What Are the Equations for a Ball Drop Experiment?

[physics88]

Homework Statement

We performed an experiment on elasticity by dropping various different balls on different surfaces.

1, Using the equations of motion firstly show that the velocity after the first bounce=e(2gH)^1/2

2, the time (t2) from the first impact to the second impact is given by

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

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


I understand that g=a and H=s when considering the equations of motion but I can't understand how to bring the e into the equations? any help would be great

 

[zhermes]

I'm not sure what 's' is, but e is the 'Coefficient of restitution' (pretty sure that's the right name), and is a way of measuring the energy loss per bounce. Actually, 'e' measures the ratio of velocities before and after the bounce.

If the object was perfectly elastic (doesn't dissipate any energy), then e=1 and the velocity before and after are the same. If e < 1, then some energy is lost, and the final velocity is a little smaller.

Knowing that e \equiv \frac{v_f}{v_i} can you derive the equation in '1' ?

 

[cupid.callin]

and is a way of measuring the energy loss per bounce.

More precisely its measure of loss of kinetic energy per bounce

Actually, 'e' measures the ratio of velocities before and after the bounce.

This is true only when other body is stationary ... when both bodies colliding are moving use the eqn: (velocity of separation) = e (velocity of approach)

 

[zhermes]

More precisely its measure of loss of kinetic energy per bounce

No, its actually the velocity (see the given equation; or the http://en.wikipedia.org/wiki/Coefficient_of_restitution" ), which can be related to the kinetic energy.

This is true only when other body is stationary ... when both bodies colliding are moving use the eqn: (velocity of separation) = e (velocity of approach)

You're giving the same equation... except replaced 'f' (final) with 'separation', and 'i' (initial) with approach... ?!
Furthermore, this is unrelated to whether the second body is stationary or not (actual, only whether it's inertial or not matters---technically). If you have a non-stationary second body (e.g. another particle instead of a floor/wall), you have to look at the ratio of differences of velocities (see the wiki linked above).

 

[cupid.callin]

we have same number of posts ... of course before i posted this reply ...

No, its actually the velocity (see the given equation; or the http://en.wikipedia.org/wiki/Coefficient_of_restitution" ), which can be related to the kinetic energy.

during any collision .. KE is used in deofrmign the objects /... this deformation is permanent for e=0 and KE is conserved if e=1 ... and of course in rest cases a part of KE is gained back.
e accounts for this loss ... though there is no direct eqn for relation b/w them but e was vorn because of loss in KE ... (here we go ... philosophy ... :D)

You're giving the same equation... except replaced 'f' (final) with 'separation', and 'i' (initial) with approach... ?!
Furthermore, this is unrelated to whether the second body is stationary or not (actual, only whether it's inertial or not matters---technically). If you have a non-stationary second body (e.g. another particle instead of a floor/wall), you have to look at the ratio of differences of velocities (see the wiki linked above).

That is what I'm saying .. you wrote \frac{v_f}{v_i} so i thought you are talking about this specific case where v of another body is 0
i just explained to to OP from where this \frac{v_f}{v_i} came so that he makes no mistake in some other question

if you look at wiki again ... they also have specified that \frac{v_f}{v_i} is the eqn for body falling on ground ...