Elastic and inelastic collision formula

1. May 11, 2010

trust_noone

i don't know how am i supposed to answer this Physics question..all i know its about generating a formula or something like that..here's the problem:

1.a)For an elastic, two-body head-on collisions, show that the general v2 - v1 = - (v2 - v1). That is, relative speed of recession after the collision is the same as the relative speed of approach before it.

1.b)In general, a collision is completely inelastic, completely elastic, or somewhere in between the degree of elasticity is sometimes expressed as the coefficient of restitution (e), defined as the relative velocity of recession and approach:
v2 - v1 = -e (v2o - v1o)
What are the values of e for elastic collision and a perfectly inelastic collision?

last question:
2. The coefficient of restitution for steel colliding with steel is 0.95. If steel is dropped from a height ho above steel plate, to what height will the ball rebound?

note: v,h are the coefficient while the next are its subscript..
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 11, 2010

rock.freak667

Well you know that in both cases, momentum is conserved. But only in the elastic collision is kinetic energy conserved.

(Two equations, two unknown )