# 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…

#### The answer

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 tunnelingÂ 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.