Why is it mentioned in the kinetic theory of gases that the time of collision with the container walls is negligible as compared with the time between collisions?
When considering impulse and hence change in momentum, aren't we interested in the the force exerted and the time for which that force is exerted(i.e. the contact time)?
In this molecular case, it seems we are concerned with the time it takes for the molecule to return back to its original position.

sulexk
 Follow
 3 followers
 6 badges
 Send a private message to sulexk
 Thread Starter
Offline6ReputationRep: Follow
 1
 30122010 14:02

 Follow
 2
 30122010 21:15
That's because the force is an average calculated from the change in momentum that occurs over the time taken between each "bounce", not over the time taken for the bounce itself.
The molecule applies a very large force on impact, but then spends a long time doing nothing.
The force is an average of this.
Think of a ball bouncing up and down on the floor with no loss of ke.
The average force the ball applies to the floor is taken over the total time the ball takes to make one complete "oscillation", not the time for which it is in contact with the floor.
The contact time would give the maximum force, but for the rest of the time the ball is applying no force to the floor. 
sulexk
 Follow
 3 followers
 6 badges
 Send a private message to sulexk
 Thread Starter
Offline6ReputationRep: Follow
 3
 30122010 22:03
(Original post by Stonebridge)
That's because the force is an average calculated from the change in momentum that occurs over the time taken between each "bounce", not over the time taken for the bounce itself.
The molecule applies a very large force on impact, but then spends a long time doing nothing.
The force is an average of this.
Think of a ball bouncing up and down on the floor with no loss of ke.
The average force the ball applies to the floor is taken over the total time the ball takes to make one complete "oscillation", not the time for which it is in contact with the floor.
The contact time would give the maximum force, but for the rest of the time the ball is applying no force to the floor.
But surely, the force has no affect before the bounce, I understand however that initial momentum could be taken, but the impulse initially would be zero, until they come into contact.
However by saying the average force, I guess then it refers to the force acting on the ball until it returns back to its initial point, Is that 1 bounce, when the ball returns to its initial point, and then would we normally time the second bounce from here?
Thank you Stonebridge.Last edited by sulexk; 30122010 at 22:05. 
 Follow
 4
 30122010 22:34
(Original post by sulexk)
It was mentioned "Between each bounce", what exactly, is this saying?
But surely, the force has no affect before the bounce, I understand however that initial momentum could be taken, but the impulse initially would be zero, until they come into contact.
However by saying the average force, I guess then it refers to the force acting on the ball until it returns back to its initial point, Is that 1 bounce, when the ball returns to its initial point, and then would we normally time the second bounce from here?
Thank you Stonebridge.
The pressure on the wall is the result of the average force applied by all the molecules.
If you divide the change in momentum of the molecule by the actual impact time, t, you get a large instantaneous force that lasts a short time. A sort of peak force.
However, if the molecule takes a time T to return to the wall, this change in momentum can be thought of as occurring over that total time.
In other words, every T seconds there is this change of momentum.
So the average force, spread out over the whole journey of the molecule, is change in momentum / T.
Imagine you and a friend both had guns that fired a stream of golf balls at a wall.
You use the same mass golf balls and fire with the same speed at the wall.
Each ball bounces straight back with speed v after impacting with speed v.
Each ball has a change in momentum of 2mv.
Your gun fires 10 balls a second and your friend's fires 5.
Which of you creates the larger force on the wall and why?
If it only depends on the change in momentum on impact (2mv) then both of you create the same force.
If it also depends on the time interval between each impact, then you create the greater force.
The average force on the wall will be 2mv / T where T is the time interval between each ball. (Your T=0.1s your friend's T=0.2s)
This is exactly what is happening with the gas molecule, except it is the same molecule coming back again and again to hit the wall. 
sulexk
 Follow
 3 followers
 6 badges
 Send a private message to sulexk
 Thread Starter
Offline6ReputationRep: Follow
 5
 31122010 01:46
(Original post by Stonebridge)
What it means is that the molecule is applying a force to the wall of the container that lasts a short time, and then spends a long time applying no force.
The pressure on the wall is the result of the average force applied by all the molecules.
If you divide the change in momentum of the molecule by the actual impact time, t, you get a large instantaneous force that lasts a short time. A sort of peak force.
However, if the molecule takes a time T to return to the wall, this change in momentum can be thought of as occurring over that total time.
In other words, every T seconds there is this change of momentum.
So the average force, spread out over the whole journey of the molecule, is change in momentum / T.
Imagine you and a friend both had guns that fired a stream of golf balls at a wall.
You use the same mass golf balls and fire with the same speed at the wall.
Each ball bounces straight back with speed v after impacting with speed v.
Each ball has a change in momentum of 2mv.
Your gun fires 10 balls a second and your friend's fires 5.
Which of you creates the larger force on the wall and why?
If it only depends on the change in momentum on impact (2mv) then both of you create the same force.
If it also depends on the time interval between each impact, then you create the greater force.
The average force on the wall will be 2mv / T where T is the time interval between each ball. (Your T=0.1s your friend's T=0.2s)
This is exactly what is happening with the gas molecule, except it is the same molecule coming back again and again to hit the wall.
So the force is "thought" to be acting throughout the journey, hence giving rise to an average force(of a single particle). The example makes great sense, if the frequency of collisions is greater then this will give rise to a greater pressure. Since there will be more collisions per second.
So when considering momentum at the particle level it would be suitable to make this assumption.
Although when a ball hits into a wall, I would consider the contact time of the ball with the wall.
Thank you 
 Follow
 6
 31122010 11:47
(Original post by sulexk)
So the force is "thought" to be acting throughout the journey, hence giving rise to an average force(of a single particle). The example makes great sense, if the frequency of collisions is greater then this will give rise to a greater pressure. Since there will be more collisions per second.
So when considering momentum at the particle level it would be suitable to make this assumption.
Although when a ball hits into a wall, I would consider the contact time of the ball with the wall.
Thank you
Here's an example I often use when thinking of this. It may help.
In the diagram on the left, a ball resting on the floor.
In the diagram on the right, the same ball bouncing perfectly elastically up and down on the floor. It rises and falls a height h.
What is the average force the ball applies to the floor in the two cases?
(That is, averaged out over a long time.)
No problem with the left diagram. The force is constant and equal to mg.
What about the right one? The force on impact is quite high, but this only lasts for a short time. I don't even know what that time is.
However, I do know that it takes the ball a time T to fall and T to rise, giving a total of 2T between each bounce.
I have worked out a formula for T in the diagram.
I also know that the change in momentum is 2mv for this elastic bounce.
I have also worked out a formula for v in the diagram.
To find the average force of the bouncing ball on the floor over a time 2T, I need
average force = change in momentum / 2T = 2mv/2T
Note that if I want the average over a longer interval, say 4T, I need to remember that the momentum changes twice (2 bounces) in this time. So I actually get the same value 4mv/4T. It doesn't matter how long an interval I take, the average force will be the same. This is the force the bouncing ball applies to the ground averaged over a long time. (It is the value the force would be if it was constant.)
What do you expect this force to be?
If you do the maths and find 2mv/2T by substituting in the values for v and T, you will find the answer. It should not surprise you.
By the way, I've seen this question (find the average force of the bouncing ball on the ground) asked at A level. It's worth making a note of.Last edited by Stonebridge; 31122010 at 11:49. 
sulexk
 Follow
 3 followers
 6 badges
 Send a private message to sulexk
 Thread Starter
Offline6ReputationRep: Follow
 7
 31122010 12:23
(Original post by Stonebridge)
The assumption works.
Here's an example I often use when thinking of this. It may help.
In the diagram on the left, a ball resting on the floor.
In the diagram on the right, the same ball bouncing perfectly elastically up and down on the floor. It rises and falls a height h.
What is the average force the ball applies to the floor in the two cases?
(That is, averaged out over a long time.)
No problem with the left diagram. The force is constant and equal to mg.
What about the right one? The force on impact is quite high, but this only lasts for a short time. I don't even know what that time is.
However, I do know that it takes the ball a time T to fall and T to rise, giving a total of 2T between each bounce.
I have worked out a formula for T in the diagram.
I also know that the change in momentum is 2mv for this elastic bounce.
I have also worked out a formula for v in the diagram.
To find the average force of the bouncing ball on the floor over a time 2T, I need
average force = change in momentum / 2T = 2mv/2T
Note that if I want the average over a longer interval, say 4T, I need to remember that the momentum changes twice (2 bounces) in this time. So I actually get the same value 4mv/4T. It doesn't matter how long an interval I take, the average force will be the same. This is the force the bouncing ball applies to the ground averaged over a long time. (It is the value the force would be if it was constant.)
What do you expect this force to be?
If you do the maths and find 2mv/2T by substituting in the values for v and T, you will find the answer. It should not surprise you.
By the way, I've seen this question (find the average force of the bouncing ball on the ground) asked at A level. It's worth making a note of.
So in the case of a bouncing ball I would consider the average force over a time T, in which case T is the time taken for a single bounce. That makes sense.
But in the case of the ball against the wall, I have seen in a question, where it gives the contact time, but I guess in reality if a ball was to collide into a wall it would not bounce of the wall again unless someone applied a force I.e. throwing the ball again.
I cannot find the centripetal force that allows a train to negotiate a bend. However, I think, that it may well perhaps be the reaction between the rails and the train track, that allows the train to negotiate the bend, however it maybe wrong and too vague. Any ideas?
what is the centripetal force that allows a train to negotiate a bend?
Thank you so much I really appreciate it Stonebridge. 
 Follow
 8
 31122010 13:13
(Original post by sulexk)
So in the case of a bouncing ball I would consider the average force over a time T, in which case T is the time taken for a single bounce. That makes sense.But in the case of the ball against the wall, I have seen in a question, where it gives the contact time, but I guess in reality if a ball was to collide into a wall it would not bounce of the wall again unless someone applied a force I.e. throwing the ball again.
In the kinetic theory case, you are considering millions of collisions by millions of molecules and the average force they exert over a long time.I cannot find the centripetal force that allows a train to negotiate a bend. However, I think, that it may well perhaps be the reaction between the rails and the train track, that allows the train to negotiate the bend, however it maybe wrong and too vague. Any ideas?
what is the centripetal force that allows a train to negotiate a bend?
Thank you so much I really appreciate it Stonebridge.
The wheels are also slightly angled ("coned") to produce a sideways component.
This acts as centripetal. Railway tracks are also banked slightly to help reduce the need for this force. If the force is too great it can cause the rails to buckle.Last edited by Stonebridge; 31122010 at 13:14. 
sulexk
 Follow
 3 followers
 6 badges
 Send a private message to sulexk
 Thread Starter
Offline6ReputationRep: Follow
 9
 31122010 13:47
(Original post by Stonebridge)
Yes, but 2T in my example as T is just the time for half the journey.
If you need to consider the contact time, you are calculating the actual force at impact over a short time.
In the kinetic theory case, you are considering millions of collisions by millions of molecules and the average force they exert over a long time.
The train's wheels have flanges that produce a sideways force.
The wheels are also slightly angled ("coned") to produce a sideways component.
This acts as centripetal. Railway tracks are also banked slightly to help reduce the need for this force. If the force is too great it can cause the rails to buckle.
When railway tracks are banked, what then acts as the centripetal force?
I understand that when a car negotiates a banked turn, then the horizontal component of the normal reaction force, between the slope and the car acts as the centripetal force. However, I have also learnt, that if the car's centripetal requirements meet that of the given(or provided) centripetal force by the road, then it can negotiate the bend without any friction. But if it goes too fast, then it needs a larger centripetal force, and then friction acts down the slope to keep it on course. But I guess, like a water ride, when a person is sliding up the side, then the angle between the person and slope is increasing, since, they require a larger centripetal force. So from F=mv^2/r, larger speed larger centripetal force.
Is it possible to negotiate a banked bend even if it was completely iced? since there would be no friction, if the car was travelling at the right speed, would this then be possible?
Sorry but I was wondering in the exams they ask questions on momentum and collisions:
I would very much like to know if this is correct?
Large mass with speed "v" collides into a small mass which is stationary or even moving in the opposite direction ===> the result would be that: both move in the direction of the large mass
Is this correct even if the small mass is moving with a tremendously greater speed than the larger mass.
Small mass colliding into a larger mass which is stationary or moving in the same direction ==> the result would be that: small mass rebounds and large mass carrys on in the same direction
Is this correct even if the small mass is moving with a tremendously greater speed than the larger mass.
There was once a 6 mark question in the jun04 paper, I have attached it.
Although In the mark scheme, it mentioned some points on the centre of mass, but I have never come across this idea in the textbook?
However that question was pretty challenging. 6 points were required.
Thank you!!Last edited by sulexk; 31122010 at 13:49. 
 Follow
 10
 31122010 14:33
(Original post by sulexk)
When railway tracks are banked, what then acts as the centripetal force?
I understand that when a car negotiates a banked turn, then the horizontal component of the normal reaction force, between the slope and the car acts as the centripetal force. However, I have also learnt, that if the car's centripetal requirements meet that of the given(or provided) centripetal force by the road, then it can negotiate the bend without any friction. But if it goes too fast, then it needs a larger centripetal force, and then friction acts down the slope to keep it on course. But I guess, like a water ride, when a person is sliding up the side, then the angle between the person and slope is increasing, since, they require a larger centripetal force. So from F=mv^2/r, larger speed larger centripetal force.
Is it possible to negotiate a banked bend even if it was completely iced? since there would be no friction, if the car was travelling at the right speed, would this then be possible?Sorry but I was wondering in the exams they ask questions on momentum and collisions:
I would very much like to know if this is correct?
Large mass with speed "v" collides into a small mass which is stationary or even moving in the opposite direction ===> the result would be that: both move in the direction of the large mass
Is this correct even if the small mass is moving with a tremendously greater speed than the larger mass.
Small mass colliding into a larger mass which is stationary or moving in the same direction ==> the result would be that: small mass rebounds and large mass carrys on in the same direction
Is this correct even if the small mass is moving with a tremendously greater speed than the larger mass.There was once a 6 mark question in the jun04 paper, I have attached it.
Although In the mark scheme, it mentioned some points on the centre of mass, but I have never come across this idea in the textbook?
However that question was pretty challenging. 6 points were required.
Thank you!!
During a collision like that, both masses experience the same change in momentum. This is because they experience the same force (Newton's 3rd Law)
 the force is mutual so acts for the same time
As momentum is mv and m doesn't change, then the larger the mass, the smaller the change in v. If one mass is very large, it will experience a very small change in its velocity while the other mass experiences a large change.
How much of a change depends on the actual values of the masses.
BTW. It's probably better to put new questions into their own thread. It stops the replies getting mixed up; but also people tend not to interrupt when they see that a thread has settled into just two people and treat it a bit as a "private" conversation. You will get more replies if you start a new thread for a new question.
Reply
Submit reply
Related discussions:
 Advanced Higher Physics 2013  2014: Discussion and Help ...
 AQA Physics Unit 2 (PHYA2) 9 June 2014
 AQA Physics PHYA5  28th June 2016 [Exam Discussion ...
 OCR Physics A (Alevel) Modelling, Exploring and Unified ...
 AQA Physics PHYA5  Thursday 18th June 2015 [Exam ...
 OCR Physics A G484  The Newtonian World  11th June ...
 Higher physics 2014/15
 EDEXCEL Unit 5 Physics Discussions  IAL only
 TSR Physics Society
 TSR Physics Society
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
 charco
 Mr M
 TSR Moderator
 Nirgilis
 usycool1
 Changing Skies
 James A
 rayquaza17
 Notnek
 RDKGames
 randdom
 davros
 Gingerbread101
 Kvothe the Arcane
 The Financier
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 Reality Check
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 LeCroissant
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Moltenmo
Updated: December 31, 2010
Share this discussion:
Tweet