PHYS 1441 – Section 002
Lecture #8
Monday, Feb. 11, 2008
Dr. Jaehoon Yu
•
•
•
Components and unit vectors
Motion in Two Dimensions
– Projectile Motion
– Maximum ranges and heights
Newton’s Laws of Motion
– Force
– Newton’s Law of Inertia & Mass
Today’s homework is homework #4, due 9pm, Monday, Feb. 25!!
Monday, Feb. 11, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
1
Announcements
• 1st term exam next Wednesday, Feb. 20
–
–
–
–
–
Time: 1 – 2:20pm
Place: SH103
Covers: Appendices, CH 1 – what we learn till this Wednesday
Style: mixture of multiple choices and essay problems
Class on Monday, Feb. 18: Jason will be here to go over the any
problems you would like to review
• Quiz Results
– Average: 3.5/8
• Equivalent to 40/100
• Quiz 1 result: 40/100
– Top score: 8/8
Monday, Feb. 11, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
2
Special Project
• Show that a projectile motions trajectory is a
parabola!!
– 20 points
– Due: Wednesday, Feb. 27
– You MUST show full details of computations to
obtain any credit
Monday, Feb. 11, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
3
Components and Unit Vectors
Coordinate systems are useful in expressing vectors in their components
y
(+,+)
Ay
A
(-,+)
(Ax,Ay)
q
(-,-)
u
r
A 

Monday, Feb. 11, 2008
(+,-)

Ax
ur
A cos q
 
2
Ax 
ur
A cos q
Ay 
ur
A sin q
x
A  Ax  Ay
2
ur
A sin q
u
r 2
A
cos 2 q  sin 2 q

PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu


2
}
Components
} Magnitude
2
u
r
 A
4
Unit Vectors
• Unit vectors are the ones that tells us the
directions of the components
• Dimensionless
• Magnitudes are exactly 1
• Unit vectors are usually expressed in i, j, k or
r r r
i, j, k
So the vector A can
be re-written as
Monday, Feb. 11, 2008
r
ur
ur
r ur
r
r
A  Ax i Ay j  A cos q i  A sin q j
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
5
Examples of Vector Operations
Find the resultant vector which is the sum of A=(2.0i+2.0j) and B =(2.0i-4.0j)

 
r
r
r
r
ur ur ur
C  A  B  2.0i  2.0 j  2.0i  4.0 j
r
  2.0  2.0  i
ur
C 
 4.0   2.0
2
r
  2.0  4.0  j

r
r
 4.0i 2.0 j  m 
2
 16  4.0  20  4.5(m)
q
2.0
tan
 tan
 27o
Cx
4.0
1
Cy
1
Find the resultant displacement of three consecutive displacements:
d1=(15i+30j +12k)cm, d2=(23i+14j -5.0k)cm, and d3=(-13i+15j)cm
ur ur ur ur
r
r
r
r
r
r
r
r
D  d 1  d 2  d 3  15i  30 j  12k  23i  14 j  5.0k  13i  15 j
r
r
r
r
r
r
 15  23  13 i   30  14  15  j  12  5.0  k  25i 59 j 7.0k (cm)

Magnitude
Monday, Feb. 11, 2008
ur
D



 25  59   7.0  65(cm)
2
2
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
2
6
Kinematic Equations in 2-Dim
x-component
y-component
vx  vxo  axt
v y  v yo  a y t
x
1
2
 vxo  vx  t
2
y
2
xo
x  vxot  ax t
1
2
Monday, Feb. 11, 2008
v
yo
 vy  t
v  v  2ay y
v  v  2a x x
2
x
y
1
2
2
2
yo
y  vyot  ayt
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
1
2
2
7
Ex.1 A Moving Spacecraft
In the x direction, the spacecraft has an initial velocity component of +22
m/s and an acceleration of +24 m/s2. In the y direction, the analogous
quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x and vx,
(b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s.
Monday, Feb. 11, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
8
How do we solve this problem?
1. Visualize the problem  Make a drawing.
2. Decide which directions are to be called positive (+) and
negative (-).
3. Write down the values that are given for any of the five
kinematic variables associated with each direction.
4. Verify that the information contains values for at least
three of the kinematic variables. Do this for x and y
separately. Select the appropriate equation.
5. When the motion is divided into segments, remember
that the final velocity of one segment is the initial velocity
for the next.
6. Keep in mind that there may be two possible answers to
a kinematics problem.
Monday, Feb. 11, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
9
Ex.1 continued
In the x direction, the spacecraft has an initial velocity component of +22
m/s and an acceleration of +24 m/s2. In the y direction, the analogous
quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x and vx,
(b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s.
x
ax
vx
vox
t
?
+24.0 m/s2
?
+22.0 m/s
7.0 s
y
ay
vy
voy
t
?
+12.0 m/s2
?
+14.0 m/s
7.0 s
Monday, Feb. 11, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
10
First, the motion in x-direciton…
x
ax
vx
vox
t
?
+24.0 m/s2
?
+22 m/s
7.0 s
2
1
v
t

a
t
x  ox 2 x
  22m s  7.0 s  
1
2
 24m s  7.0 s 
2
vx  vox  axt
  22 m s    24 m s
Monday, Feb. 11, 2008
2
2
 740 m
  7.0 s   190 m s
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
11
Now, the motion in y-direction…
y
ay
vy
voy
t
?
+12.0 m/s2
?
+14 m/s
7.0 s
y  voy t  a y t
1
2
2
 14m s  7.0 s  
1
2
12m s  7.0 s 
2
v y  voy  a y t
 14 m s   12 m s
Monday, Feb. 11, 2008
2
2
 390 m
  7.0 s   98 m s
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
12
The final velocity…
v
v y  98 m s
q
vx  190 m s
190 m s    98 m s   210 m s
A vector can be fully described when
1
q  tan  98 190   27 the magnitude and the direction are
v
2
2
given. Any other way to describe it?
Yes, you are right! Using
components and unit vectors!!
Monday, Feb. 11, 2008
v  vx i  vy j  190i  98 j  m s
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
13
If we visualize the motion…
Monday, Feb. 11, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
14