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8.1.
Model:
The model rocket and the target will be treated as particles. The kinematics equations in two
dimensions apply.
Visualize:
Solve:
For the rocket, Newton’s second law along the
y
direction is
!"
net
R
R
22
RR
11
15 N
0.8 kg
9.8 m/s
8.95 m/s
0.8 kg
y
F
F
mg
ma
a
F
mg
m
#
$
#
%&
’
#
$
#
$
#
()
Using the kinematic equation
2
1
1R
0R
0R
1R
0R
R
1R
0R
2
(
)
,
y
y
y
v
t
t
a
t
t
#
*
$
*
$
! "
2
2
1
1R
2
30 m
0 m
0 m
8.95 m/s
0 s
t
#
*
*
$
1R
2.589 s
t
’#
For the target
1T
1R
(noting
),
tt
#
2
1
1T
0T
0T
1T
0T
T
1T
0T
2
x
x
x
v
t
t
a
t
t
#
*
$
*
$
! "! "
0 m
15 m/s
2.589 s
0 s
0 m
39 m
#
*
$
*
#
You should launch when the target is 39 m away.
Assess:
The rocket is to be fired when the target is at
0T
.
x
For a net acceleration of approximately
2
9 m/s in
the vertical direction and a time of 2.6 s to cover a vertical distance of 30 m, a horizontal distance of 39 m is
reasonable.
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View Full Document 8.2.
Model:
The model rocket will be treated as a particle. Kinematic equations in two dimensions apply.
Air resistance is neglected.
Visualize:
The horizontal velocity of the rocket is equal to the speed of the car, which is 3.0 m/s.
Solve:
For the rocket, Newton’s second law along the
y
direction is:
net
R
R
()
y
F
F
mg
ma
!
"
!
# $ # $
#$
2
1
8.0 N
0.5 kg
9.8 m/s
0.5 kg
y
a
%
&
’
!
"
(
)
2
6.2 m/s
!
Thus using
2
1
1
0
0
1
0
1
0
2
( )
,
yy
y
y
v
t
t
a t
t
!
*
"
*
"
# $
2
2
1
1R
2
20 m
0 m
0 m
6.2 m/s
0 s
t
!
*
*
"
# $
22
1
20 m
3.1 m/s
t
’!
1
2.54 s
t
Since
1
t
is also the time for the rocket to move horizontally up to the hoop,
2
1
1
0
0
1
0
1
0
2
xx
x
x
v
t
t
a t
t
!
*
"
*
"
# $# $
0 m
3.0 m/s
2.54 s
0 s
0 m
7.6 m
!
*
"
*
!
Assess:
In view of the rocket’s horizontal speed of 3.0 m/s and its vertical thrust of 8.0 N, the aboveobtained
value for the horizontal distance is reasonable.
8.3.
Model:
The asteroid and the giant rocket will be treated as particles undergoing motion according to the
constantacceleration equations of kinematics.
Visualize:
Solve:
(a)
The time it will take the asteroid to reach the earth is
6
5
displacement
4.0 10 km
2.0 10 s
56 h
velocity
20 km/s
!
"
"
!
"
(b)
The angle of a line that just misses the earth is
11
6
00
R
R
6400 km
tan
tan
tan
0.092
yy
##
$$
%&
"
’
"
"
"
(
)*
!
+,
(c)
When the rocket is fired, the horizontal acceleration of the asteroid is
9
2
10
5.0 10 N
0.125 m/s
4.0 10 kg
x
a
!
""
!
(Note that the mass of the rocket is much smaller than the mass of the asteroid and can therefore be ignored
completely.) The velocity of the asteroid after the rocket has been fired for 300 s is
.
 .
2
0 m/s
0.125 m/s
300 s
0 s
37.5 m/s
x
x
x
v
v
a
t
t
"
/
$
"
/
$
"
After 300 s, the vertical velocity is
4
2 10 m/s
y
v
"!
and the horizontal velocity is
37.5 m/s.
x
v
"
The deflection
due to this horizontal velocity is
1
4
37.5 m/s
tan
tan
0.107
x
y
v
v
$
"
’
"
"
(
!
That is, the earth is saved.
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View Full Document8.4.
Model:
We are using the particle model for the car in uniform circular motion on a flat circular track.
There must be friction between the tires and the road for the car to move in a circle.
Visualize:
Solve:
The centripetal acceleration is
!"
2
2
2
25 m/s
6.25 m/s
100 m
r
v
a
r
#
#
#
The acceleration points to the center of the circle, so the net force is
! "
2
1500 kg
6.25 m/s , toward center
9380 N, toward center
r
F
ma
##
#
!
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This note was uploaded on 02/19/2010 for the course MTHSC 104 taught by Professor Simms during the Spring '10 term at Clemson.
 Spring '10
 SIMMS
 Math, Equations

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