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Phys 2101
Homework 5 Solution Spring `11
The numerical answers to the problems are calculated with the values used in the problems found
in the printed textbook; the values for your particular WebAssign assignment will be different.
1.P91.We use Eq. 95 to solve for
33
(
,
).
xy
(a) The
x
coordinate of the system
’
s center of mass is:
3
1 1
2
2
3
3
com
1
2
3
(2.00 kg)( 1.20 m)
4.00 kg
0.600 m
3.00 kg
2.00 kg
4.00 kg
3.00 kg
0.500 m.
x
m x
m x
m x
x
m
m
m
Solving the equation yields
x
3
=
–
1.50 m.
(b) The
y
coordinate of the system
’
s center of mass is:
3
1
1
2
2
3
3
com
1
2
3
(2.00 kg)(0.500 m)
4.00 kg
0.750 m
3.00 kg
2.00 kg
4.00 kg
3.00 kg
0.700 m.
y
m y
m y
m y
y
m
m
m
Solving the equation yields
y
3
=
–
1.43 m.
2. P94. We will refer to the arrangement as a
“table.”
We locate the coordinate origin at the left
end of the tabletop (as shown in Fig. 937). With +
x
rightward and +
y
upward, then the center of
mass of the right leg is at (
x,y
) = (+
L
,
–
L
/2), the center of mass of the left leg is at (
x,y
) = (0,
–
L
/2), and the center of mass of the tabletop is at (
x,y
) = (
L
/2, 0).
(a) The
x
coordinate of the (whole table) center of mass is
com
0
3
/ 2
32
M
L
M
M
L
L
x
M
M
M
.
With
L
= 22 cm, we have
x
com
= (22 cm)/2 = 11 cm.
(b) The
y
coordinate of the (whole table) center of mass is
com
/ 2
/ 2
3
0
35
M
L
M
L
M
L
y
M
M
M
,
or
y
com
=
–
(22 cm)/5 =
–
4.4 cm.
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View Full DocumentFrom the coordinates, we see that the whole table center of mass is a small distance 4.4 cm
directly below the middle of the tabletop.
3. 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
v
com
.
(a) The location of the first stone (of mass
m
1
) at
t
= 300
10
–
3
s is
y
1
= (1/2)
gt
2
= (1/2)(9.8 m/s
2
) (300
10
–
3
s)
2
= 0.44 m,
and the location of the second stone (of mass
m
2
= 2
m
1
) at
t
= 300
10
–
3
s is
y
2
= (1/2)
gt
2
= (1/2)(9.8 m/s
2
)(300
10
–
3
s
–
100
10
–
3
s)
2
= 0.20 m.
Thus, the center of mass is at
m
m
m
m
m
m
m
m
m
y
m
y
m
com
y
28
.
0
1
2
1
)
20
.
0
(
1
2
)
44
.
0
(
1
2
1
2
2
1
1
(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
2
3
2
3
3
11
1 1
2 2
com
1
2
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.
mm
m v
m v
v
m
m
m
m
4. (a)
The phrase (in the problem statement) “such that it [
particle 2] always stays directly above
particle 1 during the flight” means that the shadow (as if a light were directly above the particles
shining down on them) of particle 2 coincides with the position of particle 1, at each moment.
We
say, in this case, that they are vertically aligned.
Because of that alignment,
v
2
x
=
v
1
= 10.0 m/s.
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 Spring '07
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