homework 03 – GAUTHIER, JOSEPH – Due: Sep 21 2007, 1:00 am
1
Question 1, chap 4, sect 3.
part 1 of 2
10 points
A particle travels horizontally between two
parallel walls separated by 18
.
4 m. It moves
toward the opposing wall at a constant rate
of 9
.
9 m
/
s. Also, it has an acceleration in the
direction parallel to the walls of 4
.
1 m
/
s
2
.
It
hits the opposite wall at the same height.
18
.
4 m
4
.
1 m
/
s
2
9
.
9 m
/
s
a) What will be its speed when it hits the
opposing wall?
Correct answer: 12
.
4931 m
/
s (tolerance
±
1
%).
Explanation:
Let :
d
= 18
.
4 m
,
v
x
= 9
.
9 m
/
s
,
a
= 4
.
1 m
/
s
2
,
Basic Concepts
Kinematics equations
v
=
v
o
+
g t
s
=
s
o
+
v
o
t
+
1
2
g t
2
d
a
7
.
6202 m
/
s
9
.
9 m
/
s
12
.
4931 m
/
s
52
.
4139
◦
7
.
0814 m
The horizontal motion will carry the parti
cle to the opposite wall, so
d
=
v
x
t
f
and
t
f
=
d
v
x
=
(18
.
4 m)
(9
.
9 m
/
s)
= 1
.
85859 s
.
is the time for the particle to reach the oppo
site wall.
Horizontally, the particle reaches the maxi
mum parallel distance when it hits the oppo
site wall at the time of
t
=
d
v
x
,
so the final
parallel velocity
v
y
is
v
y
=
a t
=
a d
v
x
=
(4
.
1 m
/
s
2
) (18
.
4 m)
(9
.
9 m
/
s)
= 7
.
6202 m
/
s
.
The velocities act at right angles to each
other, so the resultant velocity is
v
f
=
radicalBig
v
2
x
+
v
2
y
=
radicalBig
(9
.
9 m
/
s)
2
+ (7
.
6202 m
/
s)
2
=
12
.
4931 m
/
s
.
Question 2, chap 4, sect 3.
part 2 of 2
10 points
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homework 03 – GAUTHIER, JOSEPH – Due: Sep 21 2007, 1:00 am
2
b) At what angle with the wall will the
particle strike?
Correct answer: 52
.
4139
◦
(tolerance
±
1 %).
Explanation:
When the particle strikes the wall, the ver
tical component is the side adjacent and the
horizontal component is the side opposite the
angle, so
tan
θ
=
v
x
v
y
,
so
θ
= arctan
parenleftbigg
v
x
v
y
parenrightbigg
= arctan
parenleftbigg
9
.
9 m
/
s
7
.
6202 m
/
s
parenrightbigg
=
52
.
4139
◦
.
Note:
The distance traveled parallel to the
walls is
y
=
1
2
a t
2
=
1
2
(4
.
1 m
/
s
2
) (1
.
85859 s)
2
= 7
.
0814 m
.
Question 3, chap 4, sect 3.
part 1 of 1
10 points
A quarterback takes the ball from the line
of scrimmage, runs backward for 11
.
0 yards,
then runs sideways parallel to the line of
scrimmage for 11
.
0 yards.
At this point, he
throws a 30
.
0 yard forward pass straight down
the field.
What is the magnitude of the football’s
resultant displacement?
Correct answer: 21
.
9545
yards (tolerance
±
1 %).
Explanation:
Basic Concepts:
Δ
x
tot
= Δ
x
Δ
y
tot
= Δ
y
1
+ Δ
y
2
The displacements are perpendicular, so
d
tot
=
radicalBig
(Δ
x
tot
)
2
+ (Δ
y
tot
)
2
Given:
Δ
y
1
=
−
11
.
0 yards
Δ
x
=
±
11
.
0 yards
Δ
y
2
= 30
.
0 yards
Solution:
Δ
y
tot
=
−
11 yards + 30 yards
= 19 yards
Δ
x
tot
=
±
11 yards
Thus
d
=
radicalBig
(
±
11 yards)
2
+ (19 yards)
2
= 21
.
9545 yards
Question 4, chap 4, sect 4.
part 1 of 1
10 points
A ball is thrown and follows the parabolic
path shown. Air friction is negligible. Point
Q
is the highest point on the path. Points
P
and
R
are the same height above the ground.
Q
R
P
How do the speeds of the ball at the three
points compare?
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 Fall '07
 Gilbert
 Acceleration, Velocity, Correct Answer, Orders of magnitude, m/s, JOSEPH

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