Head on Collision: Conservation of Momentum Explained

[gracy]

My teacher taught us conservation of momentum.She gave us an example (or way to think )as if one of the blocks has spring attached to it.As given in the image.As u1 >u2 there will be collision ,though surface is not shown it is assumed to be smooth so that friction does not act,this is necessary condition for external force to be zero and in turn for conservation of momentum.The block one will collide with block 2.So,in this process spring is compressed so in order to restore it's position it will apply force on both the blocks,force applied by spring on the left block would be in left direction and force applied by spring on the right block would be in right direction.So,left block which was having greater velocity will decelerate after collision and m2 will accelerate after collision.What I don't understand is why spring applies force on both the blocks to restore its position?Shouldn't it apply force only on m1 in left direction so that it comes to it's original position?

 

[PWiz]

No, the spring will apply a force on both of them. A body that undergoes elastic compression pushes on all objects attached to it along the axis of compression.While I don't want you to do what's not recommended by your teacher, I would suggest that you don't take this "spring" idea too far in more complex problems. Your intuition might work against you (as was in this case).

 

[gracy]

A body that undergoes elastic compression pushes on all objects attached to it along the axis of compression

But this site shows something different.
http://www.dummies.com/how-to/content/how-to-calculate-a-spring-constant-using-hookes-la.html

 

[PWiz]

A force is exerted on the wall as well.

 

[gracy]

But this site shows something different.
http://www.dummies.com/how-to/content/how-to-calculate-a-spring-constant-using-hookes-la.html

A force is exerted on the wall as well.

How this answers my question?

 

[PWiz]

The compressed string pushes both the masses.

 

[gracy]

My teacher also says that there is maximum compression when u1=u2 .I really did not understand.Because when u1=u2 there is no head on collision so there should be no compression.

 

[PWiz]

My teacher also says that there is maximum compression when u1=u2 .I really did not understand.Because when u1=u2 there is no head on collision so there should be no compression.

That does not make much sense to me either. Perhaps your teacher meant it in a different context? And for the compression thing, try this:
Use a small string and place it on your thumb. Press it in with your index finger. You'll notice that you feel the pressure on both your index finger as well as your thumb, even though it is your index finger that's pushing the string in.

 

[jbriggs444]

My teacher also says that there is maximum compression when u1=u2 .I really did not understand.Because when u1=u2 there is no head on collision so there should be no compression.

During a single collision the speeds of the two masses will be changing as the spring is compressing and then expanding. Before the moment of maximum compression, the rear mass will still be catching up to the front mass. After the moment of maximum compression, the rear mass will be falling behind. At the moment of maximum compression, the two masses will necessarily be moving at the same speed.

 

[gracy]

rear mass

Did you mean m1 in my case?

 

[gracy]

Because when u1=u2 there is no head on collision so there should be no compression.

What is wrong in it?

 

[jbriggs444]

Did you mean m1 in my case?

Yes.

 

[jbriggs444]

My teacher also says that there is maximum compression when u1=u2 .I really did not understand.Because when u1=u2 there is no head on collision so there should be no compression.

What is wrong in it?

The problem is the interpretation of u1 and u2. Your teacher is taking u1 and u2 to be the current instantaneous velocities of m1 and m2 during a single collision. You are taking them to be the velocities of m1 and m2 before a collision.

 

[gracy]

Ok,even if I take u1 and u2 as instantaneous velocities of m1 and m2 why there should be maximum compression when these instantaneous velocities are equal?

 

[PWiz]

Ok,even if I take u1 and u2 as instantaneous velocities of m1 and m2 why there should be maximum compression when these instantaneous velocities are equal?

At the moment of maximum compression, the two masses will necessarily be moving at the same speed.

 

[gracy]

place it on your thumb.

Do you want me to tie the string around my thumb?

 

[jbriggs444]

Ok,even if I take u1 and u2 as instantaneous velocities of m1 and m2 why there should be maximum compression when these instantaneous velocities are equal?

If u1 is greater than u2 then the masses are getting closer together. Compression is not yet maximum because it is still increasing.
If u1 is less than u2 then the masses are getting farther apart. Compression is not maximum because it was greater in the past.
The remaining possibility is that u1 = u2.

 

[PWiz]

Do you want me to tie the string around my thumb?

No, just hold it between your thumb and finger and then push it in with the help of your finger only.

 

[gracy]

At the moment of maximum compression, the two masses will necessarily be moving at the same speed.

That's what I want to understand why such obligation?What's the reason behind it?

 

[jbriggs444]

What is the first derivative of a continuous function at a position where it has a local maximum?

[Edit -- addition]

If you are standing on the side of a hill where the ground slopes up, are you at the top?

 

[gracy]

What is the first derivative of a continuous function at a position where it has a local maximum?

Sorry I have not taken maths.

 

[jbriggs444]

Sorry I have not taken maths.

See the edit to my prior post, if you are standing on the side of a hill where the ground slopes up, are you at the top?

 

[gracy]

if you are standing on the side of a hill where the ground slopes up, are you at the top?

No.I will be at at the top if I see ground sloping down.
if I am standing on the side of a hill where the ground slopes up,I am at the bottom most point.Right?

 

[gracy]

I think my picture of ground sloping up was not correct in my prior post.This one is correct.

 

[jbriggs444]

I think my picture of ground sloping up was not correct in my prior post.This one is correct.

Ok, good. We are making progress. The key point I wanted you to recognize is that if you are on the side of a hill where there is a slope, that you can climb higher yet. You are not at the top.

Now there is one feature of hills that is not reflected on your drawing -- they tend to be smooth. The tops will be rounded off and not pointy. If you are exactly at the top of a round hill can you see that the slope there must be zero?

[Edit: Can you at least see that it must be nearly zero so that we don't get into distracting quibbles about what it means to be exactly zero]

 

[gracy]

If you are exactly at the top of a round hill can you see that the slope there must be zero?

Yes.

 

[gracy]

comparing this analogy with my problem,who is playing the role of slope?

 

[jbriggs444]

comparing this analogy with my problem,who is playing the role of slope?

The slope is u1 minus u2 -- the rate at which the distance between m1 and m2 is decreasing.

 

[gracy]

And being on top of the hill is condition of maximum compression ,right?

 

[jbriggs444]

And being on top of the hill is condition of maximum compression ,right?

Right.

 

[gracy]

And what about bottom most point?Is it the condition prior to first collision between blocks?

 

[Delta2]

A spring pushes with equal and opposite force on each of its edges. Why this is happening? It boils down to Newton's 3rd law in classical mechanics the law of action-reaction.

 

[jbriggs444]

And what about bottom most point?Is it the condition prior to first collision between blocks?

The analogy attempts to get you to think about why u1 = u2 at the point of maximum compression, i.e. at the top of the hill. What happens at the bottom of the hill is not important.

 

[gracy]

Thanks.If you don't mind I wanted to make a point here.
Why you are not given any badges such as science advisor ,homework helper.I really wanted to know how one can become science advisor and homework helper

 

[jbriggs444]

https://www.physicsforums.com/help/medals/

In my opinion, the reward should be in the doing, not in the resulting recognition.

 

[gracy]

Take an example with front block moving with 4m/s and rear block moving with 12 m/s. As long as velocity of rear block is more it travels more compared to front and spring compresses. This compressed spring retards rear block and accelerates front block. So rear block speed decreases to 11,10,9 etc and front speed increases to 5,6,7 m/s
This decrement in rear block velocity occurs in first time or it goes to the left then again comes ,pushes the spring then spring applies force in left direction to rear block ,retards the rear blocks and it velocity decreased to 11m/s and block goes in left then again comes, pushes the spring then spring applies force in left direction to rear block ,retards the rear blocks and it velocity decreased to .10 m/s .Then again comes ,pushes the spring then spring applies force in left direction to rear block ,retards the rear blocks and it velocity decreased to 9 m/s and so on...
is it like this?

 

[jbriggs444]

Take an example with front block moving with 4m/s and rear block moving with 12 m/s. As long as velocity of rear block is more it travels more compared to front and spring compresses. This compressed spring retards rear block and accelerates front block. So rear block speed decreases to 11,10,9 etc and front speed increases to 5,6,7 m/s.

Yes.

The velocities will match at 8 m/s each. At that point the compression of the spring will no longer be increasing. It will be at its maximum. The rear block will still be slowing down. The front block will still be speeding up. The rear block will slow to 7, 6, 5, 4 while the front block speeds up to 9, 10, 11, 12. While this is happening, the compression on the spring will be reducing. Eventually it will no longer be compressed at all and will not be touching the rear block.

This decrement in rear block velocity occurs in first time or it goes to the left then again comes ,pushes the spring then spring applies force in left direction to rear block ,retards the rear blocks and it velocity decreased to 11m/s and block goes in left then again comes, pushes the spring then spring applies force in left direction to rear block ,retards the rear blocks and it velocity decreased to .10 m/s .Then again comes ,pushes the spring then spring applies force in left direction to rear block ,retards the rear blocks and it velocity decreased to 9 m/s and so on...
is it like this?

I have very great difficulty understanding what you are saying here. Possibly it is a language difficulty.

The word "decrement" implies an instantaneous reduction in velocity from one integer value to the next lower integer. That would be wrong. The phrase "it goes to the left then again comes" suggests a series of bounces. That would be wrong.

 

[sophiecentaur]

That's what I want to understand why such obligation?What's the reason behind it?

You have to be prepared to use some Maths here.