Physics Semester 1 Review and Tutorial

SEMESTER 1 REVIEW

&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.

Make sure you know how to use the following formulas.These will be given to you on a separate card the day of the final.

Solve for displacement

Dr. Fiala is traveling on hisHarley at a constant 13.67 m/s. What is the distancetraveled by Doc in 7.32seconds?

SOLUTION:

K U E

Find the distance traveled.

Vi&f = 13.67 m/s Xf

ti = 0 stf = 7.32 sXi = 0 m Xf= ma = 0 m/s2

SOLUTION:

K U E

Find the distance traveled.

Vi&f = 13.67 m/s Xf

ti = 0 stf = 7.32 sXi = 0 m Xf= 100.07 ma = 0 m/s2

Dr. Fiala notices he is nowtraveling at a constant 49.21 km/h. What is the

distancein meters traveled by Doc in 7.32seconds?


SOLUTION:

49.21 km h

Dimensional analysis.

49.21 km/h = 13.67 m/sSo the Xf remains

100.07 m

1 h 60 min

1000 m 1 km

X 1 min 60 s

X X

M K H D 0 d c m

1 x x x

Dr. Fiala jumps in his un-started car. He accelerates

at arate 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.

DisplacementVelocityAccelerationInertiaForceMomentum

Match• 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.

Matched

Time (s)

Object #1Position

(m)

Object #2Position

(m)

Object #3Position

(m)

Object #4Position

(m)

1 16 4

2 4

3 48 24 6

4 16 32

Complete the data table of an object moving at constant velocity.

Completed data table.

Time (s)

Object #1Position

(m)

Object #2Position

(m)

Object #3Position

(m)

Object #4Position

(m)

1 16 4 8 2

2 32 8 16 4

3 48 12 24 6

4 64 16 32 8

Time(s)

Object #1Velocity

(m/s)

Object #2Velocity

(m/s)

Object #3Velocity

(m/s)

Object #4Velocity

(m/s)

1 16 8 2

2 9 4

3 13 8

4 10

Complete the data table of an object moving with constant acceleration.

Completed data table.Time(s)

Object #1Velocity

(m/s)

Object #2Velocity

(m/s)

Object #3Velocity

(m/s)

Object #4Velocity

(m/s)

1 16 4 8 2

2 14.5 6 9 4

3 13 8 10 6

4 11.5 10 11 8

Analyze the graph. Compare velocities between the different time intervals as determined by the slopes.

Analyze the graph. Compare velocities between the different time intervals as determined by the slopes.

#1

#2 #3 #4

Analyzed graph.

-2 m/s 0 m/s

1 m/s

0 m/s

Analyze the graph of a projectile. Calculate the initial velocity.

Analyzed graph of a projectile.

Vy = 0 m/s

ty = 5 s

Yf = Yi + Vi t + ½ gt2Vi = 48.5

m/s

Vf2 = Vi

2 + 2g

ΔyVi = 48.5 m/s

Match

Position Graph

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

Graph Options

Match

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.

Match

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).

Match

Position Graph

Graph Options

Matched

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 fullstop and accelerates at 2.25 m/s2, how incredibly

fastwill he be traveling when

he has traveled 1530 meters?

Solve for velocity

SOLUTION:

K U E

Calculate final velocity without knowing time.

Vi = 0 m/s Vf

Xi = 0 m tf

Xf = 1530 ma = 2.25 m/s2 ti = 0 s

Vf= 82.98 m/s

If Dr. Fiala starts from a fullstop and accelerates at 2.25 m/s2, how long will it

takehim to drive 1530 meters?

Solve for time

SOLUTION:

K U E

Solve for time.

Vi = 0 m/s tf

Xi = 0 m Vf

Xf = 1530 ma = 2.25 m/s2 ti = 0 sVf = 82.98 m/s

tf= 36.88 s

Analyze the graph. Calculate the acceleration on and change in position of an object during the time period represented by each individual slope.

Analyzed graph.

0 m/s2

.875

m/s2

-2.33 m

/s 2

0 m/s2

30 m

15 m

33.75 m

92 m

Analyze each of these x-t graphs. Which one matches the v-t?

0 2 4 6 8 10 120

50

100

150

200

Time (s)

Position(m)

0 2 4 6 8 10 120

2

4

6

8

10

12

Time (s)

Position(m)

0 2 4 6 8 10 12

-3-2.5

-2-1.5

-1-0.5

00.5

11.5

Time (s)

Position(m)

0 2 4 6 8 10 12-20

0

20

40

60

80

100

120

Time (s)

Position(m)

Analyzed graph.

0 2 4 6 8 10 120

20

40

60

80

100

120

140

160

180

Time (s)

Position(m)

Match

MotionMap

Graph Options

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

Match

MotionMap

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.

Match

MotionMap

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.

Match

MotionMap

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.

MatchedPosition Graph

Motion Map

Acceleration GraphVelocity Graph

Match Graph Options

MotionMap


Match Graph Options

MotionMap

Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?

Match Graph Options

MotionMap


Position Graph

Velocity Graph

Match Graph Options

MotionMap


Position Graph

Velocity Graph


Motion Map


Match

MotionMap

Graph Options


Match

MotionMap

Graph Options

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

Position Graph

Match

MotionMap

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.

Match

MotionMap

Graph Options

Can you predict the slope shape and orientation of the acceleration graph now based on this velocity graph?

Position Graph

Velocity Graph


Motion Map


Analyze the graph. Calculate the jerk and change in velocity of an object during the time period represented by each individual slope.


18 m/s

0 m/s3


18 m/s

2 m/s17.5 m/s

0 m/s3

0 m/s3

-1 m/s 3

Add vectors. ()

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

Solve for mass

Determine the mass of a 153.08 N

object.

153

N

?

kg

SOLUTION:

K U E

Calculate mass.

W = 153.08 N mg = -9.8 m/s2

m = 15.62 kg

Solve for force of friction

Determine the force of friction on a

15.62 kg object traveling at aconstant horizontal velocity of3.62 m/s while experiencing anapplied force of 6 N.

SOLUTION:

K U E

Calculate force of friction.

m = 15.62 kg Ff

a = 0 m/s2 g = -9.8 m/s2

Fa = 6 N Ff = -6 N

MatchForce Diagra

ms

MotionMap

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

MatchForce Diagra

ms

MotionMap


MatchedForce Diagra

m

MotionMap

Position

Graph

Velocity Graph

Acceleration

Graph

Solve for force

Determine the force needed toaccelerate Dr. Fiala’s car and

itsoccupants at a rate of 3.23

m/s2 ifthe total mass of car and

occupantsis 1315 kg and there is no

frictionforce.

SOLUTION:

K U E

Find applied force.

m = 1315 kg Fa = 3.23 m/s2

F = 4247.45 N (kg)(m/s2)

Solve for force

This time, when we apply that4247.45 N force to Dr. Fiala’s car andits occupants, the resultingacceleration is actually lower. Itregisters at a rate of only 3.00 m/s2. What is the magnitude for the forceof friction causing the acceleration to bedecreased?

SOLUTION:

K U E

Find applied force.

m = 1315 kg Ff

F = 4247.45 Na = 3.00 m/s2

Ff = - 302.45 N

MatchForce Diagra

ms

MotionMap


MatchForce Diagra

ms

MotionMap


MatchedForce Diagra

m

MotionMap

Position

Graph

Velocity Graph

Acceleration

Graph

Solve for force of support

When an object is freefalling it isweightless. Prove mathematically

thata .448 kg apple is weightless

during itsfreefall from a tree. Draw a forcediagram of the apple during its

fallfrom the tree.

SOLUTION:

K U E

Find force of support.

m = .448 kg Fs

g = -9.8 m/s2

Fs = 0 N

Fg = -4.39 N

Match

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?

Matched

Force Diagra

m

Motion Map

Velocity (Vy) Graph

Acceleration Graph


Solve for force

Assuming a perfectly frictionless surface, ideal for

launching students in a game of faculty bowling,Dr. Fiala uses a brand new gizmo that

automaticallyapplies a force that results in an acceleration of 1.1 m/s2. Experimentation resulted in a student

with amass of 44.10 kg, accelerating at 1.1 m/s2. Find

theforce generated by the gizmo for that student.

SOLUTION:

K U E

Find force in the horizontal.

m = 44.10 kg Fa = 1.1 m/s2

F = 48.51 N

Analyze the graph of information collected from the class. Complete the data table of student masses based on this graph.

Mass (kg) Force (N)

0 0

42 46.2

47.56

44 48.4

48.65

49.25

49.51

45.45 50

Analyzed graph of information collected from the class.

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

Solve for force of gravity

and force of support

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

SOLUTION:

K U EFind force of gravity and force of support.

m1 = 42 kg Fg

m2 = 43.25 kg Fs

m3 = 44 kgm4 = 44.23 kgm5 = 44.77 kgm6 = 45.01 kgm2 = 45.45 kg Fg= -3025.36 Ng = -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 forcediagram of this situation including the magnitude of Fg

andFs.

SOLUTION:

K U EFind force of support.

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

Match

Force Diagram

Motion Map

Position Graph

Velocity (Vy) Graph

Acceleration Graph

Can you predict the motion map, and kinematic graphs for this elevator?

MatchedForce Diagra

m

Motion Map

Position Graph

Velocity (Vy) Graph

Acceleration Graph


This same elevator acceleratesat a rate of .50 m/s2 as it begins itsstop for the second floor. Draw aForce diagram of this situationincluding the magnitude of Fg

and Fs.

SOLUTION:

K U EFind force of support.

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

Match

Force Diagram

Motion Map

Position Graph

Velocity (Vy) Graph

Acceleration Graph

Can you predict the motion map, and kinematic graphs for the ENTIRE TRIP?

MatchedForce Diagra

m

Motion Map

Position Graph

Velocity (Vy) Graph

Acceleration Graph

Solve for acceleration

According to Newton’s 3rd law, anaction force causes an equal on oppositereaction force. It is no wonder a truckwindshield squashes a bug and not viceversa. A 2000 kg truck and a .0002 kgbug hit with a 50 N force. Take a closerlook at why the truck wins the collisionby calculating the accelerationexerienced by the bug and by the truck.

SOLUTION:

K U E

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

Solve for mass

These cables will snap if themass of the trafffic light exceeds10.1 kg. Does the traffic lightexceed 10.1 kg?

375.40

N

SOLUTION:

K U E

The cable does not break.

T1 = 375.4 N mg = -9.8 m/s2 T1y

Θ = 7.5° m= 10 kg

Solve for resultant vector

Dr. Fiala attempts to walkdue east at 5 m/s at thesame time as a 30 m/s cold,winter wind is blowing duesouth. What is themagnitude of Dr. Fiala’s velocity.

SOLUTION:

K U E

Resultant velocity magnitude.

Vi = 30 m/s Vf a2

+ b2

= c2

Vi = 5 m/s

Vf= 30.41 m/s

Vy = 30 m/s

Vx = 5 m/s

Solve for resultant vector angle

If Dr. Fiala continues hisvelocity and the windcontinues to blow steadily,at what angle, as measured frompositive “X”, is Dr. Fiala’svelocity.

Vx = 5 m/s

Vy = 30 m/s

V = 30.41 m

/s

SOLUTION:

Vy = 30 m/s

Vx = 5 m/s

tan Θ = x yΘ = 9.46°

tan Φ = y x

Φ = 80.54°

Θ (from +x) = 279.46°

Resultant velocity angle measured from positive x.

Solve for force

Because of this wind, a 15 kgpackage is blown from Dr. Fiala’sarms and onto the ground. The

15 kgpackage reaches a velocity of

30.41 m/s in a time of 4 seconds. Find

theforce acting on the box

horizontally ifthere is no friction.

SOLUTION:

K U E

Find applied force.

Yf = -15 m aYi = 0 m Fm = 15 kgg = -9.8 m/s2 Vi = 0 m/s Vf = 31.41 m/sti = 0 s a = 7.60 m/s2

ti = 4 s F = 114 N

MatchForce Diagra

ms

Motion Map


MatchedForce Diagra

m

MotionMap

Position

Graph

Velocity Graph

Acceleration

Graph

Labeled

Force Diagra

mMotion 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

Solve for time in air

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 beforestriking the ground? What does

themotion map look like?

SOLUTION:

K U E

Find time package spends in the air.

Yf = -15 m Vf

Yi = 0 m tf

m = 15 kgg = -9.8 m/s2 tf = 1.75 sVi = 0 m/s ti = 0 m/s

Match Force DiagramMotion Map

Acceleration Graph

Can you predict what the force diagram, and vertical kinematic graphs for this freefalling object?

Velocity (Vy) Graph


Matched Force Diagra

m

Motion Map Position (ΔY) Graph

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 -

Complete the data table of this object freefalling.

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

Completed data table.

Solve for velocity

componentsDr. Fiala throws a baseball in theair with an initial velocity of 27 m/s

atan angle of 27° to the horizon.

Create avelocity vector diagram and show,

byparallelogram method, the “X” and

“Y” components of the baseball’s

velocity.

SOLUTION:

K U E

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 m/s

27°

Solve for time

How much time will it take for the

baseball to reach the same height

from which it was thrown?

SOLUTION:

K U E

Find time in the air.

g = -9.8 m/s2 tf

Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/sti = 0 sYi = 0 mYf = 0 m tf = 2.5 s

Solvefor

horizontal displacement

How far will the baseball travel in

2.5 seconds?

SOLUTION:

K U EFind range.

g = -9.8 m/s2 Xf

Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/sti = 0 sYi = 0 m Yf = 0 mXi = 0 m tf = 2.5 s Xf = 60.15 m

Solve for maximum

vertical displacement

What is the maximum height the

baseball attained during its

flight?

SOLUTION:

K U EFind Δy.

g = -9.8 m/s2 ΔyΘ= 27° Viy = 12.26 m/s Vix = 24.06 m/sti = 0 sYi = 0 m Yf = 0 mXi = 0 m Δy = 7.67 mtf = 2.5 s Xf = 60.15 m

Analyze this graph of a projectile. Calculate its maximum height.

Analyzed graph of a projectile. Vy = 0 m/s

Vf = Vi + g Δt

tf = 3.06 s

Yf = 45.9 m

Yf = Yi + Vi t + ½ at2

a = -9.8 m/s 2

AREA = ½ Base x Height

Match 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

Matched

Force

Acceleration

Velocity

Position

Solve for momentum

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 E

Find momentum of apple.

m = .448 kg pVi = 30 m/sg = -9.8 m/s2

p = 13.44 kgm/s

Solve for force

What constant force is needed to

get a change in the apple’smomentum from 13.44 kgm/s

to 0In 3.06 seconds?

SOLUTION:

K U E

Find force necessary to change momentum.

m = .448 kg FVi = 30 m/sg = -9.8 m/s2

ti = 0 stf = 3.06 sΔp = -13.44 kgm/s

F = -4.39 N

Solve for velocity

After falling to the ground the .448 kg apple rolled at a

constant10.4 m/s where collided with astationary .577 kg apple. If the

twoapples stuck together, at whatvelocity would they roll?

SOLUTION:

K U E

Find the velocity of two apples stuck together.

m1 = .448 kg Vf

m2 = .577 kg pg = -9.8 m/s2 Vi1 = 10.4 m/s Vi2 = 0 m/s

p = 4.66 kgm/s2

Vf = 4.55 m/s

Solve for force

Determine the force applied if

the rolling apples strike a wall

and a come to a stop in .311

seconds.

SOLUTION:

K U EFind force needed to stop apples.

m1 = .448 kg Fm2 = .577 kgti = 0 stf = .311 sg = -9.8 m/s2 Vi1 = 4.55 m/s Vi2 = 0 m/s p = 4.66 kgm/s F = 14.98 N

This Semester Exam Cannot Harm me. My Study Habits are like a

shield of steel!

Education

Physics Semester 1 Review and Tutorial