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-PHYS 102-
Midterm 1 Study Guide Midterm Exam Guide
0 Problem Solving Tips
0.1 You can’t memorize your way through a physics
class
The ▯rst instinct of many life science students is to study for exams by
memorizing as much information as possible. This is not a bad instinct | it
is generally a successful strategy in biology courses.
However, this will not work in physics classeIntroductory physics
courses teach problem solving. The only way to get better at problem solving
is to practice | this means that you should not only do all of the homework
that your professor assigns, you should also work together with a study group
to solve or review di▯cult problems. The best way to study for a physics test
is to review the solutions and revise your answers to your corrected home-
work problems (but don’t try to simply memorize them!). Physics professors
are known to put homework problems, either verbatim or slightly modi▯ed
on exams.
0.2 How novices and experts solve problems
Education research shows that novices and experts approach solving physics
problems very di▯erently. If you would like a high grade on your physics
exam, you should emulate expert problem solvers, rather than follow your
instincts as a novice.
How novices (students in physics 101) solve problems:
1. Panic!
1 2. Hunt through the textbook (or equation sheet) until they ▯nd an equa-
tion that looks promising or has (some of) the right variables
3. Immediately plug in the numbers given in the problem
4. Plug all of those numbers into their calculator
5. Get a result
6. Relax! The problem has been solved.
How experts (physics professors) solve problems:
1. Describe the physical system (objects, forces, interactions)
2. Clarify the geometrical relationships of the system (for example by
drawing a picture or adding a coordinate system)
3. Identify the relevant physical principle (this is a Newton’s 3rd law prob-
lem, this is a 1-D kinematics problem)
4. Make simplifying assumptions (can we represent this as a particle?, is
air resistance important?)
5. Find relevant equations
6. Solve the equations algebraically ▯rst, then plug in numbers
7. Get a result
8. Check the units of the result (to verify the answer is correct) and in-
terpret the result (is this reasonable?)
0.3 The mandatory steps for solving a physics problem
You may not be an expert problem solver yet, but you can avoid some novice
mistakes by following these steps:
1. Examine the situation to determine which physical principles are in-
volved
2. Draw a picture
2 3. Write down the things that we know (knowns) and convert units if
necessary
4. Write down what we want to ▯gure out (unknowns)
5. Write down an appropriate equation(s)
6. Solve that equation using algebra so that the variable we are interested
in is by itself on one side of the equals sign
7. Plug in numbers
8. Use a calculator (or our heads) to do the arithmetic
9. Check to see if our answer is reasonable
Most di▯culties that students have when they are doing physics homework
assignments come from skipping steps or doing the steps out of order. You
will be very tempted to plug in numbers to your equation as soon as possible.
Do not fall into this trap! Wait until the end. I promise this will lead to less
mistakes and higher exam grades. Drawing a picture will help, even if you
are not very artistic, I promise, really.
0.4 Units
Another di▯erence between experts and novices at physics problem solving
is dealing with units. Novices tend to ignore them, experts are obsessed with
them.
There is a big di▯erence between 10 km and 10cm (or 10 miles and 10
inches if you are not metrically inclined)! If your units do not cancel when
you are solving a problem (10km=10cm) you will get the wrong answer.
Professors like to see if you are paying attention on exams by giving you
quantities in weird units.
One way to make sure that you are solving a problem correctly is to
plug in numbers and units into your equation and then cancel the units.
You should get an answer with the correct units. If not, you probably did
something wrong.
3 0.4.1 Example of our problem solving technique
Calculate how long it takes the radio waves to arrive at your current location
from a radio station 10 km away.
1. This problem is a straight-forward 1-D kinematics problem, which also
requires us to know a little bit about the properties of light. Since we
know the displacement of the radio waves and can look up the velocity
of light, we should be able to solve for the elapsed time.
2. A picture of our situation:
You
Tower 10 km
3. We know that we are 10 km away from the tower. Let’s convert our
distance in kilometres into a distance in meters, so that we can plug it
into an equation in later steps.
d = 10 km
1 km = 1000m
1000 m
d = 10▯km ▯
1 ▯m
d = 10000 m
We also know the speed of the wave, since all light waves, including
radio waves, travel at the speed of light.
v = c = 3 ▯ 10 m=s
4 4. We want to ▯nd the time (in seconds) that it takes the radio waves to
travel 10km.
t =?
5. The equation that shows the relationship between speed (v), distance
traveled (d) and time it takes to travel (t) for anything, including light
waves is :
d
v =
t
In other words, our speed is the distance (meters) that we cover per
unit time (seconds).
6. We need to solve this equation, so that t is by itself on one side of the
equals sign before we can plug in numbers:
d
v =
t
t = d
v
7. Now, ▯nally we can plug in numbers
10000 m
t = 3 ▯ 10 m=s
8. Using a calculator
10000 m
t = 8
3 ▯ 10 m=s
t = 0:0000333333 s
▯5
t = 3:33 ▯ 10 s = 0:03 milliseconds
9. Is our answer reasonable?
This time, 0.03 milliseconds, is much too small for us to notice, which
is good.
5 1 Chapter 2: Kinematics
1.1 Important Equations
You should be able to explain the following equations, what concepts they
represent, and what each variable represents, and situations where you are
likely to use them.
▯x = x f x 0 : displacement
▯t = tf▯ t 0 : elapsed time
▯x
v = : Average velocity
▯t
▯v
a = : Average acceleration
▯t
kinematic equations :
v0+ v
x = x0+ vtv =
2
v = v0+ at
1
x = x0+ v 0 + at 2
2
v = v 0 2a(x ▯ x )0
1.2 Important Concepts
1.2.1 Vectors, Scalars, and Coordinate Systems
A vector is a quantity with both a magnitude and direction. Examples
include displacement, velocity, acceleration and force.
A scalar is a quantity that has a magnitude, but no direction. Examples
include time, temperature, mass and speed.
Unless otherwise speci▯ed, motion to the right is usually considered pos-
itive, and motion to the left is considered negative. Motion up is usually
positive and motion down is negative. However, you are free to de▯ne what-
ever positive direction you would like for a problem. Just be very careful
that you are consistent! When solving a problem, it is usually a good idea
to indicate the positive direction with an arrow in your drawing.
6 1.2.2 Displacement
Displacement is di▯erence between the starting and ending positions of
an object. The units are meters. It is a vector quantity (you can have a
displacement of -2m).
Distance traveled is the total length of the path taken between two
points. It is a scaler quantity (you cannot travel a distance of -2m).
1.2.3 Time, Velocity, and Speed
Velocity is displacement (change in position) per unit time traveled. Its
units are m/s. Velocity is a vector quantity (you can travel -2 m/s). A
negative velocity indicates that you are traveling in the negative direction
(to the left or down, unless otherwise noted).
Speed is much like velocity, except it is a scalar quantity. This means
that your speed is always a positive number. For example, if you have a
velocity of -2 m/s, your speed is 2 m/s. if you have a velocity of 2 m/s, your
speed is also 2 m/s. Do not confuse velocity and speed.
1.2.4 Acceleration
Acceleration is a change in velocity per unit time. When you accelerate,
you are speeding up, slowing down, or changing direction. Its units are m/s .
Acceleration is a vector quantity.
Be very careful with signs when dealing with acceleration! A negative
acceleration does not mean that you are traveling in the negative direction, it
means that you are slowing down. The acceleration vector is pointing in the
opposite direction as your velocity vector.
A positive acceleration does not mean that you are traveling in the positive
direction, it means that you are spreading up. The acceleration vector is
pointing in the same direction as your velocity vector.
1.2.5 Falling Objects
If an object is falling under the in
uence of gravity, we know its acceleration.
g = 9:80m=s . The acceleration vector points down. According to our sign
2
convention (down = negative), a = ▯g = ▯9:80m=s . Be careful with your
signs when solving problems with falling objects. Things will speed up if
7 42 Chapter 2 | Kinematics
Figure 2.14 (a) This car is speeding up as it moves toward the right. It therefore has positive acceleration in our coordinate system. (b) This car is
slowing down as it moves toward the right. Therefore, it has negative acceleration in our coordinate system, because its acceleration is toward the
leFigure 1: (a) Positive velocity, positive acceleration. The car is speedingsite to its direction of motion. (c) This car is moving toward the left, but
slowing down over time. Therefore, its acceleration is positive in our coordinate system because it is toward the right. However, the car is
deup. (b) Positive velocity, negative acceleration. The car is slowing down.r is speeding up as it moves toward the left. It has negative acceleration
because it is accelerating toward the left. However, because its acceleration is in the same direction as its motion, it is speeding up (not
de(c) Negative velocity, negative acceleration. The car is slowing down. (d)
Negative velocity, positive acceleration. The car is speeding up.
Example 2.1 Calculating Acceleration: A Racehorse Leaves the Gate
A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average
acceleration?
8 dimensional kinematics.
Slopes and General Relationships
First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are
plotted against one another in such a graph, the horizontal axis is usually considered to be an independent variable and the
vertical axis a dependent variable. If we call the horizontal axiY-axis and the vertical axis tZe-axis, as in Figure 2.46, a
straight-line graph has the general form
Z = NY + C. (2.89)
Here N is the slope, defined to be the rise divided by the run (as seen in the figure) of the straight linC.is used for
the y-intercept, which is the point at which the line crosses the vertical axis.
Figure 2.46 A straight-line graph. The equation for a sZ = NY + Cne.is
Figure 2: Equation for a straight line.
Graph of Displacement vs. Time (a = 0, so v is constant)
Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement
versus time would, thus, haveY on the vertical axis anU on the horizontal axis. Figure 2.47 is just such a straight-line graph.
It shows a graph of displacement versus time for a jet-powered car on a very flat dry lake bed in Nevada.
traveling up.
1.2.6 Graphical Analysis of One-Dimensional Motion
The slope of a position vs. time graph is velocity. The slope of a velocity vs.
time graph is acceleration.
9 64 Chapter 2 | Kinematics
Figure 2.40 Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that
velocity changes linearly with time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical
position only. It is easy to get the impression that the graph shows some horizontal motion—the shape of the graph looks like the path of a
projectilegraphs for something falling under the in
uence of gravity. Note that thectual path of the rock in space is straight up, and straight down.
Discussionposition and velocity of the object are positive for some points negative for
2
The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upwaZ1, since
graph switches from positive to negative at the point where the slope of the
and W1 are both positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving
downward. At 3.00 s, bothmZ gandh Wwitare negative, meaning the rock is below its starting point and continuing to move
3 3
downward. Notice that when the rock is at its highest point (at 1.5 s), its velocity is zero, but its acceleration is still
2 2
−9.80 m/s . Its acceleration −9.80 m/s for the whole trip—while it is moving up and while it is moving down. Note
that the values for are the positions (or displacements) of the rock, not the total distances traveled. Finally, note that free-
fall applies to upward motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which
remains constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall while
arcing up as well as down, as we will discuss in more detail later.
Making Connections: Take-Home Experiment—Reaction Time 1.3 Example Exam Question
1.3.1 Ball A and Ball B roll parallel to each other on a level
table top. Below is a stroboscopic photograph, showing the
position of the balls at 1 second intervals and a length scale
for the table top. (20 points)
1. At what time to the balls have the same speed? Explain your answer.
(5 points) The magnitude of velocity is speed. Veloctiy is de▯ned as
▯x=▯t. Since the balls are shown at equal time intervals, they will
have the same speed when the distance intervals are the same. If we
de▯ne t = 0 at x = 0, this happens between t = 3 and t = 4.
2. On the axis provided, sketch a position vs time, a velocity vs time and
an acceleration vs time diagram for both balls. Make sure the points
and/or lines on your graphs are clearly labeled. (10 points) see below
3. What is the physical signi▯cance of the slopes of your graphs? (5
points) The slope of the x vs. t graph is velocity. The slope of the v
vs. t graph is acceleration. The slope of the a vs. t graph is the change
in acceleration over time.
11 x v
t t
a
t
12 2 Chapter 3: Two-Dimensional Kinematics
2.1 Important Equations
You should be able to explain the following equations, what concepts they
represent, and what each variable represents, and situations where you are
likely to use them.
vector components :
A x Acos▯
A = Asin▯
y
adding vector components :
R x A + x x
R y A + y y
q
R = R + R 2 : magnitude vector sum
x y
▯1
▯ = tan (Ry=R x : direction vector sum
Projectile motion :
x = x 0 v x
vx= v 0x = vx
y = y0+ 12(v 0y+ v yt
vy = v 0y▯ gt
1 2
y = y 0 v 0y▯ gt
2
v = v 2 ▯ 2g(y ▯ y0)
y 0y
Quadratic equation :
2
at + pt + c = 0
b ▯ b ▯ 4ac
t = ▯
2a
13 2.2 Important Concepts
2.2.1 Kinematics in Two Dimensions: An Introduction
The horizontal (x) and vertical (y) components of motion can be analyzed
separately from one another. This greatly simpli▯es problem solving.
Figure 4: A red and yellow ball fall under the in
uence of gravity. Even-
though the yellow ball has an initial velocity in the x direction and the red
ball does not, their y motions are the same. They hit the same y positions
at the same time.
2.2.2 Vector Addition and Subtraction: Graphical Methods
Graphical method for adding vectors:
Make sure that you follow these steps exactly! You should have two arrow
heads that touch. Many students will be tempted to draw an arrow from the
head of the ▯rst vector to the tail of the second vector in the last step. The
might look nicer, because no arrow heads touch, but it will get you the wrong
answer.
To subtract vectors, put the head of the vector on the other end of your
arrow. For vector B, this gets you -B. Then do vector addition, as described
above (A + (-B)).
14 Figure 3.10 To describe the resultant vector for the person walking in a city considered in Figure 3.9 graphically, draw an arrow to represent the total
displacement vectorUsing a protractor, draw a linȆ relative to the east-west axis% Tof the arrow is proportional to the
vector’s magnitude and is measured along the line with a ruler. In t%isof the vector is 10.3 units, anȆ ise direction
29.1º north of east.
Vector Addition: Head-to-Tail Method
The head-to-tail method is a graphical way to add vectors, described in Figure 3.11 below and in the steps following. The tail
of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.
Figure 3.11 Head-to-Tail Method: The head-to-tail method of graphically adding vectors is illustrated for the two displacements of the person walking
in a city considered in Figure 3.9. (a) Draw a vector representing the displacement to the east. (b) Draw a vector representing the displacement to the
north. ThFigure 5: (a)Draw an arrow to represent the ▯rst vector (b) Draw an arrowointing vector. (c) Draw a line from the tail of the east-pointing vector to the
head of the north-pointing vector to form the sum Dr. The length of thD ais proportional to the vector’s magnitude and is
to represent the second vector. Place the tail of the second vector at the
measured head of the ▯rst vector (c) Find the sum of the two vectors by drawing anmeasured with a protractor to be
29.1º .
arrow from the tail of the ▯rst vector to the head of the second vector
Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor.
2.2.3 Vector Addition and Subtraction: Analytical Methods
Usually it is much easier to put away your ruler and protractor, and use a
little trigonometry when you have to deal with vector quantities. Always
make sure your calculator is in degree (DEG) mode when using sin and cos.
You can deal with the x and y components of your motion separately,
then combine them back together at the end.
2.2.4 Projectile Motion
A projectile is something that is falling only under the in
uence of gravity.
The path it follows is called a trajectory. We generally assume that air
This OpenStax book is available for free at http://cnx.org/content/col11406/1.9
resistance is negligible for these problems (otherwise they would be way too
hard!).
Projectile motion is made up of two independent motions: Free fall mo-
tion in the vertical direction and uniform motion at constant velocity in the
horizontal direction. These motions can be analyzed independently and then
combined to understand the total motion of an object.
To solve a projectile motion problem:
1. Resolve the the motion into horizontal (x) and vertical components (y)
15 " Y " ≠Z" (3.5)
If the vectoA is known, then its magnitud" (its length) and its aȆg(its direction) are known. To "iYand " Z, its x-
and y-components, we use the following relationships for a right triangle.
" = " cos Ȇ (3.6)
Y
and
" = " sin Ȇ. (3.7)
Z
Figure 3.27 The magnitudes of the vector coApYandtA Z can be related to the resultaAt and the angȆewith trigonometric
Figure 6: How to ▯nd the X and Y components of a vector.
identities. Here we se" Y " cos Ȇ and "Z= " sin Ȇ .
Suppose, for example, tha| A =xAcos▯ and A = Asyn▯. For example, you will often be givenson walk

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