Can external forces change the momentum of an exploding system of particles?

[HWGXX7]

Hey,

question about the work done onto a system of particles. I have to include the work of the internal force because the particles itself can undergo different displacements. I don't understand this very well..Is this only possible if the body is distorted ?

ty

 

[tiny-tim]

Hey HWGXX7!

… question about the work done onto a system of particles. I have to include the work of the internal force because the particles itself can undergo different displacements. ..Is this only possible if the body is distorted ?

More or less.

Suppose you have water in a tube, and you press down on it to increase the pressure …

water is pretty nearly incompressible, so would you say that it was distorted?

But generally speaking, if the internal forces in a body change, there must be distortion, for example a beam that is under tension will bend, or will get shorter.

 

[HWGXX7]

General case, it's very logically indeed that the body will be distorted. So an explosion for example is an extreme case where the body initially fall apart and each subparticle will undergo the same internal force. But why doesn't travel all the particle the same traject?

I suppose that de internal forces are the same 'during' the explosion...Perhaps too complex to explain in this fashion..

ty

 

[jehan60188]

all of the particles don't have the same trajectory, but the sum of their momentums (mass times velocity) will be the same as the momentum of the body before the explosion

here's something to think about
work = change in kinetic energy
so, if you take a balloon at 1 ATM and 273.15 K, with zero volume, and inflate it to 22.4 liters, you've inserted 6*10^23 molecules into it. each molecule has a velocity and mass, so it has kinetic energy
Ke = .5*mv^2
assume all particles have the same m and v
sum of Ke = 6*10^23*.5*m*v^2
so when you inflate a balloon, you do work on it (since the Ke of the system has changed)

now- take a hermetically sealed rigid box, and put a CO2 cartridge in it, open the cartridge, and observe the pressure change in the box.
internal work

 

[HWGXX7]

but the sum of their momentums (mass times velocity) will be the same as the momentum of the body before the explosion

If the body was first in rest, the momentum was zero. After the explosion the sum = 0. So the the 'exploding object' will first rise in the sky, because some particle must have momentum towards earth.


But what if I can exert external force to counteract the internal forces, which creates the explosion.
Prinicple of work and energy states: that the change in kinetic energy equals the sum of work done by external forces and interal forces. So zero in this case, logic.

Momentum states that external work done on een object will change it's momentum. So an exploding body at rest/constant velocity doesn't change momentum (as you stated) .

But if I push the exploding (moving) body I do work on it so I change the momentum of the total body (now sum of particles).

Is this somehow a correct interpretation?


ty