Inelastic Collision Selfcheck Key

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Characterizing Collisions — Self-Check • INSTRUCTOR KEY
Model answers shown in red. Worked with g = 9.81 m/s² and the example data of Part 3.
1. Formulas
Flight interval: 𝑇𝑛 = 𝑡𝑛+1 − 𝑡𝑛 Rebound speed: 𝑤𝑛 = 1
2 𝑔 𝑇𝑛
Peak height: ℎ𝑛 = 1
8 𝑔 𝑇𝑛2 COR from flight times: 𝑒𝑛 = 𝑇𝑛
𝑇𝑛−1
COR from rebound heights: 𝑒𝑛 = √ ℎ𝑛
ℎ𝑛−1
Fraction of energy lost: 1 − 𝑒2
2. Calculations
Given: 𝑡1 = 0.45 𝑠, 𝑡2 = 1.25 𝑠, 𝑡3 = 1.97 𝑠
(a) 𝑇1 = 1.25 − 0.45 = 0.80 s 𝑇2 = 1.97 − 1.25 = 0.72 s
(b) 𝑒 = 𝑇2
𝑇1
= 0.72
0.80 = 0.90
(c) 𝑤1 = 1
2 (9.81)(0.80) = 3.92 m/s
(d) ℎ1 = 1
8 (9.81)(0.80)2 = 0.785 m
(e) 1 − 𝑒2 = 1 − (0.90)2 = 0.19 = 19%
(f) 𝑒 = 𝑒−0.105 = 0.90 (same e as part b — a good consistency check)
3. Explanations (model answers)
(a) Why is the peak at the midpoint of a flight interval?
Gravity has the same constant magnitude on the way up and on the way down, so the time to rise equals the time to
fall. The ball therefore reaches its highest point at the midpoint of the interval, t = 𝑇𝑛
2 .
(b) Why is the log-plot slope a more reliable value of e than one pair of times?
The slope is fit to all of the bounces at once, so random timing error in any single bounce is averaged out. A value
from just one pair of times depends entirely on the error in those two measurements, so it is far noisier.
Challenge. What would a flat (zero-slope) ln Tn vs. n graph mean?
A flat line means ln 𝑇𝑛 — and therefore 𝑇𝑛 — is constant. Then e = 𝑇𝑛
𝑇𝑛−1
= 1: no energy is lost and the collisions are
perfectly elastic; the ball would bounce forever to the same height.