Oakland Technical High School
AP PHYSICS SUMMER ASSIGNMENT
Due Monday, August 26th
I.
This packet is a review to brush up on valuable skills, and perhaps a means to assess whether you are
correctly placed in Advanced Placement Physics.
II. Physics, and AP Physics C in particular, requires an exceptional proficiency in algebra, trigonometry,
geometry and calculus. In addition to the science concepts, Physics often seems like a course in applied
mathematics. The following assignment includes mathematical problems that are considered routine in AP
Physics. This includes knowing several key metric system conversion factors and how to employ them.
Another key area in Physics is understanding the geometry of vectors.
Base Units. In the English System of measurement, the fundamental quantities are force, length, and time, which are
measured in the standard units: pound, foot, and second. This system is commonly used by American engineers. On
the other hand, the International System (abbreviated SI) uses mass, length and time, which are measured in the
standard units: kilogram, meter, and second. We will be using these units in this course.
Prefixes. Prefixes will be used with measurements to help us compare quantities. For example, if you had to
compare 4.2 gigabytes of memory space with 1700 megabytes, you could probably do it after some thought. But, it
would be even easier to compare 4.2 gigabytes with 1.7 gigabytes (which happen to describe the same two quantities
as above). Prefixes also let us avoid writing out long numbers. Although we could compare 4,200,000,000 bytes
with 1,700,000,000 bytes above, it is cumbersome.
We can talk about megabytes or megaseconds or megameters. Mega-, wherever it appears, will always mean 1
million of the base unit. A megasecond represents the time that goes by in about 11.5 days. A megameter represents
the distance from Oakland to Portland, Oregon, or Phoenix, Arizona (approximately).
You will need to memorize 8 prefixes. From large to small, they are:
Prefix
gigamega-
Factor
Power of 10
1,000,000,000
1,000,000
Symbol
10
9
G
10
6
M
3
k
kilo-
1,000
10
centi-
0.01
10-2
c
0.001
-3
m
0.000001
10-6
µ
nano-
0.000000001
-9
n
pico-
0.000000000001
10-12
p
millimicro-
10
10
Notice that there are two upper-case symbols (G and M), five lower-case symbols (k, c, m, n, p), and one symbol
that is a Greek letter (µ, pronounced mu). The symbol for the prefix is used with the symbol for the base unit: m for
meter, s for second, and g for gram.
Comparisons. The weight of 1.0 kilogram of mass is about 2.2 pounds. (That is, gravity pulls on one kilogram of
mass with a force of 2.2 pounds.) The mass of a small paper clip is about 1.0 gram, and the mass of a 5-cent coin is
about 5.0 grams. (The kilogram is the only standard SI unit that uses a prefix.)
1
The most common SI units for distance are the meter, kilometer, centimeter, and millimeter. The distance from your
nose to the end of your outstretched arm is about a meter. A kilometer is about 0.6 mile--a little over one-half. The
width of your little finger is about 1.0 centimeter. The thickness of the most common pencil lead is 0.5 millimeter,
so double it to have 1.0 millimeter.
Scientific Notation. Physicists often write very large or very small measurements using scientific notation (or
exponential notation in some books). A number in scientific notation always is written with two factors: a number
between 0 and 10, multiplied by a power of 10. The first factor will have a number of decimal places that show the
precision of the measuring instrument that was used. If you used a wall clock to measure a time interval and you
counted 51 ticks, you would write your measurement as 5.1 x 101 seconds (if you wanted to use scientific notation).
However, the same time interval might be measured using a precise stopwatch, and the time might be written 5.145
x 101 in scientific notation.
You will need to know enough about scientific notation to take a number in standard notation and write it using
scientific notation. You will also need to know how to enter a number in scientific notation into your calculator and
record a number from your calculator correctly.
Conversion of Units and Conversion Factors. Many times, before a calculation can be done with two
measurements, one of the measurements must be changed to have the same units as the other. Both the black book
and the blue book use the method of conversion factors in this situation. The math is not difficult; it is more a
method of record-keeping.
Example: Suppose you need to add 3.5004 m plus 150 µm.
1. Convert 150 µm to the same measurement in m.
2. Since 1 µm = 0.000001 m, the ratio
0.000001 m
1 µm
is exactly equal to 1. This means you can multiply any
measurement by that ratio without changing the size of the measurement. This ratio is called the conversion factor
for this problem.
3. Using the conversion factor method, multiply 150 µm times the ratio shown above. Notice that there is the unit
symbol µm in both the numerator and the denominator, so cancel it out of both places. When you multiply the
numbers together, you have 0.000150, and m is the only unit that remains. Now this measurement can be added to
3.5004 m.
4. Add 3.5004 m plus 0.000150 m, to get 3.500550 m, which is the answer to the problem.
You will need to use conversion factors to convert several measurements from one unit to another.
Derived Units. When a calculation is made with measurements, the units are combined just as if they were
variables. The final combination of units is called a derived unit, because it comes from the other units.
Example: Find the volume of a box that is 1.2 m long, 0.3 m high, and 1.1 m deep, using the formula V = lwh.
1. (1.2 m)(0.3 m)(1.1 m) = 0.396.
2. The derived unit for the answer is m3 because (m)(m)(m) = m3.
3. 0.396 m3 is the complete answer, or 3.96 x 10-1 m3 in scientific notation.
You will need to write the correct derived units to go with an answer that was calculated using a formula.
Vectors are used in the AP Physics book to show displacement, velocity, acceleration, and force. We will learn a
notation system for doing vector calculations in three dimensions that is slightly different from the black book.
However, for the summer assignment, the problems are limited to two dimensions.
2
Scalar quantities. Many measurements can be reported without using vectors, because they do not refer to a
direction. For example, time, temperature, and mass can be described completely without using direction. They are
scalar quantities.
Vector quantities. Vector quantities, on the other hand, are measurements that cannot be completely described
without using direction. For example, a car may be moving at 90 kilometers per hour, but the result after one hour
would be very different if it were moving west compared to the result if it were moving east. Therefore, to
completely describe the velocity of the car, you would need to say 90 km/h, East. Displacement (s), velocity (v), and
acceleration (a) are the vector quantities we will use to describe the motion of an object.
Force vectors also must include direction. The direction may be written as an angle, or it may be shown by writing
the two components of the force vector (horizontal and vertical, or x and y). Therefore, the force vector 20 N, 60°
(read 20 newtons, at a 60° angle) means the same thing as a force vector where Fx = 10 N, and Fy = 17.3 N. They are
two different ways of showing the direction of the vector. The magnitude of the vector remains the same: 20 N.
For the majority of situations in the blue book, the force vectors will be concurrent; that is, all vectors in a problem
will act on a single point. This is a change from the black book, in which the majority of problems had vectors that
acted on different point along an object. In that sense, the problems in the blue book will be less complex. In the
blue book, the complexity will come in the motion of the object that results from the forces.
A free-body diagram is used to represent the force vectors. An accurate free-body diagram is the first step in a
successful solution of a force problem. In the blue book, since the forces are concurrent, the free-body diagram will
be drawn so that the endpoints (or tails) of the vectors are touching. This will be the consistent method used
throughout AP Physics. When the problem involves a large object, the tails of the vectors will all meet at the center
of gravity of the object, and the arrows will point out from there. In most of the problems that we solve in AP
Physics, we can represent the object with a point mass without any loss of accuracy.
Example: In the free-body diagram shown, suppose vectors A, B, and C represent
concurrent forces, and the center of gravity of the object is at point D. To find the
resultant of the three forces, the blue book will find the x-component of the resultant
and the y-component of the resultant, and then combine them using the Pythagorean
Theorem.
Suppose force A is 50 N, force B is 50 N, and force C is 20 N. Then
Ax = (50)(cos 30°) = 43.3 N.
A
D
B
C
Bx = (50)(cos 0°) = 50 N.
Cx = (20)(cos(-100°)) = -3.5 N.
Also,
Ay = (50)(sin 30°) = 25 N.
By = (50)(sin 0°) = 0 N.
Cy = (20)(sin(-100°)) = -19.7 N.
The resultant of the forces, R, has the following x- and y-components:
∑ F = 43.3 N + 50 N + (-3.5 N) = 89.8 N.
Ry = ∑ F = 25 N + 0 N + (-19.7 N) = 5.3 N.
Rx =
x
y
Using the Pythagorean Theorem to find the magnitude of the resultant, R2 = Rx2 + Ry2.
So, R2 = (89.8)2 + (5.3)2 = 90.0 N.
The angle, θ = tan-1 (Ry/Rx) = 3.4°.
The resultant force of combining forces A, B, and C, is a force of 90.0 N at an angle of 3.4° above the x-axis. The
short form of this statement is
R = 90.0 N, 3.4°. R is called the net force, or the sum of the forces.
You must be able to calculate the net force exerted on an object by combining the individual forces correctly.
3
PROBLEMS
1.
Area For each figure, estimate the area. Your answer will have three parts: (a) the “best” answer, (b) the
greatest reasonable answer, and (c) the least reasonable answer. Briefly explain your reasoning for each
problem. Use your best judgment, and support your judgment with reasoning.
1.
2.
3.
4
4.
2.
Algebra. Often problems on the AP exam are done with variables only. Solve each problem for the
variable indicated. Don’t let the different letters confuse you. Manipulate them algebraically as though
they were numbers. Many subscripts are used as labels in physics (for example: 1, 2, c, o, i); simply keep
the subscript with the variable in your solution.
a.
v2 = v02 + 2a ( x − x0 ) , a =
c.
Tp = 2π
l
,
g
e.
mgh =
1
mv2 ,
2
g.
B=
i.
1 2
kx ,
2
b.
U=
g=
d.
Fg = G
v=
f.
x = x0 + v0t +
µ0 Ι
⋅ ,
2π r
r=
h.
xm =
pV = nRT ,
T=
j.
sinθc =
m1m2
r2
mλ L
,
d
n1
,
n2
x=
,
r=
1 2
at , t =
2
d=
θc =
5
k.
3.
qV =
1
mv2 ,
2
v=
l.
1 1 1
=
+ ,
f so si
Quadratic Equations. Frequently, equations in physics will require the quadratic formula to find a
solution. Solve each of the following quadratic equations.
a.
x2 + 8x − 2 = 0
b.
2x2 − 3x + 1 = 0
c.
3 x + 4 = x2
d.
x2 − 5 = 3 x
e.
x ( x + 5) = 12
f.
x2 − 2x + 6 = 2x2 − 6x − 26
4.
si =
Geometry Review. Solve the following geometry problems.
Line B touches the circle at one point. Line A passes
through the center of the circle.
B
a.
Describe line B in relation to the circle:
______________________________________
A
b.
What is the angle between A and B?
_______________________________________
C
c.
30º
What is the measure of angle C?
______________________________________
45º
6
30º
A
d.
If A is parallel to B, what is the measure of angle θ ?
_________________________________________
e.
What is the measure of angle θ ?
θ
B
__________________________________________
θ
30º
f.
The radius of a circle is 5.5 m.
i. What is the circle’s circumference in meters?
_________________________________________
_
ii. What is the circle’s area in square meters?
_________________________________________
__
5.
Right Triangle Trigonometry. Use the generic right triangle shown in the figure.
Solve the following, using the same units in your answer as are given in the
problem
a.
θ = 55º and c = 32 m, solve for a and b.
b.
θ = 45º and a = 15 m/s, solve for b and c.
c.
b = 17.8 m and θ = 65º, solve for a and c.
d.
a = 250 m and b = 180 m, solve for θ and c.
7
e.
a =25 cm and c = 32 cm, solve for b and θ.
6.
f.
b =104 cm and c = 65 cm, solve for a and θ.
Measurement.
Scientists use the mks system (SI system) of units. mks stands for meter-kilogram-second; these are the
preferred units for solving physics problems. They are referred to as base units. You must check to be sure all
measurements are converted to mks before beginning a solution. This is known as “agreement of units”. In
preparation for AP Physics, you must know how to make the following conversions:
kilometers (km) to meters (m)
micrometers (µm) to meters (m)
hours (h) to seconds (s)
centimeters (cm) to meters (m)
grams (g) to kilograms (kg)
days (d) to seconds (s)
millimeters (mm) to meters (m)
minutes (min) to seconds (s)
years (y) to seconds (s)
Complete each of these conversions.
1.
4008 g =
kg
2.
1.2 km =
m
3.
823 µm =
m
4.
298 K =
ºC
5.
0.77 cm =
m
6.
8.8 × 10–8 mm =
m
7.
2.23 × 104 g =
kg
8.
2.65 cm =
m
Write each measurement using the correct mks base unit.
9.
6.22 grams =
10. 105 milligrams =
11. 250 milliseconds =
12. 32.3 centimeters =
13. 18.2 days =
14. 27 years =
15. 50 micrometers =
16. 1000 minutes =
_______
Write each measurement using scientific notation and the correct mks base unit.
17. 9.2 mm =
18. 1040 mg =
19. 83 kg =
20. 4225 µg =
21. 6.33 µg =
22. 11,308 µs =
23. 11.00 s =
24. 40.5 ms =
25. 1.00 µs =
26. 7.223 kg =
8
7.
Vectors. Most of the quantities that are used in physics are vectors. Therefore, students must be proficient
in using vectors.
Definitions.
Magnitude – size or extent. Magnitude is the numerical portion of a vector’s description.
Direction – alignment or orientation of any position with respect to any other position. Direction is often
given as an angle.
Scalar – a physical quantity that is described by a single number and units.
Vector – a physical quantity that is described by a magnitude and a direction. A “directional quantity”.
Examples: force, velocity
r
A
A
Notation:
or
Length of the arrow is proportional
to the
vector’s magnitude.
Direction of the arrow is the
B
vector’s direction.
The head of the vector has the
arrow; the tail of the vector is the
other end.
r
r
A and B are identical vectors because they both have the same magnitude and direction.
Negative Vector – It is possible to have a negative vector. The negative vector has the same magnitude as
its positive counterpart, but it is pointing in the opposite direction.
–A
A
i.
ii.
Vector Addition – Add vectors by combining both the magnitude and the direction. The result of adding
two vectors is called the resultant. There are four situations to learn:
r r r
Both vectors point in the same direction (parallel). A + B = R
A
B
(Add the magnitudes)
r
r
r
R
R
B
A
The vectors point in opposite directions (anti-parallel). A + B = R
(Subtract the magnitudes. Resultant direction is same as vector with greater magnitude.)
r
r
r
iii. The vectors are perpendicular. A + B = R
(Find the resultant magnitude with the Pythagorean Theorem.)
A
B
R
A
R
A
R
B
B
These two vector diagrams show the same vector addition in two different ways.
iv. The vectors are in different directions. Either of two different methods will give the same resultant.
r r r
A +B = R
A
a.
r
R
B
r
Parallelogram Method. The tails of A and B are connected to form adjacent sides of a
r
parallelogram. The resultant R points from the point where the tails connect to the
opposite corner of the parallelogram.
R
B
A
b.
r
Head-to-Tail Method. The vectors are drawn with their correct directions, so that the tail of B is
r
r
r
r
connected to the head of A . The resultant R points from the tail of A to the head of B .
R
Many vectors can be connected using the head-to-tail method, and the resultant will
point from the tail of the first vector in the chain to the head of the last vector.
A
9
B
Vector Subtraction – Vector subtraction is exactly the same as adding the negative of the second vector.
r r r
r r
A − B = A + (−B) = R .
B
A
–B
R
A
–B
R
r
r
r
Instructions. For each problem, draw the diagram to show the resultant for the vector addition A + B = R of each of
the following. Draw the given vectors using either the parallelogram method or the head-to-tail method.
a.
A
B
A
B
b.
A
c.
B
B
A
d.
B
e.
A
r
r
r
r
r
Draw the resultant for A + B + C + D = R
f.
A
C
D
B
10