# Homework Help: Average force exerted by ball on the wall

1. Oct 6, 2012

### Tanya Sharma

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

A superball is to bounce elastically back and forth between two rigid walls at a distance d from each other. neglecting gravity and assuming the velocity of superball to be horizontal .Find the average force being exerted by the superball on each wall is ?

2. Relevant equations

3. The attempt at a solution

We consider rightwards direction to be positive .
If the ball moves at velocity v towards right wall then after collision it moves with velocity -v towards right.
The change in momentum of the ball is -2mv .The change in momentum of the wall is 2mv .
But we do not know the time of collision , so how do we calculate rate of change of momentum and hence force ?

2. Oct 6, 2012

### ehild

You need average force. What is the time interval between two subsequent collisions on the same wall?

ehild

3. Oct 6, 2012

### Tanya Sharma

2d/v .This is what i want to understand ...why time between two collisions ...the same concept comes in the Kinetic Theory of gases where a molecule collides with the wall
of the container ...Why are we not concerned with time of collision ...

How is average force on the wall different from force on the wall?

4. Oct 6, 2012

### ehild

Yes, it is the same as in the Kinetic Theory of gases when you derive the formula for pressure.

The change of the momentum is really equal to the integral of the force for the time of interaction. But we do not know usually the exact time dependence of the force. The integral of the force divided by the time of interaction is the average force, so we can say that the change of the momentum is equal to the average force multiplied the time of interaction. That is true for the ball, it experiences a great average force from the wall for the very short time of a collision, but it does not experiences any force between the collisions. .
But what does the wall feel? It is kicked by the ball again and again. Say, in 10 collisions, it gets 10*2mv momentum, in 10*2d/v time. The net momentum divided by the whole time is the average force the wall experiences: Δpwall/Δt==Fav, not in a single collision, but in the whole time.

ehild

5. Oct 6, 2012

### Tanya Sharma

Thanks ehild...I have understood what you have explained ...But somehow I am not able to distinguish properly between average force and force . Can u give an example where we have to calculate force and average force separately ? How are they mathematically different ?

6. Oct 6, 2012

### ehild

Do you feel what average is? Think about your weight. Sometimes you are slimmer, sometimes you gain some weight. But in average you are say 50 kg... You get that average by measuring your weight every day, and add the weights and divide by the number of the days.

You have a time dependent force. It is different in every second. You get the average if you measure it in every second, add the data and divide by the number of the data.

ehild

7. Oct 6, 2012

### Tanya Sharma

ehild....thanks a lot