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Example: The system shown in the
figure below consists of block A of weight wA,
a light pulley, B,
a "collar", C, of weight, wC, and a block D, of weight, wD. A
mechanical stop, E, with a hole in it
is
a distance, d, below the rider as
shown. The system is released from
rest. Find the final
distance
of block D below the stop when its motion ceases. Also, find the maximum distance that
the
rider, C, rebounds after striking the stop.
Let wC
= (2/3) wA and wD
= (2/3) wA .
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Strategy: Use the principle of work and energy to
determine the speed of A, C, and D just
prior
to impact with the stop, E. After
impact block A continues to move up
and block D
continues
to move down and finally stops. Use
work and energy to calculate where block
D ends up below the stop. The collar, C, will rebound its maximum
amount if the
coefficient
of restitution, e, equals 1. Then
use work and energy to determine the final
height
that block C ends above the stop.
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Solution: Let 1
denote the initial position of the system (A + C + D). Let
2 denote the
position
of the system just prior to impact with the stop.
W1-2 =
T2 ˗ T1 which yields: ˗ wA
d +
wC d + wD d
= (1/2) {[( wA + wC
+ wD)]/g} v22
So
in position 2 v22 = 2[ ( wC + wD ˗ wA
) / ( wA + wC
+ wD )] d g
Or v22 = 2[ ( wC + wD ˗ wA
) / ( wA + wC
+ wD )] d g =
(2/7) d g
Click
here to continue with this example.
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