# Change in internal energy of particle colliding with piston

• maxim07
maxim07
Homework Statement
A gas with particles of mass m and number density n is in a closed cylinder with a piston of area A. The cylinders initial length is L0. Th probability distribution of velocities in the x direction is p(v).

At time t the piston moves with velocity -u in the x direction. The piston ives much slower than the mean speed of the particles. The volume of the cylinder is reduced according to V= A(L0-ut). Show that the additional momentum from the piston is -2mu and the change in energy of the particles is 2muv. Where v is the velocity of a particle after colliding with the piston. Hint: consider a collision in the piston reference frame.

I am familiar with kinetic theory and know how to derive P = 1/3nmv^2 from a similar set up, but I’m not sure how to work out the momentum in this one.
Relevant Equations
p = mv Homework Helper
The question has nothing to do with the actual kinetic theory. It is just a matter of ellastic collisions. You need to apply the usual treatment of an ellastic collision and take the limits indicated (m<<M and u<<v).

maxim07 but then how can I use that to find the internal energy

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View attachment 301563
but then how can I use that to find the internal energy
Please review the "LaTeX Guide" link below the Edit window, so that you can post your future math equations using LaTeX. Thank you.

maxim07
The advice made sense I just thought it only referred to the part of the question about momentum not internal energy, does the limit u <<v mean I need to use the binomial expansion since for kinetic energy you have v2, which in this case would be (v+u)2 ?

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Apply the results for elastic collision to find the final velocity of the particle and then you can use this velocity to calculate both momentum and kinetic energy in the initial and the final state and the change in both.

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The advice made sense I just thought it only referred to the part of the question about momentum
Well, then, I suggest you take the advice. Have you reviewed the topic of elastic collisions in an introductory college-level textbook?

maxim07
So applying energy conservation in an elastic collision I’d have 1/2mv2 + 1/2Mu2 = 1/2mvf2 + 1/2Mu2 if I consider the energy of the piston too. Where vf is the final velocity of the particle. But then I’d get vf= v so that must be wrong

Homework Helper
You keep guessing rather than using the equations. The final velocity of the particle is not u (in magnitude). You need to use both momentum and KE conservation. Or just look up the formulas for final velocities.

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A few thoughts...
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The question implies that the momentum and kinetic energy of the piston are fixed - because the piston’s mass (M) and velocity (-u) are both fixed.

If so, this is not a standard elastic collision between 2 free objects (piston and particle).

The particle (reflected from the moving piston) gains additional kinetic energy from somewhere! It will increase your understanding if you can identify where the particle’s additional kinetic energy comes from (and how momentum is conserved).
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The piston’s velocity is give as ‘-u’, suggesting 'u' represents speed and the particle moves in the -x direction. The reflected particle must then also move in the -x direction. But the reflected particle’s velocity is given as ‘v’ rather than ‘-v’ which seems inconsistent. It might be best to regard the reflected particle's final velocity as -v.
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A minor point - the title “internal energy of particle” is not great. The term ‘internal energy’ (of a gas) means the sum of the kinetic (and potential) energies of all the individual particles in the gas.

But this question relates to the momentum and kinetic energy of only a single particle.
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To solve this, use the hint given at the end of the problem.

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Rather than taking the piston's velocity as "fixed" you can just see what happens in the limit m/M >>1 where m is the mass of a molecule and M is the mass of the piston. The formulas for elastic collision applies as for any other collision.

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Rather than taking the piston's velocity as "fixed" you can just see what happens in the limit m/M >>1 where m is the mass of a molecule and M is the mass of the piston.
You mean M/m>1.

The formulas for elastic collision applies as for any other collision.
Edited.

I agree it will work. But there is something I'm uncomfortable with.

Consider the overall system. To push a piston at constant speed into a cylinder of gas requires external work to be done on the piston. In fact the required power to the piston must continuously increase during the process.

Formulating the problem as a 2-body-collision between isolated objects (piston and particle) with no external forces acting, therefore isn’t correct. However, the 2-body-collision approach converges to the correct answer for M/m>>1.

In fact the mass of the piston is irrelevant. Because the piston's velocity is kept constant, the particle interacts with the piston as if the piston has infinite mass. That's why the 2-body collision approach works.

It’s also worth noting another fault in the question. The question should state that the piston has a negligible thermal capacity and/or the process is adiabatic. If this is not the case, some of the incident particle’s kinetic energy will be lost heating the piston.

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Homework Helper
You are right about the overall system. If you consider all the molecules then the mass is comparable with the mass of the piston and the piston will slow down unless acted by a force. But the question in the OP is about one single collision event and not about the overall system. It is a simple homework problem.

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You are right about the overall system. If you consider all the molecules then the mass is comparable with the mass of the piston and the piston will slow down unless acted by a force. But the question in the OP is about one single collision event and not about the overall system. It is a simple homework problem.
We agree. It’s just that my preferred approach is to use the hint given at the end of the question: “consider a collision in the piston reference frame”.

In the piston reference frame, the particle’s x-velocity is simply reversed. The initial and final particle velocities in the lab’ frame (in terms of u and v) follow from that.

maxim07
Probably going to need someone to show me the first lines of workings for this, still can’t seem to make it work

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Probably going to need someone to show me the first lines of workings for this, still can’t seem to make it work
All motion is in x-direction only.

1.You throw a 1kg ball at 2m/s to the right (+x) at a wall. The ball hits the wall and bounces elastically back.
What are:
a) the ball’s initial and final velocities?
b) the ball’s change in momentum?
c) the ball’s change in kinetic energy?
d) out of interest, explain how momentum has been conserved.

2. It turns out that the wall is mounted on motorised wheels(!). It starts to move left (towards you) with speed 3m/s.
You throw the ball again with speed 2m/s at the moving wall.
What are:
a) the ball’s initial and final velocities?
b) the ball’s change in momentum?
c) the ball’s change in kinetic energy?

3. What is the increase in the ball’s change of momentum due to the wall moving?

4. What is the increase in the ball’s kinetic energy due to the wall moving?

maxim07
All motion is in x-direction only.

1.You throw a 1kg ball at 2m/s to the right (+x) at a wall. The ball hits the wall and bounces elastically back.
What are:
a) the ball’s initial and final velocities?
b) the ball’s change in momentum?
c) the ball’s change in kinetic energy?
d) out of interest, explain how momentum has been conserved.

2. It turns out that the wall is mounted on motorised wheels(!). It starts to move left (towards you) with speed 3m/s.
You throw the ball again with speed 2m/s at the moving wall.
What are:
a) the ball’s initial and final velocities?
b) the ball’s change in momentum?
c) the ball’s change in kinetic energy?

3. What is the increase in the ball’s change of momentum due to the wall moving?

4. What is the increase in the ball’s kinetic energy due to the wall moving?
1
a) u = 2 ms-1v = -2ms-1
b) Δp = -2mu = -4 kgms-1
c) 0
d) F = dp/dt from Newton’s 2nd law, is it conserved because there’s and equal and opposite force acting on the hall by the wall ?

2
a) u = 2ms-1, v = -5 ms-1
b) -7 kgms-1
c) 22.5 J

3. -3 kgms-1
4. 22.5 J

i think v in 2 a is definitely wrong, but if I use momentum conservation I just get v= 2ms-1, because i take the walls final velocity to still be -3, so it’s initial and final momentum cancels

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1
a) u = 2 ms-1v = -2ms-1
b) Δp = -2mu = -4 kgms-1
c) 0
d) F = dp/dt from Newton’s 2nd law, is it conserved because there’s and equal and opposite force acting on the hall by the wall ?

2
a) u = 2ms-1, v = -5 ms-1
b) -7 kgms-1
c) 22.5 J

3. -3 kgms-1
4. 22.5 J

i think v in 2 a is definitely wrong, but if I use momentum conservation I just get v= 2ms-1, because i take the walls final velocity to still be -3, so it’s initial and final momentum cancels
1a, 1b, 1c – agreed.

1d: I would explain something like this: since the ball’s change of momentum is -4kgm/s, the wall+earth (they are attached) changes momentum by +4kgm/s (so the net change for the system is zero).

(Of course, since the mass of the wall+ Earth is so big, their velocity-change is imperceptibly small.)
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2a – disagree. I would think of it like this:

The expression for change of momentum (what you call ‘-2mu’) only applies in the frame of reference of the wall.

In the frame of reference of the moving wall (e.g. as measured by someone sitting on the wall) the ball appears to approach with velocity 5m/s. The wall-observer sees the ball rebound elastically, i.e. with velocity -5m/s.

Converting back to the ground frame of reference (which is what the question requires) a ground-observer sees the ball with:
initial velocity = 2m/s
final velocity = -8m/s (because (-5)+(-3) = -8).

maxim07
Okay, but then for 2 b wouldn’t I get Δp = -8 - 2 = -10 Kgm/s, am i supposed to be getting Δp = 6 Kgm/s as this is -2mu = -2(1)(-3) as the question requires?

Homework Helper
The 2mu is the additional momentum from the wall. This is what you say in the OP. With fixed wall the change in momentum is 4 units. With moving wall is 10. Or, in magnitude, initial momentum is 2 and final is 8. Either way, the effect of moving wall is 6 units extra.

maxim07
okay I see, then to get the change in internal energy ΔE = 1/2m(v+2u)2 - v2) = 2muv if you consider u << v, where (v+2u) is the velocity of the particle after the collision

Homework Helper
okay I see, then to get the change in internal energy ΔE = 1/2m(v+2u)2 - v2) = 2muv if you consider u << v, where (v+2u) is the velocity of the particle after the collision
Yes, for kinetic energy the change is 2muv as first approximation.