Average force exerted by ball on the wall

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Homework Help Overview

The discussion revolves around calculating the average force exerted by a superball on rigid walls during elastic collisions, while neglecting gravity. The problem involves understanding the relationship between momentum change and time intervals between collisions.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants explore the concept of average force and its distinction from instantaneous force, questioning the relevance of collision time in the context of momentum change. There is an attempt to relate this problem to concepts in the Kinetic Theory of gases.

Discussion Status

Some participants have provided insights into the relationship between average force and momentum change over time, while others express a desire for further clarification on the differences between average and instantaneous force. The discussion is ongoing, with various interpretations being explored.

Contextual Notes

There is a noted uncertainty regarding the time of collision and its impact on the calculation of average force, as well as the implications of this in the context of the problem. Participants are encouraged to think critically about these assumptions.

Tanya Sharma
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Homework Statement



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 ?

Homework Equations


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

ehild
 
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?
 
Tanya Sharma said:
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?

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? :smile: 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
 
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 ?
 
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
 
ehild...thanks a lot
 

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