Thread: Momentum and coefficient of restitution

We performed an experiment in which we dropped different balls onto different surfaces in order to find different values of the coefficient of restitution.

The time from the first impact to the second impact is given by:

t2 =2e(2H/g)^1/2

it can be shown that

T = ((2H/g)^1/2)((1+e)/(1-e))

I don't know where I should start with this??

I have no idea how to answer this because I have no idea what all of your symbols mean. "g" I presume is the acceleration due to gravity- about 9.81 m/s^2. Is "H" the height from which the ball is dropped? What are "t" and "T"? The "coefficient of restoration" is the fraction of energy recovered after the bounce- and since the total energy, at the ball's highest point is all potential energy which is proportional to the height, the "coefficient of recovery" is just the height from which the ball was dropped, divided by the height to which it rebounds. Unless you are trying to do other than just determine the coefficient of recovery, you don't need any other formula.

Originally Posted by HallsofIvy

I have no idea how to answer this because I have no idea what all of your symbols mean. "g" I presume is the acceleration due to gravity- about 9.81 m/s^2. Is "H" the height from which the ball is dropped? What are "t" and "T"? The "coefficient of restoration" is the fraction of energy recovered after the bounce- and since the total energy, at the ball's highest point is all potential energy which is proportional to the height, the "coefficient of recovery" is just the height from which the ball was dropped, divided by the height to which it rebounds. Unless you are trying to do other than just determine the coefficient of recovery, you don't need any other formula.

The coefficient of restitution is the ratio of speeds after to before impact (or rather the normal components of the speeds).

CB