# Tennis balls in action

#### The tennis balls

The “standard” speed of the tennis balls is 100 mph. At this speed the tennis balls bounce off the tennis rackets. If the speed is sufficiently high, the ball will go through the strings The racket can be approximated as a potential-energy barrier having a height equal to the elastic energy of one of its strings. A 100 g tennis ball that travels at 220 mph is just sufficient to go through the 2.0-mm-thick strings. What is the probability of 150 mph ball going through the racket without breaking the strings? Read about wave functions

Basically the tennis ball is described by a wave function $\psi$. Since it needs to go through a barrier of potential the wave function will decrease exponentially in the barrier with increasing the barrier thickness $x$.

$\Psi(x) = \Psi(0)*\exp(-kx)$

$k$ is the wave number so that $(kx)$ is adimensional.

$k =1/\lambda = P/\hbar= \sqrt{2m(U-E))/\hbar}$

$U$ is the height of the barrier, $E$ is the kinetic energy of the incoming tennis balls, $m$ is the ball mass

(it is eactly like when an electron of energy $E<U$ tunnels through a potential barrier $U$)

$V1 =150 mph =67.0 m/s$

$V2= 220 mph =98.35 m/s$

The height of the potential barrier is equal to the maximum energy of the incoming tennis balls that breaks the strings

$U = mv2^2/2 = 0.1*98.35^2 /2 =483.6 J$

#### The probability

The incoming ball has a kinetic energy

$Ek =mv1^2/2 = 0.1*67^2/2 =224.45 J$

$U – E =259.15 J$

The wavenumber is

$k =\sqrt{(2*0.1*259.15)/10^{-34}} =7.2*10^{34}$

The probability of tunneling  is just (d=2 mm is the thickness of the barrier)

$P = |\psi(d)/\psi(0) |^2 = \exp(-2kd) =\exp(-2*7.2*10^{34}*2*10^{-3}) =\exp(-28.8*10^{31})$

Now we know that $\exp(2.303) =10$ so that

$P = [e^{2.303}]^{-12.5*10^{31}} =10^{-12.5*10^{31}}$

This is a probability P<<1 so in reality tennis balls that go through the racket cannot be observed.