Average force by ball on floor

[vissh]

heya :)

Homework Statement

(Q)A ball of mass m is dropped onto a floor from a certain height. The collision is perfectly elastic and the ball rebounds to the same height and again falls. Find the average force exerted by the ball on the floor during a long time interval.

Homework Equations

>Change in linear momentum of body = $$\int$$Fdt
.....where F is the net force on the body during the collision
>Velocity of separation = e (Velocity of appraoch)

The Attempt at a Solution

>As the ball is abt to strike, its velocity v = (2gh)^1/2
>The ball will exert force when it hits the floor and impulse of that force will be approximately equal to [neglecting impulse by gravity] "Change in momentum of Ball".
Thus, $$\int$$Fdt = mv-(-mv) = 2mv
-- Now as we are talking abt The Average force , F_av (let) , so it can be taken out of integration for it to be const on the time interval.
Thus, F_av $$\int$$dt = 2mv
--Now,the problem is that i can't decide what will be the required time interval.
Need a bit guidance in this ^.^
The answer of prob is F_av = mg [if u need xD hehe]
Thanks in advance :)

 

[Doc Al]

If you take a long enough interval, does the average position of the ball change? So what must be the net force on the ball?

 

[vissh]

hello Doc :D
Hmm By average position , you mean the position where the ball is most of the time in the long time interval . If that so I think the average position of the ball will be somewhat in air And the net force on it is going to be "mg" [as the force on striking the floor acts for very small time interval].
How i proceed with this further oO ? Some more guidance :D
Thanks in advance again ^.^

 

[Doc Al]

Hmm By average position , you mean the position where the ball is most of the time in the long time interval . If that so I think the average position of the ball will be somewhat in air

OK. The average position of the ball is somewhere between the floor and the height it bounces to. It's not getting very far.

And the net force on it is going to be "mg" [as the force on striking the floor acts for very small time interval].

Think long term. If the average force on the ball were 'mg', then after a few years that ball would be long gone, wouldn't you think?

 

[vissh]

Think long term.

Hmm..Still getting mg towards Earth's center :P
>The ball is affected by "mg" for most of the time[When its b/w the height 'h' and just above the floor].So i think its should be "mg"[Not getting how it could be somewhat different].
--->Does the "large force (but short time interval)which acts on striking" the floor also comes into account for the average oO]

-->If mg acts as an average the ball should be always getting accelerated towards center.But the "large force" continuosly changes ball's motion's direction. So the ball doesn't go any far off.

>Thats came into my mind.
Think i getting something left off , but can't figure it out.

And how [by this] we can find average force on floor??
Thanks for ur help :)

 

[Doc Al]

Hmm..Still getting mg towards Earth's center :P
>The ball is affected by "mg" for most of the time[When its b/w the height 'h' and just above the floor].So i think its should be "mg"[Not getting how it could be somewhat different].

When it's in the air, the force on it is "mg".

--->Does the "large force (but short time interval)which acts on striking" the floor also comes into account for the average oO]

Absolutely!

-->If mg acts as an average the ball should be always getting accelerated towards center.But the "large force" continuosly changes ball's motion's direction. So the ball doesn't go any far off.

Right.

Here's an analogy. Say everyday at 9am I give you $10. And at 10am I take away $10. Over the course of a year, on average how much money do you gain per day?

 

[Doc Al]

Here's an analogy. Say everyday at 9am I give you $10. And at 10am I take away $10. Over the course of a year, on average how much money do you gain per day?

Even better: Everyday at 9am, 10am, 11am, 12noon, and 1pm I give you $2. And at 2pm I take away $10.

 

[vissh]

Hehe on an average, I will get $0 per day.
--->Hmm now i think :-
On a Long time interval , the average force on the ball wud be abt (Impulsive force - mg)
i.e. F_av = [2m(2gh)^1/2]/(dt) - mg .
But if this average force exerts on ball, sud not the ball fly off in the sky on a long time interval [lol]
But i still don't get How the average force on the ball would help in finding the average force on the floor by ball Oo [hehe]

^.^

 

[Doc Al]

But if this average force exerts on ball, sud not the ball fly off in the sky on a long time interval [lol]

Right. So what's the average net force on the ball?

But i still don't get How the average force on the ball would help in finding the average force on the floor by ball Oo [hehe]

There are two forces acting on the ball. They add up to make the net force.

 

[vissh]

ayyyyyyeeeeeeeyoooooooooo xD
I now and finally getting F_av = [2m(2gh)^1/2]/(dt) - mg
If this is not correct , i will really start to laugh.
Can u tell what will be the average force and why ?
Sometimes, i can't think of the right direction for the solution hehe
Thank you for so much hints :)

 

[Doc Al]

ayyyyyyeeeeeeeyoooooooooo xD
I now and finally getting F_av = [2m(2gh)^1/2]/(dt) - mg
If this is not correct , i will really start to laugh.

That's ok, but what's the answer?

Fave = Ave Upward force from ground - Ave Downward force from gravity

What does that equal?

Can u tell what will be the average force and why ?

Sure I can. Reread all my hints!

 

[vissh]

hmmm..On re-reading and thinking :-
The ball is at the "same" average position in a long time interval.
So, the F_av = 0

 

[Doc Al]

hmmm..On re-reading and thinking :-
The ball is at the "same" average position in a long time interval.
So, the F_av = 0

You got it! So what's the average force exerted by the floor?

 

[vissh]

hmm so :-
F_av = Av Upward force from ground - Av Downward force from gravity
And F_av = 0
As the average force by gravity is "mg"[as Earth is applying this constant force throughout]
So, Av Upward force from ground = mg .
>WoW got it :)
-->But want to ask a thing . Could there be any other way of finding this average force ??
Like -- F_av $$\int$$dt = mv-(-mv) = 2mv
And then choosing the variation of 't' and solving for F_av.
->I saw someone took $$\int$$dt = 2(2h/g)^1/2 And found the correct answer "mg". I am not sure how he got that time interval . Is it just luck ?

Btw thanks doc :)

 

[Doc Al]

hmm so :-
F_av = Av Upward force from ground - Av Downward force from gravity
And F_av = 0
As the average force by gravity is "mg"[as Earth is applying this constant force throughout]
So, Av Upward force from ground = mg .
>WoW got it :)

Good!

-->But want to ask a thing . Could there be any other way of finding this average force ??
Like -- F_av $$\int$$dt = mv-(-mv) = 2mv
And then choosing the variation of 't' and solving for F_av.
->I saw someone took $$\int$$dt = 2(2h/g)^1/2 And found the correct answer "mg". I am not sure how he got that time interval . Is it just luck ?

No, not luck at all. Here's how you do it directly.

Each time the ball contacts the floor the momentum change is Δmv = 2m(2gh)^1/2
The time it takes for the ball to do a complete bounce to its original position is Δt = 2(2h/g)^1/2
So the average force exerted by the floor is Δmv/Δt = mg.

(But, being lazy, I like to do things the easy way first.)

 

[vissh]

oK Got that too :)
Thanks again ^.^

(But, being lazy, I like to do things the easy way first.)

hehe