Momentum vs energy

1. Apr 25, 2005

Mr.V.

Hi thanks for taking the time to read this. I apologize in advance for the length. Anyway, at some point in my education I allegedly learned about energy and momentum and someone even saw fit to give me an 'A', but when thinking about the simple concepts again a couple of years later I suddenly realize, I don't actually get it...

For instance let's take this problem.
Say I have a gun that fires two different bullets.
The first bullet travels at say 100m/s and has a mass of 0.1kg.
Therefore it's momentum is $$100m/s*0.1kg = 10 kg*m/s$$
and it's energy is $$1/2*0.1kg*(100m/s)^2 = 500J$$ (at least I think that's right)

The second bullet travels at say 50m/s and has a mass of 0.3kg.
Therefore it's momentum is $$50m/s*0.3kg = 15 kg*m/s$$
and it's energy is $$1/2*0.3kg*(50m/s)^2 = 375J$$

This is the part I don't get. The first bullet has more energy than the second but less momentum.

What does that mean to, say, the moose or bear or cute baby seal that get's hit by the bullet. Let's assume the bullet hits the target and goes into it but doesn't come out the side and therefore all the energy is transferred to the target. Is the momentum what determines the damage or knockdown or is it the energy?

Let's say the target is a moose that isn't moving anywhere and is 100kg.
From the conservation of momentum formula and given that this would be an inelastic collision you would expect that the first bullet would be
$$10kg*m/s=(100kg+0.1kg)*V_2$$ you could calculate that the moose+bullet1 would end up traveling ~0.10m/s in the direction of the bullet

bullet two would be $$15kg*m/s=(100kg+0.3kg)*V_2$$ you could calculate that the moose+bullet two would end up traveling 0.15m/s.

But if we say that the bullet transfers its energy (hits the moose and transfers the kinetic energy) we can see that the moose in the first bullet instance gains 500J of kinetic energy and therefore $$500J=1/2*(100kg+0.1kg)*V^2$$ and solving for V we get V=3.16m/s
and in bullet2's case we get $$375J=1/2*(100kg+0.3kg)*V^2$$ and solving for V we get V=2.73m/s.

So now I have two problems. Using the conservation of momentum equation I see that a bullet with LESS energy but MORE momentum can cause the moose to move more
But using kinetic energy I see that the bullet with MORE energy but LESS momentum causes the moose to move more.
And the HUGE problem that the velocities don't add up.
I'm clearly doing something wrong and I also don't understand the difference between momentum and energy.

Any help would be appreciated. Sorry for the drawn out example. I thought creating a senario would be more helpful than just asking a bunch of questions.

Thanks!
Vik

2. Apr 25, 2005

dx

I didnt read the whole post, but you dont seem to understand what momentum is.
"Momentum" is another word for "Quantity of motion". So, it is proportional to both velocity and the mass (because the greater the mass, the more "motion" there would be. You should understand that velocity and motion are not the same) so momentum would be written 'm x v'. So, now you easily see that momentum is very different from energy. A stationary object (i.e v = 0) would have zero momentum, but it would still have energy(E = mc^2)

3. Apr 25, 2005

dextercioby

Momentum is conserved in both elastic & plastic collisions,while energy (KE) is conserved only in the former.U've got it right when saying that the bullet with more momentum causes more damage,simply becasue we quantify "damage" by force...And force is given by the momentum transfer in unit time.Therefore a bigger force would come from the bullet with bigger momentum...(assuming both bullets come to rest,therefore u can easily compare the momentum transfer).

Daniel.

4. Apr 25, 2005

dextercioby

U needn't go to relativistic physics for that.In fact,relativistic physics unifies energy & momentum into a 4 vector (or 4 tensor).

Inclusion of forces will do the trick.

Daniel.

5. Apr 25, 2005

Mr.V.

I see...so the "energy" of the first bullet with 500J that comes to rest is lost to other things like friction/heat/etc and its momentum knocks the target backwards but to a lesser degree than the bullet with more momentum. But that bullet would have less energy transferred to friction/heat/etc than the first...

6. Apr 25, 2005

arildno

MrV:
Your last example is wrong; because you cannot assume you have conservation of energy here.
That is, not all the bullet's initial kinetic energy will afterwards be the sum of the bullet's new kinetic energy and the moose's new kinetic energy; rather, quite a bit of that initial kinetic energy has been converted into HEAT, i.e, the temperature has increased.

7. Apr 25, 2005

Mr.V.

I see arildno. I think I understand it now. Momentum is a value that is always conserved vs kinetic energy which is mutated. I must have remembered it wrong that kinetic energy was also conserved.

8. Apr 25, 2005

arildno

Just to clear up:
MOmentum is APPROXIMATELY conserved during a collision between two objects, because the EXTERNAL forces acting upon either the moose or bullet (say, gravity) are negligible in their effects compared to the collision forces BETWEEN the bullet and moose (i.e, which are internal forces in system bullet+moose).

9. Apr 25, 2005

T@P

so thats why say a magnum will hurt more than say a 9mm? because the size of the bullet is bigger? sorry about the totally off topic post, but it has a bit to do with it

10. Apr 25, 2005

dx

Sorry, I dont know much about relativity, neither do I know what a 4-vector or a tensor it. How would "Inclusion of forces" give rise to energy of a stationary object. I thought the only energy a stationary object possessed was its mass ( which, of course is an idea that can be understood only in the frame-work of relativity)

11. Apr 25, 2005

whozum

Assuming they penetrate teh same (which they dont), the momentum will be larger if its heavier or faster. I believe a magnum fires faster than a 9mm, so has a higher momentum. I think it also is heavier.

$$p = mv$$

12. Apr 25, 2005

whozum

I think he means inclusion of all forces external to the system, aka expanding the system so that everything acting on the system is included. A stationary object, (you in your chair) still has potential energy and many other forms of kinetic energy.

13. Apr 25, 2005

T@P

thanks whozum

14. Apr 26, 2005

dx

I would have potential energy if I was in a gravitational field. But, when im in a gravitational field, im not statoinary, im accelerating. When I sit in a chair, im not stationary, im accelerating.

15. Apr 26, 2005

dx

dextercioby, could u please clarify this.