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 10Since 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 ----> V
2 > V
1Relative speed of approach = Relative speed of separation
u
1 - (-u
2) = v
2 - v
1For 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