Version 065/ABAAB – Test #1 – Antoniewicz – (56445)
1
This
printout
should
have
20
questions.
Multiplechoice questions may continue on
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before answering.
001
9.0 points
A planet with a mass of 1
×
10
23
kg travels
around a star in a nearly circular obrbit in the
xy plane as shown in the diagram. Its speed
is constant at 15000 m
/
s.
Star
C
A
D
B
Which arrow best describes the direction of
Δ
P
in going from B to C?
a
e
c
g
h
b
d
f
1. g
.
2. a
3. c
4. e
5. h
6. d
7. b correct
8. f
Explanation:
We know that direction of
Δ
P
is given by
P
B
+
Δ
P
=
P
C
Δ
P
=
P
C

P
B
Thus, from the directions of
P
C
and
P
B
it
follows that the correct answer is
b
002 (part 1 of 3) 3.0 points
A soccer ball is kicked at an angle of 60
◦
to
the horizontal, with an initial speed of 22 m
/
s.
Assume for the moment that we can neglect
air resistance.
g
=

9
.
8 m
/
s
2
.
How long is the ball in the air?
1. 4.0354
2. 3.34348
3. 3.88828
4. 3.80756
5. 3.76546
6. 3.11528
7. 3.14878
8. 3.21289
9. 3.30169
10. 3.38383
Correct answer: 3
.
88828 s.
Explanation:
To determine how long the ball is in the
air, we can use the position update formula
derived from the momentum principle:
y
f
=
y
i
+
v
i,y
Δ
t
+
1
2
a
y
Δ
t
2
.
We just need to fill in the quantities that we
know. Clearly,
y
f
=
y
i
= 0 m and
a
y
=
g
=

9
.
8 m
/
s
2
. The only other thing we need to
solve for the time is the initial velocity in the
y
direction. We obtain this from trigonometry:
v
i,y
=
v
sin 60
◦
= (22 m
/
s) sin 60
◦
= 19
.
0526 m
/
s
.
So, rearranging the above equation to solve
for the time, we obtain
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Version 065/ABAAB – Test #1 – Antoniewicz – (56445)
2
0 =
v
i,y
Δ
t
+
1
2
a
y
Δ
t
2
⇒
v
i,y
Δ
t
=

1
2
a
y
Δ
t
2
⇒
Δ
t
=

2
v
i,y
a
y
=
2 (19
.
0526 m
/
s)
9
.
8 m
/
s
2
=
3
.
88828 s
.
003 (part 2 of 3) 3.0 points
How far does the ball go (horizontal distance
along the field)?
1. 26.7471
2. 46.4093
3. 33.8383
4. 35.348
5. 42.7711
6. 37.8442
7. 32.1637
8. 45.118
9. 26.0526
10. 30.6538
Correct answer: 42
.
7711 m.
Explanation:
There is no force in the
x
direction, so
v
x
is
constant. This makes this part simple:
v
x
=
Δ
x
Δ
t
Δ
x
=
v
x
Δ
t
= (
v
cos 60
◦
)(
Δ
t
)
= (22 m
/
s) cos 60
◦
(3
.
88828 s)
=
42
.
7711 m
.
004 (part 3 of 3) 3.0 points
How high does it go?
1. 18.5204
2. 19.2513
3. 16.5698
4. 11.8886
5. 18.8898
6. 11.3615
7. 13.6941
8. 18.1435
9. 14.3545
10. 13.1235
Correct answer: 18
.
5204 m.
Explanation:
To find the height, we can use the position
update formula again, although this time we
are solving for a di
ff
erent variable.
Now we
want to find
y
f
, and we need to plug in the
correct time that it takes to reach the peak.
Since the trip up and the trip down take the
same amount of time, we can simply divide
the total flight time in half to find that
Δ
t
peak
=
Δ
t
2
= 1
.
94414 s
.
Now we just plug into the formula above:
y
f
=
y
i
+
v
i,y
Δ
t
peak
+
1
2
a
y
Δ
t
2
peak
= 0 + (19
.
0526 m
/
s)(1
.
94414 s)
+
1
2
(

9
.
8 m
/
s
2
)(1
.
94414 s)
2
=
18
.
5204 m
.
005
9.0 points
A proton of mass
m
p
is moving along the ˆ
z
direction and undergoes a change in its speed
from 0.994c to 0.998c. What is the magnitude
and direction of the impulse acting on the
proton?
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 Fall '08
 Turner
 Proton, Mass, Velocity, Correct Answer, m/s

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