practice 07 – GADHIA, TEJAS – Due: Feb 27 2007, 10:00 pm
1
Question 1
part 1 of 2
10 points
Given:
G
= 6
.
672
×
10
−
11
N
·
m
2
/
kg
2
A coordinate system (in meters) is con
structed on the surface of a pool table, and
three objects are placed on the coordinate sys
tem as follows: a 1
.
2 kg object at the origin, a
3
.
5 kg object at (0,1
.
3 m), and a 4
.
3 kg object
at (3
.
6 m,0).
Find the resultant gravitational force ex
erted on the object at the origin by the other
two objects.
Correct answer: 1
.
67927
×
10
−
10
N (tolerance
±
1 %).
Explanation:
Given :
G
= 6
.
672
×
10
−
11
N
·
m
2
/
kg
2
,
m
1
= 1
.
2 kg
,
m
2
= 3
.
5 kg
,
and
r
= 1
.
3 m
.
1.2 kg
3.5 kg
4.3 kg
F
x
F
y
3.6 m
1.3 m
θ
The force exerted on the 1
.
2 kg mass by the
3
.
5 kg mass is in the positive
y
direction and
given by
F
y
=
G m
1
m
2
r
2
=
(
6
.
672
×
10
−
11
N
·
m
2
/
kg
2
)
·
(1
.
2 kg) (3
.
5 kg)
(1
.
3 m)
2
= 1
.
65813
×
10
−
10
N
.
The force exerted on the 1
.
2 kg object by
the 4
.
3 kg object is in the positive
x
direction
and is
F
x
=
G m
1
m
2
r
2
=
(
6
.
672
×
10
−
11
N
·
m
2
/
kg
2
)
·
(1
.
2 kg) (4
.
3 kg)
(3
.
6 m)
2
= 2
.
65644
×
10
−
11
N
.
The
resultant
F
is
found
from
the
Pythagorean Theorem:
F
R
2
=
F
y
2
+
F
x
2
= (1
.
65813
×
10
−
10
N)
2
+ (2
.
65644
×
10
−
11
N)
2
= 2
.
81996
×
10
−
20
N
2
.
so that
F
R
=
radicalbig
2
.
81996
×
10
−
20
N
2
=
1
.
67927
×
10
−
10
N
.
Question 2
part 2 of 2
10 points
At what angle does it act with respect to
the positive
x
axis? Let counterclockwise be
positive, within the limits of
−
180
◦
to 180
◦
.
Correct answer: 80
.
8981
◦
(tolerance
±
1 %).
Explanation:
tan
θ
=
F
y
F
x
θ
= tan
−
1
parenleftbigg
F
y
F
x
parenrightbigg
= tan
−
1
parenleftbigg
1
.
65813
×
10
−
10
N
2
.
65644
×
10
−
11
N
parenrightbigg
=
80
.
8981
◦
above the +
x
axis.
Question 3
part 1 of 1
10 points
An object is dropped from rest from a
height 4
.
8
×
10
6
m above the surface of the
earth.
The acceleration of gravity is 9
.
81 m
/
s
2
.
If there is no air resistance, what is its speed
when it strikes the earth?
Correct answer: 7
.
32847 km
/
s (tolerance
±
1
%).
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practice 07 – GADHIA, TEJAS – Due: Feb 27 2007, 10:00 pm
2
Explanation:
Let :
h
= 4
.
8
×
10
6
m
.
g
=
G M
E
R
2
E
and the potential energy at a
distance
r
from the surface of the earth is
U
(
r
) =
−
G M
E
m
r
.
Using conservation of energy to relate the
initial potential energy of the system to its
energy as the object is about to strike the
earth (
K
i
= 0),
K
f
+
U
f
−
U
i
= 0
K
(
R
E
) +
U
(
R
E
)
−
U
(
R
E
+
h
) = 0
1
2
m v
2
−
G M
E
m
R
E
+
G M
E
m
R
E
+
h
= 0
.
Solve for
v
:
v
=
radicalBigg
2
parenleftbigg
G M
E
R
E
−
G M
E
R
E
+
h
parenrightbigg
=
radicalBigg
2
g R
E
parenleftbigg
h
R
E
+
h
parenrightbigg
=
radicalBigg
2 (9
.
81 m
/
s
2
) (6
.
37
×
10
6
m)
4
.
8
×
10
6
m + 6
.
37
×
10
6
m
×
radicalbig
4
.
8
×
10
6
m
·
1 km
1000 m
=
7
.
32847 km
/
s
.
Question 4
part 1 of 1
10 points
An object is projected upward from the
surface of the earth with an initial speed of
4 km
/
s.
The acceleration of gravity is 9
.
81 m
/
s
2
.
Find the maximum height it reaches.
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 Spring '08
 Turner
 Mass, Correct Answer

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