oldmidterm 01 – KIM, JI – Due: Dec 18 2007, noon
1
Question 1, chap 1, sect 5.
part 1 of 1
10 points
A newly discovered Jupiterlike planet has
an average radius 10
.
6 times that of the Earth
and a mass 314 times that of the Earth.
Calculate the ratio of new planet’s mass
density to the mass density of the Earth.
Correct answer: 0
.
26364 (tolerance
±
1 %).
Explanation:
Let :
R
np
= 10
.
6
R
E
and
M
np
= 314
M
E
.
The volume of a sphere of radius
R
is
V
=
4
3
π R
3
. The (average) density
ρ
of a body is
the ratio of the body’s mass to its volume,
ρ
=
M
V
.
Planets are spherical, so the (average)
density of a planet of a given mass
M
and a
given radius
R
is
ρ
=
M
4
3
π R
3
.
Comparing the newly discovered planet to the
Earth, we have
ρ
np
ρ
E
=
M
np
4
3
π R
3
np
M
E
4
3
π R
3
E
=
M
np
M
E
parenleftbigg
R
np
R
E
parenrightbigg
3
=
314
(10
.
6)
3
=
0
.
26364
.
Question 2, chap 2, sect 5.
part 1 of 1
10 points
A body moving with uniform acceleration
has a velocity of 6
.
01 cm
/
s when its
x
coordi
nate is 3
.
69 cm.
If its
x
coordinate 3
.
05 s later is

8 cm,
what is the magnitude of its acceleration?
Correct answer: 6
.
45429 cm
/
s
2
(tolerance
±
1 %).
Explanation:
Basic Concept:
Kinematic Equations
x
=
x
0
+
v
0
t
+
1
2
a t
2
Solution:
Solving for
a
, we get
a
=
2[
x

x
0

v
0
t
]
t
2
=

6
.
45429 cm
/
s
2
2
.
Question 3, chap 2, sect 6.
part 1 of 1
10 points
An object is thrown downward with an ini
tial speed of 9 m
/
s from a height of 45 m
above the ground. At the same instant, a sec
ond object is propelled vertically from ground
level with a speed of 59 m
/
s
.
The acceleration of gravity is 9
.
8 m
/
s
2
.
At what height above the ground will the
two objects pass each other?
Correct answer:
36
.
8982
m (tolerance
±
1
%).
Explanation:
Let :
v
up
= 59 m
/
s
,
v
down
= 9 m
/
s
,
and
h
= 45 m
.
Basic Concepts:
y
down
=
h

v
down
t

1
2
g t
2
y
up
=
v
up
t

1
2
g t
2
.
Solution:
The
two
objects
pass
one
another
when
y
down
=
y
up
.
h

v
down
t

1
2
g t
2
=
v
up
t

1
2
g t
2
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oldmidterm 01 – KIM, JI – Due: Dec 18 2007, noon
2
h
=
v
down
t
+
v
up
t
=
⇒
t
=
h
v
down
+
v
up
= 0
.
661765 s
.
So,
H
=
v
up
t

1
2
g t
2
= (59 m
/
s) (0
.
661765 s)

1
2
(9
.
8 m
/
s
2
) (0
.
661765 s)
2
= 36
.
8982 m
.
Question 4, chap 3, sect 4.
part 1 of 1
10 points
Consider vectors
vector
A
and
vector
B
with coordinate
components shown in the illustration below.
A
B
A perspective drawing:
Each
coordinate has a length of
±
5 units.
To indicate the coordinates of each
vector, a line is projected to the hor
izontal plane then two lines are pro
jected to the horizontal coordinates.
As well, a line is directly projected
to the vertical coordinate.
What is the scalar product
vector
A
·
vector
B
for these
two vectors?
Correct answer:

19
units
2
(tolerance
±
1
%).
Explanation:
Basic Concept:
vector
A
·
vector
B
= (
A
x
ˆ
ı
+
A
y
ˆ
+
A
z
ˆ
k
)
(1)
·
(
B
x
ˆ
ı
+
B
y
ˆ
+
B
z
ˆ
k
)
,
where ˆ
ı
·
ˆ
ı
= ˆ
·
ˆ
=
ˆ
k
·
ˆ
k
= 1 and ˆ
ı
·
ˆ
= ˆ
ı
·
ˆ
k
=
ˆ
·
ˆ
k
= ˆ
·
ˆ
ı
=
ˆ
k
·
ˆ
ı
=
ˆ
k
·
ˆ
ı
= 0
.
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 Fall '08
 Turner
 Physics, Acceleration, Force, Friction, Mass, Velocity, Correct Answer

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