Coefficient of restitution?
Note: velocity = v; kinetic energy = (1/2)mv^2
consider bouncing a ball on the earth.
Just before it hits the ground, it has velocity v1,initial, going down
when it leave the ground, it has velocity v1, initial, going up.
the coeffient of resitution eqn (in your phsyics book)
is something like:
e = (v2 – v1),final divided by (v1-v2) initial.
v1 is velocity of mass 1, v2 of mass 2.
Let the ball be mass 1, the earth be mass 2.
since we measure with respect to the earth, and the earth is so massive, its velocity before and after bounce is zero.
e = (-v1,final)/(v1,initial)
= (speed leaving ground)/(speed hitting ground)
(negative sign on v1,final is there because v1,f and v1,i are opposites… one is up (positive), the other down (neg.).
unless you have a high speed camera, balls speeds just before and just after bounce are hard to measure. however, consider energy…..
potential energy of a ball at height above ground h,i (initial)
PE,i = (1/2)mgh,i
when dropped from rest (v=0), the ball accelerates to ground. at ground…. h = 0, and all of the energy is converted to kinetic energy
KE,i = (1/2) m (v,i)^2 = PE,i = mgh,i
in the collision, energy is lost (heat, sound, deformation).
… so ball leaves bounce with less energy.
KE,f = (1/2)m (v,f)^2
and reaches a final height, h,f
PE,f = KE,f = 1/2m(v,f)^2 = mgh,f
this continues, the height after bounce getting smaller and smaller each time.
BUT, Coef of Restitution, e = (v,f/v,i) ….
so……………………….e^2 = (v,f/v,i)^2 = h,f/h,i
measure max height before bounce, h,i.
measure max height after bounce, h,f
e = sqrt (h,f/h,i) … less than one.
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