns on recoil to it's original position. I don't know how to find that given the mass and k and displacement, but I assume there is a way.

 

[Jittles]

OK, I think I figured it out. The key is that there is no friction as the bullet passes thru the block.

If I assume the block is super thin, then the bullet strikes the block, gives it some velocity, therefore giving it momentum and therefore losing momentum of the bullet.

I will calculate the energy of the spring compressed, convert to kinetic, find velocity. Using this velocity, find the momentum of the block, subtract from the original momentum of the bullet and find the new momentum of the bullet, divide by mass to find V.

I'll crunch it out now.

Thank you all for the replies! btw this is college physics, but it's the first level so essentially it's high school physics over again.


EDIT: I get 146.5 m/s now which seems much more reasonable than my first answer of 12 m/s.
EDIT2: CORRECT! THANK YOU! I was just missing the fact that the bullet just imparts momentum then exits and there is no friction/force throughout the block.

 

[Armitage]

Originally posted by: Jittles
OK, I think I figured it out. The key is that there is no friction as the bullet passes thru the block.

Yea, you definitely don't have enough information to mess around with the actual penetration of the block by the bullet.

If I assume the block is super thin, then the bullet strikes the block, gives it some velocity, therefore giving it momentum and therefore losing momentum of the bullet.

I will calculate the energy of the spring compressed, convert to kinetic, find velocity. Using this velocity, find the momentum of the block, subtract from the original momentum of the bullet and find the new momentum of the bullet, divide by mass to find V.

Why do you need the velocity or momentum of the block. You have the energy, that's all you need.

I'll crunch it out now.

Thank you all for the replies! btw this is college physics, but it's the first level so essentially it's high school physics over again.


EDIT: I get 146.5 m/s now which seems much more reasonable than my first answer of 12 m/s.
EDIT2: CORRECT! THANK YOU! I was just missing the fact that the bullet just imparts momentum then exits and there is no friction/force throughout the block.

Take a look at my posts again.
Forget about momentum. Do it all as energy and the mass & velocity of the block is irrelavent.

 

[Moonbeam]

How can you know the energy since the energy required to compress the spring also required energy to overcome the inertia of the block?

 

[HokieESM]

Originally posted by: ergeorge

Yea, you definitely don't have enough information to mess around with the actual penetration of the block by the bullet.

Why do you need the velocity or momentum of the block. You have the energy, that's all you need.

Take a look at my posts again.
Forget about momentum. Do it all as energy and the mass & velocity of the block is irrelavent.

Ergeorge.... it really depends a LOT on how you take the "bullet passes through" part. If the bullet passes through "instantaneously" (or a very short time), this would be a conservation of MOMENTUM problem, not conservation of energy. MOST impact (read: small delta-t events) problems are taken as momentum-conserving--because energy is usually dissipated. Recall all the "two cars of different masses crash at XX mph and they stick together, what is the velocity after the crash". This comes from the "impulsive" nature of the force (impulses, by definition, don't exhibit conservation of energy). And a "short period of time" just has to deal with the natural frequency of the system.... in this case, the bullet is moving very fast, so you would normally consider this conservation of momentum.

Now, if the problem was "a ball is dropped onto the plate, and then falls off", one could make a case for conservation of energy.


Jittles, in other words, I think the conservation of momentum bit is right. You can calculate how much velocity the block should gain through conservation of energy (the energy stored in the spring).... but the initial velocity is imparted through conservation of momentum with the bullet. Good luck!

 

[Jittles]

Ok I got the answer correct doing it my second way. There is a substantial loss of energy in the problem and that would account for the friction/bullet penetrating the block. The second part of the problem was calculate energy lost and I got that correct also.

 

[Alistar7]

Originally posted by: 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.

No cigar for you, a bullet fired at something LESS dense will take LESS time and energy to penetrate and create LESS friction than the same bullet fired at something MORE dense, which will require MORE energy and time creating MORE friction.

 

[Moonbeam]

Alistar, you're not looking at the situation. We're talking about a block and a particular mass. So the less dense the larger the block and the more time to traverse. But experiments would really have to be done to see what effect would occur. I don't really think they are proportional since something half as dense is not twice as thick. I was just pointing to thickness as an extra factor. Anyway we have to assume no friction to work the problem

 

[Alistar7]

LMAO Density and SIZE have nothing in common, I will stake my ASNT SNT-TCA-1A Level II Radiographic Certification on the line over this question.

Friction is a MAJOR factor to just let slide by in search of the "truth" here, bet your professor would have been impressed if you were able to quantify the value of the energy lost there.

 

[Moonbeam]

Well, several things here. We can't calculate the true exit velocity. In a real life test it would probably never be the same twice. There would probably be chaos factors involved in collisions at high velocities and vaporization and melting and eddies ans swirls that would never duplicate. The problem is idealized so that's why I answered "easy frictionless bullet". I then added that something less dense would have greater dimension and therefore while the bullet would pass through easier it might have equivalent or somewhat more equal friction due to the fact that a less dense block say of 500g is thicker than one of a more dense 500g configuration and therefore the bullet would be subjected to friction for a longer period of time. Density and sive are only related when you are talking about a specific mass. You can have one cubic foot of anything and they will have different masses, or you can have the same volume and the less density the less mass. I didn't say that density and size have nothing in common or mean to imply so. Also a block of kevlar might stop a bullet better than the same weight block of steel due to friction.

 

[Alistar7]

not if you are using a calibrated block, one determined to be of uniform DENSITY and without imperfection, quite possible, they are made everyday for testing standards. The largest variable would always be the load and the bullet itself. The second part was to determine energy lost, much of this lost energy is due to friction. Seems like his prof let him off easy on this one.... Thank you for clarifying, however your kevlar example is somewhat weak, let's say they both weigh ten pounds but the kevlar is made into a block a half inch thick and 2 feet tall, while the ten pound steel block is made into a 6 inch tall, 12 inch thick block, you still think that the kevlar would have more stopping power (friction)? There is a reason it requires certification to be able to test ailerons for the F22, aft engine mounts for the Boeing 777, the largest apassenger airplane ( tested 12, try not to think of that if you ever get to fly on one LOL), fuel delivery systems for RocketDyne, etc... and it doesn't include "idealizing" essential data in the testing process.

 

[Moonbeam]

To me a bolck implies a cube. You just use block when talking about some material like a target made out os something and cube when you refer specifically to shape. At any rate if you keep it in proportion less density has still got to lead to a thicker block since it will equally ex[pand l w and depth