1Physics
Necessary Math Skills for 1st Year Physics
st
This document describes the mathematics skills you will need to be successful in the 1 -year physics program.
Physics attempts to understand, explain, justify, and quantify the behavior of objects that you can observe (and
some objects that you can’t observe). Physics speaks in the language of mathematics: symbols, numbers, units,
and angles. As a result, it is necessary for you to be able to do algebra, rudimentary trigonometry, unit
conversions, dimensional analysis, simplification of expressions of numbers and units, scientific notation, and
graphing.
nd
It must be understood that these skills will be tested for specifically on the Friday of the 2 week of school.
HOWEVER, THESE SKILLS WILL THEN BE TAKEN FOR GRANTED FROM THAT POINT FORWARD. These skills
are absolutely required. If you are not comfortable doing this mathematics, you have two choices: get
comfortable, or fight it for the rest of the year.
1. Algebra: The algebra skills required for this course typically involve solving an expression for a particular
variable. The concepts and physical principles that we study are frequently shown as a formula or mathematical
relationship. Typically the problem statement will include some data, which may be in the form of symbols or
numbers with units, and a description of the situation. You then will be asked to solve for one (or more)
variable(s), either symbolically or numerically. In the event that the data is provided as symbols, you need to
know how to isolate a variable.
For example, we might have an equation that says
d=
(v + v ) t
i
f
and we may want to solve the equation for vf.
2
We would first multiply both sides by 2 and get 2d = ( vi + v f ) t . Then we need to distribute the t to get the vf
term out of the parentheses. We get
2d = vit + v f t , subtract the vit from both sides, and get 2d − vit = v f t .
Now we divide both sides by t and get
nd
2d − vit
2d
= v f . We could also say that v f =
− vi .
t
t
For another example, Newton’s 2 Law uses a formula that is written as follows: ΣF = ma where ΣF refers to the
vector sum of the forces, m is the mass, and a is the acceleration of the object. We might consider a particular
problem statement, apply this principle to it, and get an equation that looks like this:
FA − µFN − FT = ma
The problem may ask you to solve for µ in terms of the other variables.
−µFN = ma + FT − FA
so
µ=
ma + FT − FA
−FN
or
µ=
FA − FT − ma
FN
You are expected to know the standard rules for isolating variables. You should be able to cross-multiply, use
the quadratic formula when necessary, distribute, factor out common variables, simplify complex fractions, solve
systems of equations, and so on. All of these skills will show up in this review and on the exam; further, they will
show up for the rest of the year.
Note that in some cases the variables will have subscripts. The subscript is a simple differentiator; it does not
have an effect on the mathematics. In the example shown above, FA, FN, and FT are different forces. The
subscripts A, N, and T will refer to specific forces in the problem.
Note also that typically case sensitivity is critical. “A” is not the same variable as “a.” In physics, capital A
typically is used to refer to amplitude, while lower case a refers to acceleration. It matters.
In most cases, the simplest form of an expression is preferred. Sometimes, alternate expressions of the same
answer can be correct depending on the order of your operations. In such cases, you may see two answers for
the same question. Both would be acceptable.
Algebra represents the biggest challenge to most students. The problems chosen for this review illustrate the
most common mistakes.
Algebra Problems
1.
1
d = vit + at 2 solve for vi
2
16.
m1v1 + m2 v2 = (m1 + m2 ) v3 solve for m2
2.
v2f = vi2 + 2ad solve for vi
17.
a=
m1v2
R
18.
FA cosθ − µ (mg − FA sinθ ) = ma solve for m
3. G
m1m2
R2
=
solve for v
4π 2R
solve for T
T2
4.
1
Fdcosθ − µmgd = mv2 solve for m
2
19.
ab =
5.
1
mgh1 = mgh2 + mv2 solve for v
2
20.
mgsinθ − µmgcosθ = ma
6.
1
d = vit + at 2 solve for t
2
21.
1 2
1
Kx − µmgx = mv2
2
2
solve for m
7.
1 1 1
= −
X Y Z
22.
1 2
1
Kx − µmgx = mv2
2
2
solve for x
8.
bc − ad = 3c + r
23.
d=
9.
(a + b) t = d
24.
R 3 GM
=
T2 4π 2
25.
mgh =
2
t
x
solve for Z
solve for c
solve for b
10.
rx + y =
11.
( t − b ) = ( m − n)
p
solve for x
s
solve for b
26.
m
x +1
solve for x
(v + v ) t
i
f
solve for vf
2
solve for R
1
1
mv2 + Iω 2
2
2
15 − F = 1.5a
solve for µ
and
solve for ω
F − 12 = 1.2a
solve for F and a
12.
v2 =
(m
(m
1
− m2 )
1
+ m2 )
v1 solve for m1
27.
1 1 1
= −
f di do
and
M=−
di
do
solve for f in terms of M and do
13.
FT − mg =
mv
R
2
solve for m
28.
3x − 2y = 28 and 1.5x + 3.2y = 2.5
solve for x and y
14. FA − µmg = ma solve for µ
29. 2.15FT − 24.5 = 3.55a
15.
mgh − µmgh =
1
mv2 solve for v
2
solve for FT and a
and 210 + 2.44FT = 24.6a
Algebra Answers
2d − at 2
or
2t
1.
vi =
2.
vi = ± v2f − 2ad
vi =
d at
−
t 2
Gm2
R
3. v = ±
4. m = 2Fdcosθ
2
(2µgd + v )
5.
v = ± 2g(h1 − h2 )
6. t =
−vi ± vi2 + 2ad
a
m2 =
17.
T = 2π
FT (cosθ − µ sinθ)
(a + µg)
19. x =
m
− 1 or x = m − ab
ab
ab
20.
µ = tanθ −
µmg ± µ 2m2g2 + Kmv2
K
XY
X−Y
22.
x=
8.
c=
r + ad
b−3
23.
vf =
9.
b=
2d − at
2d
−a
or b =
t
t
24.
R=
10.
11.
b=t−
12.
m1 =
13. m =
( m − n) p
s
m2 ( v1 + v2 )
(v
1
− v2 )
FTR
(v
2
+ gR
a
gsinθ − a
or µ =
gcosθ
gcosθ
Kx 2
(v2 + 2µgx)
21. m =
Z=
−y ± y 2 + 4rt
2r
R
a
18. m =
7.
x=
m1(v3 − v1)
(v2 − v3 )
16.
2d − vit
2d
− vi or v f =
t
t
3
25. ω = ±
GmT2
4µ 2
2mgh − mv2
I
26.
F = 13.3
27.
f=
a = 1.1
Mdo
M−1
28.
x = 7.5
29.
FT = 30.5
y = -2.7
)
FA − ma
F
a
or µ = A −
mg
mg g
14.
µ=
15.
v = ± 2gh(1− µ)
a = 11.6
2. Trigonometry: Two-dimensional vector operations require you to be able to use trigonometric relations (sine,
cosine, and tangent) to find the sides and angles of a right triangle. The acronym SOHCAHTOA may ring a bell.
Trigonometric functions (we will use the abbreviation “trig” hereafter) refer to the ratios of the lengths of
particular sides of a right triangle. Usually these sides are labeled A, B, and C, where A and B are the short legs
of the triangle and C is the hypotenuse, or the longest side. The trig functions are listed with reference to one of
the acute angles of the right triangle, conventionally called θ (theta) and φ (phi). These are Greek letters.
Before we get to the trig functions, it is important to recall two facts about right triangles. Since one of the three
o
o
angles must be 90 , and the sum of all three angles must be 180 for any triangle, the relationship between θ and
φ can be shown to be:
θ + φ = 90o
A 2 + B 2 = C2
Also the Pythagorean theorem can be used at any time:
The trig functions are defined as follows, with reference to a specific acute angle:
sin =
opposite side
hypotenuse
cos =
adjacent side
hypotenuse
tan =
opposite side
adjacent side
So using the triangle shown below, we would write the trig ratios:
sinθ =
A
C
sinφ =
B
C
cosθ =
B
C
cos φ =
A
C
tanθ =
A
B
tanφ =
B
A
There are five pieces of information needed to know all there is to know about a right triangle: the three sides and
the two acute angles. In a problem of this type, you will be provided with two of the five and you will have to
solve for the other three. You might be given one angle and one side, or you might be given two sides.
In the event that you are given one angle and one side, use the angle that is given and write out the trig ratios for
it. It will be obvious which of the ratios will be useful; whatever expression has just one variable in it.
If you are provided with two sides of the triangle, again, pick an angle and write out all three trig ratios for it. It
will be seen easily which ratio will be useful. You then must use an inverse trig function, shown on your
-1
-1
-1
calculator as sin (or cos or tan ).
For the time being, it is important that your calculator is in “degrees” mode. Click the mode key and look at the
screen; the third or fourth option down on the list will say RADIANS DEGREES. Usually the default setting is
radians. Click the down arrow until that line is blinking, then click the right arrow key to switch to degrees mode.
nd
The DEGREES should now be highlighted. Click “enter.” Click “2 f” then mode to go back to the normal
screen.
Trigonometry Problems
1.
5.
A=
B=
C=
θ=
φ=
φ=
2.
6.
A=
C=
B=
θ=
φ=
φ=
3.
7.
B=
B=
C=
θ=
θ=
φ=
4.
8.
A=
A=
B=
θ=
θ=
φ=
Trigonometry Answers
1. A = 17.5 m
C = 26.3 m
2. A = 34.0 cm B = 46.1 cm
3. B = 49.5 m
C = 129 m
4. A = 67.4 km B = 41.8 km
φ = 48.3o
φ = 53.6o
θ = 22.6o
θ = 31.8o
5. B = 81.1 m
6. C = 41.3 cm
θ = 26.3o φ = 63.7o
θ = 39.5o φ = 50.5o
7. B = 178 m
θ = 30.8o φ = 59.2o
8. A = 35.9 m
θ = 24.5o φ = 65.5o
3. Unit Conversions: In order to use the formulas that govern behavior, we must have unit consistency within
the problem. We may see formulas that relate displacement, velocity, acceleration, and time; before we can use
the formula, we may have to convert units within the data to have consistency. This may require conversion
from English system units (such as miles, feet, or pounds) to Metric system units (meters or kilograms or
Newtons). It also may require us to convert within the metric system, like from millimeters to meters.
There is a preferred method for unit conversion. Let’s say we want to convert from feet to meters. We construct
a conversion factor in the form of a fraction; we allow the fraction to eliminate the units we want to convert from.
So let’s say the measurement is 6.50 feet. The conversion is 1 m = 3.28 feet.
! 1m $
6.50 feet #
& = 1.98 m
" 3.28 feet %
Note that the conversion factor equals 1 because the numerator and the denominator are equal. The feet cancel,
and the numbers we can multiply or divide as necessary.
We can even do two units at once, such as if we want to convert from miles/hour to meters/second.
85.0
$
miles ! 1610m $ !
1hour
meters
#
&#
& = 38.0
hour " 1mile % " 3600 seconds %
second
The miles cancel, the hours cancel, and we are left with meters per second. Once again, the arithmetic is easy.
Within the metric system, we may need to convert from one metric unit to another. You (hopefully) remember
that the metric system operates on multiples of 10. You will need to know the metric prefixes that are listed
below.
The prefixes are used in conjunction with a “base unit,” typically meters in our case. They can also be used for
liters, grams, bytes, tons, or any other base unit.
So let’s say we need to convert from 744 kilometers to meters. Construct the conversion factor the same way we
did before, and use the metric prefix with the exponent shown in the list above.
! 1m $
744km # 3
& = 744,000m or alternatively
" 10 km %
! 1000m $
744km #
& = 744,000m
" 1km %
As long as the numerator is equal to the denominator, the conversion factor is valid.
What if the conversion is from a unit smaller than the base unit to one that is larger? For example, we want to
convert from millimeters to megameters. We should do this in two steps, such that we first go to the base unit
(meters) and then to the desired megameters.
" 10−3 m % " 1Mm %
546000mm $
' $ 6 ' = 0.000546 Mm
# 1mm & # 10 m &
You may want to use scientific notation.
Be advised – if you decide that you are going to just “move the decimal point over,” you will invariably move it in
the wrong direction, and your answer will be either a million times too big or a million times too small.
Unit Conversion Problems (English-metric)
Use these conversion factors from English to metric units (or vice versa). You don’t have to memorize them.
1.0 m = 3.28 ft = 39.4 inches
1.0 mile = 1610 m = 1.61 km
1.0 N = 0.225 lb
1.0 mile = 5280 ft
2.54 cm = 1.0 inches
2
1.0 acre = 4047 m
1.0 horsepower = 746 Watts
1.0 hour = 3600 seconds
2
1.0 psi (lb/inch ) = 6895 Pascals (Pa)
1.0 calorie = 4.184 Joules
1.0 gallon = 3.785 liters
Now convert the following.
1. 76.9 feet = _____________m
76.9 feet = 23.4 m
2. 37.5 cm = _____________inches
37.5 cm = 14.8 inches
3. 950 Joules = ______________calories
950 Joules = 227 calories
4. 16.5 feet = ______________cm
16.5 feet = 503 cm
5. 15.0 psi = _____________Pa
15.0 psi = 103421 Pa = 103000 Pa
6. 14700 lb = ____________N
14700 lb = 65389 N = 65400 N
7. 65 miles/hour = ______________ft/sec
65 miles/hour = 95 ft/sec
8. 85 miles/hour = ______________m/sec
85 miles/hour = 38 m/sec
9. 225 horsepower = ____________Watts
225 hp = 167850 W = 168000 W
10. 2.25 liters = _____________gallons
2.25 liters = 0.594 gallons
11. 24.0 hours = _____________seconds
24.0 hours = 86400 seconds
2
2
12. 75300 m = ______________acres
75300 m = 18.6 acres
13. 1110 ft/sec = _____________km/hr
1110 ft/sec = 1218 km/hr = 1220 km/hr
14. 175 lb = _____________N
175 lb = 778 N
15. 18.0 miles/gallon = ______________km/liter
18.0 miles/gallon = 7.65 km/liter
16. 65.3 m/s = ______________miles/hour
65.3 m/s = 146 miles/hour
17. 2000. calories = _______________Joules
2000. calories = 8368 Joules
18. 555 N = _______________lb
555 N = 125 lb
19. 3.57 acres = ______________ft
2
20. 100. Watts = ______________horsepower
2
3.57 acres = 155509 ft = 156000 ft
100. Watts = 0.134 horsepower
2
Unit Conversion Problems (metric-metric)
1. 864 cm = _____________m
864 cm = 8.64 m
2. 75.0 km = _____________m
75.0 km = 75000 m
3. 65.8 g = _____________kg
65.8 g = 0.0658 kg
4. 660. nm = ____________m
660. nm = 6.60E-7 m
5. 3.50 Mbytes = ____________bytes
3.50 Mbytes = 3.50E6 bytes
6. 264 µC = ____________C
264 µC = 0.000264 C
7. 431 mm = ____________m
431 mm = 0.431 m
8. 719 l = ____________ml
719 l = 719000 ml
9. 9.22 Gm = ____________m
9.22 Gm = 9.22E9 m
10. 3.42 ml = _____________l
3.42 ml = .00342 l
11. 1215 µm = _____________km
1215 µm = 1.215E-6 km
12. 317 mg = ____________kg
317 mg = 3.17E-4 kg
13. 28.4 kC = ____________mC
28.4 kC = 2.84E7 mC
14. 2674 cm = ___________µm
2674 cm = 2.674E7 µm
15. 432 nm = ____________Mm
432 nm = 4.32E-13 Mm
4. Simplification: The general problem-solving process in physics involves a number of steps. We select a
relationship that applies to the problem in question (that’s the physics), rearrange it as necessary (that’s the
algebra), substitute in the data provided, and then simplify the resulting expression into a single number with a
single unit.
We’ll be dealing with units of meters (m), seconds (s), and kilograms (kg), or some combination of these, for
most of the year. If the data you are provided is not in these units, convert them as per the process shown on the
previous page. Units are to be treated as algebraic variables throughout the problem.
It is important to note that the only time you can add or subtract quantities is when they have the same units.
Units are like variables. You can’t add 3A + 6B. So you can’t add 3 m + 4 m/s. You can, however, multiply or
divide units of any type.
1
d = vit + at 2 , and we want to solve for vi. First we would rearrange the
2
d at
vi = − . In the problem statement, we see that d = 90.4 m, a = 3.22 m/s2,
t 2
Let’s say we are using the formula
formula so
2d − at 2
vi =
2t
or
and t = 5.79 s. We substitute into the first expression and get the following:
2(90.4 m) − (3.22 m / s2 )(5.79 s)2
vi =
2(5.79 s)
This is the expression we need to simplify. Note that the squared
2
2
symbol is outside the parentheses, so it applies to everything inside them. So (5.79 s) = 33.52 s . Therefore the
2
2
s cancels with the s term in the acceleration (a).
First we take care of the numerator and get
So
vi = 6.30
m
s
vi =
180.8m − 107.9m
11.58 s
or
vi =
72.9m
11.58 s
The numbers divide.
Meters per second is the unit we would expect for a velocity – that recognition will come later.
Try substituting the same data into the alternate expression for vi, and see that it simplifies to the same quantity.
When we report a quantitative answer in physics, the unit is as important as the number. If I ask you for a
velocity and you say “5,” that tells me nothing. Five miles per hour? Kilometers per second? Centimeters per
year? You get the idea. Get into your head now that the units always matter.
The importance of this skill cannot be overstated. If your algebra and your simplification are weak, you will
almost never get the right answer quantitatively.
These are the required math skills. We will come across some others as we go, including significant figures and
scientific notation. But if you can do the problems in this review and in the WebAssign homework for the first
nd
two weeks, you will be fine on the exam scheduled for Friday of the 2 week of school.
There are several video and educreation presentations available one the website under the math review section.
You can consult them as you please. In order to access the educreation presentations, you must register for the
class at http://www.educreations.com/sr/87DB37.
At some point before Friday, you should attempt to do one or more versions of the practice exam. One is a paper
copy of an old exam, and the other is online in WebAssign. These are the approximate duration and difficulty
level that you can expect from the real exam. The practice exams are not assignments to be turned in and will
not affect your grade. However, it is in your best interest to attempt them.
Simplification Problems
"
m
m%
$ 6.00 − 3.50 '
s
s&
#
1. d =
(5.00s)
2
2.
v f = 78.5
2
!
!
m$
m$
6. v f = # 6.20 & + 2 # 1.20 2 & ( 5.00m)
s%
s %
"
"
m !
m$
+ #12.2 2 & ( 43.0s)
s "
s %
7.
v=
(3.65kg − 1.83kg) "16.9 m %
$
'
(3.65kg + 1.83kg) # s &
2
"
m
m%
$17.3 − 11.2 '
s
s&
#
3. a =
0.422s
!
m$
2 # 865 & cos30o sin30o
s%
"
8. R =
m
9.80 2
s
(
"
m
m%
$ 6.55 − 4.13 '
s
s&
#
4. t =
m
3.47 2
s
)(
(
)
)(
2 % 6.65E24kg 1.90E30kg
"
9. F = $ 6.67E − 11 N m '
g
2
kg2 &
#
1.55E11m
(
2
5.
!
2
m$
1!
m$
d = # 6.20 & ( 4.00s) + # 2.10 2 & ( 4.00s)
s%
2"
s %
"
)
!
!
m
m$
m$
12.0 ± # 2.00 & + 2 #1.55 2 & (12.0m)
s
s%
s %
"
"
10. t =
m
1.55 2
s
Simplification Answers
1. d = 6.25 m
6. vf =
7.10 m/s
2. vf = 603 m/s
7. v = 5.61 m/s
3. a = 14.5 m/s2
8. R = 66100 m or 6.61E4 m
4. t = 0.697 s
9. Fg = 3.51E22 N
5. d = 41.6 m
10. t = 11.9 s and 3.60 s
)
5. Graphing: Often in physics, quantities are displayed graphically. We want to think in terms of independent
and dependent variables. The independent variable is the quantity that either is changed purposely by the
researcher or changes on its own (like time). The dependent variable is the quantity that depends on the
independent variable.
We will begin the year by looking at how quantities such as position and velocity change over time. In such
cases, we typically have a function that shows, for example, displacement (or position) as a function of time. We
would call displacement the dependent variable and time the independent variable.
Before showing the graphing technique, it is important to remember a few rules about graphs:
A. Graphs should have titles. Typically they would be titled as “[dependent variable] vs. [independent variable].”
B. The dependent variable is graphed on what you call the “y axis,” or the ordinate axis. The dependent variable
is graphed on what you call the “x axis,” or the abscissa. You would describe a typical graph as “y vs. x.”
C. The graph axes should have labels, with units as appropriate. If we are graphing d vs. t, the y axis should be
titled “d (m)” and the x axis would be “t (s),” assuming we are measuring time in seconds.
D. Electronic tools can be used to easily construct graphs, but making graphs manually is an excellent skill. The
graph should be neat and easily interpretable. Given a particular space to use, use as much of the space as is
feasible. Large graphs are easier to interpret for a reader than small graphs.
Graph this function of d(t) for the interval t = 0 to 4 seconds, in one second intervals, in the space below:
!
m$ 1!
m$
d(t) = # 2.50 & t + # 6.00 2 & t 2
s % 2"
s %
"
Note the range of the independent variable (0 to 4 s) and the intervals (1 s) are provided. We begin at the start of
the range, where t = 0 seconds, and we insert this value into the equation wherever t appears. We call the
dependent variable d0, or the displacement at t = 0 s. Then we proceed to t = 1 s, t = 2 s, and so on.
!
2
m$
1!
m$
d0 = # 2.50 & (0 s) + # 6.00 2 & (0 s) This solution is trivial, but it is important that d0 = 0m.
s%
2"
s %
"
!
2
m$
1!
m$
d1 = # 2.50 & (1s) + # 6.00 2 & (1s) Note that this is a simplification problem. d1 = 5.50 m.
s%
2"
s %
"
!
2
m$
1!
m$
d2 = # 2.50 & (2 s) + # 6.00 2 & (2 s)
s
2
s %
"
%
"
d2 = 17.0 m. Continue until t = 4 s.
The data should be shown in a table, as below.
d"(m)"
0"
5.50"
17.0"
34.5"
58.0"
t"(s)"
0"
1"
2"
3"
4"
Graphing Problems
1. Fill in the data table and graph the displacement vs. time function below for t = 0, 1, 2, 3, 4 seconds.
&
+ $−
%
=−
t (s)
# & #&
! + $ !$
" % "%
d (m)
#
!
"
2. Fill in the data table and graph the displacement vs. time function below for t = 0, 1, 2, 3, 4 seconds.
m # & 1 #&
m#
&
d = −8.55m + $ 5.88 !t + $ !$ − 1.33 2 !t 2
s " % 2 "%
s "
%
t (s)
d (m)
Graphing Answers
1.
2.
t (s)
d (m)
0
-2.20
1
-4.31
2
-4.06
3
-1.47
4
3.48
t (s)
d (m)
0
-8.55
1
-3.34
2
0.55
3
3.11
4
4.33