practice 04 – – Due: Feb 12 2007, noon
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 8
.
6 m
/
s. Also, it has an acceleration in the
direction parallel to the walls of 2
.
2 m
/
s
2
.
It
hits the opposite wall at the same height.
18
.
4 m
2
.
2 m
/
s
8
.
6 m
/
s
a) What will be its speed when it hits the
opposing wall?
Correct answer: 9
.
80386 m
/
s (tolerance
±
1
%).
Explanation:
Let :
d
= 18
.
4 m
,
v
x
= 8
.
6 m
/
s
,
a
= 2
.
2 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
4
70698 m
8
.
6 m
/
s
9
80386 m
61
3071
◦
5
.
03537 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)
(8
.
6 m
/
s)
= 2
.
13953 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 ±nal
parallel velocity
v
y
is
v
y
=
a t
=
a d
v
x
=
(2
.
2 m
/
s
2
) (18
.
4 m)
(8
.
6 m
/
s)
= 4
.
70698 m
/
s
.
The velocities act at right angles to each
other, so the resultant velocity is
v
f
=
r
v
2
x
+
v
2
y
=
r
(8
.
6 m
/
s)
2
+ (4
.
70698 m
/
s)
2
=
9
.
80386 m
/
s
.
Question 2, chap 4, sect 3.
part 2 of 2
10 points
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View Full Documentpractice 04 – – Due: Feb 12 2007, noon
2
b) At what angle with the wall will the
particle strike?
Correct answer: 61
.
3071
◦
(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
p
v
x
v
y
P
= arctan
p
8
.
6 m
/
s
4
.
70698 m
/
s
P
=
61
.
3071
◦
.
Note:
The distance traveled parallel to the
walls is
y
=
1
2
a t
2
=
1
2
(2
.
2 m
/
s
2
) (2
.
13953 s)
2
= 5
.
03537 m
.
Question 3, chap 4, sect 4.
part 1 of 4
5 points
A ball of mass 0
.
4 kg, initially at rest, is
kicked directly toward a fence from a point
20 m away, as shown below.
The velocity of the ball as it leaves the
kicker’s foot is 18 m
/
s at angle of 60
◦
above
the horizontal. The top of the fence is 9 m
high. The kicker’s foot is in contact with the
ball for 0
.
05 s. The ball hits nothing while in
±ight and air resistance is negligible.
The acceleration due to gravity is 9
.
8 m
/
s
2
.
b
20 m
9 m
18
m
/
s
60
◦
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
b
Determine the magnitude of the average
net force exerted on the ball during the kick.
Correct answer: 144 N (tolerance
±
1 %).
Explanation:
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 Spring '08
 Turner
 Acceleration, Force, Correct Answer, m/s

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