Animated Solution for Physics - System of Particles: A rubber ball is released from a height of $5\text{ m}$ above the floor. It bounces back repeatedly, always rising to $\frac{81}{100}$ of the height through which it falls. Find the average speed of the ball. (Take, $g=10{\text{ ms}}^{-2}$)

Imagine dropping a rubber ball from a certain height. It hits the ground, compresses, and then springs back up. However, it never quite reaches its original height. Why? Because during the collision with the floor, some of its kinetic energy is lost as heat and sound. This loss of energy is mathematically captured by a dimensionless number called the coefficient of restitution, denoted by $e$.

When a ball drops from an initial height $h_0$, it hits the ground with a velocity $v_0=\sqrt{2g{h}_0}$. After the bounce, it rebounds with a velocity $v_1=e{v}_0$. The height it reaches after the first bounce, $h_1$, is related to this rebound velocity by $h_1=\frac{v_1^2}{2g}$. Substituting $v_1$, we get $h_1=e^2{h}_0$.

In our specific problem, we are told that the ball always rises to $\frac{81}{100}$ of the height through which it falls. This immediately tells us that $e^2=0.81$, which means the coefficient of restitution $e=0.9$.

To find the average speed of the ball from the moment it is dropped until it finally comes to rest, we need two critical pieces of information: the total distance it travels and the total time it takes.

Let's map out the distance. The ball falls a distance $h_0$. Then it bounces up a distance $h_1$ and falls back down the same distance $h_1$. Then it bounces up $h_2$ and falls $h_2$, and so on. The total distance $H$ is an infinite series:

Since $h_n=e^{2n}{h}_0$, we can rewrite this as:

Factoring out $2e^2{h}_0$ from the subsequent terms, we reveal a classic infinite Geometric Progression (GP):

The sum of an infinite GP $1+r+r^2+\dots$ is $\frac{1}{1-r}$ (for $|r|<1$). Here, our common ratio is $e^2$. Applying this formula, we get the elegant result for total distance:

Plugging in our values ($h_0=5\text{ m}$ and $e^2=0.81$), we find $H\approx 47.6\text{ m}$.

Now, let's tackle the total time. The time it takes to fall from the initial height $h_0$ is $t_0=\sqrt{\frac{2h_0}{g}}$. For every subsequent bounce, the ball takes time $t_n$ to go up and another $t_n$ to come down. The total time $T$ is:

Since velocity scales by $e$ with each bounce ($v_n=e^n{v}_0$), the time for each bounce also scales linearly by $e$ ($t_n=e^n{t}_0$). This gives us another infinite GP:

Using the same GP summation logic, the total time formula emerges as:

With $h_0=5\text{ m}$ and $g=10{\text{ m/s}}^2$, the initial drop time $t_0$ is exactly $1\text{ s}$. Substituting $e=0.9$, the total time $T$ evaluates to a neat $19\text{ s}$.

Average Speed vs Average Velocity

We now have everything we need. The average speed is simply the total distance divided by the total time:

A Word of Caution: It is crucial to distinguish between average speed and average velocity. If the question had asked for average velocity, we would need to use the net displacement. No matter how many times the ball bounces, its final resting place is on the floor, exactly $5\text{ m}$ below its starting point. Therefore, the average velocity would be $\frac{5\text{ m}}{19\text{ s}}\approx 0.26\text{ m/s}$ downwards. Always read the question carefully to avoid this classic trap!