Collisions_Analysis_Guide

Characterizing Collisions — Analysis Guide • Page 1
Characterizing Collisions
From Bounce Times to Velocity, Energy, and the Coefficient of Restitution
A step-by-step analysis guide • Scientific Inquiry with AI
In this experiment you record the sound of a ping-pong ball bouncing on a hard surface. Each
impact makes a sharp click, so the audio gives you the precise clock time of every bounce. From
this one kind of measurement — a list of bounce times — you can reconstruct the ball’s speed, the
height it reaches, the energy it loses in each collision, and the coefficient of restitution that
characterizes how “bouncy” the collision is.
This guide builds the analysis up one step at a time, starting from the equations of motion you
already know. Nothing is skipped: every formula is derived from the one before it, and worked
numbers are shown so you can check your own work.
What you measure vs. what you calculate
You measure only the bounce times 𝑡1, 𝑡2, 𝑡3, …. Everything else — velocity, height, energy, and
the coefficient of restitution — is calculated from those times using the physics below.
1. The Raw Measurement: Bounce Times and the Time Between Bounces
The ball is dropped from an initial height 𝐻0. It strikes the surface, rebounds to a lower height, falls,
strikes again, and so on. We label the bounces with an index n = 1, 2, 3, … and call the clock time of
the n-th impact 𝑡𝑛.
The quantity that carries the physics is the time between successive bounces, which we call the
flight interval 𝑇𝑛:
𝑇𝑛 = 𝑡𝑛+1 − 𝑡𝑛 (1)
In words: Tₙ is how long the ball is in the air between bounce n and bounce n+1. During that interval
the ball leaves the floor, rises to a peak height ℎ𝑛, and falls back down. Figure 1 shows these
quantities.

Characterizing Collisions — Analysis Guide • Page 2
Figure 1. Height vs. time. Each red dashed line marks a bounce time tₙ. The flight interval Tₙ is the time between
two bounces; hₙ is the peak height the ball reaches during that interval. The peaks shrink because energy is lost at
each collision.
Two words that are easy to confuse
Bounce time: 𝑡𝑛 = the moment an impact happens (read directly from the audio).
Flight interval: 𝑇𝑛 = the gap of time between two impacts (a difference of two bounce times).
All of the analysis below uses the intervals 𝑇𝑛.
2. Quick Review: Equations of Motion for Constant Acceleration
While the ball is in the air, the only force on it is gravity (for now we assume no drag force), so it
moves with a constant acceleration. For any object with constant acceleration a, starting at
position 𝑦0 with velocity 𝑣0, the three standard kinematic equations are:
𝑣 = 𝑣0 + 𝑎 𝑡 (2)
𝛥𝑦 = 𝑣0 𝑡 + 1
2 𝑎 𝑡2 (3)
𝑣2 = 𝑣0
2 + 2 𝑎 𝛥𝑦 (4)
Here v is the velocity at time t, and 𝛥𝑦 is the change in height. For our bouncing ball we take
upward as positive, so the acceleration is downward: 𝑎 = −𝑔, with 𝑔 = 9.81 𝑚/𝑠2 (the
magnitude of free-fall acceleration). These three equations are the only physics we need; the rest of
the guide is algebra applied to them.

Characterizing Collisions — Analysis Guide • Page 3
3. From a Flight Interval to Speed and Height
Consider one trip through the air: the ball leaves the floor going straight up with some rebound
speed 𝑤𝑛, rises, slows to a stop at the top, then falls back. Because gravity is the same on the way
up and the way down, the motion is symmetric: the time to rise equals the time to fall. So the ball
reaches its peak exactly halfway through the interval, at time 𝑇𝑛
2 .
3a. Rebound speed from the flight interval
At the very top the vertical velocity is zero. Apply Equation (2) from launch (velocity 𝑤𝑛) to the apex
(velocity 0), over the rise time 𝑇𝑛
2 ):
0 = 𝑤𝑛 − 𝑔 𝑇𝑛
2 (5)
Solving for the rebound speed gives a clean result — the speed the ball left the floor with is set
entirely by how long it stays in the air:
𝑤𝑛 = 1
2 𝑔 𝑇𝑛 (6)
Worked example. If a flight interval is 𝑇𝑛 = 0.40 𝑠, then 𝑤𝑛 = 1
2 (9.81)(0.40) = 1.96 𝑚/𝑠.
3b. Peak height from the flight interval
The ball rises for a time 𝑇𝑛
2 . Using Equation (3) for the rise (starting speed 𝑤𝑛, acceleration −𝑔), the
peak height is
ℎ𝑛 = 𝑤𝑛
𝑇𝑛
2 − 1
2 𝑔 𝑇𝑛
2
2
(7)
Substitute 𝑤𝑛 = 1
2 𝑔 𝑇𝑛 from Equation (6) and simplify. The two terms combine to give the
compact form:
ℎ𝑛 = 1
8 𝑔 𝑇𝑛
2 (8)
(Equivalently, since ℎ𝑛 = 𝑤𝑛
2
2𝑔 from Equation (4) with 𝑣 = 0 at the top — both routes give the same
answer.)
Worked example. With 𝑇𝑛 = 0.40 𝑠: ℎ𝑛 = 1
8 (9.81)(0.40)2 = 0.196 𝑚.
4. Energy of the Bounce
Just after leaving the floor, all of the ball’s mechanical energy is kinetic. At the top of its flight, all of
it is potential. Energy conservation in the air means these are equal, so the mechanical energy
associated with flight interval n can be written either way:

Characterizing Collisions — Analysis Guide • Page 4
𝐸𝑛 = 1
2 𝑚 𝑤𝑛
2 = 𝑚 𝑔 ℎ𝑛 (9)
Because both the kinetic energy (∝ 𝑤2) and the potential energy (∝ ℎ) are fixed by the flight
interval, you do not even need the ball’s mass to compare one bounce to the next — the ratios of
energy depend only on the measured times. That is exactly what the coefficient of restitution
captures.
5. The Coefficient of Restitution, e
The coefficient of restitution e measures how much speed survives a collision. It is defined as the
ratio of the speed leaving the surface to the speed arriving at it:
𝑒 = 𝑟𝑒𝑏𝑜𝑢𝑛𝑑 𝑠𝑝𝑒𝑒𝑑
𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑠𝑝𝑒𝑒𝑑 (10)
A perfectly elastic bounce has 𝑒 = 1 (no speed lost); a dead, fully inelastic impact has 𝑒 = 0. Real
balls fall in between. There are two complementary ways to get e from your bounce times.
5a. Method 1 — e for each bounce, from approach and rebound speeds
Here is the key insight that links speeds to your measured intervals. When the ball comes back
down to the floor, it arrives at the same speed it left with (the flight is symmetric). So the speed
approaching bounce n is just the rebound speed from the previous interval, 𝑤𝑛−1, and the speed
leaving bounce n is 𝑤𝑛. Therefore:
𝑒𝑛 = 𝑤𝑛
𝑤𝑛−1
(11)
Now substitute the speed–interval relation 𝑤𝑛 = 1
2 𝑔 𝑇𝑛 from Equation (6). The factor of 1
2 𝑔
cancels top and bottom, leaving a remarkably simple result:
𝑒𝑛 =
1
2 𝑔 𝑇𝑛
1
2 𝑔 𝑇𝑛−1
= 𝑇𝑛
𝑇𝑛−1
(12)
Result
The coefficient of restitution for each bounce is simply the ratio of consecutive flight intervals.
You can read e straight off the bounce times — no mass, no height, no calibration required:
𝑒𝑛 = 𝑇𝑛 / 𝑇𝑛−1.
5b. Energy lost per collision
Since energy goes as speed squared (Equation 9), the fraction of kinetic energy kept in a bounce is
𝑒𝑛
2, and the fraction lost is whatever is left over:
𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 1 − 𝑒𝑛
2 = 1 − 𝑇𝑛
𝑇𝑛−1
2
(13)

Characterizing Collisions — Analysis Guide • Page 5
Worked example. If 𝑒𝑛 = 0.96, the ball keeps (0.96)2 = 0.92, i.e. 92% of its energy, and loses
about 8% in that single collision.
5c. Relating e to the rebound heights
You can also express e through the peak heights, but be careful: height scales with speed squared,
not speed itself. Since ℎ𝑛 ∝ 𝑤𝑛
2, the height ratio is the square of the speed ratio:
ℎ𝑛
ℎ𝑛−1
= 𝑤𝑛
𝑤𝑛−1
2 = 𝑒𝑛
2 (14)
so taking the square root gives the coefficient of restitution from heights:
𝑒𝑛 = √ ℎ𝑛
ℎ𝑛−1
(15)
Watch the square root
A common slip is to write 𝑒 = ℎ𝑛 / ℎ𝑛−1 without the square root. That would be the energy
ratio, not the speed ratio. The speed-based definition of e requires the square root, because
energy and height scale as speed squared.
6. The Best Estimate of e: The Log Plot Method
Method 1 gives a value of e from every neighboring pair of bounces, but each one is a bit noisy. If
the surface and ball behave the same way at every bounce, then e is roughly constant, and we can
use all the bounces at once to pin it down far more precisely. Here is how.
If e is constant, each interval is just e times the one before it (from Equation 12, 𝑇𝑛 = 𝑒 · 𝑇𝑛−1).
Applying that repeatedly:
𝑇2 = 𝑒 𝑇1, 𝑇3 = 𝑒 𝑇2 = 𝑒2 𝑇1 , …
which gives a geometric pattern — the intervals shrink by the same factor each time:
𝑇𝑛 = 𝑇1 𝑒𝑛−1 (16)
Geometric patterns become straight lines when you take a logarithm. Taking the natural log of both
sides of Equation (16) and using 𝑙𝑛(𝐴 𝐵) = 𝑙𝑛 𝐴 + 𝑙𝑛 𝐵 and 𝑙𝑛(𝑒𝑘) = 𝑘 · 𝑙𝑛 𝑒:
𝑙𝑛 𝑇𝑛 = 𝑙𝑛 𝑇1 + (𝑛 − 1) 𝑙𝑛 𝑒
Grouping the terms by powers of n puts this in the form of a straight line, 𝑦 = (𝑠𝑙𝑜𝑝𝑒) · 𝑛 +
(𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡):
𝑙𝑛 𝑇𝑛 = (𝑙𝑛 𝑒) 𝑛 + (𝑙𝑛 𝑇1 − 𝑙𝑛 𝑒) (17)
So if you plot 𝑙𝑛 𝑇𝑛 on the vertical axis against the bounce number n on the horizontal axis, the data
fall on a straight line whose slope equals 𝑙𝑛 𝑒. Figure 2 shows this.

Characterizing Collisions — Analysis Guide • Page 6
Figure 2. Plotting ln Tₙ against bounce number n gives a straight line. Fitting the line gives one precise slope from
all the data at once; the coefficient of restitution is then e = e^(slope).
To recover e itself, undo the logarithm by exponentiating the slope:
𝑒 = 𝑒𝑠𝑙𝑜𝑝𝑒 (18)
Worked example (from real bounce data)
A line fit to bounces 1–45 gives a slope of −0.0398. Then 𝑒 = 𝑒−0.0398 = 0.961. This single
number summarizes all 45 bounces and is far less sensitive to noise in any one bounce than the
pair-by-pair values from Method 1.
A note on which logarithm. The method works with any base of logarithm. If you use base-10
instead of natural log, you determine e with 𝑒 = 10𝑠𝑙𝑜𝑝𝑒. The data here were fit with the natural
log, so we use 𝑒 = 𝑒𝑠𝑙𝑜𝑝𝑒.
7. Putting It Together: From Times to Physics
Every quantity in this experiment flows from the single list of bounce times. The chain of reasoning
is:
• Measure the bounce times 𝑡𝑛 from the audio.
• Subtract to get the flight intervals 𝑇𝑛 = 𝑡𝑛+1 − 𝑡𝑛.
• Compute rebound speed 𝑤𝑛, peak height ℎ𝑛, and energy 𝐸𝑛 for each interval.
• Characterize the collisions with e — either bounce-by-bounce (ratio of intervals) or, more
precisely, from the slope of the log plot.
• Interpret: the fraction of energy lost per collision is 1 − 𝑒2.

Characterizing Collisions — Analysis Guide • Page 7
Quick-reference equation sheet
Quantity Formula (from bounce times)
Flight interval 𝑇𝑛 = 𝑡𝑛+1 − 𝑡𝑛
Rebound speed 𝑤𝑛 = 1
2 𝑔 𝑇𝑛
Peak height ℎ𝑛 = 1
8 𝑔 𝑇𝑛
2
Mechanical energy 𝐸𝑛 = 1
2 𝑚 𝑤𝑛
2 = 𝑚 𝑔 ℎ𝑛
COR (per bounce)
𝑒𝑛 = 𝑇𝑛
𝑇𝑛−1
= √ ℎ𝑛
ℎ𝑛−1
COR (log-plot slope) 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑙𝑛 𝑇𝑛 𝑣𝑠. 𝑛 = 𝑙𝑛 𝑒 ; 𝑒 = 𝑒𝑠𝑙𝑜𝑝𝑒
Energy lost / collision 1 − 𝑒𝑛
2
Every equation above is a native Word equation — click any of them to edit in the equation editor.