# Shwoing the sensitivity of many body systems oi small perturbations

1. Oct 8, 2011

### XCBRA

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

Imagine we are observing the motion of frictionless, elastically colliding billiard balls.
Using sensible assumptions about the size of things, show that, if someone enters the
room, small deflections of the balls due this visitor's gravitational pull will completely
alter their trajectories after just about 10 collisions.

Hint: For simplicity, consider all balls but one fixed in place and then analyse the
motion of one ball.

2. Relevant equations

3. The attempt at a solution

Not entirely sure how to approach this problem. I chose to consider a one ball rebouinding between a wall and another fixed ball, such that the distance between the wall and th3 fised ball is 2m, and the the bal was moving at 0.1 m/s. Then i assumed the person is 2 m away perpendicular to the direction of motion. the person is 70kg and the ball is 1kg.

For image
http://img802.imageshack.us/img802/1383/smallperturbationsprobl.png [Broken]

$$F = \frac{Gm_pm_b}{r^2}$$
$$F=\frac{6.6726*10^-11*70*1}{2^2}$$
$$F=1.1*10^-9$$

Then if we assume tyen collsions, the ball will travle 20m horizontally and hence will travel for 200s.

If we consider only the vertical force and assume that it is constant.

since $F=ma$
$$a=1.16*10^-9$$

$$s=ut+\frac{at^2}{2}$$

therefore s = 2.3366*10^-5 m, whcih is clearly not enough to be considered to have completly altered the trajectory of the ball.

I dont think this approach is right, so please could you advise how it would be best to tackle this problem. Thanks for the help!

Last edited by a moderator: May 5, 2017
2. Oct 11, 2011

### 256bits

You have a correct approach. It is a first time approximation to see if the person has an effect upon the ball. The centre of mass has moved s = 2.3366*10^-5 m ( 2.3366*10^-3 cm ). If your ball radius is 2.336 cm that is s/r = 0.001 or 0.1%. Or the ball centre is off sin(theta) = s/r; where theta = 0.057 degrees, from its head on contact point. Or the ball has moved sin(alpha) = s/L; where L= 2m, and alpha = 0.0003 degrees from the original straight line contact with the cushion.

Next step is to do each of the 10 iterations.
Have the ball bounce to the cushion and back to the stationary ball. When the ball hits a flat surface at an angle alpha, it rebounds at the same angle. What happens when the ball hits another ball of the same angle - it hits at a tangent from the head on so the angle alpha doubles.
Do that for 10 hits and see what you get.