# Moving wall and ball collision

1. Jan 26, 2013

### PhizKid

1. The problem statement, all variables and given/known data

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

2. Relevant equations
Energy

3. 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!

2. Jan 27, 2013

### 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.

3. Jan 27, 2013

### haruspex

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.

4. Jan 27, 2013

### 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!

5. Jan 27, 2013

### haruspex

I think you mean V/v through all that.
Yes, the average force.
$\Delta t = \frac{2x}{v} + O(\frac{V}{v})$
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.

6. Jan 27, 2013

### 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.

7. Jan 27, 2013

### haruspex

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.
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.

8. Jan 27, 2013

### 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.

9. Jan 28, 2013

### 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

10. Jan 28, 2013

### 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

11. Jan 28, 2013

### haruspex

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.

12. Jan 29, 2013

### 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?

Last edited: Jan 29, 2013
13. Jan 29, 2013

### ehild

It is less confusing if you consider a finite-mass wall (mass M) travelling 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-->∞
2m|v-V|
and also what is the exchanged energy with the same limit.
2m|v-V|V
.

ehild

14. Jan 29, 2013

### 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.

15. Jan 29, 2013

### 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

16. Jan 29, 2013

### voko

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.

17. Jan 29, 2013

### 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?

Last edited: Jan 29, 2013
18. Jan 29, 2013

### ehild

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

19. Jan 29, 2013

### haruspex

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.

20. Jan 29, 2013

### voko

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.