Momentum Question

[Dania]

Hi, can someone please explain to me fully, with and example [or any other method] what is meant by:

"Relative speed of approach = Relative speed of separation".

[I know it's silly, but I'm having problems trying to apply this principle]

Thanks

[astarmathsandphysics]

Not really to do with momentum more restitution. I have done a page on my website. Will post a link first thing in morning.

[Deadly_king]

Hi, can someone please explain to me fully, with an example [or any other method] what is meant by:

"Relative speed of approach = Relative speed of separation".

[I know it's silly, but I'm having problems trying to apply this principle]

Thanks

In other words it also implies that total kinetic energy is conserved.

Relative speed of approach is the difference between the speed of two particles which will be colliding in a matter of time.

Relative speed of separation is the difference between the speed of the two particles after collision.

Example : Nov 08 P1 No 10
Since collision is elastic the two particles will not merge after collision. Each will move separately at a certain speed such that total k.E is conserved.
For the two spheres not to merge ----> V₂ > V₁
Relative speed of approach = Relative speed of separation
u₁ - (-u₂) = v₂ - v₁

For this question....Answer is therefore D.

NOTE : u2 is taken to be negative since it moves in opposite direction.

If you have time......take a close look at this link.....it might help clear your doubts
http://thatlaureltree.blogspot.com/2008_03_01_archive.html

[Dania]

Thank you, that was really helpful

[Deadly_king]

Thank you, that was really helpful

You are most welcome

And don't worry it was not a silly question since many students get misguided by the question

I was one of them too 

[astarmathsandphysics]

http://www.astarmathsandphysics.com/a_level_maths_notes/M2/a_level_maths_notes_m2_restitution_momentum_collisions.html