"Superball" bouncing between two walls

[cpgp]

Homework Statement

A "superball" of mass m bounces back and forth between two surfaces with speed v0. Gravity is neglected and the collisions are perfectly elastic.
a. Find the average force F on each wall.
Ans. F = mv0^2 /l
b. If one surface is slowly moved toward the other with speed V <<v0, the bounce rate will increase due to the shorter distance between collisions, and because the ball's speed increases when it bounces from the moving surface. Find F in terms of the separation of the surfaces, x.
(Hint: Find the average rate at which the ball's speed increases as the surface moves.)
Ans. F =(mv0^2/l)(l/x)^3

Relevant Equations

Newtons laws, conservation of momentum and energy.

I have been able to solve part a correctly, but am having some trouble with part b.
I started out calculating values of v and x at various points in time when the ball collides with either wall. I found:

t = 0, v = v0, x = l

t = (l/v0), v = v0, x = l(v0-v)/v0

t = 2l/(v0+V), v = v0 + 2V, x = l(v0-V)/(v0+V)

t = 3l/(v0+2V), v = v0 + 2V, x = l(v0-V)/(v0+2V)

t = 4l/(v0+3V), v = v0 + 4V, x = l(v0-V)/(v0+3V)

t = 5l/(v0 + 4V), v = v0 + 4V, x = l(v0-V)/(v0+4V)

t = 6l/(v0+5V), v = v0 + 6V, x = l(v0-V)/(v0+5V)

Extrapolating from this, one might write for v = v0 + nV, x = l(v0-V)/(v0+(n-1)V), and t = nl/(v0+(n-1)V). Manipulating x = l(v0-V)/(v0+(n-1)V) yields
n = (l-x)(v0-V)/Vx. Substituting this into the expression for v, we have v = v0 + (l-x)(v0-V)/x.

Differentiating this with respect to x should yield a(x), so we have $$a(x) = \frac d {dx} (v_0 + \frac{(l-x)(v_0-V)}{x}).$$

$$a(x) = (v_0 - V) \frac d {dx} (\frac lx -1)$$
$$a(x) = \frac{(V - v_0)l} {x^2}$$

Now, F = ma = m(V - v0)l/x^2.

Letting V - v0 = v0, we have F = -mv0(l/x^2). But this is clearly not the answer indicated. What is the flaw in my solution?

 

[PeroK]

Are you trying to calculate accleleration as a function of $x$ here?

 

[cpgp]

I thought that might be the way to calculate the force in terms of x.

 

[haruspex]

t = 2l/(v0+V)

How so?

 

[cpgp]

How so?

Assuming the ball is released at r = 0, it first collides with the moving wall at t = 2l/(v0+V).

 

[haruspex]

Assuming the ball is released at r = 0, it first collides with the moving wall at t = 2l/(v0+V).

Judging from your v=v0 for the second transit, the ball started at the moving end. You got that the first transit takes l/v0. How far is the second transit? How long does that take? What is the total time for those two transits?

 

[cpgp]

Judging from your v=v0 for the second transit, the ball started at the moving end. You got that the first transit takes l/v0. How far is the second transit? How long does that take? What is the total time for those two transits?

The total time for those two transits is 2l/(v0+V). The distance traveled is l(v0-V)/(v0+V).

 

[haruspex]

The total time for those two transits is 2l/(v0+V). The distance traveled is l(v0-V)/(v0+V).

For the second transit you have velocity v0 and distance l(v0-V)/v0. That takes time l(v0-V)/v0². Adding the time of the first transit we have l/v0+l(v0-V)/v0²=(l/v0)(1+(v0-V)/v0)=l(2v0-V)/v0²

Edit: that was using your second line
t = (l/v0), v = v0, x = l(v0-v)/v0
But I just realized that x is wrong. l(v0-v)/v0 is the distance when it starts the second transit, but it will not have to go that far.

 

[cpgp]

For the second transit you have velocity v0 and distance l(v0-V)/v0. That takes time l(v0-V)/v0². Adding the time of the first transit we have l/v0+l(v0-V)/v0²=(l/v0)(1+(v0-V)/v0)=l(2v0-V)/v0²

Edit: that was using your second line
t = (l/v0), v = v0, x = l(v0-v)/v0
But I just realized that x is wrong. l(v0-v)/v0 is the distance when it starts the second transit, but it will not have to go that far.

Yes, x is l(v0-V)/(v0+2V).

 

[haruspex]

Yes, x is l(v0-V)/(v0+2V).

I don't get that 2.
First transit: dist L, time L/v0
In this time the barrier has advanced VL/v0
Second transit time t, V(L/v0 + t) + v0 t = L
t(v0+V) = L - L(V/v0)
x = v0 t = L(v0-V)/(v0+V)

So what is the total time for the first two transits?

To solve it analytically, define sequences of variables like
v_n velocity after n bounces off the moving barrier
t_n time between n^th and (n+1)^th bounce off moving barrier.
Then obtain equations relating successive terms.

At some point, presumably, you should use V<<v0, but avoid doing that too soon as it can lead to a wrong answer.