


MECHANICS

USER MANUAL

VERSION 1.0

















SPARTAN SOFTWARE

PROVIDER OF MATHEMATICAL AND SCIENTIFIC SOFTWARE FOR EDUCATION




TABLE OF CONTENTS

Introduction									3
Metrics										4
Significant Figures 								5
Temperature Conversion								5
Statistics										6
Trigonometry Values								7
Areas/Volumes									8
Pythagorean Theorem								8
Equations/Graphs									8
Motion										9
Acceleration									11
Vectors										14
Newton's Laws									16
Friction										18
Energy										19
Equilibrium										20
Momentum										22
Gravitation										24
Rotational Motion									25
Circular Motion									28
Harmonic Motion									28
Archimede's Principle								29
Pascal's Principle								30



INTRODUCTION

Mechanics was designed to help high school and college students solve problems in an 
introductory physics class. The program is easy to learn and run. If you have used other
 Windows programs before, Mechanics will not cause you any problems. You can use your mouse 
or the keyboard. The menu functions operate the same as the mouse functions. 

Mechanics is not a physics tutorial. Rather it is a problem solving assistant. Its primary 
purpose is to help students with many of the calculations encountered in a typical physics 
class. Students can use Mechanics as a homework helper, a study guide, or as a review for 
a test. One of the strong points about Mechanics is that it shows the work needed to get 
the answer to the problem. The answer alone would not help much. With the work shown, students 
can see how a problem is addressed and use that in solving other similar problems.

Teachers can use Mechanics as an instructional tool in the classroom. With a computer hooked 
up to an overhead projector, a teacher can demonstrate many types of chemical principles and 
problems. You can experiment with a particular variable in order to illustrate a particular 
concept. For example, you can change the velocity of an object and show the kinetic energy. 
Then have the students see if they can identify the relationship that exists. The 
possibilities are limited only by one's imagination.

In the media center, groups of students or an individual can receive extra assistance with a 
particular problem or concept. Students can use Mechanics to study for a quiz or a test. It 
can be used in group competitions in the classroom. 
Mechanics is a tool that can be used by both students and teachers. 


TUTORIAL
In this tutorial, you will receive a brief introduction to all sections of Mechanics. 
Hopefully, this will help you in your understanding of physics and make using the program 
easier.

In each section of the program, there are two sources of help in addition to this manual. 
First you have the help button, illustrated by the question mark. Press that button and you 
are given a brief explanation of the concept or problem type. Secondly, by placing the cursor 
over certain data entry boxes, you will be given an additional prompt or reminder. 
Many steps have been taken to prevent the entry of invalid data. If you attempt to enter 
invalid data or fail to enter data where it is required, a message box will notifying you of 
the error. You can then enter the correct data and the program will continue normally.

At the top, you have two rows of menu items that can be accesses by using the Alt key in 
combination with the underlined letter. For example, when you see G, you would press the 
Alt key and the G at the same time. By using the Tab key, you can move through all the menu 
items. Below the menu items are two rows of buttons. Each button corresponds to a major 
section of Mechanics. By clicking a particular button, you can access a particular section 
of the program. At the bottom of the Main screen is a single row of buttons that corresponds 
to data tables. When you click the Periodic Table button the Periodic Table of the Elements 
appears. 

In each section of Mechanics, you have the Print option, If you want to print a copy of your 
calculation, simply click the Print button on the form you are working with and it will print. 
On each section, there is also the  ?  button. This is the Help button and when you click on 
it, you will be given a brief, additional explanation of the problem type. You can also Copy 
and Paste results from one calculation to another. You can have several forms open at the same 
time. However, only one is visible at a time. The others will be minimized at the bottom of 
the screen. If you do have several forms open at the same time, be sure to close all of them 
when you exit the program. That way the forms will be removed from your computers memory.

General

The General section contains problems that are found throughout physics. It does not refer to 
a particular physics concept but rather it addresses some of the ?math? that is used 
throughout the program and throughout physics

1.  Metrics.  In science, you cannot use the English system. All measurements must be in the 
metric system. Many times you have to convert a number from one metric unit to another. What 
is involved is basically multiplying or dividing by 10 or a multiple of 10.  Mechanics does 
this for you.

Example 1.  How many kilometers are there in 500 meters? In the General section, click on the 
Metrics button. The Metric Conversion form then appears. At the top of the form is the data 
box where you enter the number you want to convert. At this point enter 500. Select the 
distance option by clicking on it. In the From area, you select the label of the number you 
want to convert. In this case, select the meters option by clicking on it. Next you select the 
label that you want to convert the number to in the To area. Click on kilo. Finally, you simply
click  Ok and the answer appears in the answer box. 500 meters is equal to 0.5 kilometer.

Example 2.  How many milligrams are there in 5 kilograms? Click the Clear button to clear the 
previous entries. In the data entry box at the top of the Metric Conversion, enter 5. Select 
the mass option by clicking on it. In the From area, select the kilogram option by clicking on 
kilo. In the To area select milligrams and press Ok. There are 5,000,000 milligrams in 5 
kilograms.

2.  Significant Figures.  This causes students a lot of confusion that is really not necessary 
once you understand what it is all about. Calculations in physics require numbers to be very 
accurate. There is no room for numbers that are guesses or just estimates. Each number that is 
used must be an accurate, dependable number. So rules have been established to guarantee that 
the numbers used are just that - dependable and reliable. You may have been introduced to these
rules and have noticed that several of the rules are concerned with zeroes.

Example 1:  How many significant figures are there in 123456789? In the General section, click 
on the Significant Figures button. At the top of the Significant Figures forms is the entry 
box. In that box type in the number 123456789 and press Ok. 123456789  contains 9 significant 
figures. Each non-zero number is assumed to be an accurate, reliable number.

Example 2: How many significant figures are there in 100. Press the Clear button to erase the 
previous entries. Enter 100 into the data entry box and press Ok. 100 has only 1 significant 
figure. The zeroes are there only to place the 1.

Example 3:  How many significant figures are there in 100.00?  Press the Clear button to erase 
the previous entries. Enter 100.00 into the data entry box and press Ok. 100.00 has 5 
significant figures. Zeroes between a non-zero number and the decimal point are significant as 
well as zeroes after the decimal point.

Example 4: How many significant figures are there in the sum of 10.65 and 12.3?  If you add 
the numbers 10.65 and 12.3, the final answer has only three significant digits. An answer can 
have only as many significant digits as the number with the least number of significant digits. 
12.3 has three and 10.65 has four. So the final answer can only have three.

3.  Temperature Conversion.  Temperatures in physics cannot be in the Fahrenheit scale. They 
must be in either the Celsius scale or the Kelvin scale. You must be able to convert 
temperatures from one scale to another. This section of Mechanics does all the conversions for 
you.  At the top of the Temperature Conversion form you will see one data entry box. This is 
where you enter the temperature that you want to convert. Once you have entered the 
temperature, you simply click the button corresponding to the conversion you want to make.



Example 1: What is 32 degrees in the Celsius scale and in the Kelvin scale? First you enter 
32 in the data entry box and the top of the Temperature Conversion form. Next you click the 
button labeled 'F To C'. The new temperature is 0 degrees. The work used to reach that answer 
is displayed at the bottom of the form. 

To determine the new temperature in the Kelvin scale, simply click the button labeled 'F To K'. 
The new Kelvin temperature is 273 degrees. You will notice that to convert from Celsius to 
Kelvin, you add 273 to the Celsius temperature.

Example 2: What is the Fahrenheit equivalent of 0 degrees Kelvin? Enter 0 in the data entry 
box and click the button labeled "K To F'. The new temperature is -459 degrees.

4.  Statistics.  The Statistics section provides you with mathematical means of analyzing data. 
This is very to scientists since they must be able to determine if the results of their work 
is accurate and dependable. There are six statistical tests available to you. 

The first test is the Mean or average. Many times if you have numerous results for an 
experiment or test you will want a single value to represent the result. This is where the 
Mean is valuable. In the Statistics section, you click on the Mean button to go to the Mean 
section. Though determining the Mean is quite easy, Mechanics helps you with it.

Example 1: What is the Mean of the following values: 10.2, 13.3, 9.9, 11.4, and 12.99? To 
start you must have the sum of these values. You can determine this by using your personal 
calculator or using the calculator provided with Mechanics. Add the numbers together and enter 
the sum (57.79) in the first data entry box in the Mean form. Next you enter the number of 
measurements (5) into the second data box and press Ok. The Mean is 11.558. What about 
significant figures? As a reminder, if you are considering significant figures, the correct 
answer is 12. Why? 9.9 has only 2 significant figures while the rest have 3 or 4. So the 
answer can have only two. You round off the 11.558 to the nearest two significant figures 
which is 12. Keep this in mind for future reference. The rest of the tutorial will give the 
actual calculator value.

The second statistical test is the Median. Instead of working out an actual example, this 
section just gives you the method of how to determine the Median. It is the number that is in 
the middle of the range of measurements that you have. If you have 13 measurements, the Median 
is the 7th measurement.

The third test is the Mode. Like the Median test, you are just given the method for determining 
the Mode. The Mode is the number that is used the most. If a certain number occurs 10 times 
while the rest of the numbers occur only 5 times, the Mode is 10.

The fourth test is a very important statistical one. This is the Standard Deviation. It is a 
measure of variability that indicates by how much all of the measurements vary from the Mean 
(average). It always uses the Mean and never the Median or Mode. Since a discussion of the 
Standard Deviation is beyond the scope of Mechanics, just remember that approximately 
two-thirds of all measurements fall within 1 Standard Deviation of the Mean. The smaller the 
Standard Deviation, the closer most of the measurements are. The larger the value, the more 
spread out they are. 

Example 1.  What is the Standard Deviation for the following measurements of displacement 
(10,8,6,4,2)? Select the Standard Deviation button on the Statistics form.   In the first data 
entry box you enter the number of measurements. In this case, you enter 5. Now you add together 
the measurements. They add up to 30 so you enter 30 into the second data entry box. Finally, 
you square each number and then add them together. This number is 220 so you enter 220 in the 
third data entry box. Press Ok and the Standard Deviation is 2.828. Six is the mean. This means 
that roughly two-thirds of the measurements fall between 3.172, which is 6 - 2.828, and 8.828, 
which is 6 + 2.828. One value for using the Standard Deviation is to determine how accurate and 
precise the results of your experiment are.

Variation is closely related to Standard Deviation. Variation is simply the square root of the 
Standard Deviation. You enter the same information and Mechanics does the calculation for you. 

Example 1.  In the Variation data boxes, enter the same data that you entered for the Standard 
Deviation and click Ok. The variation is 8.

The last statistical function, and it is not really statistical, is Percent of Error. It is 
included in this section merely for grouping purposes. The Percent of Error lets you know how 
far off you are from the correct answer percentage wise. The actual magnitude of the difference 
between your results and the actual results can be a lot different that the  percent that you 
are off. For example, your answer may be 50 off of the correct answer but may be only 2% off. 
The Percent of Error just gives you another way to view your results.

Example 1: Your result was 9500 and the actual value was 10000. What was your Percent of Error? 
You were actually 500 lower than the actual value. That may seem like a lot. On the Statistics 
form, select Error. On the Error form, you enter your value in the first data entry box. So 
enter 9500 in the first box. Next enter 10000 in the second box for the actual value. Press Ok. 
Even though you were 500 off in actual value, you were only -5% off. The minus sign means that 
your answer was lower than the actual value. 5% may or may not be an acceptable Percent of 
Error. It depends upon the work you are doing. However, 5% sounds a little better than 500. You 
may require substantially lower error.

5. Trigonometry Values.  In physics, trigonometry is a very important tool. It is used whenever 
an angle is involved. You will see it involved in most kinds of motion problems. Since 
trigonometry values are so important, it is provided here for even though you may have a good 
calculator available. Many times you need the Sin, Cos, or Tan value for a particular angle. 
Also, you may need to calculate the angle from one of these values. Mechanics provides both 
options.

Example 1.  What is the sin of 60 degrees? On the Statistics form, select Trig Values.  Enter 
60 in the top data entry box. Then click the Sine button. The answer appears in the bottom data 
box. The sin of 60 degrees is 0.86603. To see what the tangent of 60 degrees is, just click on 
the Tangent button. You do not have to clear the data.

Example 2.  If the tangent value is 3.73205, what is the corresponding angle? In the middle box 
enter 3.73205 and click the button labeled 'Angle From Tangent" and press Ok. The angle is 75 
degrees.

6. Areas / Volumes.  On occasions in physics, you need to know the area of a certain figure or 
its volume. Mechanics contains the formulas for the most common areas and volumes that you will 
probably need. After clicking on the Areas/Volumes button, you select the formula you need by 
selecting it and clicking Ok.

Example 1.  What is the area of a triangle that has a height of 10 and a base of 15? In the 
Areas/Volumes form, select Area of triangle. You will notice that the prompts change for each 
formula. When you select Area of triangle, the prompts ask you for the height and the base. In 
this example, enter 10 in the first data entry box for the height of the triangle. Then enter 
15 in the second data entry box for the base and press Ok. The area is 75. The work for the 
problem is shown at the bottom of the form. All you have to provide is the correct label.

Example 2.  What is the volume of a cone that has a radius of 5 and a height of 15? Select the 
Area of cone option. Enter 5 in the first data entry box for the radius and 15 in the second 
data entry box for the height. Press Ok and the volume is 392.699. 

7.  Pythagorean Theorem.  The Pythagorean Theorem function is provided for your convenience. It 
is a basic function from Geometry and Trigonometry. When you have a right triangle, and know 
any two of the sides, you can easily calculate the length of the third side. This function also 
has an important role in many physics calculations.

Example 1.  What is the hypotenuse of a right triangle with sides of 12 and 15? On the 
Pythagorean theorem form select the 'find the hypotenuse' option. Enter 10 in the first data 
entry box and 15 in the second box. Click Ok. The hypotenuse is 19.209.

Example 2.  In a right triangle with a hypotenuse of 25 and one side of 13, what is the length 
of the second side? Select the 'Find the length of a side' option. Enter 25 in the first data 
entry box for the hypotenuse and 13 in the second box for the length of one side. Press Ok and 
the length of the second side is 21.354

8.  Equations/Graphs.  Analyzing large amounts of data can be quite difficult. That is also 
true of an equation. Physics has many equations especially for motion type problems. Simply 
having an equation is often not enough to really analyze what is happening. To really see what 
is happening, it is often beneficial to be able to visualize an equation. In addition to seeing 
an equation, you often need to analyze what is happening at certain points in the equation. In 
Mechanics you have the ability to graph an equation and then to analyze it. For a specific 
value of X, you can calculate the corresponding value of Y and you can determine the slope. The 
value of Y tells you how high the graph is at a certain point and the slope tells you how fast 
the graph is increasing or decreasing.

The graphing function in Mechanics is limited to 100 values of X and these values must be 1 or 
2 digit integers. The values of X are to be a range of consecutive values like 1 to 100. 

Example 1.  Plot the equation X^2 + 10. Use the range of 1 to 100 for the values of X. What is 
the value of Y when X = 25? What is the slope of the equation when X = 25? After you have 
selected the Equations/Graphs, you will notice the equation entry box at the top of the form. 
Correct entry of an equation is very important. To assist you, an example equation is given to 
you in the form in which an equation is to be entered. In addition, remember that you can only 
use 1 or 2 digit numbers and you can only have a range of 100 consecutive numbers. 

In the equation entry box at the top of the form, enter the equation exactly as shown above. 
Next enter 1 in the 'Starting value for X' box. Enter 100 in the 'Ending value for X'  box. 
Press Ok and the equation is plotted for you. As you can see, the line rapidly curves upward.  
Enter 25 in the 'Enter a value for X' box. Click on the 'Calculate Y' button. The value of Y 
at X = 25 is 635. Now press the 'Slope at X' button. The slope at X=25 is 50.000. This means 
for that value of X, Y is increasing at a rate of 50.000.  (Not shown)

Example 2.  Plot the equation  2X^2  - X + 5. Use the range 1 to 100. What is the Y value for 
X = 50?  What is the slope of the equation at X =90?  Enter the equation carefully exactly as 
shown and press Ok. The line is curving rapidly upward. Enter 50 in the 'Enter a value for X' 
box and press the 'Calculate Y' button. The value of Y is 4955. To see the slope at X=90, enter 
90  in the 'Enter a value for X' box. DO NOT CLICK ON CLEAR! That will clear the graph and all 
of the X and Y values. After entering 90, simply click on the 'Slope at X' button. The slope 
is 359.

Motion

In Mechanics, all of the problems associated with motion have been categorized. In this 
particular section only displacement and velocity have been included. Other topics related to 
motion have been grouped together in other sections.

1.  Displacement.  The first of the topics addressed here is displacement. This sometimes can 
be confusing. Displacement refers to the distance between the starting point and the ending 
point. If they are the same, as in running around a track, the displacement is zero. 
Displacement is not the same as distance. You can have a large distance but a very small 
displacement.

In the Main screen click on the Motion button. You will then see two additional buttons for 
displacement and velocity.  Click the Displacement button. In the Displacement form, you will 
notice that there are only two data entry boxes. You will enter the starting point in the first 
box and the ending point in the second box. Before you enter your data, you must decide on a 
measurement system like the X axis on a grid or simply meters. 

Example 1.  A runner runs four laps of a 400 meter track. What is the displacement? First you 
decide that the starting point is zero. Also since the finish line is at the same spot as the 
starting point, it is also labeled as zero. On the Displacement form, enter zeroes in both of 
the data boxes and press Ok. As you obviously knew, the displacement was zero even though the 
runner went four times around the track. This concept is important in many physics problems.

Example 2.  During a vacation, a family decided to drive across the country from New York to 
Los Angeles. They drove to Los Angeles and back to Detroit which is 800 kilometers from 
New York. What was their displacement? With zero for the starting value in New York and -800 
for the location of Detroit from New York, enter 0 for the starting point and- 800 for the 
ending point. Press Ok. The displacement is 800. They drove a lot further than 800 kilometers 
but they were only 800 kilometers away from the starting point.

Displacement is essentially a subtraction problem. All you need to know is the starting and 
ending points.

2.   Velocity.  When you click on the Velocity button on the Motion form, you are presented 
with two additional choices. These are Average Velocity and Instantaneous Velocity.  While 
Average Velocity determines the velocity over the entire trip, Instantaneous Velocity 
determines the velocity at a particular instant in time. 

Example 1.  Average Velocity is the total displacement divided by the total time. This does not 
take into account times when the velocity is 0 and other times when it is very high. If it 
takes a race car driver 3.45 hours to drive the Indianapolis 500, what was his average velocity?  
Select Average Velocity in the Velocity form. You notice that there are four data entry boxes 
with prompts. In the first box, you enter 500 which is the final position. Put 0 in the second 
data box as the initial starting position. Then you enter the final time of 3.45 in the third 
data box and a zero in the last box for the starting time. When you press Ok, you see that the 
average velocity is 144.928 miles per hour. This is one of the exceptions to the use of the 
metric system. 

Example 2.  Instantaneous Velocity is the velocity at a particular point in time. It does not 
cover the entire trip. Select the Instantaneous Velocity option at the top of the Motion form. 
You will also notice that when you select different options in Mechanics, the prompts also 
change to match the type of problem you are working on. 

Instantaneous Velocity can be solved in a couple of ways. If you have the equation that 
describes the motion of the object in question, you can take the derivative of the equation. 
Or you could graph the equation with the graphing section earlier in Mechanics. Another method 
is to determine the starting velocity, the acceleration and the time involved and use these to 
determine the Instantaneous Velocity. For our purposes, we will use the last method. The space 
shuttle blasts off and accelerates at 11.2 meters per second per second. What is its 
Instantaneous Velocity after 10 seconds? In the first data box enter a 0 for the initial 
velocity. Enter 11.2 for the acceleration and 10 for the time. Press Ok. The Instantaneous 
Velocity is 112 meters per second.

Acceleration

Acceleration is the rate of increase or decrease in velocity. If you fall from a height, you 
will go downward faster and faster because of acceleration. Think of it as a force that pulls 
on you as you fall. If you throw an object upward, it gradually slows down and begins to fall 
downward faster and faster until it hits the ground.

When you click on the Acceleration button, you notice that there are four additional choices. 
Each deals with a different aspect of acceleration.

1.  Average Acceleration.  Average Acceleration deals with acceleration over the length of time 
the object is moving. Like average velocity, it does con show increases or decrease in 
acceleration, but what the acceleration is overall. 

Example 1.  A car comes to a stop at a red light. It slowed down from 9.00 meters per second 
to 0 meters per second in 5.00 seconds. What was the average acceleration? After clicking on 
the Acceleration button click on the Average Acceleration button. In the first data box enter 
9.00 as the initial velocity. Enter 0 in the second data entry box for the final velocity. 
Finally, enter 5.00 in the third data box for the change in time. Press Ok. The Average 
Acceleration is -1.8 meters per second per second. The negative sign means that the velocity 
of the object is slowing down. The 1.8 is how much the object is slowing down each second. In 
this case it means that for every second that passes, the velocity slows down by 1.8 meters 
per second. Your label for acceleration must include a velocity change (-1.8 meters per second) 
and a time change (per second).

Example 2.  A rocket is launched and reaches a velocity of 16 meters per second in 2 seconds. 
What is the Average Acceleration? Click the Clear button to erase the precious entries. Enter 
0 for the starting velocity and 16 for the final velocity. Enter 2 for the time change and 
click Ok. The Average Acceleration is 8.00 meters per second per second. The sign is positive 
since the velocity is increasing. 8.00 meters per second is how much the velocity increases 
each second.

2.  Constant Acceleration.  Constant Acceleration implies that the rate of increase or decrease 
stays the same. For example, the force of gravity on Earth causes a constant downward 
acceleration of 9.81 meters per second per second. Knowing the value of acceleration, you can 
calculate several different values like displacement, velocity and time.

On the Acceleration form, click the Constant Acceleration button. You are then presented with 
four different options all related to constant acceleration. 

Example 1.  If a car is traveling at 42 meters per second and then hits the brakes stopping 
in 5.5 seconds, how far did the car travel while stopping? The distance covered is actually 
displacement. So select the displacement option on the Constant Acceleration form. Enter 42 
in the first data box as the initial velocity. Enter 0 in the second data box since the car 
came to a complete stop. Finally, enter 5.5 in the third box as the time needed to come to a 
complete stop. Press Ok. The displacement is 115.5 meters.

Example 2.  A plane waiting at the end of a runway starts to accelerate at a rate of 4.8 
meters per second per second for 15 seconds.  What is its velocity at takeoff?  Press the 
Clear button to erase the previous entries. Select the velocity option. Enter 0 for the 
initial velocity, 4.8 for the acceleration and 15 for the time. Press Ok. The plane's velocity 
is 72 meters per second.

Example 3.  Another plane also waits to take off on the runway after the first plane has left. 
If this plane accelerates at 1.6 meters per second per second for a distance of 1600 meters, 
what is the time required for takeoff? Clear the previous entries and select the time option. 
Only two entries are needed for the time, acceleration and displacement. In the first data box 
enter 1.6 as the acceleration. Enter 1600 in the second data box for the displacement. Press 
Ok. The time is 44.721 seconds. If you are considering significant digits, the answer is 45 
seconds since 1.6 has only 2 significant digits and that is all your answer can have. You 
simply round 44.721 to 45.

Example 4.  A sports car can accelerate from rest to 87 miles per hour in 8 seconds. What is 
the acceleration of the car? Clear the entries and select the Acceleration option. Since the 
car started at rest, enter 0 in the first data box. Enter 87 in the second and 8 in the third. 
Press Ok. The acceleration is 10.875 miles per hour per second. For each second that passes, 
the car increases its velocity by 10.875 miles per hour.

3.  Falling Objects.   In the Falling Objects section, you analyze various factors relating to 
objects falling under the influence of Earth's gravity.  Factors such as velocity, time and 
distance are calculated.  All objects accelerate at the same rate downward unless affected by 
some other force. Without any outside influence, objects will accelerate downward at a rate of 
9.81 meters per second per second. Objects thrown upward will decelerate at an identical rate 
of 9.81 meters per second per second and gradually begin to accelerate downward at a rate of 
9.81 meters per second per second.

Example 1.  A rock sitting at the edge of a 1000  meter  cliff loosens and falls to the bottom 
of the cliff. What was the rock's velocity when it hit the ground?  On  the Acceleration form, 
click on the Falling Objects button. You will then see three options: velocity, time and 
distance. Select the Velocity option. Since the rock was sitting at rest, enter a 0 for the 
velocity. Acceleration on Earth is constant at 9.81 meters per second per second. However, how 
about the Moon or some other planet? This is why the acceleration box is not filled in for you. 
You may want to calculate velocity on the Moon. So for this problem, enter 9.81 for 
acceleration. Finally, enter 1000 for the distance and click Ok. The velocity is 140.071 meters 
per second. Realistically, an object might not fall that fast due to upward air pressure. As an 
object falls, the force of the air pushes upward harder and harder until a terminal velocity is 
reached. It will not continue to accelerate.

Example 2.  How long will it take the rock, in example 1, to hit the ground? Clear the entries 
from the previous problem and select the Time option. The entries as you see are exactly the 
same. Enter 0 for the initial velocity, 9.81 for acceleration, and 1000 for distance. Click on 
Ok. The rock will hit the ground in 14.278 seconds.

Example 3.  A ball is dropped from the top of a tall tower. Neglecting air resistance, what is 
the displacement after 3 seconds? Clear all entries and select the distance option. The initial 
velocity of the ball is 0 meters per second. The acceleration is 9.8 meters per second per 
second. Finally, enter 3 for the time and press enter. The displacement is 44.1 meters.

4. Projectile Motion.  In projectile motion problems, you have two categories to consider. One 
is when projectiles are launched horizontally. The other is when projectiles are launched at an 
angle. You will notice when you select the Projectile Motion option, the first choice you have 
to make is which type of projectile motion you are dealing with. Once you have selected the 
type of motion you have, you can then select the option you wish to find. All projectile motion 
problems involve the force of gravity. However, that value is not entered for you since you may 
wish to consider motion someplace else other than Earth.

Example 1. A low-flying airplane is traveling at 22.5 meters per second and at a height of 10 
meters. The pilot drops a dummy, practice bomb. How far horizontally will the bomb travel 
before it hits the ground? First select Launched Horizontally as the type of motion problem. 
This is an example of a two step problem. One of the first things you need to know is the time 
it will take for the bomb to hit the ground.  So select the Time option. The only value you 
have to enter is the height. Enter 10 in the data box and press enter. The time required for 
the bomb to hit the ground is 1.429 seconds. Now clear the entry and select the Horizontal 
distance option. Enter 1.429 in the first data box as the time. Enter 22.5 in the second data 
box as the velocity and press enter. The bomb will travel  32.153 meters before it hits the 
ground.

Example 2.  Using the problem in example 1, what is the velocity of the bomb when it hits the 
ground? Select Launched Horizontally as the problem type and the Initial forward velocity 
option. Enter 32.153 as the distance. Enter 10 as the height and press enter. The initial 
forward velocity is 22.507 meters per second.

Example 3.  A ball rolls off a table with a velocity of 1 meter per second. It hits the floor 
.3 meters from the edge of the table. How high is the table? Select Launched Horizontally and 
the Height option. Enter .3 as the distance. Enter 1 as the velocity in meters per second and 
press enter. The height of the table is .441 meters. 

Example 4.  A rock fell off of the edge of the Grand Canyon. If the rock fell 800 meters 
(approximately 1/2 mile), how long would it take to hit the bottom of the canyon? Select 
Launched Horizontally as the problem type and the Time option. Time is based only on height. 
Enter 800 as the height and press enter. The time is 12.778 seconds.

Projectiles launched at an angle require a little more work. You now have to calculate 
components of forces.

Example 5.  A long jumper jumps at an angle of 20 degrees and at a velocity of 11 meters per 
second. How far will he go? Select Launched at an angle and then select Horizontal distance. 
Enter 20 as the angle at which the jumper left the ground. Enter 11 as the initial velocity 
and press enter. The horizontal distance is 7.936 meters.

Example 6.  Continuing with example 5, how long was the jumper in the air? Select the Time 
option. Enter 20 as the angle and 11 as the initial velocity and press enter. The time the 
jumper was in the air was 0.768 seconds. The time at which the jumper was at his highest point 
in the jump was 0.384 seconds or 1/2 the total time.

Example 7.  Again using example 5, how high did the jumper go vertically? The Height option 
uses the same values as the previous two examples did. Enter 20 for the angle and 11 for the 
initial velocity. Pressing enter gives 0.722 meters as the height.

Example 8.  A ball is thrown a distance of 55.231 meters and at an angle of 30 degrees. What 
was the initial velocity? Clear the entries and select Initial forward velocity. Enter 55.231 
as the distance in the first data box. Enter 30 in the second data box as the angle and press 
enter. The Initial forward velocity is 25.000 meters per second.

Example 9.  A ball is thrown with an initial velocity of 20 meters per second. After 2.886 
seconds, it lands 40.816 meters away. At what angle was the ball thrown? Clear the entries and 
select the Angle option. Enter 20 in the first data box as the initial velocity. Enter 2.886 
seconds as the time. Enter 40.816 meters as the horizontal distance. Press enter. The angle is 
44.997 degrees.

Vectors

Vectors are forces that have a numeric value and a direction. Vector calculations can be done 
algebraically or graphically.  Drawing vectors out can be very useful in seeing the overall 
picture. Once you understand the overall situation, you can tackle the vectors algebraically. 
There are usually just two categories of vector operations: resolving a vector into its 
component parts and determining the resultant vector from several other vectors. In Mechanics, 
vectors have been broken down into three sub-sections. The first is to determine the resultant 
of two perpendicular vectors. The second takes one resultant vector and breaks it down into its 
X and Y components. The third sub-section adds together two or three non-perpendicular vectors. 
In all cases, the problems are handled algebraically. However, if you draw out or can picture 
the individual vectors or their components, the problems are a lot easier.

1.  Determine The Resultant.  In this part, you add together two perpendicular vectors. In 
order to do so, they must be broken down into their respective X and Y components. When you 
are asked to enter a vector, you only enter its magnitude. Mechanics handles the algebra.

Example 1.  Determine the magnitude and direction of the resultant of adding together two 
vectors whose magnitudes are 100 and 50. Remember that these vectors are perpendicular to each 
other. If you are drawing these, you can draw two perpendicular arrows pulling on a common point. 
From the Main screen, you select the Vectors button. Once the Vectors selections are presented 
to you, select the first button Determine Resultant. Once you see the Determine Resultant form, 
you will notice that there are only two data entry boxes for you to fill. You can enter the 
magnitudes of the vectors in any order. Since they are perpendicular, the order does not matter 
much. For this problem, enter 100 into the first data box and 50 into the second. Press return. 
The resultant force is 111.803 and the direction is 63.435 degrees. Common sense tells you that 
the resultant has to be more than 100 and the direction will be towards the stronger vector. 
That is why the direction is not in the middle between the two vectors.

2. Resolve Vectors  The opposite of determining the resultant of adding two vectors is to break 
a resultant vector down into its components. 

Example 1.  A car is traveling 95 km/h in a direction of 35 degrees east of north. What are the 
component velocities? Select the Resolve Vectors button from the Vectors form. On the Resolve 
Vectors form you enter the magnitude and angle of the resultant. Enter 95 into the first data 
box for the magnitude. Enter 35 into the second box for the angle and press enter. The X 
component is 77.819 and the Y component is 54.490. This means that even though the car is 
traveling 95 km/h at a direction of 35 degrees east of north, the car is also traveling north 
at 77.819 km/h and east at 54.490 km/h at the same time. This may seem confusing and even 
impossible but if you sketch it out to scale, it begins to make sense.

3. Add Vectors  Many times the vectors you want to add together are not perpendicular but at 
different angles. This may seem quite difficult but Mechanics handles the problems quite 
easily. You are limited, though, to no more than three vectors. Basically, what happens is 
that each vector is resolved into its X and Y components taking into consideration their 
angles.  All the X components are added together and all the Y components are added together. 
The resultant and its angle are then determined.

Example 1.  Add the following three vectors. Vector 1 has a magnitude of 8 in a northerly 
direction (0 degrees). Vector 2 has a magnitude of 3.5 and has a direction of 35 degrees east 
of north. Vector 3 has a magnitude of 5 and the direction is east (90 degrees). Select Adding 
Vectors from the Vectors form. For each of the three vectors, you will enter the magnitude and 
the angle. Enter 8 in the first data box for the magnitude of vector 1. Enter 0 for its 
direction. North is considered 0 degrees. Enter 3.5 for the magnitude of vector 2 and 35 for 
its direction. Enter 5 for the magnitude of vector 3 and 90 for its direction (east is 
considered 90 degrees). Press enter. The resultant of these three vectors has a magnitude of 
12.930 and its direction is 32.816 degrees east of north.

Newton's Laws

This section contains calculations that deal with all of Newton's laws. The options presented 
address the major values of importance for motion calculations.

When you select Newton's Laws in the Main screen, the form that opens up contains 7 options. 
This is because there are basically 7 calculations involved with Newton's Laws. You will see 
that the first two options are really the same. Both force and weight require the same input 
and give the same answer. The reason is simply to avoid confusion for some students. The weight 
of an object is a force downward caused by gravity. However, some students feel more 
comfortable with weight instead of force. Force, though, is the better concept to use.

Example 1.  A person has a mass of 73 kilograms. What is that person's weight and force? Since 
these two options are essentially the same, they can be handled at the same time. Select either 
the Weight option or the Force option. For the mass, enter 73, and for acceleration, enter 9.81. 
This value is not entered for you  so you can determine your weight or force on some other 
planet or on the moon. Press enter and your weight / force is 716.130 Newtons. 

Example 2.  An astronaut has a mass of 70 kilograms. This same astronaut exerts a downward 
force of 114.45 Newtons. What is the acceleration of the astronaut? Clear the entries and 
select the Acceleration option. For the force, enter 114.45 and for the mass enter 70. When 
you press enter the acceleration is 1.635 meters per second per second. With this acceleration, 
the astronaut is probably on the moon since its acceleration is about 1/6 of the Earth's 
acceleration.

Example 3.  A car is being pushed to get it started. A force of 8250 Newtons is applied to the 
car and it accelerates at a rate of 1.5 meters per second per second. What is the mass of the 
car? In the first data box enter 8250 for the force being applied to the car. Enter 1.5 in the 
second data box as the acceleration. Press enter. The mass of the car is 5500 Kilograms.

The External Force option refers to different forces acting upon an object. This is not like 
the vector problems addressed earlier. Though similar in that you resolve the forces into X 
and Y components, the forces are not connected together. They are acting separately on the 
same object. Imagine being pulled on by three people who are pulling in different directions.

Example 4.  Three men are trying to pull a tree stump out of the ground. Each has a rope 
attached to the tree stump. The first man is pulling with a force of 500 Newtons in a direction 
30 degrees west of north. The second man is pulling due north with a force of 800 Newtons. The 
third man is pulling with a force of 700 Newtons in a direction of 45 degrees east of north. 
When, and if, the stump comes out, in what direction will it go and with what force? Clear all 
entries and select Net External Force. You will notice that you can enter up to 4 forces.  The 
order of the forces does not matter so we will just enter them in the order given in the 
problem. Enter 500 and -30 (or 330) for the magnitude and angle for the first man. Enter 500 
and 0 (north is 0) for the second man. Enter 700 and 45 for the magnitude and angle of the 
third man. Leave the magnitude and angle for a fourth force empty. Press enter.  The magnitude 
of the resultant is 1,745.266 at an angle of 8.069 degrees. If you had a fourth force and 
magnitude, just enter it like the others.

Caution!! How you enter the angles is important. If you assign 0 degrees to north, any angle 
to the left of north will be negative or 360 minus your angle. For example, if your angle is 
45 degrees west of north, you would enter it as -45 or 315 degrees. Either way would work. 
Likewise, when you are given a negative angle as your answer, you must consider the negative 
sign when interpreting it. If your resultant angle was -10 degrees, it is really 10 degrees 
west of north or 350 degrees.

Example 5.  A crate is pulled to the right with a force of 85 Newtons, to the left with a 
force of 115 Newtons, to the front with a force of 565 Newtons, and to the back with a force 
of 236 Newtons. Find the magnitude and direction of the net external force on the crate. Enter 
the magnitudes and angles the same as in the last example. Enter 85 and 90 for the magnitude 
and angle for the first force, 115 and 270 (or -90) for the second force, 565 and 0 for the 
third force and 236 and 180 (or -180) for the last force. Press enter. The magnitude is 330.365 
Newtons and the angle is -5.210 degrees. The angle can be interpreted as 5.210 degrees west of 
north or 354.79 degrees. 

The next option is similar to one of the vector problems. It is included again for your 
convenience. If you are given the magnitude of a force and its angle, you can break it down or 
resolve it into its X and Y components. 

Example 6.  A man is pulling on a sled with a force of 70 Newtons at an angle of 30 degrees. 
What are the X and Y components of this force?  Select the XY-Components option. Enter 70 for 
the magnitude of the force and 30 for the angle. The X component is 60.622 Newtons and the Y 
component is 35.000 Newtons.

The final part of the Newton's Laws section is the Normal force. If an object is resting on a 
slanted surface, there are several forces acting upon it. One is the force that is pulling the 
object down the slanted surface. Another is the force of friction that is keeping the object at 
rest and countering the force pulling it down the slanted surface. Another force is the force 
of gravity pulling the object downward towards the center of the Earth. Finally, there is what 
is called the Normal force. This is the force that is perpendicular to the surface. This force 
is not necessarily opposite to the force of gravity. For example, consider a television sitting 
on a tv stand. The force of gravity is pulling the television downward. The normal force is the 
force perpendicular to the table top and is pushing upward against the television. 

Example 7.  A box with a mass of 5.5 kilograms is resting on a ramp which has an angle of 25 
degrees. What is the Normal force acting upon the box? Select the Normal Force option. Enter 
5.5 as the mass of the object and 25 as the angle of the ramp. Press enter. The Normal force is 
48.850 Newtons.

Friction

Friction is a force caused by two things rubbing against each other. If both objects were 
perfectly smooth, there would be no friction and thus no force. This is fine is some cases, but 
in many cases, friction is necessary for movement to occur. Friction allows you to walk and run  
but also provides resistance to motion. It has both good and bad points. Friction between the 
road the tires on your car is good. Friction within the engine of your car can be very bad.

Many friction problems are related to Newton's second law of motion, F=MA. Another important 
value is the Coefficient of Friction. This is the ratio of the force of friction to the normal 
force acting between two objects.

Example 1.  A 75 kg. Crate is resting on a horizontal surface. It requires a 200 N horizontal 
force to move it. What is the coefficient of friction?  Select the Friction option from the 
Main screen. Once the Friction form is shown, select the Coefficient of friction option. Enter 
200 as the moving force in the first data box and 75 in the second data box for the mass. Press 
enter. The Coefficient of friction is .272. There are two kinds of Coefficients, static and 
kinetic. The present problem addressed the static type which involves non-moving objects. The 
kinetic type addresses objects once they are moving. They are solved the same way.

Example 2.  Michelle is rearranging the furniture in her apartment. She pushes an 82 kg hutch 
across the floor. The coefficient of friction is 0.39. What force must she apply to the hutch 
in order to move it? Clear the entries and select the Force option. Enter 0.39 in the first 
data box as the coefficient of friction. Enter 82 as the mass of the hutch and press enter. 
The force needed is 313.724 Newtons.

Example 3.  Mike is trying to move his St. Bernard so he can clean his room. The coefficient 
of friction is 0.50 and he is pushing with a force of 350 Newtons. What is the mass of this 
massive dog? After clearing the entries, enter 350 as the force and 0.50 as the coefficient of 
friction. Press enter. The mass is 71.356 kg.

Example 4. What is the acceleration of a 1400 kg car when a force of 750 Newtons is applied to 
it and the coefficient of friction is 1.0? Enter 750 as the force, 1 as the coefficient of 
friction and 1400 as the mass.  The acceleration is 0.536 meters per second per second.

Energy

Energy is a force or potential to do work. It is the stuff that makes things go. It is a hard 
term to clearly define so scientists have classified it instead. There is kinetic energy - the 
energy of motion. There is stored energy or potential energy.  However it is described, it is 
essentially a force, in motion or stored. When you click on the Energy button in the Main 
screen, you are presented with 4 additional choices: kinetic energy, potential energy, work, 
and power.

Kinetic Energy depends upon mass and velocity. The formula is 1/2 mv^2.  m is the mass and v 
is the velocity. As you can see from the v^2, the velocity is a very important factor in  
calculating kinetic energy.

Example 1.  What is the kinetic energy of a 7.00 kg object moving at a rate of 3 meters per 
second? Click on the Kinetic Energy Button. Enter 7 as the mass of the object and 3 as the 
velocity. Click OK. The kinetic energy is 31.500. Kinetic Energy is expressed in Joules.

Example 2.  What is the mass of an object that has a kinetic energy value of  1.12E9 Joules 
and traveling at a velocity of 600 Km/H? The velocity must be in meters per second. 600 Km/H 
is equal to 167 meters per second. Enter 1.12E9 for the kinetic energy and 167 for the velocity. 
Press Ok. The mass is 80,318.410 Kg.

Example 3.  What is the velocity of a bullet with a mass of 3.0 grams and a kinetic energy of 
9.6 Joules? Enter 9.6 for the kinetic energy. The mass must be in Kg. 3 grams is equal to 0.003 
Kg. Enter 0.003 for the mass and press Ok. The velocity is 80 meters per second.

Potential energy is stored energy. The stored energy is usually due to position. An object 
held high in the air (high potential energy) and dropped has more kinetic energy while falling 
than an object held low and then dropped. This type of potential energy is referred to as 
gravitational potential energy.  Another type of potential energy, called elastic potential, 
is energy stored in a compressed spring or stretched spring.

Example 1.  An apple with a mass of 0.20 kg falls 7 meters and hits a guy named Isaac Newton 
on the head. What was the potential energy before it fell? This is a case of gravitational 
potential energy and depends upon the distance or height. The potential energy (PE) is 
mass x gravitational force x height. From the Main screen, click the 'Energy' button and then 
click the 'Potential Energy' button. On the screen that appears, enter 0.20 as the mass and 7 
as the height. Press Ok. The potential energy is 13.734 Joules. 

Example 2.  A spring with a force constant of 4.3 N/m has a relaxed length of 3.2 meters. When 
a mass is attached to the end of the spring, the length becomes 5.3 meters. What is the elastic 
potential energy stored in the spring? Select the Elastic Potential option. The first value to 
calculate is the difference between the relaxed length and the stretched length. This is 2.1 
meters (5.3 - 3.2). Enter 4.3 as the spring constant in the first data box. Enter 2.1 in the 
second data box as the distance. Press Ok. The elastic potential energy is 9.482 Joules.

The third Energy option is Work. Work is defined as the product of the magnitude of the force 
and the displacement. There can be no Work without a force and likewise there can be no Work 
without displacement. 

Example 1.  How much work is done on a wagon pulled 10 meters by a force of 75 Newtons at an 
angle of 30 degrees above the horizontal? Select the Work option from the Energy Choices form.  
Select the Work option. Enter 75 as the force, 10 as the displacement and 30 as the angle. 
Press Ok. The work done is 649.519 Joules.

Example 2.  Calculate the displacement of an object if the work is 130 Joules, the force is 
50 Newtons and the angle is 45 degrees. Clear all entries. Enter 130 as the work, 50 as the 
force and 30 as the angle. Press Enter. The displacement is 3.677 meters.

Example 3.  What is the force exerted if the work is 260 Newtons, the displacement is 6 meters 
and the angle is 45 degrees? Clear the entries. Enter 260 for the force, 6 for the displacement 
and 45 for the angle. The force is 61.283 Newtons.

Power is the last energy option. It is described as the rate at which work is done. It is 
simply Work / Time. The key element is time since it will give us the rate.

Example 1.  A 500 Kg. crate needs to be raised 20 meters in about 10 seconds. How much power 
is needed to raise the crate? Select Power from the Energy Choices form. The first entry needed 
is force. Force is equal to mass times acceleration. This means we multiply 500 times 9.81 
giving us a force of 4905 Newtons. Enter 4905 as the force. Enter 20 as the distance and 10 as 
the time. Press Enter. The Power needed is 9810 Watts or 9.81 Kilowatts. 

Equilibrium

In order for an object to remain stationary, the sum of all the forces acting upon that object 
must be equal. For a traffic light to remain stationary, the two opposing vertical forces must 
be equal, the left and right forces must be equal and the forward and backward forces must be 
equal. When all the forces are equal and an object is stationary, you have what is called 
equilibrium.

In the Equilibrium problems in Mechanics, you must first imagine the setting of the problems. 
Think of an object being suspended by a single wire or cable. Now imagine that wire or cable 
being held in turn by two other wires or cables. This is like the capital letter Y with the 
object attached at the very bottom. It is the upper two wires or cables that can vary in 
problems of this type. The two wires may be parallel up and down. They could also be like the 
upper case Y in that the two wires form identical angles with respect to the horizontal. They 
could also form unequal angles with respect to the horizontal. Each of these three variations 
is presented in this section of Mechanics.

In the first problem type, there are two parallel wires or cables attached to the bottom cable 
which is attached to the object in question. Each of these two cables provide an equal force to 
help suspend the cable with the object.

Example 1. An acrobat with a mass of 72 Kg. is hanging from a trapeze with both arms. What 
force is being exerted in each arm to hold the gymnast up? From the Main screen, click the 
Equilibrium button. Having done that, select the 2 equal parallel forces option. The only value 
you have to enter is the weight of the gymnast. A mass of 72 Kg. will have a weight (or 
downward force ) of 706.32 Newtons (72 x 9.81). Enter this value in the data box. Press Enter. 
The force exerted by each arm is 353.160 Newtons. Since both arms are considered parallel and 
vertical, angles are not important in this problem.

In the second type of problem, the two forces are not parallel but form equal angles like the 
letter Y. The two forces in this case are also equal since the angles are equal.

Example 2.  A traffic light with a mass of 50 Kg. is suspended by two cables each making an 
angle of 45 degrees with the horizontal.  What is the force exerted up each of these cables? 
An object with a mass of 50 Kg. will exert a downward force of 490.5 Newtons. Enter 490.5 into 
the first data box. Enter 45 into the second data box as the angle formed by the cables. Press 
Enter. The force exerted by each cable is 346.836 Newtons.

In the third type of problem, the two supporting cables do not form equal angles. For whatever 
reason, the cables make different angles. Therefore, the force exerted by each cable is 
different due to the different angles they form.

Example 3. A 90 Newton boy hangs from a rope stretched between to trees. The rope forms a 10 
degree angle on the left and a 5 degree angle on the right. What are the forces exerted on each 
part of the rope? Select the third option (2 unequal concurrent forces). Enter 90 in the first 
box as his weight. Enter 10 as the angle of force 1 (on the left) and 5 for the angle on the 
right (on the right). Press Ok. The force on the rope on the left of the boy is 346.410 Newtons. 
The force on the rope on the right of the boy is 342.451 Newtons.

Momentum

Momentum is described as mass x velocity.  Mass is given in kilograms and velocity is usually 
in meters / second.  To solve Momentum problems, click on the Momentum button on the main 
screen. You are then shown three additional buttons. Click on the button labeled Momentum.

Example 1.  Calculate the  momentum of a car with a mass of 2250 kg and a velocity of 25 m/s 
to the east. In momentum problems, a direction must be supplied since velocity is a part of 
the solution. In the Momentum screen, select the Momentum option. Enter 2250 for the mass and 
25 for the velocity. Press Ok. The momentum is 5.63E4 kg m/s.

Example 2. What mass must another vehicle have traveling at 15 m/s to have the same momentum 
as the vehicle in example 1? Clear all entries. Select the mass option. Enter 5.63E4 for the 
momentum and 15 for the velocity. Press Ok. The mass has to be 3,753.333 kg.

Example 3. What must the velocity be for a 2.5 kg baseball bat to provide 100 kg m/s of 
momentum? Clear the entries. Select the velocity option. Enter 100 for the momentum and 2.5 
for the mass. Press Ok. The velocity must be 40 m/s.

In the last type of problems in this section, force is an important quantity to be able to 
calculate. Sometimes you are asked how much force is needed to stop a moving object or to 
start it moving. You may also see Newton's 2nd law of motion as well as the concept of impulse 
involved. Newton's 2nd law is Force = Mass x Acceleration. Impulse is force x time. An example 
of impulse might be the padded dash of your car. It gives a little and spreads out the force 
over a slightly longer period of time thus reducing the severity of injury. 

Example 4.  A 2000 kg car is moving with a velocity of 20 m/s. (Remember, you must include 
direction with velocity.) It hits a parked car and stops completely in 0.4 seconds. What was 
the force exerted on the car during the collision? Select the Net Force option. Enter 2000 as 
the mass of the car, 20 as the initial velocity, 0 as the final velocity and 0.4 as the time 
change. Press Ok. The Net Force is -100,000 kg m/s in the same direction as the car or 100,000 
kg m/s in the opposite direction. A picture of the situation is sometimes helpful. Imagine the 
X axis with west to the left of the origin which would give it a negative value. Thus the force 
of the car is negative and the opposing force to the east is positive.

Elastic Collisions

In an elastic collision, two objects collide and rebound away from each other. They retain 
their original shape and the total kinetic energy stays the same. Think of playing pool. When 
the cue ball hits one of the other balls, there is no change of shape. They simply rebound away 
from each other in various directions. The direction depends upon the angle in which the cue 
ball hits the other ball.

The formula for elastic collisions is fairly easy. The initial kinetic energy of the first 
object plus the initial kinetic energy of the second object equals the final kinetic energy of 
the first object plus the final kinetic energy of the second object. The formula looks as 
follows:

m1v1i + m2v2i = m1v1f + m2v2f

The total kinetic energy on the left side of the = sign is exactly the same as the total 
kinetic energy on the right side.

Example 1. A marble with a mass of .015 kg moves to the right with a velocity of .225 m/s. It 
collides with another marble whose mass is .030 kg and is moving to the left with a velocity 
of .180 m/s. After the collision, the first marble moves to the left with a velocity of 
.315 m/s. What is the final velocity of the second marble?

Select the Elastic Collisions button. Then select the Final Velocity of Object 2 option. One 
important consideration to make is that of signs. Objects moving to the left have a negative 
velocity while objects moving to the right have a positive velocity. This must be considered 
when entering velocities. Enter .015 for the mass of object 1 and .030 for the mass of object 
2.  Enter .225 for the initial velocity of object 1. For the initial velocity of  object 2, 
enter -.180 since object 2 was going to the left. Enter -.315 for the final velocity of object 
1 since it was also going to the left. Press Ok. The final velocity of object 2 is .09 m/s. 
Since this is a positive number, it is assumed that object 2 is moving to the right.

Example 2.  What was the total kinetic energy involved in the collision in example 1? Clear 
the entries. Select Total Kinetic Energy of Both Objects. Enter .015 and .030 for the masses 
of the two objects. Enter .225 for the velocity of the first object and -.315 for the velocity 
of the second object. Press Ok. The total kinetic energy is .001868 Joules.

Inelastic Collisions

In inelastic collisions, two objects collide and stick together. They continue moving as one 
object.  One difference is that the total kinetic energy is not conserved. The  amount of 
kinetic energy may be less due to some of the energy being converted to hear or some stored 
potential energy.

Example 1.  A car is at a complete stop at a red light. Another car with a mass of 975 kg 
moving at 22 m/s runs into the back of the stopped car. The two cars become entangled and move 
together at 7.59 m/s.  What is the mass of the stopped car? From the Momentum / Collisions 
form select Inelastic Collisions. Once the Inelastic Collisions screen appears select the Mass 
of Object 1 option. Object 1 was the stopped car. So its velocity is 0. Enter 0 into the first 
data box. The mass of object 2 is 975 and its velocity is 22. Enter 975 and 22 into the 2nd 
and 3rd data boxes. Finally, the combined mass of the two cars moves at 7.59 m/s. Enter 7.59 
into the final data box and press Ok. The mass of object 1 is 1,851.087 kg.

Example 2.  A car with a mass of 1500 kg collides with a 4500 kg truck that is at rest at a 
stoplight. The car and truck stick together and move at 3.8 m/s. What was the velocity of the 
car? Clear all entries. The  mass of the first object is 1500. Enter 1500 into the first data 
box. Enter 4500 into the second data box as the mass of the truck. Enter 3.8 as the final 
velocity of the combined objects. The truck was at rest so its velocity is 0. Enter 0 into the 
final data box and press Ok. The velocity of the car was 15.2 m/s.

Example 3.  A pool ball with a mass of .5 kg and a velocity of 1.5 m/s hits another identical 
pool ball with a mass of .5 kg going in the opposite direction with a velocity of 2.0 m/s. 
What is the combined kinetic energy of the two pool balls? Clear all entries. Enter .5 for the 
mass of object 1 and object 2. Enter 1.5 for the velocity of object 1 and 2.0 for the velocity 
of object 2. Press Ok. The total kinetic energy is 1.563 Joules.

Example 4.  A 1.5E4 kg railroad car moving at 7.00 m/s collides with and sticks to another 
railroad car of the same mass that is moving in the same direction at 1.50 m/s? What is the 
velocity of the joined cars? Enter 1.5E4 for the masses of the two objects. Enter 7 for the 
velocity of object 1 and 1.5 for the velocity of object 2 and press Ok. The final velocity of 
the two joined railroad cars is 4.250 m/s.

Gravitation

Gravitation is the mutual force of attraction between particles of matter. In other words, it 
is the force of attraction between any two objects. This force of attraction is dependant on 
the distance between the objects and their respective masses. The larger the mass the larger 
the force of attraction. Also the closer an object the larger the force of attraction. It is 
possible for a small object that is real close to exert greater gravitational attraction than 
a large object that is far away.  The force of attraction between you and the person sitting 
next to you may be greater than the force of attraction between you and the Moon. The Moon is 
a lot larger, but the person sitting next to you is a lot closer. 

To solve gravitation problems, click the Gravitation button on the main screen. You are then 
presented with three options: Gravitational Force, Mass and Distance.

Newton's Universal Law of Gravitation is:  F = G * (M1 + M2)/ R^2.

G is the constant of universal gravitation and has a value of 6.673E-11 Nm^2/kg^2. The two M 
values are the masses of the objects and R is the distance between the two centers of the 
masses.

Example 1. What is the force of gravity between a 67.5 kg person standing on the surface of 
Earth? Select the Gravitational Force option. Enter 67.5 for mass 1. Enter 5.98E24 for the 
mass of the Earth. The distance between the center of the Earth and the person is 6.37E6 
meters so enter 6.37E6 for the distance. Press Ok. The Gravitational Force is 664 Newtons.

Example 2.  The gravitational force between two football players just before the play starts 
is 2.77E-3 Newtons. The distance between then is 0.025 meters. If one player has a mass of 
144 kg, what is the mass of the other player? Select the Mass option. Enter 144 for the mass 
of player 1.  For the distance, enter 0.025. In the final box enter 2.77E-3 for the 
gravitational force. Press Ok. The mass of player 2 is 180.167 kg.

Example 3.  What is the distance between the Earth and the Moon? The mass of the Earth is 
5.98E24 kg. The mass of the Moon is 7.36E22 kg. The force of attraction between the Earth and 
the Moon is 1.99E20 Newtons. Press Ok. The distance is 3.84E8 meters.

Rotational Motion

Rotational Motion deals with the motion of a body that spins around an axis. The Earth exhibits 
rotational motion as does a tire on a car.  Some of the factors considered in rotational motion 
are the radius of the body, the angle of change and the speed of an object at the outside of 
the body. One must also be familiar with the concept of radians in dealing with this type of 
problem.

Angular Displacement

Angular Displacement refers to the angle through which a point is rotated in a particular 
direction and around a particular axis. A person riding on a Ferris wheel exhibits angular 
displacement when the wheel rotates. 

Example 1.  A child sits on the edge of a merry-go-round at a distance of 1.5 meters from the 
center. If the child moves through an arc length of 2.5 meters, what was the angular 
displacement? To work with rotational motion problems, click the Rotational Motion button on 
the Main screen. On the Rotational Motion form that appears, click the Angular Displacement 
button. On the Angular Displacement form, select the Displacement option. Enter 2.5 for the 
change in arc length. Enter 1.5 for the distance from the axis and press Ok. The angular 
displacement is 1.667 radians. 

Example 2.  Calculate the arc length that a passenger cart on a Ferris Wheel moves if the 
angular displacement is .343. The carts are 35 meters from the axis of rotation. Select the 
Arc Length option. Enter .343 for the angular displacement. Enter 35 for the distance from the 
axis. Press Ok. The arc length is 12.005 meters.

Example 3.  What is the radius of a tire if the angular displacement is 3 and the change in 
arc length is .9.  Enter .9 into the first data box as the change in arc length and 3 in the 
second data box for the angular displacement. Press Ok. The radius is .300.

Angular Speed

Angular speed refers to the rate at which a body rotates around an axis. This is usually given 
in radians per second. The basic formula for angular speed is angular displacement / time 
interval.

Example 1.  A truck tire rotates counterclockwise 7 times in 1.5 seconds. What is the angular 
speed of the tire? From the Rotational Motion form click on the Angular Speed button. In the 
Angular Speed form, select the Sped option. You are requested to enter the angular displacement in radians. Since there are 2 pi radians in a circle, you would multiply the 7 rotations times 2 pi. This is 43.98. Enter this number into the first data box. Now enter 1.5 as the time into the second data box and press enter. The angular speed is 29.32 radians per second.

Example 2. A truck tire moves at an angular speed of 31.41 radians per second. What will the 
distance of a mark on the tire if the tire rotates for 5 seconds? Select the Displacement 
option on the Angular Speed form. Enter 31.41 for the angular speed and 5 for the time 
interval. Press Ok. The displacement (actually distance in this case) is 157.05 pi radians.

Example 3. What time is required for a gear rotating at 2 pi radians per second to cover 1.5 
pi radians? Select the Time Interval option on the Angular Speed form. For the speed enter 
6.283 (2 x Pi). For the displacement, enter 4.712 ( 1.5 x Pi). Press Ok. The time interval is 
0.75 second.

Angular Acceleration

Angular acceleration is the time rate of change of angular speed expressed in radians per 
second per second.  It is the change in angular speed divided by the time interval. 

Example 1.  A tire rotates at an angular speed of 21.5 radians per second. The tire is 
accelerated for 3.5 seconds and achieves an angular speed of 28.0 radians per second. What was 
the tire's average angular acceleration? Click on Angular Acceleration from the Rotational 
Motion form. Select the Acceleration option. Enter 21.5 into the first data box, 28.0 into the 
second data box and 3.5 for the time in the last data box. Press Ok. The average angular 
acceleration is 1.857 radians per second per second. 

Example 2. A truck tire rotates at an initial speed of 10 radians per second. It accelerates 
at 5 radians per second per second until its speed is 20 radians per second. What is the time 
required for the change in  speed to occur? Select the Time Change option. Enter 10 for the 
initial speed, 20 for the final speed and 5 for the acceleration. Press Ok. The time required 
is 2 seconds.

Example 3. A turbine starts at rest and accelerates for 10 seconds at a rate of 2.5 radians 
per second per second. What was the turbine's final speed? Select the Final Speed option. Enter 
0 for the initial speed since the turbine started at rest. Enter 10 for the time and 2.5 for 
the acceleration. Press Ok. The final speed is 25 radians per second.

Torque

Torque refers to the ability of a force to rotate an object around an axis. A wrench applying 
an effort to loosen a nut would be an example. Also a door on a hinge would illustrate torque. 
Torque is force x the lever arm x the sine of the angle. The angle may or may not be important 
to the problem. The length of the lever arm is, however, very important. Consider trying to 
loosen a nut on a bolt with a large wrench as compared to a small wrench. The larger wrench 
(or lever arm) serves to magnify the force.

Example 1.  What is the torque produced by a 7.0 Newton force at an angle of 60 degrees on a 
door .33 meter from the hinge? Click the Torque button on the Rotational Motion form. Select 
the Torque option on the Torque form. Enter 7.0 for the torque in the first data box. Enter 
.33 for the distance in the second data box and 60 for the angle in the third data box. Press 
Ok. The torque is 2.001 Newton-Meters.

Example 2. A worker applies a force of 12 Newtons to a wrench at an angle of 90 degrees. The 
torque was 4.8 Newton* Meters. What was the distance over which the force was applied? Click 
the Torque button on the Rotational Motion form. Select the Distance option. Enter 4.8 in to 
the first data box for the torque. Enter 12 for the applied force into the second data box. 
Finally, enter 90 for the angle. Press Ok. The distance is 0.40 meter.

Example 3. What force must be applied to a lever to create a torque of 9.6 Newton*Meters? The 
distance is .5 meter and the angle is 90 degrees. Select the force option. Enter 9.6 for the 
torque, .5 for the distance, and 90 for the angle. Press Ok. The force is 19.2 Newtons.

Example 4. If a force of 6 Newtons is applied over a distance of .75 meters, it creates a 
torque of 2.4 Newton*Meters. At what angle was this force applied? Select the angle option. 
Enter 2.4 for the torque, 6 for the force and .75 for the distance. Press Ok. The angle is 
32.231 degrees/.

Angular Momentum

Angular Momentum is the product of a rotating object's moment of inertia and the angular speed 
about the same axis. Angular momentum is calculated by multiplying the mass x the velocity x 
the radius. It can also be considered how difficult it might be to stop an object from 
rotating.

Example 1.  Each tire on a four-wheel vehicle has a mass of 30 kg. The radius of each tire is 
30 cm. If the wheels are turning at 5.0 revolutions per second, what is the angular momentum 
of each tire?  Click on the Angular Momentum button on the Rotational Motion form. Select the 
Angular Momentum option. Enter 30 into the first data box for the mass. The radius is required 
in meters. 30 cm is equal to .30 meters. So enter .30 into the second data box for the radius. 
The final entry is the speed. This needs to be in radians per second. Since 1 revolution is 2 
pi radians, 5 revolutions is 10 pi radians or 31.41 radians per second. Enter 31.41 into the 
third data box as the speed. Press Ok. The angular momentum of each tire is 283-kg *m^2/s.

Example 2. In example 1, what is the moment of inertia? Select the Moment of Inertia option. 
Copy the values from example 1 into the boxes for this example. 283 is the momentum, 31.41 is 
the speed and .30 is the radius of the tire. Press Ok. The moment of inertia is 2.70.

Example 3. In another problem, Jupiter's mass was given as 1.90E27 kg. Jupiter's momentum was 
calculated to be 2.82E38 kg*m^2/s. If, for some reason, Jupiter's distance from the sun 
increased to 80,000,000 meters, and its momentum was conserved, what would be the new angular 
speed of Jupiter? Select the Angular Speed option. Enter 2.82E38 in the first box as the 
momentum, 1.90E27 as the mass, and 80000000 as the radius.  Press Ok. The new speed would be 
1,855.263 m/s. The old radius was 71398000 and the speed was 2079 m/s. As you can see, as the 
radius increases the speed decreases in order to conserve the momentum.

Kinetic Energy

Kinetic energy is usually thought of as an object moving in a straight line. The formula in 
that case is 1/2 mv^2 where m is the mass and v is the velocity. In rotational kinetic energy, 
things are a little bit different. The formula used here is 1/2 IW^2 where I is the moment of 
inertia and W is the speed. The first calculation you must perform is to determine the value 
of I. This is mass times the radius of the object. This quantity is then substituted into the 
rotational kinetic energy formula.

Example 1.  Calculate the rotational kinetic energy of a 25 kg wheel rotating at a speed of 6 
revolutions per second when the radius of the wheel is .22 meters. Click on the Kinetic Energy 
button on the Rotational Motion form. Select the Kinetic Energy option. Enter 25 for the mass 
in data box one. Enter .22 for the radius in data box two and 37.669 for the speed. Where did 
this come from? The speed must be given in radians. Since the wheel rotates at 6 revolutions 
per second, we have 6 times 2 pi (12 pi = 37.699). Press Ok. The kinetic energy is 859.835 
Joules.

Example 2.  What is the moment of inertia of a large roulette wheel with a radius of 0.60 
meters and a mass of 3.0 kg? Select the Moment option. Enter 3.0 in box one for the mass and 
0.6 in box 2 for the radius of the wheel. Press O. The moment of inertia is 0.54 kg*m^2.

Example 3. What is the angular speed of the wheel in example 2? Select the Angular Speed 
option. Enter 0.54 in box one for the kinetic energy as calculated in the last example. Enter 
.6 for the radius and 3 for the mass. Press Ok. The angular speed is 1.0.

Circular Motion

Two forms of circular motion are addressed:  centripetal acceleration and the force that 
causes circular motion. Centripetal acceleration is considered because as an object moves in a 
circular path, there is a continuous change in direction. Acceleration is a quantity that 
includes a numerical quantity and direction. So when the direction is changed, you have 
acceleration. The force that causes that change in direction is called centripetal and tends 
to pull the object toward the center of a circle. It is countered  by the velocity of the 
object and its tendency to go in a straight path. This is the force that causes circular 
motion.

Example 1. A test car moves at a constant speed of 19.7 m/s around a circular track. If the 
distance from the car to the center of the circle is 48.2 meters, what is the centripetal 
acceleration of the car? Click the Circular Motion button on the Main screen. Select the 
Centripetal Acceleration option on the Circular Motion form. Enter 19.7 in data box one for 
the speed and 48.2 in the second data box for the radius. Press Ok. The centripetal 
acceleration is 8.05 meters/per second/per second.

Example 2. A pilot with a mass of 75 kg is flying a small plane at 50 m/s in a circular path. 
This path has a radius of 500 meters. What is the force that produces the circular motion of 
the pilot (not the airplane)? Clear all entries. Enter 75 in the first data box for the mass 
of the pilot. Next enter 500 for the radius of the circular path. Finally enter 50 for the 
speed of the pilot. Press Ok. The force that causes circular motion is 375 Newtons.

Harmonic Motion

Simple harmonic motion is a back and forth motion about an equilibrium point. The force needed 
to restore equilibrium is proportional to the displacement from the equilibrium point. If you 
have a mass hanging from a string and it is at rest, you have equilibrium. If you pull the mass 
back a little ways and let it go, it will swing back and forth and eventually reach a state of 
rest or equilibrium. The force required to bring the mass to rest depends upon how far back you 
pulled the mass from the rest or equilibrium point. The farther back you pulled it, the more 
force is necessary.

In a mass-spring system, you have a similar setup. Instead of a mass attached to a string, you 
have a mass attached to a spring. The only basic difference is that the spring has a constant 
that must be considered. Not all springs are made of the same material and behave the same way. 
Hooke's law calculates the force needed to bring the mass attached to a spring back to a state 
of rest or equilibrium. 

Example 1.  If a mass of 0.55 kg is attached to a vertical spring, the spring stretches 2.0 cm 
from its equilibrium position. What is the spring constant? On the Main screen click the 
Harmonic Motion button. You will then be presented with the Harmonic Motion screen which 
presents you with two further choices. Click the Mass-Spring button. Select the Spring Constant 
option. You are asked to enter the force in Newtons. Multiply the mass of 0.55 kg times 
9.81 m/s/s. The force is 5.3955 Newtons. Enter 5.3955 in the first data box. Since the mass 
has been moved down from its equilibrium point by 2 cm  or by .02 m. the displacement is 
-.02 mm. It has to go back .02 m to reach its rest or equilibrium point. Enter -0.02 in the 
second data entry box and press Ok. The spring constant is 269.775 N/m.

Example 2. If a force of 45 Newtons is applied to a spring with a spring constant of 320, what 
would be the displacement? Select the Displacement option. Enter 45 in the first box as the 
force and 320 in the second box as the spring constant. Press Ok. The displacement is 0.141 
meter.

Example 3. The pull handle on a pinball machine has a spring constant of 200 N/m. If the 
handle is pulled backl5.0 centimeters, what is the force applied to the spring?  Select the 
Force option. Enter 200 for the spring constant in box 1 and .05 as the displacement 
(in meters!) in the second box. Press Ok. The force is 10 Newtons.

Simple Pendulum

A pendulum is simply a mass at the end of a string or rope. Like the mass-spring system, the 
mass is brought away from its equilibrium position and released. The mass then moves back and 
forth. A force gradually brings the mass back to its state of rest or equilibrium. The pendulum 
illustrates several properties associated with simple harmonic motion: frequency, period, 
length and acceleration. 

Example 1. Calculate the period of a pendulum if the length of the pendulum is 36 meters. From 
the Main screen click the Harmonic Motion button. On the Harmonic Motion form click the Simple 
Pendulum button. On the Simple Pendulum form select the Period option. Enter 36 into data box 
one for the length of the pendulum. Enter 9.81 in the second data box for the acceleration. 
Press Ok. The period is 12.036 seconds. This means that it takes 12.036 seconds for the mass 
to make a round trip thus returning to its starting position.

Example 2. If the period of the pendulum in example 1 is 12.036 seconds, what is the frequency 
of the pendulum? Clear all entries. Select the Frequency option. The only piece of information 
you need to enter is the period. Enter 12.036 in the only data box and press Ok. The frequency 
is 0.083. This means that in one second, the pendulum travels 0.083 of the distance the 
pendulum travels in one complete back and forth path. 

Example 3.  A spider hangs from a thread that has a period of 0.6 seconds. How long is the 
spider's strand of thread? From the Simple Pendulum form, select the Length option. Enter .6 
in box 1 as the period of the thread which is acting like a pendulum. The acceleration is the 
downward force of gravity. Enter 9.81 in the second box. Press Ok. The length is 0.089 meter.

Example 4. An astronaut on the moon makes a simple pendulum with a piece of string and a small 
rock. The string has a length of 9 centimeters. The period of this simple pendulum is 1.48 
seconds. What is the force of gravitational acceleration? Select the Acceleration option. Enter 
1.48 in the first box as the period. Enter .09 for the length of the string in meters. Press 
Ok. The acceleration is 1.622 or about 1/6 that of Earth's gravitational acceleration.

Archimede's Principle

Archimede's principle determines the amount of a buoyant force. An object placed in water 
displaces an amount of water equal to the volume of the object that is in the water. A ship 
displaces an amount of water equal to the volume of the ship submerged.  The magnitude of the 
buoyant force is equal to the weight of the fluid displaced.

Example 1. An ice cube is placed in water. The ice cube has a volume of 1.2E-5 cubic meter. 
Ice also has a density of 917 kg/cubic meter. What is the buoyant force causing the ice cube 
to float? On the Main screen click the Miscellaneous button. Next click the Archimede's 
Principle button. On the Archimede's Principle form select the buoyant force option. Enter 917 
in data box one for the density of ice. Enter 1.2E-5 in the second data box for the volume. 
Press Ok. The buoyant force is 0.108 Newtons.

Example 2. A bar of soap floats in a tub of water. A buoyant force of 1.61 Newtons acts upon 
the bar causing it to float. If the volume of the bar is 1.62E-4 cubic centimeters, what is 
the density of the soap? Select the Density option. Enter 1.61 for the buoyant force and 
1.62E-4 for the volume. Press Ok. The density is 1,014.109 kg/cubic meters.

Example 3.  A piece of an iceberg breaks off a larger iceberg. The buoyant force supporting 
the piece of ice is 1200 Newtons. The density of the ice is 1025 kg/m^3. What is the volume of 
the smaller piece of the iceberg? Select the Volume option. Enter 1200 for the buoyant force 
and 1025 for the density of water. Press Ok. The volume is 1.19E-1  meter.

Pascal's Principle

Pascal's Principle deals with pressure. If you have ever seen a car raised with a hydraulic 
lift, you have seen Pascal's Principle in action. A small force exerted over a small area will 
translate into a large force exerted over a large area. In a hydraulic lift a small pressure 
is applied to a small piston. The pressure per square inch on the small piston is applied to 
the same pressure per square inch on a larger piston. This results in a larger force capable 
of lifting a heavy object. 

Example 1.  A car weighing 1.2E4 Newtons is sitting on a hydraulic press piston which has an 
area of 0.90 square meters. What is the pressure being exerted? From the Miscellaneous form 
click on Pascal's Principle. On the Pascal's Principle form select the Pressure option. In the 
first data box enter 1.2E4 for the force of the car. In the second data box, enter 0.90 for 
the area of the piston that the car is sitting on. Press Ok. The pressure is 1.33E4 Pascals. 

Example 2. A car with a mass of 1200 kg is placed on a hoist. If the area of input pressure is 
12 square centimeters and the area of output pressure is 700 square centimeters, what is the 
force exerted on the input area?  Select the force option. Enter the mass of the car (1200) in 
box one, 12 for the input area in box two and 700 for the output area in box three. Press Ok. 
The force is 202 Newtons.

Example 3.  2.4E4 Newtons are required to lift a small car on a hoist. If the area of the 
cylinder the car is sitting on is 1.8 square meters and the force applied to the initial 
cylinder is 5.4E3 Newtons, what is the area of the initial cylinder? Select the Area option. 
Enter 5.4E3 as the force on the initial cylinder, 2.4E4 as the force on the lifting cylinder, 
and 1.8 as the area of the lifting cylinder. Press Ok. The area of the initial cylinder is 
.405 square meters.

Data Tables

At the bottom of the Main screen are buttons referring data tables that may be of use in 
problem solving. These data tables provide a variety of information. 

The Constants table provides the most commonly needed values for physics constants.

The Periodic table illustrates the Periodic Table of Elements. If you click on any of the 
element boxes, a data form for that particular element appears showing the most important data 
for that element. 



The Coefficient of Friction table shows you the various values for static and kinetic friction. 
The most commonly used one are given.

The Conversion table gives you the various conversions from one unit to one or two other units.

The Refraction Indices is for a future version.

The Help button is a reminder that for help, you have this user manual to refer to for 
assistance in running the program. You also have the ? button on each problem section.

The Calculator button brings up the calculator provided by the Windows operating system. It 
contains a scientific calculator which will handle all the mathematical operations you need to 
perform.













