Physics I
Review 1
Review Notes
Exam 1
Rev. 26-Sep-04 GB
R1-1
Definitions
Scalar:
Magnitude:
Vector:
A number – positive, negative, or 0.
Absolute value – positive or 0.
Magnitude (or length) and direction
in space.
Time:
Position:
Displacement:
Time interval:
t(scalar)
x (vector)
  
x  x  x 0
t  t  t 0
R1-2
Definitions (Continued)
Average or mean velocity is defined as follows:

v avg
 

x  x 0 x


t  t0
t
Instantaneous velocity or just “velocity”:


x d x

v  lim

t 0 t
dt
R1-3
Definitions (Continued)
Average acceleration is defined as follows:

a avg
 

v  v 0 v


t  t 0 t
Instantaneous acceleration or just “acceleration”:


2 
v d v d x

a  lim


t  0  t
dt dt 2
R1-4
Class #1
Take-Away Concepts
1D Equations of Motion for Constant Acceleration
Basic Equations
1. v  v 0  a  t  t 0 
2. x  x 0  v 0 ( t  t 0 )  2 a ( t  t 0 )
1
2
Derived Equations
1
x

x

( v 0  v)(t  t 0 )
3.
0
2
1
2
x

x

v
(
t

t
)

a
(
t

t
)
4.
(compare with 2.)
0
0
0
2
5. v  v 0  2a x  x 0 
2
2
R1-5
Problem Solving Strategy for
Projectile Motion
Make a table, see what you know and what you need to find.
a
v0
x0 or y0
vf
xf or yf
t-t0
X
Y
SAME
SAME
The common factor in both the X and Y equations is the
time at which something happens (last row).
R1-6
Hit the Falling Target Diagram
The target will drop at the instant
the ball leaves the launcher.
h

d
The objective is to adjust the angle  so that the ball
hits the falling target.
R1-7
Hit the Falling Target
Table of Kinematic Quantities
a
v0
x0 or y0
vf
xf or yf
t-t0
X ball
0
v0 cos()
0
v0 cos()
d
?
Y ball
-g
v0 sin()
0
?
?
SAME
Y target
-g
0
h
?
Same as ball.
SAME
We have all data in the “X ball” column except time.
Solve for that first.
R1-8
Hit the Falling Target
Solving for Time
d
x f  x 0  v 0 cos( ) ( t  t 0 )  ( t  t 0 ) 
v 0 cos( )
We don’t care what vf is for the ball or target (Y).
a
v0
x0 or y0
vf
xf or yf
t-t0
X ball
0
v0 cos()
0
v0 cos()
d
d/[v0 cos()]
Y ball
-g
v0 sin()
0
(don’t care)
?
SAME
Y target
-g
0
h
(don’t care)
Same as ball.
SAME
Next, use the kinematics equation to find yf.
R1-9
Hit the Falling Target
Solving for Final Y Position
Y ball:
Y target:

d
1 
d
y f  0  v 0 sin( )
 g
v 0 cos( ) 2  v 0 cos( ) 

1 
d
yf  h  0  g 
2  v 0 cos( ) 
2
2
Setting the two expressions equal, the “g” terms cancel and we are left with
d
h
sin( )
 h OR tan( ) 
cos( )
d
R1-10
Class #2
Take-Away Concepts
1.
2.
3.
4.
X and Y motions are independent.
In projectile motion problems, the acceleration is
constant = 9.8 m/s2 down (normally -Y direction).
Strategy for solving projectile motion problems:
Create a table, fill in known quantities, work on finding
unknown quantities.
Use time to connect information from one column to
another.
R1-11
Using Newton’s Second Law to
Solve Problems
1.
2.
3.
4.
5.
6.
Identify all forces acting on the object.
Pushes or Pulls
Friction (if specified)
Gravity
Normal (Surface) Forces
Choose a coordinate system.
If you know the direction of acceleration, one
coordinate axis should be in that direction.
Draw a “Free-Body Diagram.”
We will show you how in the next slide.
Express the force vectors in components.
This may require trigonometry.
Use Newton’s Second Law to write one
equation for each direction considered.
One equation for each unknown.
Solve the equation(s).
R1-12
Free-Body Diagrams
1.
2.
3.
4.
Draw the object as a box or a circle, detached
from everything else. (“free-body”)
Draw and label the force arrows acting on the
object, with all tails on the object. The arrows
should point in the correct directions relative to
your choice of coordinate system. If you have
some indication of the relative magnitudes of
the forces, you can adjust the lengths of the
arrows, but that is not critical.
It helps to put the coordinate axes in the
diagram to remember which direction is +.
It helps if you know the direction of
acceleration to align it with the + direction of
one of the axes.
a
Y
X
N
F
P
W = mg
R1-13
Example Problem in 1D
An Elevator Going Down
Consider an elevator moving downward and speeding
up with an acceleration of 2 m/s2. The mass of the
elevator is 100 kg. Ignore air resistance. What is the
tension in the elevator cable?
1.
Forces: Weight (W) down and Tension (T) up.
2.
Coordinates: +X down. (Why?)
3.
Free-body diagram:
4.
X Components: (W) and (–T). (Why – for T?)
5.
Second Law: (W) + (–T) = m a.
6.
Solve: T = W – m a = m g – m a = m (g–a)
T = 100 (9.8–2) = 780 N.
a
X
T
W = mg
R1-14
Newton’s Third Law Pairs:
How to Recognize Them
A pair of forces qualifies as a Newton’s Third Law Pair if (all of)
1.
2.
3.
They act on two different objects.
They are the same type of force.
Each object is a target for one force and a source for the other.
A pair of forces is not a Newton’s Third Law Pair if (any of)
1.
2.
3.
They act on the same object.
They are two different types, like normal force and gravity.
They are only equal and opposite for a certain combination of
accelerations and/or other conditions in the problem.
R1-15
Class #3
Take-Away Concepts
1.
2.
3.
Newton’s First Law: Nonet force, no change in motion.

Newton’s Second Law: Fnet  m a
Newton’s Third Law: All forces come in pairs.
4.
Solve force/acceleration problems with Newton’s
Second Law and free-body diagrams.
R1-16
A Common Example:
Atwood’s Machine
a
W = mg
X
W = Mg
T
T
m
w = mg
a
X
M
W = Mg
R1-17
Solution to Atwood’s Machine
T  mg  ma
Mg  T  Ma
Add these to eliminate T .
Mg  mg  Ma  ma
( M  m) g  ( M  m)a
M  m

a
g
 M  m
R1-18
Inclined Plane
R1-19
Coordinate Systems and
Free-Body Diagrams
Y
X
a
X
Use trigonometry to determine X & Y components
of forces not aligned with coordinate system.
R1-20
Solving for Acceleration
For mass 1:
T  m1g sin( )  m1 a
X:
Y:
N  m1g cos( )  0
For mass 2:
m 2g  T  m 2 a
X:
To solve for a, add the two X equations:
T  m1g sin( )  m 2 g  T  m1 a  m 2 a
m  m1 sin( )
a 2
g
m1  m 2
What would it mean if we found a < 0 after plugging in the values?
If  = 0, does the “inclined” plane resemble something in class?
R1-21
Class #4
Take-Away Concepts
1.
2.
3.
4.
Keep using the six-step process for doing Newton’s Second Law
problems for complex cases, it will help you keep things straight.
For each dimension and each object, you will get one equation. You
may or may not need to know the forces in the “normal” direction.
You should have the same number of unknowns as equations.
The easiest way to solve is usually to add the two equations resulting
from opposite ends of a rope or string, or opposite sides of a contact
surface where two objects push on each other.
R1-22
Important Facts About
Velocity and Acceleration Vectors
v
a
v
Same direction: speeding up.
a
v
a
Opposite directions: slowing down.
Right angles: changing direction, same speed.
R1-23
Uniform Circular Motion
“Uniform” circular motion means that the
object moves in a circle at a constant speed.
Some definitions and equations:
T = period = time to go around exactly once
r = radius of circle
v = speed (scalar, not vector)
circumfere nce 2  r
v

period
T
2r
T
v
R1-24
What is the Direction of
Acceleration?
Since the speed is not changing, only the
direction of velocity, acceleration must be always
at right angles to velocity. The acceleration
vector points inward, toward the center of the
circle. This is called centripetal acceleration
from Latin for “to go to or seek the center.”
Like the direction of the velocity vector, the
direction of centripetal acceleration is constantly
changing as the object moves around the circle.
v2
The magnitude of centripetal acceleration is given by a 
.
r
R1-25
Centripetal Force
 

Since  F  Fnet  m a , the net force on any object in uniform
circular motion must be given by

v2
Fnet  m a  m
r
and its direction is the same as acceleration: toward the center.
The net force in this case is called the centripetal force. It is not a
separate physical force in its own right, but only a name that we
give to the total or net force on an object in uniform circular
motion. It may, in fact, be the sum of several forces of several
different types.
R1-26
Example:
A Rock on a String
Twirl a 1 kg rock attached to a string in a 1 m radius vertical
circle. The speed is 4 m/sec. What forces act on the rock and
what are the directions of those forces?
R1-27
Case A:
Rock at the Top of the Circle
The center of the circle is below the rock, so acceleration is down.
1.
2.
3.
4.
5.
Forces: Weight (W) down and Tension (T) down.
Coordinates: +X down. (Why?)
Free-body diagram:
X Components: (W) and (T).
Second Law: (W) + (T) = m a.
 v2

 g
6. Solve: T  m a  W  m a  m g  m
 r

a
T
X
W = mg
T = 1 (16/1–9.8) = 6.2 N.
W = 9.8 N.
R1-28
Case B:
Rock at the Bottom of the Circle
The center of the circle is above the rock, so acceleration is up.
1.
2.
3.
4.
5.
Forces: Weight (W) down and Tension (T) up.
Coordinates: +X up. (Why?)
Free-body diagram:
X Components: (–W) and (T).
Second Law: (–W) + (T) = m a.
 v2

 g
6. Solve: T  m a  W  m a  m g  m
 r

T = 1 (16/1+9.8) = 25.8 N.
W = 9.8 N.
X
a
T
W = mg
R1-29
Class #5
Take-Away Concepts
1.
2.
Acceleration (or net force) at a right angle to velocity causes a change
of direction but not a change of speed.
As an object moves around a circle at a constant speed, it accelerates
toward the center with magnitude given by
v2
a
r


3. By Newton’s Second Law, Fnet  m a , the magnitude of the net force
on such an object must be given by

v2
Fnet  m a  m
r
4.
This net force is called centripetal force. It is not a separate physical
force but a name that we give the net force in this situation.
R1-30
Momentum of an Object
Definitions
We define momentum
for an object to be:

p  mv
Momentum is a vector. It is in the same direction as velocity.
SI units for momentum: kg m/s.
R1-31
Change of Momentum
Change of momentum is the difference between the final value
and the initial value.
Beware: We are subtracting vectors!
 



 p  p final  p initial  m v final  v initial 
final
initial
+3

+3
= 0 kg m/s
+3

-3
=
+6 kg m/s
R1-32
Impulse and the ImpulseMomentum Theorem
Impulse is defined to be the time integral of force. SI units = N s.
Like force, it is a vector. (Net or total force is implied.)


J   Fnet dt
Using the Fundamental Theorem of Calculus:



J   Fnet dt   p
In Physics, this is known as the Impulse-Momentum Theorem.
R1-33
Example Problem Using
Impulse-Momentum
An object of mass 0.5 kg is subjected to a
F
force in the +X direction that varies as shown 5N
in the graph from 0 to 7 seconds. Its initial X
velocity is zero. What is its final X velocity?
Doing this with F=ma would be hard. Doing 0
0
it with impulse-momentum is much easier.
7s t
J = area = 2+3+3.5+3+2+3+1.5 = 18 N s.
18 N s  J   p  p final  p initial
p final  p initial  p final  0  p final  18 kg m / s
v final  p final  m  36 m / s
R1-34
Class #6
Take-Away Concepts
1.
Momentum

 defined for an object:
2.
A new way to write Newton’s Second Law:
3.
Impulse defined:
4.
Impulse-Momentum Theorem:
5.
Average Force over time interval t (one dim.):
p  mv

 
dp
 F  Fnet  d t


J   F dt



J   F dt   p
Favg  t  J   F dt   p
R1-35
Internal and External Forces
Our system here consists of Objects A and B.
Forces between A and B are internal forces.
Forces on A or B from sources outside the system are external forces.
If we change the definition of the system, could that affect which forces
are internal and which are external?
F on A from C
F on B from C
External Forces
F on A from B F on B from A
Object A
Internal Forces
Object B
R1-36
The Momentum of a System
The momentum of a system is the sum of all the individual parts:
 N 
P   pi
i 1
Newton’s Second Law for each object:


 d pi
Fnet ,i  m i a i 
dt
Newton’s Second Law for the system:
 N 

dP
d pi N 

  Fnet ,i   F
d t i 1 d t i 1
all system
R1-37
Conservation of Momentum
(in a Nutshell)
Only external forces can change the momentum of a system.


dP
  Fext
dt
If the external forces cancel and/or can be neglected, then momentum
is constant (zero time derivative), or as physicists say, conserved.

dP
0
dt
R1-38
Class #7
Take-Away Concepts
1.
2.
Systems; internal/external forces in systems.
Momentum defined for a system:
 N 
P   pi
i 1
3.
Newton’s
 Second Law for a system:

dP
  Fext
dt
4. Conservation
of momentum when


dP
  Fext  0
dt
Pafter  Pbefore
R1-39
Conservation of Momentum in
Multiple Dimensions
 Each direction of motion is independent.
 Conservation of momentum occurs (or not) separately in each direction.
d Px
  Fext , x
dt
d Py
dt
  Fext , y
d Pz
  Fext ,z
dt
R1-40
Collisions in Multiple Dimensions
Y
X
Before:
After:
Px ,before  m1v1, x ,before  m 2 v 2, x ,before
Px ,after  m1v1, x ,after  m 2 v 2, x ,after
Py ,before  m1v1, y ,before  m 2 v 2, y ,before
Py ,after  m1v1, y ,after  m 2 v 2, y ,after
R1-41
Center of Mass
Center of mass defined for a system:
N
M   mi
i 1
x cm
y cm
1 N
  mi x i
M i 1
1 N
  mi yi
M i 1
R1-42
Class #8
Take-Away Concepts
1.
2.
3.
Momentum is conserved (or not) separately for each direction
if Fext for that direction is negligible (or not).
Conserved components of momentum do not mix with each
other.
Center of mass defined (x equation for example):
x cm
4.
1 N
  mi x i
M i 1
Velocity of the center of mass and system momentum:


P  M v cm
R1-43