Moving wall and ball collision

[PhizKid]

Homework Statement

I'm only interested in the second part where V << v

Homework Equations

Energy

The Attempt at a Solution

First I boosted to the frame of the moving wall to find the change in velocity of the ball after a collision with it. In the frame of the moving wall, the ball first comes towards it with some velocity $v + V$ and leaves with some velocity $v' - V$ so by conservation of momentum, $v + V = v' - V$ (I don't need to insert a negative sign on the left hand side expression right since I already took into account the direction by indicating its velocity would increase by V in the frame of the wall?) therefore $\Delta v = 2V$. Now focusing on the stationary wall, let's say the ball bounces off the wall and travels a distance $x$ to the moving wall and then of course travels the same distance x back to the stationary wall. The time for this round trip is $\Delta t = \frac{x}{v} + \frac{x}{v + 2V} = \frac{2(vx + Vx)}{v^{2} + 2vV} = \frac{2vx(1 + \frac{V}{v})}{v^{2}(1 + 2\frac{V}{v})} = \frac{2x(1 + \frac{V}{v})(1 + 2\frac{V}{v})^{-1}}{v} \approx \frac{2x}{v}(1 + \frac{V}{v})(1 - 2\frac{V}{v}) = \frac{2x}{v}(1 - \frac{V}{v} - 2(\frac{V}{v})^{2}) \approx \frac{2x}{v} + O(\frac{V}{v})$ where I have used a binomial expansion. Then we have that $\lim_{\Delta t\rightarrow 0}\frac{\Delta v}{\Delta t} = \frac{\mathrm{d} v}{\mathrm{d} t} = \frac{vV}{x} = \frac{vV}{l - Vt}$ where the last part comes from the diagram. Integrating this and applying initial conditions gives $v = \frac{v_{0}l}{l - Vt} = \frac{v_{0}l}{x}$. The average force on the wall is $-\frac{\Delta p}{\Delta t} = \frac{2mv}{\Delta t} = \frac{mv^{2}}{x} = \frac{m(\frac{v_{0^{2}l^{2}}}{x^{2}})}{x} = \frac{mv_{0}^{2}l^{2}}{x^{3}} = \frac{mv_{0}^{2}}{l}(\frac{l}{x})^{3}$. This agrees with the book but I have three questions: was the way I found the change in velocity correct (like I said in the parenthesis at the start of the post I'm unsure about the signs), I made many approximations is that ok, and finally this is the average force on the stationary wall but the problem doesn't specify which one, will the one on the moving wall be the same? Thanks!

 

[voko]

You cannot assume simultaneously that the moving wall is unaffected by the collision with the ball and that momentum is conserved. This is most obvious when the wall is stationary, then the total momentum is that of the ball, and each collision changes the total momentum to its opposite.

Yet you have derived the correct result that the change in the speed is 2V. You should just do it correctly.

 

[haruspex]

so by conservation of momentum, $v + V = v' - V$

As voko pointed out, this does not follow from conservation of momentum. It does, however, follow from conservation of energy. And this explains why you can ignore the sign.
Your approximations are fine, but arguably you should show this right through the integration to the final result.
Each bounce off the moving wall will have a constant amount more impact than the adjacent bounces off the stationary wall. That difference becomes insignificant as v increases.

 

[PhizKid]

Hi Haruspex thanks for responding. I redid it using energy. I just wasn't sure about the approximations because I was unsure if I was allowed to drop the v / V term with respect to 1 - v/V and my only justification was that v/V << 1. Can you elaborate on your last point about integration of the final result? I was lost in that part. By final result did u mean the average force on the wall? Thanks haruspex!

 

[haruspex]

I was unsure if I was allowed to drop the v / V term with respect to 1 - v/V and my only justification was that v/V << 1.

I think you mean V/v through all that.

Can you elaborate on your last point about integration of the final result? By final result did u mean the average force on the wall?

Yes, the average force.
You had, correctly,
$\Delta t = \frac{2x}{v} + O(\frac{V}{v})$
but then took the limit and dropped the O() term before integrating. It would be more rigorous to continue to include a bound for the error all the way through; otherwise there may be doubts as to whether it continues sufficiently bounded through the integration process. But I'm not at all sure you need to do that here. That's a judgment for your tutor.

 

[PhizKid]

Oh ok. So the solution was fine then (after changing to the energy conservation)? So regarding the integration we keep the error bound and basically since v constantly increases as time goes on the error bound gets smaller and smaller so that after performing the integral we still have a finite error bound? If it is ok can you answer these two final questions: why can't we use conservation of momentum in the frame of the moving wall and in general when can I justify dropping the V / v in a 1 +- V / v term if V << v. Thank you so much.

 

[haruspex]

why can't we use conservation of momentum in the frame of the moving wall

For the same reason you can't use it with a fixed wall, or a ball bouncing off the ground. The change in momentum of the ball is matched by that in the wall and whatever else it is attached to - perhaps the Earth, ultimately. Those objects having great mass, their change in velocity is minuscule, but the momentum change is not. (Indeed, the ball's change in momentum is 2v+2V.) Their change energy depends on the square of that tiny velocity, so can be ignored.

in general when can I justify dropping the V / v in a 1 +- V / v term if V << v. Thank you so much.

The general rule is that you should only drop small terms right at the end, when you can show they are small in comparison with the leading term. At any point before that there's the risk that what you thought was the leading term will cancel, and the term you dropped becomes the important one. E.g. if in this problem you had an equation like y = ln(1-V/v), it might be inadvisable to drop the V/v. In practice, it's common to drop smaller terms earlier, when it's 'obvious' no such problem will arise.

 

[PhizKid]

So momentum conservation won't work when hitting something like a wall of very large mass because if the collision is elastic then the ball's momentum will increase? In the case of bouncing off the stationary wall it is still -2mv as I wrote above though right? So is it safe to say in physical situations like this where I just get a regular expression in the relevant quantities similar to the above with (2x/v)*(1 - V/v - 2(V/v)^2) I can usually drop a V / v term in something like 1 +- V / v ,and of course drop the second order terms if they occur too, if V << v? Thanks a ton haurspex.

 

[ehild]

Conservation of the momentum holds for the whole system, (wall+ball). The ball gets some impulse and the wall gets the same, with opposite sign. Only the mass of the wall is infinite, its velocity does not change.
If energy is conserved, you find a relation between initial and final velocities of two colliding bodies. You get the final velocity of the ball with infinite mass of the wall.
If M is the mass of the wall, and V is its velocity, those for the ball are m and v respectively, and V' and v' are the velocities after collision,

$$v'=\frac{2MV-(M-m)v}{M+m}$$
With the limit M-->∞ v'-->-v+2V. v is negative, so the speed increased by 2V.

You can also use a frame of reference fixed to the moving wall. Here the ball has -v-V velocity before the collision, and it turns to v+V after, relative to the wall.
The velocity is (v+V)+V=v+2V in the rest frame of reference, so the speed of he ball increased by 2V.

ehild

 

[PhizKid]

But I did apply it to the whole system so that's why I got confused. I was in the rest frame of the moving wall so it would be at rest. Thats why I only looked at the momentum of the ball in this frame.

Thank you very much

 

[haruspex]

But I did apply it to the whole system so that's why I got confused. I was in the rest frame of the moving wall so it would be at rest. Thats why I only looked at the momentum of the ball in this frame.

The trouble with that approach is that it presupposes the wall moves at constant velocity (otherwise it is not an inertial frame). In reality, it must be accelerated, however slightly, by the impact with the ball.

 

[PhizKid]

so can we not do it because the changes in momentum occur to first order in velocity and mass whereas the changes to kinetic energy occur to first order in mass and second order in velocity so if we assume large mass and very small velocity then kinetic energy terms can be dropped but not changes in momentum?

 

[ehild]

It is less confusing if you consider a finite-mass wall (mass M) traveling with the initial velocity V. The velocities after the collision became

$$v′=\frac{2MV−(M−m)v}{M+m}$$

$$V′=\frac{2mv+(M−m)V}{M+m}$$

At the limit of infinite mass of the wall, v'=2V-v and V'=V.

You can derive how much momentum is exchanged between the wall and ball, and what is it at the limit when M-->∞

Spoiler

2m|v-V|

and also what is the exchanged energy with the same limit.

Spoiler

2m|v-V|V

. ehild

 

[PhizKid]

Thanks haruspex and ehild. Haruspex you mentioned that each time the ball bounces off the wall, the wall's own velocity will change accordingly so that when calculating the average rate of change of the velocity of the ball by solving $\int_{v_{0}}^{v}\frac{dv}{v} = \int_{0}^{t} \frac{Vdt}{l - Vt}$, I cannot treat V as a constant as I did in the above solution because it changes over time in conjunction with the average rate of change of the ball. Wouldn't I have to change it to a term like $V + \delta {\xi(t)}$ where $\delta {\xi(t)} << V$ and V is the constant initial velocity of the wall before collisions start and $\delta \dot{\xi (t)}$ encodes how much the wall decelerates due to the recoil of the bouncing ball.

 

[ehild]

The wall can be treated as having infinite mass. Its own velocity does not change in time. It is V, although it gets some finite impulse during each collision.

ehild

 

[voko]

Wouldn't I have to change it to a term like $V + \delta {\xi(t)}$ where $\delta {\xi(t)} << V$ and V is the constant initial velocity of the wall before collisions start and $\delta \dot{\xi (t)}$ encodes how much the wall decelerates due to the recoil of the bouncing ball.

If you do this, consider the other wall as well. Treat the moving wall and the fixed wall as parts of one system, then the time-averaged change in the system's momentum should be zero. Which means you can pretty much ignore the effect of the ball on the moving wall even without assuming anything infinite.

 

[PhizKid]

ehild that is confusing me. If the wall receives an impulse how can it's velocity not change? Voko that makes sense to me in the case of both walls receiving the same impulse from the ball which in this case is thankfully true. Thanks both.

Actually voko I've confused myself again. I understand you can take the time average but how will that integral get rid of the perturbation term in the integral I typed before which only contains the velocity of the moving wall?

 

[ehild]

"Ehild that is confusing me. If the wall receives an impulse how can it's velocity not change? Voko that makes sense to me in the case of both walls receiving the same impulse from the ball which in this case is thankfully true. Thanks both."

The impulse is finite, 2m(v-V). The change of momentum is MΔV, so the change of the velocity of the wall is ΔV=2m/M (v-V). If M is infinite what is ΔV?

The first wall gets 2m(v-V) impulse, where v is the velocity of the ball, (negative). The impulse points to the left.
The ball reaches the stationary wall with velocity 2V-v, which changes to the opposite in the collision. So that wall receives 2m(2V-v) impulse (positive) different from that the first wall got.

ehild

 

[haruspex]

"Ehild that is confusing me. If the wall receives an impulse how can it's velocity not change? Voko that makes sense to me in the case of both walls receiving the same impulse from the ball which in this case is thankfully true. Thanks both."

It often happens that we treat part of a system as 'being of infinite mass'. In this case, the wall may be connected via some sequence to the Earth. Of course, it's not really infinite, but what we mean is that we can treat that part of the system as not changing in velocity. But note what this means for momentum: m δv = ∞ × 0. It's indeterminate, so it may have a change in momentum. Concealed in the 'being of infinite mass' approximation is the assumption that the change in momentum is finite. So the change in energy is zero.

 

[voko]

Actually voko I've confused myself again. I understand you can take the time average but how will that integral get rid of the perturbation term in the integral I typed before which only contains the velocity of the moving wall?

Assume that the walls are connected (so that they can transfer to each other any impulse they get) and have some finite total mass. Work out the the total impulses the system gets during "one cycle". You don't have to do any integrals, you need to consider just two collisions.

 

[PhizKid]

ehild, I'm confused as to how you got those numbers. Let's say the ball has velocity v right before colliding with the moving wall. Pi =-mv and Pf = mv + 2mV so Iwall = -Iball = -2m(v + V). It reaches the stationary wall with velocity v + 2V and rebounds approximately with the same velocity v + 2V in the opposite direction so Iwall = -Iball = -m(-v - 2V - v - 2V) = 2m(v + 2V). I really don't like this business of infinite mass is there no other way to do this? Also, haruspex I don't quite understand either of your last two sentences regarding why the change in energy of the system is zero and why the infinite mass limit implies a finite change in momentum when it was shown it would be indeterminate. Voko, in the period of one collision for the moving wall and one collision for the stationary wall, assuming they are somehow connected, the net impulse is 2m(v + 2V) - 2m(v + V) = 2m(v + 2V - v - V) = 2mV.

 

[haruspex]

haruspex, I don't quite understand either of your last two sentences regarding why the change in energy of the system is zero and why the infinite mass limit implies a finite change in momentum when it was shown it would be indeterminate.

I didn't say that the infinite mass limit implies a finite change in momentum; I said that was also being assumed. If you had two infinite masses interacting you would not be able to deduce anything, but typically you don't. Since the change in momentum of the finite part of the system is finite, the change in momentum of the infinite mass must also be finite.
If a constant mass has a change of momentum, Δp = mΔv, then its change in energy ΔE = Δ(mv²/2) = mvΔv = pΔp/m. If p and Δp remain finite in the limit as m→∞ then ΔE = 0.

 

[PhizKid]

Thanks haruspex. I understand the last point. So for the first point, the change in momentum of the "infinite mass" walls cannot be itself infinite otherwise it would feel an infinite impulse which would imply the ball also has an infinite impulse (the ball is the finite part of the system u were talking about right?)

 

[haruspex]

Thanks haruspex. I understand the last point. So for the first point, the change in momentum of the "infinite mass" walls cannot be itself infinite otherwise it would feel an infinite impulse which would imply the ball also has an infinite impulse (the ball is the finite part of the system u were talking about right?)

Yes.

 

[ehild]

ehild, I'm confused as to how you got those numbers. Let's say the ball has velocity v right before colliding with the moving wall. Pi =-mv and Pf = mv + 2mV so Iwall = -Iball = -2m(v + V). It reaches the stationary wall with velocity v + 2V and rebounds approximately with the same velocity v + 2V in the opposite direction so Iwall = -Iball = -m(-v - 2V - v - 2V) = 2m(v + 2V). I really don't like this business of infinite mass is there no other way to do this? A

I denoted the velocity vector by v. It is negative with respect to the velocity of the wall.

In post #13, I have shown the velocities obtained for finite mass of the wall.

$$v′=\frac{2MV−(M−m)v}{M+m}$$

$$V′=\frac{2mv+(M−m)V}{M+m}$$

The expressions have been derived from conservation of momentum and conservation of energy, and true in general for head-on elastic collisions. With the limit M-->∞, m/M-->0 , you get the velocities after collision for this situation.

$$v′=\frac{M(2V−(1−\frac{m}{M})v)}{M(1+\frac{m}{M})}=\frac{2V−(1−\frac{m}{M})v}{1+\frac{m}{M}} \rightarrow 2V-v$$

$$V′=\frac{M(2 \frac{m}{M} v+(1−\frac{m}{M})V)}{M(1+\frac{m}{M} 9}=\frac{2 \frac {m}{M} v+(1−\frac {m}{M}) V}{1+\frac {m}{M}} \rightarrow V$$



ehild