What is the force exerted on an elastic object, on elastic collision?

Oldor

1. The problem statement, all variables and given/known data

An elastic object (F = k e) is thrown horizontally at a wall, so it is travelling at constant velocity, which then rebounds elastically (energy conserved). I'm only interested in when the object first makes contact and decelerates to 0ms-1 which will be when the object is at maximum compression. I thought horizontal motion would make it a bit simpler.
The object will come to rest compressed (before rebounding), due to the force of the collision. The deceleration it undergoes will be an average one, half the maximum which is at maximum compression. At rest/compression, all of the object's kinetic energy will be converted into elastic potential energy (assuming the object doesn't break).

I'd like to know the average, or maximum, force that the object experiences, due to the object's elasticity/spring constant, and due to the object's mass. I can then see quantitatively, for example, if a less-stiff (lower k) object absorbs more energy on impact, so the force that it experiences is less but extension greater (if this is the case).


2. Relevant equations

F,max = - k x e
F,avg = m x a,avg
F,avg x t = -m x u
KE = 0.5 x m x v^2
PE = 0.5 x F x e


3. The attempt at a solution

Not sure if it's best to go about it in terms of energy or momentum.
If I could work out the time it took to be compressed/stop, or the distance it took (i.e. the compression), I think it would help.

The following didn't get me far:

PE = 0.5 x F,max x e
F,max = 2PE / e

e = F,max / k
= 2PE / k e

e^2 = 2PE / k

PE = KE = 0.5 x m x u^2

e^2 = m x u^2 / k

e^2 / u^2 = m / k
e/u = sqrt( m/k )
t = sqrt( m/k) Is this anything?

Substituted t into
F,avg x t = -m x u

To get
F,avg^2 = k x m x u^2

But using that doesn’t agree with the equations of motion (v^2 = u^2 + 2as) I don't think.

Any advice or an alternative method would be good (preferably along the lines of the above).

Dadface

An elastic collision is not one where there is zero loss of energy,energy is conserved in all collisions, it is one where there is zero loss of kinetic energy.Perfectly elastic collisions can happen with atomic scale objects ,for example gas atoms colliding with a wall, but not with macroscopic objects.The smaller the distortion at impact the smaller the amount of kinetic energy converted and the more elastic the collision is.A good example of this is given by Newton's cradle,the distortion of the steel balls on collision being much smaller than it would be if say tennis balls were used.

PhanthomJay

You were doing good until you started messing with the time, and incorrectly assumed that the time, t, was equal to the displacement, e, divided by the initial speed, u, when actually the time, t, is the displacement, e, divided by the average speed, u/2. But you don't need to calculate the time. You have already solved for e in your prior equation (e^2 = mu^2/k). Then F_max = ke, or F_avg =ke/2.

Oldor

So a more massive object would distort more, and a stiffer object would distort less.
For two objects of the same mass, would the stiffer experience a greater maximum force on collision/deceleration, to compensate for the lesser distortion in order to convert the same kinetic energy to elastic potential energy?

Also, is the average speed u/2 even though deceleration is not constant (acc. increases with F increasing with e)?

PhanthomJay

Yes, provided u is constant. Yes, although the acceleration is not constant, it is linear with the displacement, thus the average acceleration is a_max/2, and the average speed is u/2. Or look at it this way: Since the work done by the spring is 1/2ke^2, and for a force varying linearly with distance (as per F = kx), work is F_avg(e), then equate the two to get F_avg = F_max/2.

Oldor

I see, thanks, that helps alot