2007
DOI: 10.1119/1.2731280
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Debouncing a Superball

Abstract: Superballs can be purchased in local toy stores. They are described as being highly elastic. For bounces on a wooden bench top, the coefficient of restitution, defined as the ratio of the velocity after collision to the velocity before collision, can be determined from the heights reached on successive rebounds. The value obtained is typically about e = 0.8.

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Cited by 5 publications
(4 citation statements)
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“…Even if the contention were on the dissimilar materials, it will fail to account for the reported values around the 0.6 range found in previous studies conducted by rebounding steel balls off massive steel plates [19]. The experiments described to demonstrate the debounce of a superball [20] appear to have some parallels to the situation here in that there is mass sensitivity involved and that the movement of one ball is stopped. However, the physics based on two mass direct interaction will not provide an adequate exposition.…”
Section: Resultsmentioning
confidence: 64%
“…Even if the contention were on the dissimilar materials, it will fail to account for the reported values around the 0.6 range found in previous studies conducted by rebounding steel balls off massive steel plates [19]. The experiments described to demonstrate the debounce of a superball [20] appear to have some parallels to the situation here in that there is mass sensitivity involved and that the movement of one ball is stopped. However, the physics based on two mass direct interaction will not provide an adequate exposition.…”
Section: Resultsmentioning
confidence: 64%
“…does not bounce at all [2]. More relevant experiments have been done by colliding gliders on an air track [1,3], but not with one of the gliders at rest against an end stop and not with gliders of variable mass and stiffness.…”
Section: R Crossmentioning
confidence: 99%
“…Considering equation ( 5) for the ball with mass m 1 , according to which the ball with mass m 1 will be stopped after the collision (v 1 = 0) and it will give the maximum energy to the ball with mass m 2 , it must be valid that em 1 − (1 + e + e 2 )m 2 = 0; thus [4]…”
Section: Collision Between the Ball With Mass M 1 And The Ball With Mmentioning
confidence: 99%