Physics Semester 1 Review and Tutorial

The slides used in this tutorial are color coded. If
you are experiencing difficulty with one aspect of
your understanding than another you might find
this coding useful.
Slides with
red
backgrounds
involve word
problems.
Slides with
tan
backgrounds
involve
matching
concepts.
Slides with olive
backgrounds
involve reading
data tables.
Slides with
green
backgrounds
involve
graphing.

Dr. Fiala is traveling on his
Harley at a constant 13.67 m/s.
What is the distance
traveled by Doc in 7.32
seconds?

SOLUTION:
K U
Find the distance traveled.
Vi&f = 13.67 m/s Xf
ti = 0 s
tf = 7.32 s
Xi = 0 m Xf= m
a = 0 m/s2

SOLUTION:
K U
Find the distance traveled.
Vi&f = 13.67 m/s Xf
ti = 0 s
tf = 7.32 s
Xi = 0 m Xf= 100.07 m
a = 0 m/s2

Dr. Fiala notices he is now
traveling at a constant
49.21 km/h. What is the distance
in meters traveled by Doc in 7.32
seconds?

SOLUTION:
49.21 km
h
Dimensional analysis.
49.21 km/h = 13.67 m/s
So the Xf remains
100.07 m
1 h
60 min
1000 m
1 km
X
1 min
60 s
XX
M K H D 0 d c m
1 x x x

Dr. Fiala jumps in his un-
started car. He accelerates at a
rate of 4 m/s2 for 8 seconds.
How far did Doc travel?

vi = 0 m/s vf =
ti = 0 s Xf =
tf = 8 s
a = 4 m/s2
Xi = 0 m
Xf = 128 m
Doc’s final position.

Displacement
Velocity
Acceleration
Inertia
Force
Momentum
• The change in the rate or
direction of motion.
• The resistance to a change in an
object’s current state of motion.
• A change in position.
• A push or a pull that tends to
accelerate an object.
• The movement of an object in a
specific direction over time.
• The product of mass times
velocity.

Displacement is a change in position.
Velocity is the movement of an object in a
specific direction over time.
Acceleration is the change in the rate or direction
of motion of an object.
Inertia is the resistance to a change in an object’s
current state of motion.
Force is a push or a pull that tends to accelerate an
object.
Momentum is the product of mass times velocity.

Time
(s)
Object #1
Position (m)
Object #2
Position (m)
Object #3
Position (m)
Object #4
Position (m)
1 16 4
2 4
3 48 24 6
4 16 32

Time
(s)
Object #1
Position (m)
Object #2
Position (m)
Object #3
Position (m)
Object #4
Position (m)
1 16 4 8 2
2 32 8 16 4
3 48 12 24 6
4 64 16 32 8

Time
(s)
Object #1
Velocity (m/s)
Object #2
Velocity (m/s)
Object #3
Velocity (m/s)
Object #4
Velocity (m/s)
1 16 8 2
2 9 4
3 13 8
4 10

Time
(s)
Object #1
Velocity (m/s)
Object #2
Velocity (m/s)
Object #3
Velocity (m/s)
Object #4
Velocity (m/s)
1 16 4 8 2
2 14.5 6 9 4
3 13 8 10 6
4 11.5 10 11 8

Vy = 0 m/s
ty = 5 s
Yf = Yi + Vi t + ½ gt2
Vi = 48.5 m/s
Vf
2 = Vi
2 + 2g Δy
Vi = 48.5 m/s

Position Graph
Can you predict
the slope shape
and orientation of
both the velocity
and acceleration
graphs?
Graph Options

Position Graph
Can you predict
the slope shape
and orientation of
the velocity graph?
Graph Options
V = 0 m/s
V = 0 m/s
These two graphs
begin with positive
velocity that is
decreasing over
time.

Position Graph
Can you predict
the slope shape
and orientation of
the acceleration
graph?
Graph Options
V = 0 m/s
This graph both decreasing positive velocity
and increasing negative velocity over time
caused by constant negative acceleration
(yellow arrow).

Velocity Graph
Position Graph
Acceleration
Graph
Vy = 0 m/s
Vy = 0 m/s
-V
+V
g = -9.8 m/s2

If Dr. Fiala starts from a full
stop and accelerates at
2.25 m/s2, how incredibly fast
will he be traveling when he
has traveled 1530 meters?

SOLUTION:
K U
Calculate final velocity without knowing time.
Vi = 0 m/s Vf
Xi = 0 m tf
Xf = 1530 m
a = 2.25 m/s2
ti = 0 s
Vf= 82.98 m/s

If Dr. Fiala starts from a full
stop and accelerates at
2.25 m/s2, how long will it take
him to drive 1530 meters?

SOLUTION:
K U
Solve for time.
Vi = 0 m/s tf
Xi = 0 m Vf
Xf = 1530 m
a = 2.25 m/s2
ti = 0 s
Vf = 82.98 m/s
tf= 36.88 s

0
m/s2
0 m/s2
30 m 15 m33.75 m 92 m

0
50
100
150
200
0 2 4 6 8 10 12
Position
(m)
Time (s)
0
20
40
60
80
100
0 2 4 6 8 10
Position
(m)
Time (s)
-3
-2
-1
0
1
2
0 2 4 6 8 10 12
Position
(m)
Time (s)
-20
0
20
40
60
80
100
120
0 2 4 6 8 10 12
Position
(m)
Time (s)

0
20
40
60
80
100
120
140
160
180
0 2 4 6 8 10 12
Position
(m)
Time (s)
0
2
4
6
8
10
12
14
16
0 2 4 6 8 10 12 14 16
Time (s)
Velocity
(m/s)

Motion
Map
Graph Options
Can you predict
the slope shape
and orientation of
the position,
velocity, and
acceleration
graphs based on
this motion map?

Motion
Map
Graph Options
Can you predict
the slope shape
and orientation of
the position
graph?
These three graphs illustrate an object
moving to the left over time.
This graph can be eliminated because
it illustrates an object that begins
moving back to the right over time.

Motion
Map
Graph Options
Can you predict
the slope shape
and orientation of
the velocity graph
now based on this
position graph?
This is the only graph that illustrates and
object moving to the left with changing
velocity (curved slope, fast to slow to
fast) over time.

Motion
Map
Graph Options
Can you predict
the slope shape
and orientation of
the acceleration
graph now based
on this velocity
graph?
This is the only graph
that illustrates negative
velocity (moving to the
left) the whole time. It
is under the influence
of constant positive
and then constant
negative (yellow
arrows) acceleration.
V = 0 m/s
V = 0 m/s
This graph can be
eliminated because it
illustrates an object
that is moving slowly
at the beginning.

Position Graph
Motion Map
Acceleration GraphVelocity Graph

Graph Options
Motion
Map
Can you predict the
slope shape and
orientation of the
position, velocity,
and acceleration
graphs based on
this motion map?

Graph Options
Motion
Map
Can you predict the
slope shape and
orientation of the
position, velocity,
and acceleration
graphs?

Graph Options
Motion
Map
Can you predict the
slope shape and
orientation of the
position, velocity,
and acceleration
graphs?
Position Graph
Velocity Graph

Motion
Map
Graph Options
Can you predict the
slope shape and
orientation of the
position, velocity,
and acceleration
graphs based on
this motion map?

Motion
Map
Graph Options
Can you predict the
slope shape and
orientation of the
velocity, and
acceleration graphs
now based on this
position graph?
Position Graph

Motion
Map
Graph Options
Can you predict the
slope shape and
orientation velocity
graph now based
on this position
graph?
Position Graph
These four graphs illustrate
positive velocity over time. The
ones circled in orange can be
eliminated because they indicate
changing acceleration which we
will not study in this class.
The one circled
in green can
be eliminated
because the
velocity does
not change.

Motion
Map
Graph Options
Can you predict the
slope shape and
orientation of the
acceleration graph
now based on this
velocity graph?
Position Graph
Velocity Graph

18 m/s 2 m/s17.5 m/s
0 m/s3
0 m/s3

2.25 + 3.25 =
2.25 + 3.25 =
2.25 + 3.25 =

SOLUTION:
Adding vectors.
+2.25 + +3.25 = +5.5
+2.25 + -3.25 = -1.00
+2.25 + +3.25 = +3.95

Determine the mass of a 153.08 N
object.
153
N
?
kg

SOLUTION:
K U
Calculate mass.
W= 153.08 N m
g = -9.8 m/s2
m = 15.62
kg

Determine the force of friction on a
15.62 kg object traveling at a
constant horizontal velocity of
3.62 m/s while experiencing an
applied force of 6 N.

SOLUTION:
K U
Calculate force of friction.
m = 15.62 kg Ff
a = 0 m/s2
g = -9.8 m/s2
Fa = 6 N
Ff = -6 N

Force
Diagrams
Motion
Map
Match the force
diagram to the motion
map. Can you also
predict the slope
shape and orientation
of the position,
velocity, and
acceleration graphs?

Force
Diagram
Motion
Map
Position
Graph
Velocity
Graph
Acceleration
Graph

Determine the force needed to
accelerate Dr. Fiala’s car and its
occupants at a rate of 3.23 m/s2 if
the total mass of car and occupants
is 1315 kg and there is no friction
force.

SOLUTION:
K U
Find applied force.
m = 1315 kg F
a = 3.23 m/s2
F = 4247.45 N (kg)(m/s2)

This time, when we apply that
4247.45 N force to Dr. Fiala’s car and
its occupants, the resulting
acceleration is actually lower. It
registers at a rate of only 3.00 m/s2.
What is the magnitude for the force
of friction causing the acceleration to be
decreased?

SOLUTION:
K U
Find applied force.
m = 1315 kg Ff
F = 4247.45 N
a = 3.00 m/s2
Ff = - 302.45 N

When an object is freefalling it is
weightless. Prove mathematically that
a .448 kg apple is weightless during its
freefall from a tree. Draw a force
diagram of the apple during its fall
from the tree.

SOLUTION:
K U
Find force of support.
m = .448 kg Fs
g = -9.8 m/s2
Fs = 0 N
Fg = -4.39 N

Force Diagram
Motion Map
Position (ΔY) Graph
Velocity (Vy) Graph
Acceleration Graph
Can you predict the
motion map, and
kinematic graphs for
this freefalling object?

Force
Diagram
Motion Map
Velocity (Vy) Graph
Acceleration Graph

Assuming a perfectly frictionless surface, ideal for
launching students in a game of faculty bowling,
Dr. Fiala uses a brand new gizmo that automatically
applies a force that results in an acceleration of
1.1 m/s2. Experimentation resulted in a student with a
mass of 44.10 kg, accelerating at 1.1 m/s2. Find the
force generated by the gizmo for that student.

SOLUTION:
K U
Find force in the horizontal.
m = 44.10 kg F
a = 1.1 m/s2
F = 48.51 N

Mass (kg) Force (N)
0 0
42 46.2
47.56
44 48.4
48.65
49.25
49.51
45.45 50

Mass (kg) Force (N)
0 0
42 46.2
43.25 47.56
44 48.4
44.23 48.65
44.77 49.25
45.01 49.51
45.45 50

All of the students from the
previous problem (combined mass)
step into an elevator at the same time.
Draw a force diagram of this situation
including the magnitude of Fg and Fs.

SOLUTION:
K U
Find force of gravity and force of support.
m1 = 42 kg Fg
m2 = 43.25 kg Fs
m3 = 44 kg
m4 = 44.23 kg
m5 = 44.77 kg
m6 = 45.01 kg
m2 = 45.45 kg Fg= -3025.36 N
g = -9.8 m/s2 Fs= 3025.36 N

This same elevator accelerates
at a rate of .75 m/s2 towards the
second floor. Draw a force
diagram of this situation
including the magnitude of Fg and
Fs.

SOLUTION:
K U
m = 308.71 kg Fs
Fg = -3025.36 N
g = -9.8 m/s2
a = .75 m/s2
Fs= 3256.89
N
Fg = 3025.36 N
Fs = 3256.89 N

Force Diagram
Motion Map
Position Graph
Velocity (Vy) Graph
Acceleration Graph
Can you predict the
motion map, and
this elevator?

Force
Diagram
Motion Map
Position Graph
Velocity (Vy) Graph
Acceleration Graph

This same elevator accelerates
at a rate of .50 m/s2 as it begins its
stop for the second floor. Draw a
Force diagram of this situation
including the magnitude of Fg
and Fs.

SOLUTION:
K U
m = 308.71 kg Fs
Fg = -3025.36 N
g = -9.8 m/s2
a = .-50 m/s2
Fs= 2871.01
N
Fg = 3025.36 N
Fs = 2871.01 N

Force Diagram
Motion Map
Position Graph
Velocity (Vy) Graph
Acceleration Graph
Can you predict the
motion map, and
the ENTIRE TRIP?

According to Newton’s 3rd law, an
action force causes an equal on opposite
reaction force. It is no wonder a truck
windshield squashes a bug and not vice
versa. A 2000 kg truck and a .0002 kg
bug hit with a 50 N force. Take a closer
look at why the truck wins the collision
by calculating the acceleration
exerienced by the bug and by the truck.

SOLUTION:
K U
Why the bug doesn’t survive.
mt = 2000 kg at
mb = .0002 kg ab
g = -9.8 m/s2
F = -50 N
at = -.025 m/s2
ab = -250,000 m/s2

These cables will snap if the
mass of the trafffic light exceeds
10.1 kg. Does the traffic light
exceed 10.1 kg?

SOLUTION:
K U
The cable does not break.
T1 = 375.4 N m
g = -9.8 m/s2 T1y
Θ = 7.5°
m= 10 kg

Dr. Fiala attempts to walk
due east at 5 m/s at the
same time as a 30 m/s cold,
winter wind is blowing due
south. What is the
magnitude of Dr. Fiala’s
velocity.

SOLUTION:
K U
Resultant velocity magnitude.
Vi = 30 m/s Vf a
2
+ b
2
= c
2
Vi = 5 m/s
Vf= 30.41 m/s
Vy = 30 m/s
Vx = 5 m/s

If Dr. Fiala continues his
velocity and the wind
continues to blow steadily,
at what angle, as measured from
positive “X”, is Dr. Fiala’s
velocity.
Vx = 5 m/s
Vy =
30 m/s

SOLUTION:
Vy = 30 m/s
Vx = 5 m/s
tan Θ = x
y
Θ = 9.46°
tan Φ = y
x
Φ = 80.54°
Θ (from +x) = 279.46°
Resultant velocity angle measured from
positive x.

Because of this wind, a 15 kg
package is blown from Dr. Fiala’s
arms and onto the ground. The 15 kg
package reaches a velocity of 30.41
m/s in a time of 4 seconds. Find the
force acting on the box horizontally if
there is no friction.

SOLUTION:
K U
Find applied force.
Yf = -15 m a
Yi = 0 m F
m = 15 kg
g = -9.8 m/s2
Vi = 0 m/s Vf = 31.41 m/s
ti = 0 s a = 7.60 m/s2
ti = 4 s F = 114 N

Force
Diagrams
Motion Map
Match the force
diagram to the motion
map. Can you also
predict the slope
shape and orientation
of the position,
velocity, and
acceleration graphs?

Force
Diagram
Motion Map
Position
Graph
Velocity
Graph
Acceleration
Graph
Fa = 117.79 N
Fg = 147 N
Fs = 147 N
Ff = 0 N
62.8
m
4 s 4 s 4 s
31.4
m/s
7.85
m/s2

If the package is blow horizontally at
30.41 m/s off a ledge onto a parking
lot that is 15 meters below how much
time will it spend in the air before
striking the ground? What does the
motion map look like?

SOLUTION:
K U
Find time package spends in the air.
Yf = -15 m Vf
Yi = 0 m tf
m = 15 kg
g = -9.8 m/s2 tf = 1.75 s
Vi = 0 m/s
ti = 0 m/s

Force Diagram
Motion Map
Acceleration Graph
Can you predict what the
force diagram, and
vertical kinematic graphs
for this freefalling object?
Velocity (Vy) Graph

Force
Diagram
Motion Map
Velocity (Vy) Graph
Acceleration Graph

Time
(s)
Vertical
Position (m)
Horizontal
Position (m)
Vertical Velocity
(m/s)
Horizontal Velocity
(m/s)
0.11 31.41
0.27 -0.36 8.48 -2.65
0.49 15.39 -4.80
0.63 -1.94 -6.17
1.07 -5.61 33.61
1.22 38.32 -11.96
1.35 -8.93 -13.23
1.46 -10.44 45.86
1.75 -

Time
(s)
Vertical
Position (m)
Horizontal
Position (m)
Vertical Velocity
(m/s)
Horizontal Velocity
(m/s)
0.11 -0.06 3.46 -1.08 31.41
0.27 -0.36 8.48 -2.65 31.41
0.49 -1.18 15.39 -4.80 31.41
0.63 -1.94 19.79 -6.17 31.41
1.07 -5.61 33.61 -10.49 31.41
1.22 -7.29 38.32 -11.96 31.41
1.35 -8.93 42.40 -13.23 31.41
1.46 -10.44 45.86 -14.31 31.41
1.75 -15.01 54.97 -17.15 31.41

Dr. Fiala throws a baseball in the
air with an initial velocity of 27 m/s at
an angle of 27° to the horizon. Create
a
velocity vector diagram and show, by
parallelogram method, the “X” and “Y”
components of the baseball’s velocity.

SOLUTION:
K U
Resolve velocity vector into “x” and “y”
components just like force or any other vector.
V= 27 m/s Viy
Θ= 27° Vix
g = -9.8 m/s2
Viy = 12.26 m/s
Vix = 24.06 m/s
Vx = V
Vy = V
27°

How much time will it take for the
baseball to reach the same height
from which it was thrown?

SOLUTION:
K U
Find time in the air.
g = -9.8 m/s2 tf
Θ= 27°
Viy = 12.26 m/s
Viy = 24.06 m/s
ti = 0 s
Yi = 0 m
Yf = 0 m tf = 2.5 s

How far will the baseball travel in
2.5 seconds?

SOLUTION:
K U
Find range.
g = -9.8 m/s2 Xf
Θ= 27°
Viy = 12.26 m/s
Viy = 24.06 m/s
ti = 0 s
Yi = 0 m Yf = 0 m
Xi = 0 m
tf = 2.5 s Xf = 60.15 m

What is the maximum height the
baseball attained during its flight?

SOLUTION:
K U
Find Δy.
g = -9.8 m/s2 Δy
Θ= 27°
Viy = 12.26 m/s
Vix = 24.06 m/s
ti = 0 s
Yi = 0 m Yf = 0 m
Xi = 0 m Δy = 7.67 m
tf = 2.5 s
Xf = 60.15 m

Vy = 0 m/s
Vf = Vi + g Δt
tf = 3.06 s
Yf = 45.9 m
Yf = Yi + Vi t + ½ at2
AREA = ½ Base x Height

Force
Vector Arrows for this
Projectile
Acceleration
Using these vector arrows can
you predict what the position,
force, velocity and
acceleration vector arrows
would look like for this
projectile at the start and at
the top?
Velocity
Position


Force
Acceleration
Velocity
Position

 
 
 


If it was a .448 kg apple that was
thrown into the air at 30 m/s what
was the apple’s intial momentum?

SOLUTION:
K U
Find momentum of apple.
m = .448 kg p
Vi = 30 m/s
g = -9.8 m/s2
p = 13.44 kgm/s

What constant force is needed to
get a change in the apple’s
momentum from 13.44 kgm/s to 0
In 3.06 seconds?

SOLUTION:
K U
Find force necessary to change momentum.
m = .448 kg F
Vi = 30 m/s
g = -9.8 m/s2
ti = 0 s
tf = 3.06 s
Δp = -13.44 kgm/s
F = -4.39 N

After falling to the ground the
.448 kg apple rolled at a constant
10.4 m/s where collided with a
stationary .577 kg apple. If the two
apples stuck together, at what
velocity would they roll?

SOLUTION:
K U
Find the velocity of two apples stuck together.
m1 = .448 kg Vf
m2 = .577 kg p
g = -9.8 m/s2
Vi1 = 10.4 m/s
Vi2 = 0 m/s
p = 4.66 kgm/s2
Vf = 4.55 m/s

Determine the force applied if
the rolling apples strike a wall
and a come to a stop in .311
seconds.

SOLUTION:
K U
Find force needed to stop apples.
m1 = .448 kg F
m2 = .577 kg
ti = 0 s
tf = .311 s
g = -9.8 m/s2
Vi1 = 4.55 m/s
Vi2 = 0 m/s
p = 4.66 kgm/s F = 14.98 N

Physics Semester 1 Review and Tutorial

Physics Semester 1 Review and Tutorial

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Physics Semester 1 Review and Tutorial

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