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y
my
mm
com
m
=
+
+
=
+
+
=
11
2 2
12
044
2
020
2
028
..
bg
(b) The speed of the first stone at time
t
is
v
1
=
gt
, while that of the second stone is
v
2
=
g
(
t
– 100
×
10
–3
s).
Thus, the centerofmass speed at
t
= 300
×
10
–3
s is
()
(
)
( )
23
3
com
1
1
9.8 m/s
300 10 s
2
9.8 m/s
300 10 s 100 10 s
2
2.3 m/s.
mv
v
m m
−−
−
×+
×
−
×
+
==
++
=
11. We use the constantacceleration equations of Table 21 (with +
y
downward and the
origin at the release point), Eq. 95 for
y
com
and Eq. 917 for
G
v
com
.
(a) The location of the first stone (of mass
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Acceleration, Mass

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