Elastic and inelastic collision formula

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In the discussion, the focus is on understanding the formulas for elastic and inelastic collisions in physics. For elastic collisions, it is established that the relative speed of recession equals the negative of the relative speed of approach, expressed as v2 - v1 = - (v2 - v1). The coefficient of restitution (e) quantifies the elasticity of a collision, with values of e equal to 1 for elastic collisions and 0 for perfectly inelastic collisions. Additionally, the coefficient of restitution for steel colliding with steel is given as 0.95, leading to a question about the rebound height when dropped from a certain height. The key takeaway is that while momentum is conserved in all collisions, kinetic energy is conserved only in elastic collisions.
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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..
thanks in advance..
 
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Well you know that in both cases, momentum is conserved. But only in the elastic collision is kinetic energy conserved.

(Two equations, two unknown :wink: )
 
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