"Superball" bouncing between two walls

  • Thread starter cpgp
  • Start date
In summary, the first transit takes 2l/(v0+V) and the second transit takes t(v0+V) = L - L(V/v0).Solving for t yields t=L/((v0+V)/v0).
  • #1
cpgp
14
1
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?
 
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  • #2
Are you trying to calculate accleleration as a function of ##x## here?
 
  • #3
I thought that might be the way to calculate the force in terms of x.
 
  • #4
cpgp said:
t = 2l/(v0+V)
How so?
 
  • #5
haruspex said:
How so?
Assuming the ball is released at r = 0, it first collides with the moving wall at t = 2l/(v0+V).
 
  • #6
cpgp said:
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?
 
  • #7
haruspex said:
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).
 
  • #8
cpgp said:
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.
 
Last edited:
  • #9
haruspex said:
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).
 
  • #10
cpgp said:
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
vn velocity after n bounces off the moving barrier
tn time between nth 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.
 
Last edited:
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FAQ: "Superball" bouncing between two walls

1. What is the "Superball" and how does it work?

The "Superball" is a type of rubber ball that is known for its high bounce. It works by utilizing the material properties of rubber, which is highly elastic and can store and release energy. When the ball is dropped, it compresses and stores energy. As it bounces off the ground, the stored energy is released, causing the ball to bounce back up with great force.

2. Why does the "Superball" bounce higher between two walls compared to a regular ball?

The "Superball" bounces higher between two walls because of its high elasticity. When the ball hits the wall, it compresses and stores more energy than a regular ball, resulting in a higher bounce. Additionally, the walls act as a support, preventing the ball from losing energy to the sides, allowing it to maintain its energy and bounce higher.

3. How does the distance between the walls affect the bouncing of the "Superball"?

The distance between the walls affects the bouncing of the "Superball" by changing the amount of time the ball spends in contact with the walls. The closer the walls are, the shorter the time the ball spends in contact with the walls, resulting in a higher bounce. On the other hand, a wider distance between the walls will result in a lower bounce as the ball spends more time in contact with the walls, losing energy to friction.

4. Can the "Superball" bounce indefinitely between two walls?

No, the "Superball" cannot bounce indefinitely between two walls. The ball will eventually lose energy due to friction and air resistance, resulting in a lower bounce each time. Eventually, the ball will come to a stop.

5. How does the surface of the walls affect the bouncing of the "Superball"?

The surface of the walls can affect the bouncing of the "Superball" by changing the amount of energy lost during each bounce. A rougher surface will result in more energy loss due to friction, resulting in a lower bounce. A smoother surface will result in less energy loss and a higher bounce. Additionally, the material of the walls can also play a role in the bouncing, with harder surfaces resulting in a higher bounce compared to softer surfaces.

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