Question

Trading Momenta in a Collision

Answer 1

Two particles mover perpendicular to each other until they collide. particle 1 has mass m and momentum of magnitude 2p, and particle 2 has mass 2m and momentum of magnitude p.

Answer 2

1) Suppose that after the collision, the particles "trade" their momenta and each particle is now moving in the direction in which the other had been moving. How much kinetic energy is lost in in the collsion?

2) Consider an alternative situation: this time the particles collide completely in-elastically. How much kinetic energy is lost in this case?

Answer 3

well, if m1v1 + m2v2 doesnt equal m1v'1 + m2v'2 then somethings been lost right?

Answer 4

m(2p) + 2m(p) = 4mp

m(p) + 2m(2p) = 5mp

Answer 5

(m1+m2)v3

3m = 4mp when p=3/4

and 3m = 5mp when p=3/5

kinetic before and kinetic after ... is what im thinking

1/2 3m (3/4)^2 = 1/2 3m (3/5)^2 + K

1/2 3m (3/4)^2 - 1/2 3m (3/5)^2 = K

1/2 3m [(3/4)^2 - (3/5)^2] = K

if anyting, what is wrong with my thought process

Answer 6

http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html

Answer 7

It is fairly straightforward to work out the answers, just remember that the kinetic energy of a particle of mass m and momentum p can be written as \[\frac{ p^2 }{ 2m }\]

Answer 8

thank you everyone. i went through this and understand it a little better. thanks again.