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marshall4
Nov2-03, 01:36 PM
If i have the equation

m1v1 + m2v2 = m1v1' + m2v2'

would i have to use any calulus to find either velocity of the second object?

What is that the equation for? Is it inelastic collisions???

pmb
Nov2-03, 01:54 PM
Originally posted by marshall4
If i have the equation

m1v1 + m2v2 = m1v1' + m2v2'

would i have to use any calulus to find either velocity of the second object?

What is that the equation for? Is it inelastic collisions???
You don't need calculus. Simple algebra. Just solve for v2.

This is the equation for conservation of momentum. It holds in both elastic and inelastic collisions. It's also valid in relativity if the m's are what some people call 'relativistic mass.' In such cases the m's are conserved. I.e. the sum of the m's before the collision is the sum of the m's after the collision.



Pete

marshall4
Nov2-03, 02:33 PM
Originally posted by pmb
You don't need calculus. Simple algebra. Just solve for v2.

This is the equation for conservation of momentum. It holds in both elastic and inelastic collisions. It's also valid in relativity if the m's are what some people call 'relativistic mass.' In such cases the m's are conserved. I.e. the sum of the m's before the collision is the sum of the m's after the collision.



Pete

Doesn't the ' mean prime???

What is the ' there for?

What is the equation for completely inelastic collisions?

pmb
Nov2-03, 02:53 PM
Originally posted by marshall4
Doesn't the ' mean prime???

What is the ' there for?

What is the equation for completely inelastic collisions?

Actually I meant to write a quote, i.e. "relativistic mass"

I want to applogize. I made an error in that last post. I made a web page on all of this a few weeks ago. See

www.geocities.com/physics_world/sr/inertial_mass.htm

The equation I posted above should have read

m1v1 + m2v2 = m3v3 + m4v4

This will hold in all collisions both eleastic and inelastic, and in both Newtonian mechanics and relativistic mechanics.

Particle N with mass mN and velocity vN. The mass may change in the collision.

Pete

nautica
Nov2-03, 05:45 PM
The ' is only there to differeniate the original value from the final value.

Nautica