A $1.0$-mm-diameter sphere bounces back and forth between two walls at $x=0\mathrm{mm}$and $x=100\mathrm{mm}$. The collisions are perfectly elastic, and the sphere repeats this motion over and over with no loss of speed. At a random instant of time, what is the probability that the center of the sphere is

a. At exactly $x=50.0\mathrm{mm}$?

b. Between $x=49.0\mathrm{mm}$and $x=51.0\mathrm{mm}$?

c. At $x\ge 75\mathrm{mm}$?

A $1$$\mathrm{mm}$diameter sphere bounces back and forth between two walls at $x=0\mathrm{mm}$and $x=100\mathrm{mm}$.

The collisions are perfectly elastic, and the sphere repeats this motion over and over with no loss of speed.

As the collisions are perfectly elastic and the sphere has diameter of $1\mathrm{mm}$it equally probable to find the center of the sphere between $x=0.5\mathrm{mm}$and $x=99.5\mathrm{mm}$.

Thus, the probability density function of finding the center of the sphere is given by $p\left(x\right)=\frac{1}{99}$for $0.5\le x\le 99.5$

$=0$otherwise

Thus, probability of finding the center of sphere exactly at $\mathrm{x}=50\mathrm{mm}$is ${\int }_{50}^{50}p\left(x\right)dx=\frac{1}{99}(50-50)=0$.

The center of the sphere is between $x=49\mathrm{mm}$and $x=51\mathrm{mm}$

The probability of finding the center between$\mathrm{x}=49\mathrm{mm}$and $\mathrm{x}=51\mathrm{mm}$ is

The center of the sphere is at $x\ge 75\mathrm{mm}$.

The probability of finding the center at $x\ge 75\mathrm{mm}$is

${\int }_{75}^{100}p\left(x\right)dx={\int }_{75}^{99.5}p\left(x\right)dx+0=\frac{1}{99}(99.5-75)=\frac{245}{990}=\frac{49}{198}$