Short answer: the coefficient of restitution (COR), written e, describes how strongly two objects rebound along their line of impact. It is the relative separation speed after impact divided by the relative approach speed before impact. For a ball dropped vertically onto a stationary rigid floor, e ≈ √(rebound height/drop height). A larger value means a stronger rebound under those test conditions.

A tennis ball hitting acrylic, a rubber ball hitting wood, and a steel sphere hitting foam do not share one universal “bounciness number.” COR belongs to the collision pair and the conditions. Speed, temperature, inflation pressure, moisture, surface texture, spin, and measurement method can all matter.

Start with the collision definition

For a head-on collision between bodies 1 and 2, use velocities measured along the impact line:

e = relative speed of separation / relative speed of approach

With a consistent sign convention, one common one-dimensional form is:

e = (v₂ − v₁)/(u₁ − u₂)

Here, u values are velocities before impact and v values are velocities after it. The numerator and denominator both have velocity units, so COR has no unit. Readers who want a broader map of the subject can place this experiment within mechanics and the major fields of physics.

A two-cart example

Cart A approaches at 3.0 m/s while cart B moves in the same direction at 1.0 m/s. Their approach speed is 2.0 m/s. After impact, B moves at 2.4 m/s and A at 1.2 m/s, so their separation speed is 1.2 m/s.

e = 1.2/2.0 = 0.60

This result says nothing by itself about how momentum is shared; masses are still needed. It only compares relative motion along the impact line. OpenStax distinguishes elastic and inelastic collisions and defines COR through the speed ratio in its College Physics collision chapter.

How bounce height reveals the coefficient of restitution

For a vertical drop, the speed just before impact is approximately √(2gH), where H is the release height measured from the ball’s center. The upward speed just after impact is approximately √(2gh), where h is the rebound height of the center. Dividing gives:

e = √(h/H)

This shortcut assumes a stationary, much more massive surface; negligible air drag; no important sideways motion; and height measurements referenced to the same point. The distinction between force, mass, and weight also matters when explaining the experiment, so review mass and weight if those ideas are being mixed.

Worked example: a moderate rebound

A ball is released from 1.44 m and reaches 0.81 m after its first impact.

• Height ratio: 0.81/1.44 = 0.5625.
• Take the square root: √0.5625 = 0.75.
• The estimated coefficient of restitution is 0.75.

Notice the two different percentages. The rebound height is 56.25% of the drop height, while the speed ratio is 75%. Calling the height ratio “the coefficient” skips the square root.

What happens over several bounces?

If e remains roughly constant, each new peak height is e² times the preceding peak:

h_n = e²ⁿh₀

Suppose e = 0.70 and the release height is 2.00 m. The model predicts 0.98 m after the first bounce, 0.480 m after the second, and about 0.235 m after the third. A University of Alabama laboratory guide develops this repeated-height relationship and suggests plotting the logarithm of height ratio against bounce number: General Physics II bouncing-ball lab.

A better phone-video experiment

Plan the measurement

1. Choose a safe ball and a flat, dry surface.
2. Fix a meter rule vertically in the same plane as the ball’s center.
3. Place the camera far enough away to reduce perspective distortion, with its lens level and square to the motion.
4. Release the ball without pushing or deliberately spinning it.
5. Record in slow motion and identify the center of the ball at release and at the rebound peak.
6. Repeat at least five times for each ball–surface pair.

Use metres consistently; the SI units reference can help students organize a clean data table. The final ratio is unitless, but the raw measurements should still carry units.

Sample analysis table

If the ruler can be read only to the nearest centimetre, reporting e = 0.75642 would imply false precision. State the measurement uncertainty and round sensibly. Use consistent notation; this physics symbols guide gives useful context for classroom writing.

Coefficient of restitution is not an energy percentage

For the simple ball–rigid-floor model, translational kinetic energy is proportional to speed squared. Therefore, the vertical kinetic-energy ratio immediately after and before impact is approximately:

K_after/K_before ≈ e²

If e = 0.75, then e² = 0.5625. About 56% of the vertical translational kinetic energy reappears immediately after impact in this simplified description—not 75%.

The remainder is not destroyed. It becomes internal deformation, heat, sound, vibration of the floor, rotation, and sideways motion. A mass–spring–damper model published through Marquette University connects bounce behavior with stiffness, damping, and contact time: A Mass-Spring-Damper Model of a Bouncing Ball.

Why the measured value changes

NASA experiments involving spheres and moisture found that added viscous liquid can reduce restitution as its viscosity or layer thickness increases. That specialized result supports a general lesson: COR depends on the contact system, not merely the label on the ball. See the NASA technical record.

Common reasoning mistakes

• Using h/H instead of √(h/H). Height follows speed squared.
• Measuring the top of the ball at one point and the center at another. Keep the reference point consistent.
• Assuming e is a permanent material constant. Report speed, temperature, surface, and inflation where relevant.
• Ignoring spin and oblique impact. The simple formula treats vertical center-of-mass motion.
• Calling lost translational energy “gone.” Energy changes form and spreads through the system.
• Using one trial. Repetition reveals release and reading variability.

The University of Iowa’s classroom demonstration compares several balls on a steel plate and provides a useful qualitative starting point before measurement: coefficient of restitution demonstration.

Quick quiz with explained answers

1. A ball drops from 1.00 m and rebounds to 0.64 m. What is e?
   √0.64 = 0.80.
2. If e = 0.50, what fraction of the previous height is predicted?
   e² = 0.25, so 25%.
3. Why is COR unitless?
   It divides one speed by another speed with the same units.
4. Does e = 0 mean all energy vanishes?
   No. It means no relative separation speed along the impact line; energy has changed form or entered other motion.
5. Why record the ball’s center?
   The derivation tracks the center-of-mass height before and after impact.
6. What camera error appears when the scale sits behind the trajectory?
   Parallax can make the rebound look too high or too low.
7. Can one ball have different e values on wood and concrete?
   Yes. COR describes the pair and test conditions.
8. What extra information is needed to solve two-body velocities fully?
   Masses plus momentum conservation and a consistent sign convention.
9. Why might repeated peaks depart from a perfect geometric sequence?
   COR can vary with impact speed, and measurement uncertainty becomes larger for small bounces.

Conclusion

The coefficient of restitution turns a familiar bounce into a careful study of collision physics. Use the relative-speed definition for general impacts and e = √(h/H) for a controlled vertical drop. Separate speed ratio from energy ratio, measure the center consistently, repeat trials, and describe the ball–surface conditions so another learner can reproduce the result. For historical context on experimental physics, see the site’s profile of Galileo and measurement.

Companion Indonesian guide: Koefisien Restitusi: Mengukur Seberapa Tinggi Bola Memantul.