Exploration of Quantum Effects
In a Bouncing Ping Pong Ball
A ping pong ball, with its center separated from the center of
another ping pong ball by 10 radii, and dropped onto the other
ping pong ball, will make a maximum of 10 bounces, if the
collisions are elastic. (The location of the first bounce is
determined by the Heisenberg uncertainty relations.)
Below are given, for each bounce, the horizontal displacement of the center of the bouncing ball, as measured from the vertical line drawn through the center of the bottom ball, as well as the increase in horizontal velocity, and the angle between the line joining the centers of the two balls and the vertical axis:
Bounce - Displacement - Horizontal Velocity - Angle from Vertical
(xi centimeters) - (vxi cm/sec) - (θi degrees)
1 1.7822e-014 1.5869e-012 2.5672e-013
2 5.9134e-013 5.3946e-011 8.4703e-012
3 2.0088e-011 1.8326e-009 2.8773e-010
4 6.8239e-010 6.2254e-008 9.7745e-009
5 2.3181e-008 2.1148e-006 3.3204e-007
6 7.8748e-007 7.1841e-005 1.1280e-005
7 2.6751e-005 20.0024405 0.00038318
8 0.00090875 0.082904 0.013017
9 0.030871 2.8162 0.44219
10 1.0487 92.022 15.199
(11) 34.306 92.022 90
Here is a figure showing the details for the i'th bounce:
φi = collision angle with respect to the vertical, for the i'th bounce
φi + θi = angle if incidence = angle of reflection, for the i'th bounce φi = 2*θi-1 + φi-1
...
φ2 = 2*θ1 + φ1
φ1 = ε = epsilon = initial collision angle
xi = horizontal displacement, for the i'th bounce
Here is the Matlab code which generates these results:
% Bouncing ping pong ball; three dimensional analysis
clear
clc
m = 2.4;
r = 2.0;
g = 980;
hbar = 1.0546e-27;
% s = initial separation between the surfaces of the two balls
% s should be expressed in centimeters
% IF: Distance between centers is 10 radii;
% THEN: Distance, s, between surfaces, is 8 radii;
% AND THEN: Ball falls a distance of only 8 radii.
s = 8*r;
t0 = sqrt(2*s/g); % t0 = time to fall distance s
v0 = g*t0; % maximum vertical speed
x1 = 2*sqrt(hbar*t0/m); % location of first bounce
% epsilon = initial collision angle with respect to the vertical axis
epsilon = sqrt(hbar/m*t0)/v0;
x(1) = x1;
theta(1) = asin(x(1)/(2*r));
theta(1) = theta(1) + epsilon/2;
% Angle of incidence = angle of reflection, for each bounce.
vx(1) = v0 * sin(2*theta(1));
i=1;
while x(i) is less than 2*r
i=i+1;
x(i) = x(i-1) + vx(i-1)*2*t0; % Displacement
theta(i) = asin(x(i)/(2*r)); % Angle
vx(i) = v0 * sin(2*(sum(theta))); % Horizontal Velocity
end
N=i-1;
vx(i)=vx(N);
theta = rad2deg(theta);
fprintf('\n\n Number of bounces = %g\n\n',N)
disp([blanks(10) 'x(cm)' blanks(10) 'vx(cm/sec)' blanks(10) 'theta(deg)'])
disp([x' vx' theta'])
semilogy(1:N+1,x,'or','MarkerFaceColor','r')
fname = ['bounce; "bounce" ' num2str(N+1) ' is off the table'];
xlabel(fname)
ylabel('horizontal displacement (cm)')
title('Bouncing ping pong ball')
Errors will begin to appear in the results for the final bounces if the initial separation, s, between the surfaces of the balls, is significantly less than one radius, owing to the approximations made in these calculations that the maximum speed, v0, and the time between collisions, 2*t0, are the same for each bounce.
Radius of ball = r = 2.0 cm
Mass of ball = m = 2.4 grams
Acceleration of gravity = g = 980 cm/sec2
Planck's constant/2*pi = hbar = 1.0546e-27 erg-sec
The mass of the ball enters into the calculations only through the momenta px and py in the Heisenberg uncertainty relations: ΔxΔpx = hbar/2, ΔyΔpy = hbar/2. (Momentum = mass multiplied by velocity.)
The location of the first bounce, x1, is obtained by minimizing:
Δx + Δpxt0/m = Δx + hbar*t0/2mΔx
by differentiating with respect to Δx, setting the result equal to zero, and solving for Δxmin; doing the same to find Δymin; and then setting:
x1 = sqrt(Δxmin^2 + Δymin^2) + sqrt(Δpx min^2 + Δpy min^2)*t0/m,
where the minimum uncertainty relations are used:
ΔxminΔpx min = hbar/2 and ΔyminΔpy min = hbar/2.
The initial collision angle is:
tanε ~ ε = epsilon = sqrt(Δpx min^2 + Δpy min^2)/mv0.
These formulas yield:
Δxmin = Δymin = sqrt(hbar*t0/2*m)
x1 = 2*sqrt(hbar*t0/m)
epsilon = sqrt(hbar/m*t0)/v0
When s = 8*r = 16 cm, as in the above example, x1 = 1.78e-014 cm. Compare to the classical electron radius: e^2/mc^2 = 2.82e-013 cm. The minimum distance of the first bounce from zero displacement under ideal conditions is less than 1/10 th the size of a classical electron radius!
IT CAN BE NOTED HERE THAT IF THE INITIAL SEPARATION BETWEEN THE CENTERS OF THE BALLS IS 12 RADII (SO THAT THE INITIAL DISTANCE, s, BETWEEN THE SURFACES OF THE BALLS, IS 10 RADII), THEN THE MAXIMUM NUMBER OF BOUNCES IS 9. THE HORIZONTAL DISPLACEMENT OF THE 9 'TH BOUNCE IS 0.178 CM.
ANALYTIC SOLUTION
Using the small argument expansions for angles, and keeping only the leading power in the expansion parameter:
2*v0*t0/r = 4*s/r,
and using the formulas:
x(i) = x(i-1) + vx(i-1)*2*t0
theta(i) = asin(x(i)/(2*r))
vx(i) = v0 * sin(2*(sum(theta)))
with
theta(1) = asin(x(1)/(2*r)) + ε/2,
the following expression for the position of the N 'th bounce is obtained:
xN = (4*s/r)^(N-1)[x1 + r*ε].
SPECIAL CASE: Initial separation of the centers of the balls is 10 radii (the surfaces are separated by 8 radii): then
s = 8*r
4*s/r = 32
To find N, let:
xN = 2*r
Then
2*r = 32^(N-1)[ x1 + r*ε]
1 = 32^(N-1)[ x1/2r + ε/2]
0 = (N-1)log(32) + log[ x1/2r + ε/2]
When r = 2 cm and s = 8*r,
x1 = 1.78 * 10-14 cm
ε = 5.03 * 10-17
and
N-1 = - log[ x1/2r + ε/2]/log(32)
N = 1 – log(4.48*10-15)/1.505
N = 1 – log(4.48)/1.505 + 15/1.505
N = 1 – 0.43 + 10 = 11 - 0.43
N = 10.57
Since N is less than 11 when xN = 2*r, and since N must be an integer, the total number of bounces must be 10. This result agrees with the previously determined value for N.
To find the position of the last bounce, which should be less than 2*r, the formula
xN = (4*s/r)^(N-1)[x1 + r*ε]
is used again, this time with N = 10. With or without using logs, it is found from
x10 = 32^9 * 1.79 * 10^-14 cm
that
x10 = 0.628 cm,
which is somewhat less than the previously computed value:
x10 = 1.05 cm.
ADDITIONAL CONSIDERATIONS
The experiment should be done in vacuum at low temperatures to avoid air currents and thermal effects (vibrations). Thermal effects are expected to always dominate quantum effects in an experiment like this.
Factors reducing the number of bounces include the Coriolis effect, rotation or spin of the bouncing ball, imperfections in the balls, human error, etc.
Inelasticity (energy loss per bounce) actually serves to increase the number of bounces by reducing the increase in horizontal velocity gained with each bounce.
