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After
impact block A continues up and block D continues down a distance, c, to
its final position, 3.
W2-3 =
T3 ˗ T2 which yields: ˗ wA
c +
wD c = 0
˗ (1/2) {[( wA + wD)]/g}
v22
c =
[ (1/2) {[( wA + wD)]/g} (2/7)dg ] / (wA
˗ wD ) =
(5/7) d (result)
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Also,
find the maximum distance, h, that
the rider, C, rebounds above the stop after striking it.
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In
this example the coefficient of restitution, e, relates the relative speed
of the collar, C, to the
stop
after impact to that before impact as follows: (denote by '
the condition immediately after
impact)
e
= ˗ v rel
after / vrel
before For this example: e
= ˗ (v2' ˗ 0) / (v2 ˗ 0) =
˗ v2' / v2
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For
maximum rebound no energy is lost so
e = 1 and v2' =
˗ v2
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Denote
by position 4 the maximum height, h, of the collar above the stop. Then apply the
principle
of work and energy again to the collar.
W2'-4 =
T4 ˗ T2' which yields: ˗ wC
h = 0 ˗
(1/2) {[( wC
)]/g} v22
Recall v22 =
(2/7) d g so h
= (1/2) (2/7) dg / g =
(1/7 ) d (result)
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