Head on Collision: Conservation of Momentum Explained

[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.