Download AP Physics 1 - Summer Assignment Test Packet

Name _____________________________________________
AP Physics 1 - Summer Assignment
Test Packet
Skills you should have before starting AP Physics 1.
If you don’t have these skills, learn them!
(There will be another Test on this material in the first week of school.)
Test Packet due Tuesday 9/6/16 at the beginning of your class period.
It is worth 100 points (a Test Grade).
Make sure this packet is in its original order when you hand it in.
Show all work neat and clear for full credit.
Try to impress me with your work.
Must have Composition Lab Book with graph paper only.
No substitutes!
Dimensional Analysis Test . . . . . . page 1
Algebra Test . . . . . . . . . . . . . . . . . . . page 2
Trigonometry / Vector Test . . . . . .page 3
Unit Conversion Test . . . . . . . . . . . page 4
DIMENSIONAL ANALYSIS and UNITS
Name________________________________
I ask you a question . . . Do you want a million?
What is your answer?
Yes? No? Maybe?
Before you answer . . . do you want to ask an important question?
Like . . . A million what?!!
A million grapefruits? A million atoms? A million mosquito bites? A million dollars?
You’d like to know more than just a number . . . right?!!
You want to know what unit or quantity I’m talking about.
You Are Expected To Know All Of The Following
1) In physics units are extremely important. The numbers will always change, but the dimensions/units are essential
for understanding and sharing information. We have dimensions that describe different aspects of this physical
world. We also have units that measure those dimensions. Units represent the dimensions of this universe!
2) Beware! Be aware! We will use many different symbols in physics class. There are symbols for dimensions and
symbols for units. Beware! Be aware!
3) There are many different dimensions in this universe; length, time, mass, speed, weight, force, torque, pressure,
electric charge, capacitance, magnetic flux. There are many others
4) We must indicate what units we are using when measuring a dimension. When measuring the dimension of length,
do we measure in feet? centimeters? yards? miles? light-years? When measuring the dimension of time, do we
measure in hours? seconds? years? It is very important to always indicate the units being used.
5) You will hear me say throughout the year, “It’s all relative.” One example is; any number you calculate will be
relative to the units you’re using.
Example 1- Four different quantities of a given length; 6 inches, 0.5 feet, 0.00009 miles, or 1.5x10−17 light years.
These are equal lengths, but measured in four different units, therefore, they have four very different numbers.
Example 2- Four different quantities of a given time interval; 2,520,000 seconds, 700 hours, 29.2 days, 0.0800 years.
These are equal times, but measured in four different units, therefore, they have four very different numbers.
6) Units are extremely important. Always look for them and always give answers with proper units to receive full
credit. “Take care of your units, and your units will take care of you”. Knowing your units is so powerful, it will
allow you to derive equations needed to solve problems.
7) For the most part, we will use System International (SI) units, otherwise known as the metric system.
Base Units
1) These are the most basic units. You can not break these down into smaller units. All additional units will be built
from these base units.
2) Base units are used to describe and measure the basic dimensions of this physical world. There are 6 base
dimensions; length, mass, time, electric current, temperature, and amount of substance.
Dimension (symbol)
length (x, y, z, d, h, s)
mass (m)
time (t)
electric current (I)
temperature (T)
amount of substance
Unit (symbol)
meter (m)
kilogram (kg)
seconds (s)
ampere (A)
Kelvin (K)
mole (mol)
3) Length, distance, height, and displacement are all essentially the same thing. They are measured in meters.
They can be represented by x, y, or z depending on what direction you are traveling in our 3-dimensional world.
We will also use; distance (d), height (h), and displacement (s).
4) Be very careful of the “m” for mass in a physics equation and the “m” for meters a unit of measure given with a
number.
If “m” is in an equation it is the dimension of mass.
If “m” is after a number it is the unit of meters.
Derived Units
1) Derived units describe more complex dimensions in our universe like; velocity, acceleration, force, work, torque,
pressure, electric potential, resistance, specific heat capacity, and magnetic flux. All derived units are a combination
of base units. These derived units sometimes get their own name.
2) Numerical factors like ½ and 2 in equations do not affect the units of the answer. They only change the numbers.
3) Variables in equations that are squared or cubed do affect units! The unit becomes squared or cubed. This changes
the dimension. For example, length is measured in (meters), but area is (meters2), and volume is (meters3). Length,
area, and volume are very different dimensions.
4) Rules of algebra apply when analyzing units.
5) We will focus on 7 derived dimensions.
From this time forward you are responsible for knowing all of the following dimensions,
equations, units, and symbols.
Dimension (symbol)
1) average velocity (v)
Equation
v = Δx
Δt
Unit (symbols)
meters
second
m
s
To create derived units, substitute the appropriate base unit for each corresponding dimension used in the equation. The equation
for average velocity (v) equals the change in displacement (Δx) divided by the change in time (Δt). Displacement has units of
meters (m) and time has units of seconds (s). Making the substitutions, we end up with the derived unit for velocity which is
meters per second m .
s
2) average acceleration (a)
a = Δv
Δt
meters/second
second
m
s2
Average acceleration (a) equals the change in velocity (Δv) divided by the change in time (Δt). Substituting units for velocity
(m/s) and units for time (s) into the acceleration equation, we get the derived unit for acceleration, meters per second per second
or meters per second2 m
s2
3) Force (F)
F = ma
kilogrammeter
second2
kg m = N
s2
Force (F) equals mass (m) times acceleration (a). Substitute the units into the force equation; kilogram (kg) for mass and m for
acceleration, the result is kgm . This is known as a Newton (N).
s2
2
s
Dimension (symbol)
4) Work (W)
Equation
W = Fdcosθ
Unit (symbols)
kilogrammeter2
second2
kgm2 = J
s2
Work (W) equals force (F) times displacement (d). Substitute the units into the work equation, a Newton or kgm for force and
meters (m) for displacement, the result is kgm2 . This is known as a Newton·meter (Nm) or a Joule (J).
s2
2
s
5) Kinetic Energy (KE)
KE = ½mv2
kilogrammeter2
second2
kgm2 = J
s2
Kinetic Energy (KE) equals ½ the mass (m) times the velocity squared (v2). Remember, the ½ doesn’t change the units, but the
squared velocity will change the units. Substitute the units into this energy equation; kilogram (kg) for mass and meters per
second m for velocity. Squaring the units for velocity yields m2 results in kgm2 . This is a (Nm) or (J).
s
s2
s2
6) Gravitational Potential Energy (Ug )
Ug = mgh
kilogrammetermeter
second2
kgm2 = J
s2
Gravitational Potential Energy (Ug) is mass (m) times the acceleration due to gravity (g) times the height of the object (h) above
zero. Notice (g) is an acceleration and height (h) is a displacement. Substituting units into this equation; kilogram (kg) for mass,
meters per second2 m for acceleration due to gravity, and meters (m) for height, results in kgm2 . This is a (Nm) or (J).
s2
s2
7) Power (P)
P = W
t
Joules
second
J = W
s
Power (P) is work (W) or change in energy (E) done per time (t). Important note, Work = Energy. They have the same units!
Substituting units in the equation; Joules (J) for work and seconds (s) for time, results in J . This is known as a Watt (W). What
is a Watt in terms of base units only?
s
Using Dimensional Analysis
1) Dimensional analysis can be used to derive physics equations or determine if they are true. We need to determine if the units
on the left side of the “equal sign” are the same as the units on the right side of the “equal sign”.
2) When using dimensional analysis, do not move units from one side of the “equal sign” to the other side.
3) Always reduce derived units into base units.
4) Coefficients like (½ /  / 2) do not affect the final units. They will be ignored and immediately dropped from the original
equation.
5) Exponential powers (variables being squared or cubed) affect the units! They will square or cube the units.
6) Terms being added or subtracted on one side of the “equal sign” in an equation, must have the same units to do so.
You can’t add or subtract distances (m) and velocities (m/s) from each other. You can add/subtract (m) from (m) or
add/subtract (m/s) to (m/s).
7) Rules of algebra apply.
Examples on the next page.
EXAMPLE 1
 Starting with the equation
v2 = vo2 + 2ax
 Drop the “2” then substitute units for each dimension; “v” has units of (m/s), “a” has units of (m/s2), “x” has units of (m).
m
s
2
= m
s
2
+ m m
s2
reduce
m2 = m2 + m2
s2
s2
s2
reduce
m2 =
s2
m2
s2
 The left side has units of (m2/s2). On the right side you can add (m2/s2) to (m2/s2) to get more (m2/s2).
Remember, the coefficient 2 does not change the units and is removed at the beginning.
We have the same units on both sides of the equal sign m2 = m2 . This is a “true” equation. It is dimensionally correct.
s2
s2
EXAMPLE 2
 Start with the equation
v = vot + ½at3
 Drop the “½” then substitute units for each dimension; “v” has units of (m/s), “a” has units of (m/s2), “t” has units of (s).
m = m s + m s3
s
s
s2
reduce
m = m + ms
s
 The left side has units of (m/s). On the right side the first term reduces to (m) because the (s) on top and bottom cancel.
The second term reduces to (ms) because the (s2) on the bottom cancels two of the (s3) on top.
Here we have a problem. We cannot add (m) and (ms) together. They are two different dimensions.
As a result, the left and right sides of the equation do not have the same units; velocity on the left with distance and an
unknown unit (ms) on the right side. This equation is dimensionally “false”. It does not represent a dimensional relationship
that exists in our known physical world.
Dimensional Analysis Test (5 points each)
Directions – Show all work neat and clear for full credit. Use dimensional analysis to prove if the following equations
are true or false as described in previous packet. Reduce all equations to base units, then state whether the equation is
True or False. Box your final answers.
1) v
2ax
v
4) ½at3 + at = v
2) x
vt
at
5) ½mv = mgh + ½kx2
“k” has units of (N/m)
3) mv
mv
6) P = Fv
F∙t
Algebra & Trigonometry Test (6 points each)
Directions Use algebra (and trigonometry where needed) to solve for the stated variable in each of the following
equations. Show all work neat and clear for full credit. Box your final answers.
1) Solve for time (t) in:
x = vot + ½ at2
2) Solve for initial velocity (vo) in:
3) Solve for the mass (m) in:
v2 = vo2 + 2ax
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4) An object on an incline plane obeys Newton's Law ∑F = ma where the net force acting on the object
is ∑F = mg(sinθ) − μmg(cosθ). Derive an equation to solve for the coefficient of friction (μ) between the object
and the incline in terms of only; the acceleration due to gravity (g), the acceleration of the mass (a), the
trigonometric functions (sin) and (cos), and the angle of incline (θ).
5) Two equations for an object hanging in equilibrium are Tcosθ = mg and Tsinθ = qE. Derive an equation to solve
for the angle (θ) of the applied tension (T) in terms of only; the acceleration due to gravity (g) the mass of the
object (m), the charge on the object (q), the external electric field (E), and one trigonometric function.
Trigonometry & Vectors Test (6 points each)
Directions– Properly draw all vectors (components and resultants) on the frame of reference provided. All vectors
have magnitude and direction. Determine the magnitude (how much of each) and what direction it is pointing (arrow
head) relative to the frame of reference. Show all work neat and clear for full credit. Box your final answers.
1) Determine the magnitude and direction of the x and y components for the velocity vector v = 18 m/s which is
directed at an angle  = 33 West of due South.
+y (N)
+x (E)
2) Find the magnitude and direction of the resultant acceleration vector created by an x-acceleration vector
ax = 4.0 m/s2 and a y-acceleration vector ay =  3.5 m/s2.
+y
+x
3) You are driving west into St. Louis, Missouri. In the distance you see the famous Gateway-to-the-West arch. This
monument rises to a maximum height of 192m. You estimate your line of sight to the top of the arch to be 2.0
above the horizontal (the flat road). Approximately how far away are you from the arch in kilometers?
N
E
Unit Conversion Test (5 points each)
Directions– Use proper dimensional analysis, unit conversion techniques, and significant figures to solve each of
the following questions. Show all work neat and clear for full credit. Box your final answers.
1) A jet is cruising at a velocity of 520 miles per hour. What is the jet’s velocity in meters per seconds?
(1.0 mile = 1.609 km)
2) A dollar bill is about 15.5cm long. If eight trillion dollar bills were laid end-to-end like a long rope, how many
times could you wrap that rope around the Earth’s equator? The radius of the Earth at the equator is 6378 km.
3) A given pyramid has a base that covers an area of 13.0 acres and a height of 481 feet. Determine the volume V of
this pyramid in cubic meters. The volume of a pyramid is V = Bh, where B is the area of the base and h is the
height of the pyramid. (1.00 acres = 43,560 ft2) and (1.00 ft = 0.3048 m)
4) If (1.0 ederims = 0.23 zeebops) and (1.0 zeebop = 3.3x104 googliesnoots) where (1.0x106 googliesnoots =
1.0x10-3 namahas) and (777 namahas = 44 tatlims) how many tatlims equal 1.0x104 ederims?