Physics Honors Notebook
2020-2021 School Year
Table of contents
3-4 ​Formula sheet
5-6 ​Describing motion with equations (distance, displacement, speed, velocity and
acceleration
7-9 ​Describing motion with graphs
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8/18/2020
Distance versus Displacement
Distance:​ Amount of ground that is covered A scalar quantity (magnitude ONLY)
Displacement:​ Overall change in position; how far out of place an object moves A
vector quantity (magnitude AND direction)
Speed:​ A measure of how fast an object is moving. Distance covered per unit of time. A
scalar quantity (magnitude ONLY).
Velocity:​ The rate at which the velocity changes. A vector quantity
(magnitude AND direction). At any given instant, velocity = the speed +
the direction.
Instantaneous Speed:​ the speed at any given instant in time
Average Speed:​ the time-based average of all instantaneous speeds.
Ave. Speed = distance/time
Ave. Velocity = displacement/time
Velocity​ Speed + Direction (at any given instant in time)
Accelerating Objects​ are changing their velocity … either their speed or their direction.
Three ways to accelerate:
1. Speed up 2. Slow down 3. Change directions
Acceleration​ The rate at which the velocity changes. Acceleration Equation:
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Acceleration Units:
1. (mi/hr)/s
2. (km/hr)/s
3. (m/s)/s
4. m/s2
An acceleration of 5.0 m/s/s means … … the velocity of the object changes by 5.0 m/s
every 1.0 second of travel.
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Constant positive velocity →
Constant leftward velocity ←
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Positive velocity - positive acceleration →
Positive velocity - negative acceleration →
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Negative Velocity - Negative Acceleration ←
Negative velocity - Positive acceleration ←
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Newton’s Laws of Motion
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Newton’s Laws of Motion
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Newton’s Second Law of Motion
Sample Questions
1)
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2)
An applied force of 50 N is used to accelerate an object to the right across a
frictional surface. The object encounters 10 N of friction. Use the diagram to
determine the normal force, the net force, the mass, and the acceleration of the
object. (Neglect air resistance.)
Note: To simplify calculations, an approximated value of g is often used - 10
m/s/s. Answers obtained using this approximation are shown in parenthesis.
F​norm​ = 80 N; m = 8.16 kg; F​net​ = 40 N, right; a = 4.9 m/s/s, right
( F​norm​ = 80 N; m = 8 kg; F​net​ = 40 N, right; a = 5 m/s/s, right )
Since there is no vertical acceleration, normal force = gravity force.
The mass can be found using the equation F​grav​ = mg.
The F​net​ is the vector sum of all the forces: 80 N, up plus 80 N, down equals 0 N.
And 50 N, right plus 10 N, left = 40 N, right.
Finally, a = F​net​ / m = (40 N) / (8.16 kg) = 4.9 m/s/s.
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Virtual Lab: Atwood’s Machine
Objective: ​Investigate how force, mass and acceleration are related according to
Newton’s Second Law of Motion.
The link of the interactive:
https://www.physicsclassroom.com/Physics-Interactives/Newtons-Laws/Atwoods-Machine/Atwo
ods-Machine-Interactiv​e
​The interactive
Equations
Data Table 1
Net Force
9.8 N
19.6
29.4 N
Mass (m=F/a)
1.98 kg
2.98 kg
3.99 kg
Acceleration
(a=2h/t^2)
4.93 m/s/s
6.57 m/s/s
7.36 m/s/s
Net Force
29.4 N
29.4 N
29.4 N
Mass (m=F/a)
3.99 kg
4.99 kg
5.93
Acceleration
(a=2h/t^2)
7.36 m/s/s
5.89 m/s/s
4.95 m/s/s
Data Table 2
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Graph for data table 1
Write a conclusion​ …
Graph for data table 2
Write a conclusion ...
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PROJECTILE MOTION
●
●
●
●
●
Horizontal velocity is not changing
Vertical velocity is changing by the gravitational acceleration
Horizontal acceleration is 0 m/s^2
Vertical acceleration is 9.81 m.s^2 (Gravitational Acceleration)
When the object is in the air, there is only gravitational force acting on it.
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● The green and the purple ball will touch the ground at the same time.
● 45 degree angle will give us the maximum range
● According to this diagram the greatest angle (90 degree) will spend the
longest time in the air.
● 30 degree and 60 degree will give you the same range because
30 + 60 = 90
● 75 degree and 15 degree will give you the same range because
15 +75 = 90
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MOMENTUM AND COLLISIONS
Momentum​ is a commonly used term in sports. A team that has the momentum is on
the move​ and is going to take some effort to stop. A team that has a lot of momentum is
really ​on the move​ and is going to be ​hard to stop​. Momentum is a physics term; it
refers to the quantity of motion that an object has. A sports team that is ​on the move
has the momentum. If an object is in motion (​on the move)​ then it has momentum.
Momentum can be defined as "mass in motion." All objects have mass; so if an object is
moving, then it has momentum - it has its mass in motion. The amount of momentum
that an object has is dependent upon two variables: how much ​stuff​ is moving and how
fast the ​stuff​ is moving. Momentum depends upon the variables ​mass​ and ​velocity​. In
terms of an equation, the momentum of an object is equal to the mass of the object
times the velocity of the object.
Momentum = mass • velocity
p = mv
● Momentum is a vector quantity
Sample Problems
1. Determine the momentum of a ...
a. 60-kg halfback moving eastward at 9 m/s.
b. 1000-kg car moving northward at 20 m/s.
c. 40-kg freshman moving southward at 2 m/s.
Solution:
a. m = 60 kg, V = 9 m/s, p = mv, p = __540__ kgm/s, eastward
b. m = 1000 kg, V = 20 m/s, p = mv, p = __20000__ kgm/s, northward
c. m = 40 kg, V = 2 m/s, p = mv, p = __80__ kgm/s, southward
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2. A car possesses 30 000 units of momentum. What would be the car's new
momentum if ...
a. its velocity was doubled.
b. its velocity was tripled.
c. its mass was doubled (by adding more passengers and a greater
load)
d. both its velocity was doubled and its mass was doubled.
Momentum = Mass x Velocity
Solution:
a.
b.
c.
d.
__ 60000 kgm/s__
__ 90000 kgm/s__
__ 60000 kgm/s__
__ 120000 kgm/s__
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Impulse and Momentum
Impulse = Change in momentum
F•t
=
●
●
●
●
m • ∆v
The impulse experienced by an object is the force•time.
The momentum change of an object is the mass•velocity change.
The impulse equals the momentum change.
Unit for impulse is same as momentum which is kg
Impulse - Momentum calculations
F x t = m x ​Δ​V
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Momentum Problems
1) A large fish is in motion at 2 m/s when it encounters a smaller fish that is at rest.
The large fish swallows the smaller fish and continues in motion at a reduced
speed. If the large fish has three times the mass of the smaller fish, then what is
the speed of the large fish (and the smaller fish) after the collision?
m(before)*V(before) = m(after)*V(after)
(3m) * ( 2 m/s ) = (4m) * V(after)
6 = 4 * V(after)
V(after) = 1.5 m/s
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2) A railroad diesel engine has five times the mass of a boxcar. A diesel coasts
backwards along the track at 4 m/s and couples together with the boxcar (initially
at rest). How fast do the two trains cars coast after they have coupled together?
m(before)*V(before) = m(after)*V(after)
5m * 4m/s = 6m * V(after)
20 = 6*V(after)
V(after) = 3.3 m/s
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3) In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 92.8
cm/s collides with a 1.50-kg cart (Cart B) moving leftward with a speed of 21.6
cm/s. The two carts stick together and move as a single object after the collision.
Determine the post-collision speed of the two carts.
m(before)*V(before) = m(after)*V(after)
m1*V1(before) + m2*V2(before) = m(combined)*V(combined)
0.5kg(92.8m/s) + 1.5kg(-21.6m/s) = 2kg * V(combined)
46.4 - 32.4 = 2 * V(combined)
14 = 2 * V(combined)
V(combined) = 7 cm/s
4) 2. A 25.0-gram bullet enters a 2.35-kg watermelon and embeds itself in the
melon. The melon is immediately set into motion with a speed of 3.82 m/s. The
bullet remains lodged inside the melon. What was the entry speed of the bullet?
m(before)*V(before) = m(after)*V(after)
m1*V1(before) + m2*V2(before) = m(combined)*V(combined)
1000 g = 1 kg
25.0 g = 0.025 kg
(0.025)*V1(before) + 2.35 * (0 m/s) = (2.375)(3.82)
0.025*V1(before) = 6.195
V1(before) = 6.195 / 0.025
V1(before) = 247.8 m/s - velocity of the bullet -
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5) A 25.0-gram bullet enters a 2.35-kg watermelon with a speed of 217 m/s and
exits the opposite side with a speed of 109 m/s. If the melon was originally at
rest, then what speed will it have as the bullet leaves its opposite side?
(CAUTION: Be careful of the units on mass.)
m(before)*V(before) = m(after)*V(after)
m1*V1(before) + m2*V2(before) = m1*V1(after) + m2*V2(after)
1000 g = 1kg
25 g = 0.025 kg
(0.025)*(217) + (2.35)*(0) = (0.025)*(109) + (2.35)*(V2after)
5.425 + 0 = 2.725 + 2.35*(V2after)
5.425 - 2.725 = 2.35*(V2after)
2.35*(V2after) = 2.7
V2after = 2.7/2.35
V2after = 1.14 m/s (watermelon post collision speed)
6) In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s
encounters a ​magnetic collision​ with a 1.50-kg cart (Cart B) that is initially at rest.
The 0.500-kg cart rebounds with a speed of 45 cm/s in the opposite direction.
Determine the post-collision speed of the 1.50-kg cart.
m(before)*V(before) = m(after)*V(after)
m1*V1(before) + m2*V2(before) = m1*V1(after) + m2*V2(after)
(0.5)*(129) + (1.5)*(0) = (0.5)*(-45) + (1.5)*(V2after)
64.5 + 0 = -22.5 + (1.5)*(V2after)
64.5 + 22.5 = (1.5)*(V2after)
(1.5)*(V2after) = 87,
V2after = 87/1.5 = 58 cm/s
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7) A 3000-kg truck moving rightward with a speed of 5 km/hr collides head-on with a
1000-kg car moving leftward with a speed of 10 km/hr. The two vehicles ​stick
together​ and move with the same velocity after the collision. Determine the
post-collision velocity of the car and truck. (CAREFUL: Be cautious of the +/- sign
on the velocity of the two vehicles.)
m(before)*V(before) = m(after)*V(after)
m1*V1(before) + m2*V2(before) = m(combined)*V(combined)
3000(5) + 1000(-10) = 4000*V(combined)
15000 - 10000 = 4000*V(combined)
5000 = 4000*V(combined)
V(combined) = 5000/4000 = 1.25 km/hr
8) During a goal-line stand, a 75-kg fullback moving eastward with a speed of 8 m/s
collides head-on with a 100-kg lineman moving westward with a speed of 4 m/s.
The two players collide and ​stick together,​ moving at the same velocity after the
collision. Determine the post-collision velocity of the two players. (​CAREFUL​: Be
cautious of the +/- sign on the velocity of the two players.)
m(before)*V(before) = m(after)*V(after)
m1*V1(before) + m2*V2(before) = m(combined)*V(combined)
...
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Exploring Momentum and Impulse: Cart Push Off
● Velocity of Daemon : 1.24 m/s, Velocity of Christine: V = d/t, V = 0.3 / 0.2167 =
1.38 m/s
● Momentum of Christine: p = mv, p = 82 x (-1.38) = - 113.1 kgm/s
● Momentum of Dameon: p = mv, p = 88.5 x 1.24 = 109.7 kgm/s
● -3.8 kgm/s. Ideally this should be zero.
● Both people apply the same amount of force to each other.
● The duration (time) of Christine's force on Daemon is equal to the duration of
Daemon's force on Christine.
● The magnitude of the impulse Christine's applies on Daemon is equal to than the
magnitude of the impulse Daemon applies on Christine.
● The magnitude of the change of momentum of Christine and her cart is equal to
than the magnitude of the change in momentum of Daemon and his cart.
● Change in momentum is same for both of them just like force and time and
impulse
● Total momentum before push is 0 kgm/s.
● Total momentum after push is -3.8 kgm/s (Ideally this should be zero as well)
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Momentum Video Analysis Labs
Billiard Ball Collision Video
Is linear momentum of the two-ball system the same after the collision as it is
before? Describe your process and results in the space below.
The linear momentum should be the same as before ideally. Momentum before
collision; p = mv, p = m(0.64) | Momentum after the collision; p = mv, p =
m(0.64). Momentum before the collision is the same as momentum after the
collision.
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Glider Explosions
Momentum before the collision is zero.
That’s why, momentum after the collision should be zero. (lets see)
After the collision:
Momentum of the yellow cart; p = (0.140)( - 0.57) = - 0.0800 kgm/s.
Momentum of the red one; p = mv, p = (0.280)(0.27) = 0.0756 kgm/s.
Percent Error = | (0.0800 - 0.0756)/(0.0800) | x 100 = 5.5 % error
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