How Does the Rocket Equation Apply to Average Force in Elastic Collisions?

[neelakash]

Consider the following question:(this 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?

 

[variation]

Hello,

Do you connect your question with how we find the pressure or average force on the box by the gas molecules in it.
Here, the average force is what you want to find.
The first ideal comes to me is
[tex]\bar{F}=\frac{\Delta P}{\Delta t}.[/tex]
But you want to find the average force lasting for a long time.
The problem become the same as the average force in the gas example.

In elastic situation, the average force is
[tex]\bar{F}=\frac{2mv}{2v/g}=mg[/tex]
In inelastic situation, the method still works easily except different initial velocity.

Hope these helpful.

 

[neelakash]

there was nothing helpful...
this should be better---

Force at a time on the floor,F(t)=dp(t)/dt+Mg
Supposing that the initial time moment is t=0, the average force on the ground can be found by

Average force <F>=Limit(T-->0) (1/T) integration(0 to T) F(t) dt
=Limit(T-->0)[{p(t)-p(0)}/T +Mg]

Since the motion of the ball is confined in a finite region, p(T)-p(0) is always bounded, and so the limit is zero

 

[neelakash]

I did a mistake...I wrote Limit T--->0.I meant that T--->infinity.I say this as time of bounce is infinite w.r.t. the time of collision...However,I think I will have a more convincing arguement

 

[neelakash]

I started correctly but then,possibly,I made a mess of everything...

Please note my reasoning carefully...

The formula for averaging any quantity is-
<f(t)>=(1/T) integration (0 to T) [f(t) dt]

now,what does T mean here?T is the "period".In our problem T may be identified as the time in which the ball falls and rises up.That is I am denoting the moment the ball starts to fall as t=0 and the moment the ball comes to rest after rebound as t=T.Then p(0)=p(T)=0

So,(1/T)[p(T)-p(0)]=0

So,my earlier reasoning was wrong and this time it is better and does the job as well.

 

[variation]

Ha Ha !

At first, I have a mistake due to my quick thinking.
But your last reply do catch my original key point and something i drop without mention on purpose.

The average force exerted on the ball by the floor can be calculated by [tex]\bar{\mathbf{F}}=\frac{\Delta\methbf{p}}{\Delta t}.[/tex]

The mistake i make is that i forget the gravitational force on the left hand side, which is a sum of all forces exerted on the ball. Therefore the left hand side is [tex]\bar{\mathbf{N}}+m\mathbf{g}.[/tex], where [tex]\bar{\mathbf{N}}[/tex] is what you want.

I also dropped how i took average over time. As your last reply, i average each raising and falling over time and obtain [tex]\frac{\Delta\mathbf{p}}{\Delta t}=0[/tex]. Surely, i do another mistake in my first reply when calculating [tex]\Delta P.[/tex] Fortunately, two negative sign give a positive sign. My apology, mistake is mistake.

Finally, [tex]\bar{\mathbf{N}}=\frac{\Delta\methbf{p}}{\Delta t}\quad\Rightarrow\quad\bar{\mathbf{N}}+m\mathbf{g}=0\quad\Rightarrow\quad\bar{\mathbf{N}}=-m\mathbf{g}[/tex]


Goodluck

 

[neelakash]

Did my last post makes everything OK?

 

[variation]

It seems OK.


Best regards