Momentum Conservation in Ideal Gas & Piston Collision

AI Thread Summary
In the collision between a gas atom and a piston, momentum conservation depends on the defined system. If the atom is considered the system, momentum is not conserved due to the external impulse from the piston. However, if the system includes the piston, cylinder, and atom, momentum is conserved because internal forces cancel out. The key to understanding this is recognizing the impulse-momentum relationship, which indicates that external forces affect momentum conservation. Thus, the conservation of momentum is contingent on the system's boundaries.
qazxsw11111
Messages
95
Reaction score
0

Homework Statement


The cylinder and piston are made of a thermal insulator. An atom of a gas collides with the piston at an angle and bounces off at an angle. State with a reason whether momentum of the atom is conserved in this collision.


Homework Equations





The Attempt at a Solution


I think that momentum of atom is not conserved as it is a vector quantity and hence change in momentum when there is a change in direction of the atom velocity.

However, the answer says that it is how you consider what is the "system". If atom is the "system", momentum is changed. If "piston+cylinder+atom"=system, momentum is conserved.

Im really confused! Which one is right?
 
Physics news on Phys.org
Hi qazxsw11111,

qazxsw11111 said:

Homework Statement


The cylinder and piston are made of a thermal insulator. An atom of a gas collides with the piston at an angle and bounces off at an angle. State with a reason whether momentum of the atom is conserved in this collision.


Homework Equations





The Attempt at a Solution


I think that momentum of atom is not conserved as it is a vector quantity and hence change in momentum when there is a change in direction of the atom velocity.

However, the answer says that it is how you consider what is the "system". If atom is the "system", momentum is changed. If "piston+cylinder+atom"=system, momentum is conserved.

Im really confused! Which one is right?

I'm not sure what you are asking here; you have the same answer as the answer key. The momentum of the atom changes.

If you are examining just the atom by itself (which is what is meant by defining the system to be the atom), then during the collision there is a large external impulse on the system--from the piston. With a large external impulse, the momentum changes (is not conserved).

If your system is (piston+cylinder+atom), then there is no external impulse due to the collision, because the collision forces are between system objects (and so action-reaction forces will have cancelling impulses). So for that system momentum is conserved.

Remember that the test for conservation of momentum is the impulse momentum relationship, which for average forces is:

<br /> \vec F_{\rm ext}\Delta t=\Delta \vec p<br />

(and there is also the integral form if your problem requires it).
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top