Consider the following questionthis is not a homework problem,but it is one I want to discuss regarding its feature)
A ball is falling onto a ground from a certain height and collides elastically and reaches the same height and the process continues.What is the averages force on the ground?
There are plenty of methods available to do this.What I like most is to use the ROCKET EQUATION!!!
m(dv/dt)=F-u(dm/dt) F is the external force to the system.
It first may look strange.But if you think about it,there is nothing wrong in it.After all,it is meant for a system of particle and is quite general in application.
[If you doubt,you apply it correctly to the famous problems asking the average force in water jet impinging on a wall/chain hanging on a flat platform suddenly starts to fall freely and falls a distance x/conveyer belt prblems etc...Just you will have to consider that LHS m(dv/dt) denotes the force on the concerned body...may be it is not accelerating---like rain falls on a roof and we want to find the average force exerted on the roof.It will begiven by the m(dv/dt) term but individually (dv/dt) will have no meaning here]
Applying that in this problem you will see that the required average force is the external force Mg.And this is the correct answer.
Now,the current problem that is haunting me.I refer to the same problem I quoted first in this thread,by assuming not perfectly elastic collision.That is I am given a co-eff. of restitution e<1.Suppose,I try to find the average force.My method still predicts it is Mg.But,the calculation works are not easy.Because,each time you have to find (del p/del t) where each of these are changing.Then,you have to take the average...and it is by no meanss easy to see what is the result...
Can anyone tell me what is the correct result?By my tool,I am almost sure that it is Mg.But,is there any other means to verify that?