IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: Dania on September 25, 2010, 08:41:58 pm
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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 :)
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Not really to do with momentum more restitution. I have done a page on my website. Will post a link first thing in morning.
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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
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Thank you, that was really helpful :)
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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 :P
I was one of them too ;)
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http://www.astarmathsandphysics.com/a_level_maths_notes/M2/a_level_maths_notes_m2_restitution_momentum_collisions.html