Solution 6: Bouncing Ball Collisions

Solution 6

Logic of Motion Segments:
The ball rebounds “elastically” from a fixed height $H$. This means: 1. No energy is lost. The ball always rebounds to the exact same height $H$. 2. The motion consists of identical segments of length $H$.

Drop: Travels $H$ downwards. Time $T_0$.
Rebound: Travels $H$ upwards. Time $T_0$.

We are given Total Distance $S = 20$ m and Total Time $t = 8$ s. We must find the number of segments ($k$) that fit these totals to determine the number of collisions.

1. Formulating the Equations

Let $k$ be the number of one-way segments (either up or down) the ball completes.

Total Distance: $S = k \times H = 20$ m.
Total Time: $t = k \times T_0 = 8$ s.

The time for one segment ($T_0$) is the free-fall time for height $H$:

$$ T_0 = \sqrt{\frac{2H}{g}} $$

2. Solving for H

We divide the distance equation by the time equation to find the average speed:

$$ \frac{S}{t} = \frac{kH}{kT_0} = \frac{H}{T_0} = \frac{20}{8} = 2.5 \text{ m/s} $$

Now substitute $T_0 = \sqrt{2H/g}$:

$$ \frac{H}{\sqrt{2H/g}} = 2.5 $$ $$ \sqrt{\frac{gH}{2}} = 2.5 $$

Square both sides ($g=10$):

$$ \frac{10H}{2} = (2.5)^2 \implies 5H = 6.25 \implies H = 1.25 \text{ m} $$

3. Counting Collisions

First, find the number of segments ($k$):

$$ k = \frac{\text{Total Distance}}{H} = \frac{20}{1.25} = 16 \text{ segments} $$

Now, trace the path to count collisions (Impacts with the ground):

Segment 1 (Down) $\to$ Collision 1
Segment 2 (Up)
Segment 3 (Down) $\to$ Collision 2
Segment 4 (Up)
…

A collision occurs at the end of every odd-numbered segment (1, 3, 5…).

Since there are 16 segments total, the last segment is 16 (Up). The last collision happened after segment 15.

Total collisions = $\frac{16}{2} = 8$.