Original HW
Percent
Original HW
Grade
In Class Test
Grade
New HW Grade
76
15
83
15
76
15
70
15
76
15
62
12
100
15
73
0

Prepare for the Regents Exam

1.35 weeks from today
◦ June 17th @12:15pm

I need my rulers and protractors back.

Get a calculator

Review Classes

Return them.


Final Homework due next Wednesday
Mark any questions that confuse you, and
then see me as soon as possible.
Who will be remembered forever in the
Physics Hall of Fame?

2015: 71%

2014: 74%


Must see me after school at least twice next
week.
Mod 2:
◦ Franco, Maddie, Molly, Mike, Chris, Ana, Shannon,
Olivia

Mod 8:
◦ Adam, Kendra, Ian, Brenna, Steven, Nick, Megan,
Colin, John, Kaylee
1.
2.
3.
4.
Must correct every question you lost credit on.
Must show your work for every correction.
When complete you must identify the three
areas/topics that you need the most
improvement on and how you plan on
improving them.
You must hand them in to me during one of
my free mods or after school.


A scalar quantity is a unit that has a
magnitude but no direction.
A vector quantity is a unit that has both a
magnitude and direction

The Big 3 of Kinematics
◦ 𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
◦ 𝑑 = 𝑣𝑖 𝑡 +
1
𝑎𝑡 2
2
◦ 𝑣𝑓 2 = 𝑣𝑖 2 + 2𝑎𝑑

What is the acceleration due to gravity on
Earth?

What is an object’s initial velocity if it is
dropped from rest?

What is an object’s velocity at the top of its
path when it is thrown straight up?


If a skydiver in free fall has an initial velocity of
-15m/s what will her final velocity be after 13
seconds?
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
𝑚
−15
𝑠
𝑚
+ (−9.81 2 )(13𝑠)
𝑠
𝑚
−142.53
𝑠

𝑣𝑓 =

𝑣𝑓 =

Or 142.53m/s towards Earth

You are drag racing a funny-car that can
accelerate 13m/s^2. If it takes you 5.3 seconds
to finish the race, what is your final velocity?
68.9m/s

A foul ball is hit perfectly straight up into the air.
It has an initial upward velocity of 15m/s and is
in the air for 3.0581 seconds. What is it’s final
velocity right before it hits the ground?
-15m/s

How far would a skydiver fall if he had an initial velocity of zero
meters per second and was in free fall near the surface of the
Earth for 36 seconds? (Neglect air resistance)
1

𝑑 = 𝑣𝑖 𝑡 + 2 𝑎𝑡 2

𝑑=0

𝑑 = 0𝑚 − 6356.88𝑚
𝑑 = −6356.88𝑚
So the skydiver would fall 6356.88m towards Earth


𝑚
𝑠
1
𝑚
36𝑠 + 2 (−9.81 𝑠2 )36𝑠 2


How long would it take a 1997 Acura Integra LS
to travel 100m if its initial velocity is 0m/s and
its maximum acceleration of 3m/s^2?
8.2 seconds; fast for a human,
but really slow for a car…or even
an 18 wheeler
What is the rocket’s initial velocity if it
accelerates constantly at 8m/s^2 and covers a
distance of 57,750m in 120s?
1.25m/s

What is the final velocity of hail ball that fell 90 meters straight down
from the heavens with an initial velocity of -3m/s?

𝑣𝑓 2 = 𝑣𝑖 2 + 2𝑎𝑑

𝑣𝑓 2 = (−3 )2 + 2 −9.81

𝑣𝑓 2 = 9

𝑣𝑓 2 = 1774.8

𝑣𝑓 = ± 1774.8

𝑣𝑓 = −42.13
𝑚
𝑠
𝑚2
𝑠2
+ 1765.8
𝑚2
𝑠2
𝑚
𝑠
𝑚2
𝑠2
𝑚2
𝑠2
𝑚
𝑠2
(−90𝑚)


What is the final velocity of hail ball that fell 90
meters straight down from the heavens with an initial
velocity of 3.0m/s?
𝑚
−42.13
𝑠
You hit a volleyball so that it moves with an initial
velocity of 6m/s straight upward. If the volleyball
starts from 2 meters above the floor, what will the
volleyball’s velocity be right before it strikes the
floor?
-8.7m/s
𝑚
 −9.81 2
𝑠


The acceleration due to gravity is constant and
the only force acting on an object in free fall.
The time it takes an object to travel from the
ground to it’s highest point, is the same time as
it takes the object to fall back down from the
same height.

All objects fall with the same acceleration.

Mass does not effect how fast an object falls!


Just kinematics in two dimensions (x and y)
Remember only the time variable can be used
in both dimensions
Dimension
X
Y
Displaceme
nt
(m)
Initial
Velocity
(m/s)
Final
Velocity
(m/s)
Acceleratio
n (m/s^2)
Time
(s)

A bullet is fired horizontally at 300m/s from
1.5m above the ground. How far down range
does the bullet land? Ignore friction.
Dimension
X
Y
Displaceme
nt
(m)
Initial
Velocity
(m/s)
Final
Velocity
(m/s)
Acceleratio
n (m/s^2)
Time
(s)


A bullet is fired horizontally at 300m/s from
1.5m above the ground. How far down range
does the bullet land? Ignore friction.
We need to find the hang time of the bullet and
then use that to figure out the distance it
traveled in that time:

We don’t have enough information to find the hang
time in the x-direction so we will have to find it in
the y-direction instead.
1
𝑚 2
+ (−9.81 2 )𝑡
2
𝑠
2

−1.5𝑚 = 0

0.3058𝑠 2 = 𝑡
t=.55s

Dimension
Displacement
(m)
Initial Velocity
(m/s)
Final Velocity
(m/s)
Acceleration
(m/s^2)
Time
(s)
X
?
300
300
0
0.55
Y
-1.5
0
?
-9.81
0.55

Now use this time to see how far the bullet traveled
horizontally. Total hang time is the same for the x
and y directions

Use:

So the bullet travels 165.7m before hitting the ground

A cannonball is shot horizontally off of a cliff at an
initial speed of 150m/s. The cannon is 45m above
the level ground at the bottom of the cliff. How far
down range will the cannonball land? Ignore
friction.

Try this one on your own!

Range :

A punter punts the football with an initial
velocity of 18m/s and at an angle of 58
degrees off of the field. Ignoring friction, what
is the hang time of the this punt? FILL IN THE
CHART!!!
Dimension
Displacement
(m)
Initial Velocity
(m/s)
X
Y
?
0
?
?
Final
Velocity
(m/s)
?
?
Acceleration
(m/s^2)
Time
(s)
0
-9.81
?
?

First we must find the Y-component of the
initial velocity!

A punter punts the football with an initial
velocity of 18m/s and at an angle of 58
degrees off of the field. Ignoring friction, what
is the hang time of the this punt? FILL IN THE
CHART!!!
Dimension
Displacement
(m)
X
Y
?
0
Initial Velocity
(m/s)
Final Velocity
(m/s)
Acceleration
(m/s^2)
?
?
0
15.3 -15.3 -9.81
Time
(s)
?
?

With the initial y-velocity we can calculate the
total hang time of the punt using:
Dimen
sion
X
Y
Displacem
ent
(m)
?
Initial Velocity
(m/s)
Final Velocity
(m/s)
Acceleration
(m/s^2)
Time
(s)
?
?
0
3.11
0 15.3 -15.3 -9.81 3.11

What is the total horizontal range of the punt?
◦ Use the chart
Dimens
ion
X
Y
Displacem
ent
(m)
?
0
Initial Velocity
(m/s)
Final Velocity (m/s)
Acceleration
(m/s^2)
Time
(s)
?
?
0
3.11
15.3 -15.3 -9.81
3.11

Need to find the initial X-velocity first!

Dimens
ion
X
Y
Plug into the chart
Displacem
ent
(m)
?
0
Initial Velocity
(m/s)
Final Velocity (m/s)
Acceleration
(m/s^2)
Time
(s)
9.5
9.5
0
3.11
15.3 -15.3 -9.81
3.11
Now find the total range of the
football

A.
B.
C.
An airplane is carrying supplies for people
stranded on an island. If the pilot wants the
supplies to land on the island when should he
drop the package? Why?
Before he is above the island
When he is above the island
After he flew over the island

Which angle will maximize the height of a projectile?
◦ 90

Which angle will maximize the range of a projectile?
◦ 45

Which angle will maximize the hang time of a
projectile?
◦ 90

Which angle will minimize the hang time of a
projectile?
◦ 0

Work on Regents Exam

Complete alone.

Highlight any questions you needed to use
your notes for.

See me for help!

A.
B.
C.
An airplane is carrying supplies for people
stranded on an island. If the pilot wants the
supplies to land on the island when should he
drop the package?
Before he is above the island
When he is above the island
After he flew over the island

Review Forces, Friction, and Kinematics

Corrections are due today

Final homework is due tomorrow

Net Force: All the force vectors added
together.

Equilibrium: Net force is zero Newtons

Static Equilibrium: the object is not moving

Inertia: A property of mass

Netwon’s Three Laws
1. An object in motion remains in motion unless
acted upon by an outside force
2. F=ma
3. Every action has an equal and opposite reaction

𝐹𝑓 = 𝜇𝐹𝑁
◦ Kinetic vs. Static

What
happens
to the
normal
force if
you add
more
mass?
Fric.
32.1N
50°
mg
38.3N
𝐴𝑥 = 𝐴𝑐𝑜𝑠𝜃
= 50𝑁𝑐𝑜𝑠50°
𝐴𝑥 = 32.1𝑁
Normal
If Gary pushes the lawn mower a constant velocity with
a force of 50N down the handle what is the magnitude
of the force moving the mower forward? The handle is
50 degrees above the level ground.

Draw a circuit that has a battery, two resistors
wired in parallel and an ammeter that
measures the current going through only one
of the resistors.

Get better at physics


Centripetal- “Center Seeking”
Centripetal Acceleration = speed²/radius
◦ Or:


v = speed of the object
r = radius of curvature

Centripetal Force is a force that is directed
towards the center of a circle that keeps an
object in circular motion.
◦ Why do you hit the car door when you turn?

What supplies the centripetal force that
allows your car to go around turns?

If your speed is constant
while you’re moving in a
circle, how can you be
accelerating?

My jeep and I have a mass of 1300 kg. If I
drive around a turn with a radius of 100 m
at a speed of 17.5 m/s calculate the:
1. The Centripetal Acceleration
2. The Centripetal Force
3. What is providing the centripetal force?
ac = v2 / r
= (17.5 m/s)2 / 100m
= 3.06 m/s2
 2. Fc = mv2 / r
= (1300 kg)(17.5 m/s2) / 100 m
= 3981.25 N
 1.


If your tires can supply a maximum
centripetal force of 10,000N how can you
increase your velocity around a turn?
Increase your radius





Page 162 of the textbook
12
13
14
15



Gravitation is one of the four fundamental
forces.
It is only attractive.
It pulls on both parties equally.


Calculate the force of gravitational attraction
between a 75 kg person and a 62 kg person
that are 1.5 meters apart!
IS THIS TRUE LOVE????!?!?!?


Is this what
love is?
What would
the force be
if the two
people were
3m apart?





𝐹𝑔 =
𝐺𝑚1 𝑚2
𝑟2
OR…
𝐹𝑔 = 𝑔𝑚
Where 𝑔 is the gravitational constant for a
certain planet!
𝑔=
𝐹𝑔
𝑚






Page 185
8
10
11
13a
13b



Why is there gravity?
Einstein said it was because mass bends the
four dimensional space-time that our
universe is made of.
Sheet Demo

What is the speed of sound in a vacuum?

Get better at physics



𝑝 = 𝑚𝑣
No special units just 𝑘𝑔 ∗
𝑚
𝑠
Unless otherwise stated, momentum is
conserved in collisions
pbefore = pafter
pAi + pBi = pAf + pBf
mAvAi + mBvBi = mAvAf + mBvBf


If you have a mass of 62 kg and you are
running at a constant velocity of 8 m/s what
is your momentum
If your car has a mass of 2000kg and is
parked what is it’s momentum?

A 90 kg running back is running up field with
a constant velocity of 7 m/s. A 105 kg
linebacker is running down field with a
velocity of 6.5 m/s as he tackles the running
back. After they collide find their total
velocity.
pbefore = pafter
pRi + pLi = pRf + pLf
mRvRi + mLvLi = mRvRf + mLvLf

Basketball
 How
does the force on an
object relate to its
change in momentum?

Why do we have crash test dummies?
http://www.youtube.com/watch?v=8adfGDQ-uRA



What is a car’s momentum before a collision
with the wall?
What is a car’s momentum after a collision
with the wall?
Is momentum of this system conserved?
 How
does the time of a
collision relate to the
change of momentum?

Is it possible to jump out of a twenty foot tall
tree without hurting yourself?

Is it possible to jump out of a twenty foot tall
tree and hurt yourself?

Do both systems have the same change in
momentum?

http://www.youtube.com/watch?v=aAhPaiajwDY

http://www.youtube.com/watch?v=WCQEYXG-3Nw


What is the difference? Why do hardcore parkour(ers?)
not injure themselves when they fall long distances?
Does gravity effect them differently?
The time of impact changes the force they feel.

Are eggs easy to break?


The impulse of a collision is the change in
momentum of the collision.
The change in momentum is equal to the
force during the collision multiplied by the
time of the collision.

Does a larger impulse always mean a larger
force?
◦ No! Time also controls the amount of force.

Which is a greater change in momentum on a
2kg object?
A. Accelerating from 2m/s to 6m/s
B. Having a net force of 12N for 0.75 seconds

Kinetic energy
1
2
◦ 𝐾𝐸 = 𝑚𝑣 2

Gravitational potential energy
◦ 𝐺𝑃𝐸 = 𝑚𝑔ℎ

Internal energy
◦ 𝑄

Spring energy
1
2
◦ 𝑃𝐸𝑠 = 𝑘𝑥 2

Units are Joules



The total amount of mass-energy in a system
is always conserved.
Mass-Energy can be transformed into a
different form, but cannot be destroyed.
𝐸𝑇 = 𝐾𝐸 + 𝑃𝐸 + 𝑄

When is GPE maximized?

When is KE maximized?

When is Q maximized?

How do you find the KE between its min and
max?

If a 6.8kg bowling ball is 1.0m above its
lowest point in its swing, what is its
maximum velocity?



𝐸𝑇𝑏𝑒𝑓𝑜𝑟𝑒 = 𝐸𝑇𝑎𝑓𝑡𝑒𝑟
𝑃𝐸 + 𝐾𝐸 + 𝑄 = 𝑃𝐸 + 𝐾𝐸 + 𝑄
𝑚𝑔ℎ + 0𝐽 + 0𝐽 = 0𝐽 +
𝑚
 6.8𝑘𝑔(9.81 2 )(1.0𝑚)
𝑠
𝑣 2 𝐽 + 0𝐽

𝑣 =4.4m/s
1
𝑚𝑣 2
2
+ 0𝐽
+ 0𝐽 + 0𝐽 = 0𝐽 +
1
(6.8𝑘𝑔)
2






Roller Coasters
Magic Can
Dropper Poppers
Balloons
Gasoline
Nuclear Power Plants


Hooke’s Law states that a spring exerts a
force directly proportional to the distance it
is stretched.
𝐹𝑠 = −𝑘𝑥
◦ 𝐹𝑠 = 𝐹𝑜𝑟𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝𝑟𝑖𝑛𝑔
◦ 𝑘 = 𝑆𝑝𝑟𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
◦ 𝑥=
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡ℎ𝑒 𝑠𝑝𝑟𝑖𝑛𝑔 𝑖𝑠 𝑠𝑡𝑟𝑒𝑡𝑐ℎ𝑒𝑑 𝑓𝑟𝑜𝑚 𝑖𝑡𝑠 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚

Springs can also store potential energy

𝑃𝐸𝑠 = 𝑘𝑥 2


1
2
What happens to the potential energy of a spring
as you compress or stretch it more and more?
It increases, until it becomes too much and
breaks…

Draw bar graphs of
the Total Energy,
Potential Energy,
and Kinetic Energy
of the red block –
spring system for all
four stage shown.


1.
2.


To understand this phenomena we need to know
more about work.
Work can be calculated two ways
𝑊 = ∆𝐸𝑇
𝑊 = 𝐹𝑑
F=Force (Newtons)
d=displacement (meters)




A hockey player uses a stick to exert a constant
4.50N force forward to a 105g puck sliding on ice
over a displacement of 0.150m forward. How
much work does the stick do on the puck? Assume
friction is negligible.
W=0.675J
Do you need the mass for this problem?
Nope

Calculate the total work done using the graph
below.


https://www.youtube.com/watch?v=Z0iEUPF
GLXI
Your thoughts?

Power is equal to the change in energy
divided by the time required for the
change.
∆𝐸𝑇
𝑡

𝑃=

∆𝐸𝑇 = 𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦
𝑡 = 𝑡𝑖𝑚𝑒
The units are J/s or Watts (W)


◦ 1 Watt = 1 Joule of energy transferred in 1 second

It is written differently on your reference table:

𝑃=

𝑃=

𝑃=

𝑃=
𝑊
; 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑊 = ∆𝐸𝑇
𝑡
𝐹𝑑
; 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑊 = 𝐹𝑑
𝑡
𝑑
𝐹 𝑣 ; 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑣 =
𝑡
𝑊
𝐹𝑑
= = 𝐹𝑣
𝑡
𝑡





Either way I did 100J of work on the stack of
books. How much power did I exert if I
completed the work in 2.0 seconds?
𝑊
𝑡
100𝐽
2.0𝑠
𝑊
𝑡
100𝐽
45𝑠
𝑃= =
= 50𝑊
What about if it took me 45 seconds?
𝑃= =
= 2.2𝑊
Doing work faster requires more power.




A forklift can lift a 900N weight at 4 m/s.
How much power is it exerting?
𝑚
4
𝑠
𝑃 = 𝐹 𝑣 = 900𝑁 ∗
= 3,600 𝑊𝑎𝑡𝑡𝑠
Your car engine exerts a force of 5,000N over
a distance of 30m in 5.0 seconds. How much
power is it exerting?
𝑃=
𝐹𝑑
𝑡
=
5000𝑁∗30𝑚
5.0 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
= 30,000 𝑊𝑎𝑡𝑡𝑠




The big guy can squat 180kg a distance of
0.8m in 2.2 seconds. The small guy can
squat 85kg a distance of 1.0m in 1.2
seconds. Who is more powerful?
𝑃=
𝐹𝑑
𝑡
=
𝑚𝑔𝑑
𝑡
Big Guy = 642 Watts
Small Guy = 695 Watts

Pick up old tests/homework/answer sheets
from the back table.

Master physics



Find a partner that has an internet machine.
Take out a calculator, reference table, and
scrap paper.
Pick up a question booklet.