Physics Chapters 8 and 9 Study Guide
Sections: Chapter 8 – Sections 1 – 5
Chapter 9 – Sections 1 – 7
Your Equation Sheet
You are required to be able to work with all equations on the equation sheet. The equations, and only
the equations, will be provided for you on the test. So, you are required to know the units, whether the
quantity is a vector, and whether the quantity is conserved.
Equations
Momentum
Impulse
Work
Power
Kinetic Energy
Gravitational Potential Energy
Impulse-Momentum
Work-Energy Theorem (KE)
Work-Energy Theorem (GPE)
p=mv
I=Ft
w=Fd
p = w/t
KE = ½ m v2
PE = m g h
F t = Δ mv
F d = ½ m v2
work = mgh
These are just the equations from this unit.
You may always need to use a prior
equation, such as F = ma.
Plug and Chug
There will be several questions that require you solve problems using the equations from this unit. Most
will require the use of only one equation, but a few will require the use of two.
You will need to:
1) Recognize which equation to use.
2) Know where to plug in your given quantities.
3) How to solve for the remaining unknown.
Example: A football player pushes a practice dummy for 10 m, doing 9000 Joules of work. With how
much force does the player push the practice dummy?
You will need to recognize the symbols for the important quantities: work, force, and distance (w, F, and
d). From this, select the equation that has these quantities: w = F d.
w=Fd
9000 = F * 10
Solving for F:
F = 900 N
Two-Step Plug and Chug
If you know what equation to use, but you are not given the values of all the variables that you need,
you are probably working on a two-step problem. You will need to perform a calculation to find the
missing value. Consider this example:
Calculate the power required to exert a force of 10 N on an object over 3 m, taking 10 seconds. The
power equation is p = w / t. But where is the work? It is not given. Well, we are given force and
distance, so we have to calculate it, using the work equation.
w = F d = 10 * 3 = 30 N m
Then, use this result in our power equation:
P = w / t = 30 / 10 = 3 W
Your turn:
Calculate the kinetic energy of a 10-kg object that is accelerated from rest at 5 m/s2 for 12 seconds.
Impulse-Momentum
An impulse is required to change momentum. This fits with what we learned before – in order to
change an object’s velocity, we need to apply a force. The exact amount required is given by the
impulse-momentum equation (see above). So, the momentum change is equal to the impulse.
Example:
An 80-kg shopping cart is rolling at 3 m/s. How much force is required to slow the shopping cart to 1
m/s in 5 seconds?
1) Pick your equation: F t = Δ m v
(remember: Δ = “the change in”).
2) Plug in what we know. (The cart slows down 2 m/s, so that goes in for v.)
F * 5 = 80 * 2
F * 5 = 160
F = 32 N
Your turn:
A 800-kg vehicle is traveling at 25 m/s. If the friction force between tires and road is 1200 N, how many
seconds does it take to stop?
Some Definitions and Notes
Potential Energy: The energy of position. An object that is raised off the surface of the Earth gains
potential energy, as does an arrow in a drawn bow and a block compressing a spring.
Kinetic Energy: The energy of motion. All objects in motion have kinetic energy.
Note: An impulse off a bounce is twice as great as an impulse off of an object that sticks.
Relationship Problems
You should understand the basic relationships between quantities in this unit. As one quantity triples,
what happens to the other quantities?
Example 1:
If you triple the speed of an object, what happens to:
a) momentum
b) kinetic energy
p = mv - They are directly related, so momentum also triples.
KE = ½ mv2 – Since v is squared, we have to square our multiplier. 32 = 9 The
kinetic energy triples when the speed doubles.
Example 2:
If you cut the mass of an object in half, what happens to the momentum?
Ans: Since the mass and momentum are directly related, the momentum is cut in half as well.
Your turn:
Holding the momentum constant, if you double the mass of an object, what happens to the velocity?
Compared with a car moving at 10 m/s, the same car moving at 40 m/s will have how much KE?
The Work-Energy Theorem
The work-energy theorem states that whenever work is done, energy changes. In fact, the amount of
work done equals the energy change, giving us the equation: F d = ½ m (Δv)2. Though there is not
technically a WET for potential energy, this concept can also be used with potential energy, with the
equation: Work = mgh. We can use these equations to solve problems.
Example 1:
How much work is done when raising a 20-kg object to a height of 5 m?
work = mgh = 20 * 10 * 5 = 1000 J
Example 2:
How much force is required to stop a 10,000-kg garbage truck in 50 m, if the truck is travelling at 20
m/s?
F d = ½ m (Δv)2
F * 50 = ½ (10,000)(20)2
50 F = 2,000,000
Solving for F, F = 40,000 N
Your turn:
What is the stopping distance of a 1500-kg car, traveling at 40 m/s, if the friction force between the road
at tires is only 5000 N, due to icy conditions?
Conservation of Energy
Energy is never lost – it merely changes form. In the case of free-fall, energy changes from PE to KE, but
the total amount never changes. Consider this situation: a 25-kg object starts at 50 m above the ground.
What kinds of energy does it have as it falls? Fill in the missing values.
PE
KE
0
2500 J
7500 J
2500 J
TE
Free-fall Conservation of Energy Problems
We can use the conversion of potential to kinetic energy to solve free-fall problems.
Example:
A rock is released from a height of 40 m and falls freely. How fast is it going when it hits the ground?
To solve, keep in mind that the amount of potential energy that the rock has at the top is equal to the
amount of kinetic energy it has just as it hits the ground. So, set those equations equal to each other:
PE
=
KE
½ mv2
=
mgh
½ v2
=
gh
(the masses on either side cancel)
½ v2
=
(10)(40)
(plug in your known values)
½ v2
=
400
v2
=
800
v = 28.3 m/s
Your Turn:
A rock is dropped, falls freely, and hits the ground with a velocity of 35 m/s. From how high was it
dropped?
Conservation of Momentum
The total momentum of two objects before and after a collision are equal. You can start these problems
by finding the total momentum of the system before the collision.
Example 1:
A 5-kg brick, traveling at 3 m/s, collides with a stationary 7-kg mass of clay. The two objects stick
together. What is their speed after the collision?
Momentum Before Collision: Since the second object is not moving, it has no momentum. So, we need
only calculate the momentum of the first object: p = mv = 5 * 3 = 15 kg m/s.
By the law of conservation of momentum, the system will have 15 kg m/s of momentum after the
collision. The total mass of the object is now 5 + 7 = 12 kg.
p = mv
15 = 12 v
Solving for v, v = 0.8 m/s
Example 2: In this example, we will have to keep track of each object individually. A 15-kg object,
traveling East at 8 m/s, collides elastically with a 10-kg object, which was traveling East at 4 m/s. If the
15-kg object slows down to 6 m/s (still going East) after the collision, what is the speed of the 10-kg
object after the collision?
Momentum Before Collision:
Object 1: p = mv = 15 * 8 = 120 kg m/s
Object 2: p = mv = 10 * 4 = 40 kg m/s
Total Momentum Before Collision = 120 + 40 = 160 kg m/s
After Collision:
Object 1: p = mv = 15 * 6 = 90 kg m/s
So, if we started with a total momentum of 160, and object 1 has 90 after the collision, that leaves 70
left over for object 2.
Object 2:
p = mv
70 = 10 * v
Solving for v, v = 7 m/s
Your turn:
A 12-kg ball of putty is thrown at 20 m/s at a stationary 30-kg desk. The two objects stick together.
What is the speed of the putty-desk after the collision?
A 12-kg bowling ball, moving at 5 m/s, collides with a 2-kg bowling pin. After the collision, the bowling
pin travels straight ahead at 15 m/s, with the bowling ball following it. What is the speed of the bowling
ball?