PHYS101 Study Guide

1a. Explain the difference between a theory and a law

• Define a scientific law, scientific theory, and scientific model.
• Give an example of a scientific law and a scientific theory.
• What are some similarities between scientific laws and theories? What is the main difference between them?

A scientific law briefly and succinctly describes an observed natural phenomenon or pattern. We often describe scientific laws as a single equation. For example, we describe one of Newton's Laws of Motion as $F = ma$. Because this is a brief, single equation, it is a law. Laws are supported by multiple, repeat experiments performed by different scientists over time.

A scientific theory also describes an observed natural phenomenon or pattern, but in a less succinct manner. We cannot describe theories as a single simple equation. Rather, they explain the phenomenon or pattern. Charles Darwin's Theory of Evolution in an example of a scientific theory. The Theory of Evolution describes natural patterns, but cannot be described by a single equation. Like laws, theories must be verified by multiple, repeat experiments performed by different scientists.

A scientific model is a representation of an object or phenomenon that is difficult or impossible to actually observe. Models provide a mental image to help us understand things we cannot see. An example of a model is the Bohr (planetary) model of the atom. This is a representation of an object (the atom) that is far too small for us to see. It allows us to develop a mental image so we can think about atomic structure.

Review the section Models, Theories, and Laws; The Role of Experimentation.

1b. Perform unit conversions using both metric and traditional U.S. units

• Define physical quantity. What makes something a physical quantity?
• Define SI fundamental units. How are they related to SI derived units?
• List the metric prefixes and their order of magnitude for commonly used units.
• Perform unit conversions between different metric units using metric prefixes and orders of magnitude.
• Perform unit conversions between metric units and U.S. units.

We define a physical quantity by how it is measured or by how it was calculated from measured values. It is either something that can be measured, or something that can be calculated from measured quantities. For example, the mass of an object in grams is a physical quantity because it is measured using a scale. The speed of a moving object in meters per second is also a physical quantity because it is based on two measured quantities (distance in meters, and time in seconds).

SI units are standardized units of measure for different measured quantities. The fundamental SI units are the following:

Review Table 1.1 Fundamental SI Units.

Derived SI units are based on the fundamental SI units. An example is speed, which is length per unit time.

The metric system is a standardized system of units used in most scientific applications. The SI units are based on the metric system. The metric system is based on series of prefixes that denote factors of ten. We call these factors of ten orders of magnitude. The prefixes tell us the relative magnitude of the measurement with respect to the base unit. Because the metric system is based on these powers of ten, it is a convenient system for describing measurements in science.

Review an extensive list of the metric prefixes in Table 1.2 Metric Prefixes for Powers of 10 and their Symbols.

Using the metric system to convert between units requires a knowledge of the prefixes and their order of magnitude. For example, consider length. The base unit for length is the meter (m). In Table 1.2, we see that one centimeter (cm) is 10^-2 m.

If we want to know how many cm are in 5.0 m, we can use dimensional analysis to convert between meters and centimeters. To do this, we use the prefix's order of magnitude as a unit conversion factor. Unit conversion factors are fractions showing two units that are equal to each other. So, for our conversion from 5.0 m to cm, we can use a conversion factor saying 1 cm = 10^-2 m (again see Table 1.2). To determine how to write this equivalence as a fraction, we need to determine what should be the numerator and what should be the denominator. That is, we could write 1 cm/ 10^-2 m or we could write 10^-2 m/ 1 cm.

We can determine the proper way to write the fraction based on the given information. When performing dimensional analysis, always begin with what you were given. Then, write the unit conversion factor as fraction with the unit you want to end up with in the numerator, and the unit you were given in the denominator. This will result in the answer being in the unit you want:

$\mathrm{what\ you\ were\ given\times\frac{unit\ you\ want}{unit\ you\ were\ given}=unit\ you\ want}$

For our example, we want to determine how many cm are in 5.0 m. The given is 5.0 m. The unit we want is cm, and the unit we were given was m. So, we would set up the conversion in the following way:

$\mathrm{5.0\ m\times\frac{1\ cm}{10^{-2}\ m}=500\ cm}$

The meter unit cancels out in the above calculation. Because the meter unit cancels out, we are left with cm as the unit of the answer.

The same use of dimensional analysis also applies for non-metric units used in the United States. For example, we know that one foot equals 12 inches. These length measurements are not part of the metric system. We can determine how many inches are in 5.5 feet using the same dimensional analysis technique as above:

$\mathrm{5.5\ feet\times\frac{12\ inches}{1\ foot}=66\ inches}$

1c. Write numbers expressed in decimals as scientific notation and write numbers expressed in scientific notation as decimals

• Convert numbers to scientific notation. What do the different parts of scientific notation mean?
• Convert numbers written in scientific notation to decimal form.
• Perform calculations with numbers written in scientific notation.

Often in science, we deal with measurements that are very large or very small. When writing these numbers or doing calculations with these physical quantities, you would have to write a large number of zeros either at the end of a large value or at the beginning of a very small value.

Scientific notation allows us to write these large or small numbers without writing all the "placeholder" zeros. We write the non-zero part of the value as a decimal, followed by an exponent showing the order of magnitude, or number of zeros before or after the number.

For example, consider the measurement: 125000 m.

To write this measurement in scientific notation, we first take the non-zero part of the number, and write it as a decimal. The decimal part of the number above would become: 1.25

Then, we need to show the order of magnitude of the number. We count the number of decimal places from where we placed the decimal to the end of the number. In this case, there are five places between the decimal we put in and the end of the number. We write this as an exponent: 10⁵.

To put the entire scientific notation together, we write: 1.25 × 10⁵ m.

We can also do an example where the measurement is very small. For example, consider the measurement: 0.0000085 s.

Here, we again begin by making the non-zero part of the number into a decimal. We would write: 8.5

Next, we need to show the order of magnitude of the number. For a small number (less than one), we count the number of places from where we wrote the decimal back to the original decimal place. Then, we write our exponent as a negative number to show that the number is less than one. For this example, the exponent is: 10^-6.

To put the entire scientific notation together, we write: 8.5 × 10^-6 m.

We can also convert values written in scientific notation to decimal notation. Consider the number: 5.0 × 10³ m.

We can write this as "normal" notation by adding the appropriate number of decimal places to the number, past the decimal written in scientific notation. Here, the order of magnitude (number of decimal places) is three, as we see from the exponent part of the number. Because the exponent is positive, we add the decimal places to the right of the number to make it a large number. The value in "normal" notation is: 5000 m.

We can also do this for small numbers written in scientific notation. Consider the example: 4.2 × 10^-4 m.

We can write this as "normal" notation by adding the appropriate number of decimal places to the left of the number to make it a small number. Here, we need to have four decimal places to the left of the decimal in the scientific notation. The value in "normal" notation is: 0.00042 m.

1d. Solve problems with the values of the most common metric prefixes

• Perform unit conversions between different metric units using metric prefixes and orders of magnitude.

If we want to know how many cm are in 5.0 m, we can use dimensional analysis to convert between meters and centimeters. To do this, we use the prefix's order of magnitude as a unit conversion factor. Unit conversion factors are fractions showing two units that are equal to each other. So, for our conversion from 5.0 m to cm, we can use a conversion factor saying 1 cm = 10^-2 m (see Table 1.2 Metric Prefixes for Powers of 10 and their Symbols). To determine how to write this equivalence as a fraction, we need to determine what should be the numerator and what should be the denominator. That is, we could write 1 cm/10^-2 m or we could write 10^-2 m/1 cm.

We determine the proper way to write the fraction based on the given information. When performing dimensional analysis, always begin with what you were given. Then, write the unit conversion factor as fraction with the unit you want to end up with in the numerator and the unit you were given in the denominator. This will result in the answer being in the unit you want:

$\mathrm{what\ you\ were\ given\times\frac{unit\ you\ want}{unit\ you\ were\ given}=unit\ you\ want}$

For our example, we want to determine how many cm are in 5.0 m. The given is 5.0 m. The unit we want is cm, and the unit we were given was m. So, we would set up the conversion in the following way:

$\mathrm{5.0\ m\times\frac{1\ cm}{10^{-2}\ m}=500\ cm}$

The meter unit cancels out in the above calculation. Because the meter unit cancels out, we are left with cm as the unit of the answer.

For another example, we can determine the number of seconds in 424 nanoseconds (ns). In Table 1.2 we see that 1 ns = 10^-9 s. Our given is in ns, so we need to write our unit conversion factor with ns in the denominator to cancel the unit out. We perform the calculation the following way:

$\mathrm{424\ ns\times\frac{10^{-9}\ s}{1\ ns}=4.24\times 10^{-7}\ s}$

Unit 1 Vocabulary

• Derived SI units
• Dimensional analysis
• Fundamental SI units
• Metric prefix
• Metric system
• Order of magnitude
• Physical quantity
• Scientific law
• Scientific model
• Scientific notation
• Scientific theory
• SI fundamental units
• Unit conversion factor
• Units

2a. Compare and contrast distance and displacement

• Define distance.
• Define displacement.
• Give an example of motion when the distance and displacement are the same.
• Give an example of motion when the distance and displacement are different.

Distance describes how much an object has moved. It depends on how the object has moved, that is, the path the object took to get from the starting point to the ending point. The units for distance are length units, such as meters. Distance is called a scalar quantity. A scalar quantity describes the magnitude of the measurement, but not a specific direction.

Displacement describes the overall change in position of an object. It only depends on the starting and ending point of the object. It does not depend on the path taken to get between the two points. Like distance, the units for displacement are length units, such as meters. Displacement is a vector quantity, which means it has a magnitude and a specific direction associated with the measurement. So, the complete units for displacement also include a direction.

For an example, consider a four-story building. A person needs to travel on the elevator from the first to the third floor. To accomplish this, the person could take an elevator directly from the first floor to the third floor. In this case, the distance and displacement are the same, because the person went directly from the starting to the ending point.

However, this is not the only way the person could travel from the first to the third floor. They could accidentally hit the fourth floor button when they got on the elevator. In this case, they would travel from the first floor to the fourth floor, and back down to the third floor. In this instance, the displacement is still the first floor to the third floor. But, the distance is longer, because the person took a detour to the fourth floor.

2b. Define and distinguish between vector and scalar physical quantities

• What is a scalar physical quantity? Give an example of a scalar physical quantity.
• What is a vector physical quantity? Give an example of a vector physical quantity.
• What is the major difference between scalar and vector physical quantities?

A scalar physical quantity is a measurement of quantity that has a magnitude (amount), but not a direction. Examples of scalar quantities include mass and temperature. These are scalar quantities because there is no direction associated with these measurements. In the previous learning outcome, we saw that distance is a scalar quantity because it has no direction associated with it.

A vector physical quantity is a measurement that has a magnitude (amount) and direction. Vectors are often depicted as an arrow. The length of the arrow shows the magnitude of the quantity, and the direction of the arrow shows the direction of the vector.

For simple one-dimensional systems, a vector is often written as the magnitude with a (+) or (-) to indicate direction. As we saw in the previous learning outcome, displacement is a vector quantity. Velocity is also a vector quantity, for example, 5.5 km/s east. This measurement shows the magnitude of the velocity (5.5 km/s) and the direction (east).

The major difference between scalar and vector quantities is that scalar quantities only have a magnitude and vector quantities have a magnitude and direction.

Review Vectors, Scalars, and Coordinate Systems.

2c. Explain the relationship between instantaneous and average values for physical quantities

• What is an instantaneous value? Give an example of an instantaneous value in physics.
• What is an average value? How is an average value calculated? Give an example of an average value in physics.

An instantaneous value is a value measured at a given instant, or time. For example, we can measure the velocity of an object at a given time as 5.5 km/s east. This is an instantaneous value because it was measured at a given instant. The velocity may not be constant over time, but at the instant it was measured, that was the velocity.

An average value is calculated over a period of time. For example, to calculate average speed, divide the distance traveled by time traveled. For example, if you drive 30 miles in two hours, your average speed is 15 miles/hour. However, as we know from driving, we rarely drive exactly the same speed for two hours. So, the instantaneous value of your speed could vary at any given time, but the average value is still 15 miles/hour.

2d. Compare and contrast speed and velocity

• Define elapsed time and show how it is calculated.
• Define average velocity and show how it is calculated.
• Why is velocity a vector quantity?
• Define instantaneous speed and average speed.
• Describe the velocity and speed for a given system.

Elapsed time, $\delta t$, is the change in time. Elapsed time is calculated in the following way:

$\Delta t=t_{f}-t_{i}$, where $t_f$ is final time and $t_i$ is initial time

The Greek letter delta, $\Delta$, means change. So, $\Delta t$ means change in time. You will see this frequently in this course. When calculating elapsed time, we often assume the initial time is zero, to make the subtraction easier.

Average velocity is the displacement divided by the elapsed time:

$\overline{v}\: =\: \frac{\Delta x}{\Delta t}\: =\: \frac{x_{f}-x_{i}}{t_{f}-t_{i}}$

Here, the line you see above the v shows that it is an average quantity. This is common notation for average quantities. To calculate the average velocity, divide the change in displacement by the elapsed time.

Review equations 2.5 and 2.6 in Average Velocity.

The average velocity is a vector quantity. This is because displacement is a vector quantity. Because we calculate average velocity from a vector quantity, it itself is a vector quantity. This means that average velocity has a direction associated with it. In one dimensional systems, this means that the average velocity is written with a (+) or (-) sign, depending on the direction of the displacement.

Instantaneous speed is the magnitude of the instantaneous velocity, measured at a given time or instant. Unlike velocity, instant speed is a scalar quantity, so it does not have a direction associated with it. For example, if the instantaneous velocity of an object is -2 m/s, the object's instantaneous speed is 2 m/s.

The average speed of an object, however, is not simply the magnitude of the average velocity. We define the average speed of an object as the distance divided by the elapsed time. Recall from learning outcome 2a that distance is a scalar quantity that describes how much an object moved and that it can be very different from the vector displacement. Therefore, the average speed of an object is also a scalar quantity, and it can differ from the average velocity.

Take a look at Figures 2.10 and 2.11. Figure 2.10 shows a diagram of the displacement and distance between a home and a store, which took 30 minutes total. For a roundtrip from home, to the store, and back home, the distance and displacement are quite different. The total distance traveled is 6 km (3 km to the store, and 3 km from the store back home). However, the total displacement is 0 because displacement is a vector. The person went +3 km to the store, and then -3 km back home. So, the total vector displacement is 0.

Based on the distance and displacement, we can calculate the average speed and velocity of the trip to the store and back. To determine the average speed, divide the distance by the elapsed time: 6 km/30 min = 2 km/min. However, to determine the average velocity, divide the displacement by the elapsed time: 0 km/30 min = 0 km/min. Figure 2.11 shows graphs of distance, average speed, and average velocity in this example.

2e. Solve one-dimensional kinematics problems

• Define acceleration, instantaneous acceleration, and average acceleration.
• Calculate displacement given average velocity and time.
• Calculate final velocity given initial velocity, acceleration, and time.
• Calculate the displacement of an accelerating object given acceleration and time.
• Calculate the final velocity of an accelerating object given acceleration and time.

Acceleration ($a$) is the rate of change of velocity. We can calculate the average acceleration using the following equation:

$\bar{a}=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}$

Because velocity is a vector, acceleration is also a vector quantity.

Instantaneous acceleration is acceleration measured at a specific instant in time. In most kinematic problems, we assume average acceleration is a constant value.

We need to derive the equations of motion before we can calculate displacement given average velocity and time.

Review equations 2.28 and 2.29 in Solving for Displacement and Final Position from Average Velocity when Acceleration is Constant:

$x=x_0+\bar{v}t$

$\bar{v}=\frac{v_0+v}{2}$

where $x_0$ is initial displacement, $v_0$ is the initial velocity, and $v$ is the final velocity.

Review an example of using these equations to solve for displacement, given average velocity and time, in Example 2.8: Calculating Displacement: How Far Does the Jogger Run?.

We need to derive the equations of motion before we can calculate the final velocity given initial velocity, acceleration, and time.

Review equation 2.35 in Solving for Final Velocity:

$v=v_0+at$

where $v$ is final velocity, $v_0$ is initial velocity, $a$ is constant acceleration,
and $t$ is elapsed time.

Review an example of using these equations to solve for final velocity given initial velocity, acceleration, and time in Example 2.9: Calculating Final Velocity: An Airplane Slowing Down after Landing.

To calculate the final displacement of an accelerating object, we first need to derive the equation necessary to solve these problems.

Review equation 2.40 in Solving for Final Position when Velocity is Not Constant:

$x=x_0+v_0 t +\frac{1}{2}at^2$

where $x$ is final displacement, $x_0$ is initial displacement, $v_0$ is initial velocity, $t$ is elapsed time, and $a$ is acceleration.

Review an example using this equation to solve for final displacement given acceleration and time in Example 2.10: Calculating Displacement of an Accelerating Object: Dragsters.

To calculate the final velocity of an accelerating object, we need to derive the equation necessary to solve these problems.

Review the necessary equation 2.46 in Solving for Final Velocity when Velocity is Not Constant:

$v^2 = v_0^2 + 2a \left (x-x_0 \right )$

where $v$ is final velocity, $x$ is final displacement, $v_0$ is initial velocity, $a$ is acceleration, and $x_0$ is initial position.

In these problems, we first use equation 2.40 to solve for $x$, and then input that result into equation 2.46 to solve for final velocity.

Review an example using this equation to solve for final velocity in an accelerating object in Example 2.11: Calculating Final Velocity: Dragsters.

Review a list of the important kinematics equations used in this section in the box Summary of Kinematics Equations.

Review more examples of using kinematics equations in Example 2.12: Calculating Displacement: How Far Does a Car Go When Coming to a Halt? and Example 2.13: Calculating Time: A Car Merges into Traffic.

2f. Describe the effects of gravity on an object in motion

• Define gravity, free-fall, and acceleration due to gravity?
• How does air resistance affect the falling motion of objects on earth?

Gravity is a force that attracts objects toward the center of the earth, or another large object or planet. In the absence of friction or air resistance, all objects fall with the same acceleration toward the center of the earth. This is known as free-fall. The acceleration due to gravity, $g$, of an object in free-fall is $g=9.80\mathrm{\frac{m}{s^2}}$.

In reality, air resistance affects the acceleration of falling objects. Air resistance opposes the motion of an object in air, and causes falling lighter objects to accelerate less than heavier objects. This is why a feather falls to earth slower than a heavier object like a brick. If there was no air resistance, a feather and brick would fall to earth with the same acceleration due to gravity.

2g. Calculate the position and velocity of an object in free fall

• Calculate the position and velocity of a falling object given initial velocity and time.
• Use position and velocity data of an object in free fall to determine the acceleration due to gravity.

To perform calculations involving objects in free fall, we first need kinematic equations for objects in free fall. Fortunately, these equations are essentially the same as those found in learning outcome 2e above. The main difference is that in free fall, acceleration ($a$) equals the acceleration due to gravity ($g$). For an object falling, we use $-g$ to show the vector direction of free fall.

Review the relevant equations in the box Kinematics Equations for Objects in Free Fall where Acceleration = -g:

$v=v_0-gt$

$y=y_0+v_0 t-\frac{1}{2} gt^2$

$v^2 = v_0^2 -2g \left (y-y_0 \right )$

Note that because the motion is free fall, it is in the $y$ direction rather than the $x$ direction. Here, $g$ is acceleration due to gravity, $g=9.80\mathrm{\frac{m}{s^2}}$.

When calculating the position and velocity of a falling object, we need to consider two different conditions. First, the object can be thrown up and then enter free fall. For example, you could throw a baseball up and watch it fall back down.

Review this case in Example 2.14: Calculating Position and Velocity of a Falling Object: A Rock Thrown Upward. After reviewing the solution, pay special attention to the graphs in Figure 2.40 in the example.

The other case is when an object is thrown directly downward. For example, you could throw a baseball directly down from a second-floor window.

Review this case in Example 2.15: Calculating Velocity of a Falling Object: A Rock Thrown Down. After completing this example, review Figure 2.42, which compares what is happening in Examples 2.14 and 2.15. It is important to understand the difference between an object that is thrown up and enters free fall, versus an object that is directly thrown down.

We can often use experimental data to calculate constants, such as gravity. In Example 2.16: Find G from Data on a Falling Object the acceleration due to gravity constant ($g$) is determined from experimental data.

2h. Draw and interpret graphs for displacement and velocity as functions of time, and determine velocity and acceleration from them

• Define the dependent and independent variable in a graph.
• Describe the graph of a straight line.
• Analyze a graph of position versus time when acceleration is zero.
• Analyze a graph of position versus time when acceleration is constant (not zero).

In graphing two variables against each other, we define the dependent variable as the variable on the vertical axis (y–axis) and the independent variable as the variable on the horizontal axis (x–axis). When plotting a line, we use the equation $y = mx + b$, where $m$ is the slope and $b$ is the y–intercept of the line.

We define slope as:

$m=\mathrm{\frac{rise}{run}}=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$

The y–intercept is the point where the line crosses the y–axis of the graph.

Review Figure 2.46.

An example of a linear graph is the graph of position versus time when acceleration is zero.

Review an example of this type of graph in Figure 2.47. In this graph, we can determine the slope by picking two different points and calculating slope using the equation above. In this case, the unit for slope is m/s, which is the unit for velocity. Therefore, the slope for a graph of position versus time with zero acceleration is the average velocity of that object.

Review how to calculate the average velocity of an object from this type of graph in Example 2.17: Determining Average Velocity from a Graph of Position versus Time: Jet Car.

When acceleration is a non-zero constant, the graph of position versus time is no longer linear. Review an example of this type of graph in Figure 2.48.

Note that while the position versus time graph is not linear, the velocity versus time graph is linear. In the position versus time graph, the slope at any given point is the instantaneous velocity of the object. The instantaneous slope can be determined by drawing tangent lines at various points along the graph, and using the tangent lines to determine slope.

Review tangent lines drawn in Figure 2.48 (a).

To determine instantaneous velocity at a given time when acceleration is a non-zero constant, review Example 2.18: Determining Instantaneous Velocity from the Slope at a Point: Jet Car. 

We can determine instantaneous velocity at multiple points along a position-time graph with constant non-zero acceleration. Then, we can plot velocity versus time, seen in Figure 2.48 (b). We see that this is a linear graph. The slope has units of m/s², which are acceleration units. Therefore, the slope of the velocity versus time graph is acceleration.

Unit 2 Vocabulary

• Acceleration
• Acceleration due to gravity
• Air resistance
• Average acceleration
• Average speed
• Average value
• Average velocity
• Delta
• Dependent variable
• Displacement
• Displacement of an accelerating object
• Distance
• Equations of motion
• Elapsed time
• Final velocity of an accelerating object
• Free fall
• Gravity
• Independent variable
• Instantaneous acceleration
• Instantaneous speed
• Instantaneous value
• Position
• Scalar physical quantity
• Scalar quantity
• Slope
• Speed
• Tangent line
• Vector physical quantity
• Vector quantity
• Velocity
• Velocity of a falling object
• Y–axis
• Y–intercept

3a. Add and subtract vectors

• Describe the properties of vectors. How do you identify a vector in a problem? 
• Define the commutative and associative properties of vectors. 
• Use the head-to-tail graphical method to add vectors.
• Use the head-to-tail graphical method to subtract vectors.

A vector is a quantity that has both a magnitude (amount) and direction. Often in texts, vectors are denoted by being bolded, and/or by having a small arrow written above the vector name.

For example, a vector called A can be written as A or as $\overrightarrow{A}$. The magnitude, or amount, of the vector A equals the value of A. We can think of vectors as arrows, with the length being the magnitude of the vector, and the arrow pointing in the direction of the vector.

Review an example of how a vector can be represented as an arrow in Figure A.1.1 Vectors as Arrows.

When adding or subtracting vectors, we can follow many of the rules we learned in math class about non-vector numbers.

Vector addition follows the commutative property, which means the order of addition does not matter. Vector addition also follows the associative property, which means it does not matter which vector is first when vectors are being added.

Review a complete list of vector addition, subtraction, multiplication, and division rules at the beginning of page A-3 of Vector Analysis. Note that the commutative and associative properties are the most useful.

One way to add or subtract vector is to do so graphically. The graphical method for adding and subtracting vectors is called the head-to-tail method.

When adding vectors using this method:

1. Draw the first vector starting from the tail, or starting point of the vector, to its head, or ending point (arrow) of the vector.
2. Begin the second vector by putting its tail at the head of the first vector.
3. Finally, draw a line from the tail of the first vector to the head of the second vector.

The vector that results is the resultant vector, or the solution to the vector addition problem. To determine the magnitude of the resultant vector, measure it with a ruler. To determine the direction of the resultant vector, use a protractor to determine the angle.

Review an example of using the head-to-tail method to add two vectors, and the step-by-step figures below it, in Figure 3.14.

Review a worked example of using the head-to-tail method to add multiple vectors in Example 3.1: Adding Vectors Graphically Using the Head-to-Tail Method: A Woman Takes a Walk.

When subtracting vectors graphically, consider the vector that is being subtracted as negative. That means the direction of the vector being subtracted is flipped so it points in the opposite direction. The head-to-tail process is the same as it is for addition.

Review a worked example of using the head-to-tail method to subtract vectors in Example 3.2: Subtracting Vectors Graphically: A Woman Sailing a Boat.

3b. Determine the components of a vector given its magnitude and direction, and determine the magnitude and direction of a vector given its components

• Explain the advantages of analytical methods for adding and subtracting vectors, as compared to graphical methods.
• Define component vectors.
• Explain how to use right triangles to resolve a vector into its components. What equations do you use to obtain the x and y coordinate component vectors?
• If you are given component vectors, what equations do you use to determine the magnitude and angle of the resultant vector?

The other method for adding and subtracting vectors is by using analytical methods. Analytical methods use trigonometry to solve vector addition and subtraction. While we still use arrows to represent vectors, analytical methods reduce the measurement errors that can happen with graphical (head-to-tail) methods. 

To use analytical methods to solve vector problems, we need to resolve vectors into their component vectors in the x–y coordinate system.

Review an example of a vector that has been resolved into its x and y components in Figure 3.26. Here, the vector A has a magnitude A and an angle 𝛳. The vector can be broken up into its two components: A_x and A_y. We know that A_x + A_y = A. However, we must use our knowledge of trigonometry to determine how the scalar or magnitude part of each vector relates to each other. The magnitudes of the component vectors relate to the resultant vector the following way:

$A_x=A\cos\theta$

$A_y=A\sin\theta$

Review an example of a vector that has been resolved into its component vectors and shows the magnitudes of the component vectors in Figure 3.27.

Sometimes you are given the component vectors and need to determine the magnitude and angle of the resultant vector. To do this, we again use the trigonometry of right triangles:

$A=\sqrt{A_x^2+A_y^2}$

$\theta=\tan^{-1}\left( \frac{A_y}{A_x} \right )$

Review an example of a resultant vector that is calculated from its component vectors in Figure 3.28.

3c. Separately analyze the horizontal and vertical motions in projectile problems

• Define projectile motion, projectile, trajectory. What is the assumption we make when we calculate projectile motion?
• What are the steps for solving a projectile motion problem?
• Solve projectile motion problems.

We define projectile motion as the motion of a thrown object that only feels the acceleration of gravity. The projectile is the object being thrown, and the trajectory is the path the object takes when it is thrown. When performing projectile motion calculations, we assume there is no air resistance, so gravity is the only force acting on the projectile.

Projectile motion problems are vector problems. We need to know the vector of the trajectory, and then resolve the vector into its components. The steps for solving these problems are the following:

• Step 1: Resolve the trajectory vector into its x and y component vectors.
• Step 2 and 3: Calculate the equations of motion for each component separately. Review the equations we use for the x and y components of the trajectory vector in equations 3.34, 3.35, 3.37, 3.38, 3.39, and 3.40 in Projectile Motion.
• Step 4: Recombine the component vectors into a new resultant vector to calculate total displacement and velocity. Review the equations we use to recombine the component vectors in equations 3.41, 3.42, 3.43, and 3.44.

Review worked examples of performing these calculations using the four-step process listed above in Example 3.4: A Fireworks Projectile Explodes High and Away and Example 3.5: Calculating Projectile Motion: Hot Rock Projectile.

Unit 3 Vocabulary

• Analytical methods
• Angle
• Associative property
• Commutative property
• Component vector
• Graphical methods
• Head-to-tail method
• Magnitude
• Projectile
• Projectile motion
• Resultant vector
• Right triangle
• Trajectory
• Vector
• Vector head
• Vector tail

4a. Compare and contrast mass and inertia

• Define mass and inertia.
• State Newton's First Law of Motion or the Law of Inertia.

We define mass as the amount of matter in an object. We measure mass in units such as grams. Mass does not depend on gravity and therefore does not depend on the location where it is being measured. Inertia is the property that describes the fact that an object at rest (not moving) will stay at rest unless an outside force acts upon it. Likewise, an object in motion will stay in motion with constant velocity unless an outside force acts upon it.

Newton's First Law of Motion is also called the Law of Inertia. This law states the definition of inertia: An object at rest will remain at rest unless an outside force acts upon it. Also, an object in motion with constant velocity will remain in motion with constant velocity unless an outside force acts upon it.

4b. Determine the net force on an object

• Define external force and system.
• Explain how acceleration changes motion.
• State Newton's Second Law of Motion.
• What is the unit for force? What is the meaning of this unit?

The system is whatever we are interested in when calculating a physics problem. The external force is any force that acts upon the system, but is not part of the system. For example, picture pushing a rock up a hill. The system is the rock, and the external force is you pushing the rock.

As we saw in Newton's First Law of Motion, an object at rest stays at rest unless acted upon by an external force. Also, an object in motion at constant velocity remains in motion unless acted upon by an external force. This is inertia. The only way to overcome inertia is to accelerate the object. Applying a net force to the system to induce acceleration.

Acceleration is proportional to the net external force on a system. That is, the higher the applied force, the bigger the acceleration. We also know that acceleration is inversely proportional to mass. That is, large objects accelerate at a slower rate than smaller objects. We know this from our everyday observations. It is easier to accelerate a light ball than a heavier bowling ball.

Newton's Second Law of Motion relates net external force to acceleration and mass of the system: $F_\mathrm{net} = ma$, where $F_\mathrm{net}$ is the net force, $m$ is mass, and $a$ is acceleration.

Review a derivation of the law in the box Newton's Second Law of Motion. 

The unit for force is the Newton, N. The definition of the Newton is 1 N = 1 kg m/s².

Review examples of using Newton's Second Law of Motion to calculate acceleration and force in objects in motion in Example 4.1: What Acceleration Can a Person Produce When Pushing a Lawn Mower? and Example 4.2: What Rocket Thrust Accelerates This Sled?.

4c. Draw and interpret free-body diagrams representing the forces on an object

• Describe the structure of a free body diagram.
• For a given system and force, draw a free body diagram.

A free body diagram shows all the forces acting upon a system within a given coordinate system. It is a simplified way to visualize what is happening in a physics problem. We draw the forces as vector arrows in the direction of the force from the center of mass of the system. This can help us to figure out how we need to add or subtract force vectors when determining the net force on an object.

Review examples of free body diagrams drawn for specific examples in Figures 4.5 and 4.6.

Review more examples of how to draw a free body diagram for a given example in Example 4.1: What Acceleration Can a Person Produce When Pushing a Lawn Mower? and Example 4.2: What Rocket Thrust Accelerates This Sled?.

4d. Identify the correct use of normal and tension forces in terms of Newton's Third Law of Motion

• State Newton's Third Law of Motion or Law of Action and Reaction. 
• For a given system and applied force, determine the opposing force. 
• Define weight, normal force, tension.

Newton's Third Law of Motion states that for every force exerted by an object, there is an equal magnitude force exerted on that object in the opposite direction. This is often called the Law of Action and Reaction. That is, for every action (exerted force), there must be an equal and opposite reaction. This law tells us that forces are always paired.

Review an example of how we can apply Newton's Third Law of Motion to a swimmer in a pool in Figure 4.9. When the swimmer kicks off the wall of the pool to begin swimming, they exert a force toward the wall. Because of the Third Law, the wall also exerts an opposing force on the swimmer. The force of the wall on the swimmer is equal in magnitude, but opposite in direction of the force exerted by the swimmer on the wall. Because the force of the swimmer's feet on the wall does not exert a force on the swimmer themself, this force does not impact the swimmer. Gravity exerts a force toward the earth on the swimmer, but buoyancy exerts an equal magnitude force away from the earth. This keeps the swimmer floating in the water.

Review another example of determining the forces in a given system in Example 4.3: Getting Up to Speed: Choosing the Correct System.

We define weight as the force of gravity on an object of a given mass. Because it comes from gravity, the weight force is directed toward the earth. Now, consider a coffee cup sitting on a table. What keeps the table from collapsing under the weight force of the cup? The coffee cup is experiencing the weight force in the direction of the earth. This "pushes" down on the table. Because of Newton's Third Law of Motion, there must be an equal magnitude force in the opposite direction also acting on this table to balance the forces.

The opposing force is the normal force. Here, "normal" means perpendicular. That is, the normal force is perpendicular to the surface of the table. It balances the weight force from the coffee cup and keeps the table from collapsing. The normal force is abbreviated as N.

Review an example of how the normal force works when an object is placed on a table in Figure 4.12.

Tension is the force along the length of an object. We normally think of tension as a force in an object, such as a rope or string. Objects, such as ropes, can only exert forces in the same direction as their length. If a rope is attached to a hanging object, the object the weight of the object exerts a force toward the earth. The tension of the rope acts as a normal force in the opposite direction of the weight. Review an example of tension in Figure 4.15.

4e. Use Newton's Second Law of Motion to analyze dynamic problems

• Identify forces in x and y directions for a given system.
• Use Newton's Second Law of Motion to solve problems.

Whenever a problem involves forces, we must use Newton's Second Law of Motion to solve the problem. There are four steps to solving these types of problems:

1. Sketch the system described in the problem.
2. Identify forces and draw the forces on the sketch. 
3. Draw a free body diagram of the forces acting on the system.
4. Use Newton's Second Law of Motion to solve the problem. 

Review worked examples of solving dynamics problems using this problem-solving strategy in Example 4.7: Drag Force on a Barge, Example 4.8: Different Tensions at Different Angles, and Example 4.9: What Does the Bathroom Scale Read in an Elevator?.

4f. Give examples of the effects of friction on the motion of an object

• Define friction, kinetic friction, and static friction.
• Give examples of how friction impacts an object's motion.

Friction is the force between two surfaces that opposes motion between them. Kinetic friction is the friction between adjacent surfaces that are moving relative to each other. Static friction is the friction that occurs when two adjacent surfaces are not in motion. Static friction is generally higher than kinetic friction. 

There are many examples of the action of friction in our everyday lives. For example, if you slide a box across a room, the box's motion will eventually stop due to the friction that occurs between the surface of the box and the surface of the floor. A box will slide relatively well across a smooth tile floor because the smooth tile floor has lower friction. It will slide less well across a floor with a rough carpet because the carpet has higher friction.

When we walk on a sidewalk our shoes do not generally slip because the friction between our shoes and the sidewalk opposes the forward force of our shoes. However, we know that icy surfaces are "slippery" when the ice exerts less friction on our shoes than concrete. 

4g. State Hooke's Law

• State Hooke's Law.

Hooke's Law describes oscillations or vibrations. Figure 16.1 shows oscillatory motion. When an object is deformed, there is a restoring force in the opposite direction as the deformation that works to bring the object back to its original position. Hooke's Law describes the forces of this type of motion with the following equation:

$F=-kx$

where $F$ is the restoring force, $k$ is a force constant, and $x$ is the displacement from the equilibrium position of the system.

The force constant describes how stiff the system is, or how difficult it is to deform the system.

4h. Solve problems involving springs

• Identify the parts of Hooke's Law ($F$, $k$, and $x$) in a given problem.
• Use Hooke's Law to determine the force, displacement, or force constant for a given system.
• Calculate the potential energy of a spring using Hooke's Law.

Review a worked example of using Hooke's Law to determine force constant in Example 16.1: How Stiff are Car Springs?. 

We can also use Hooke's Law to determine the potential energy, or stored energy in a spring. Any deformed system, such as a pulled spring, has stored energy. We can calculate the potential energy of a spring using the following equation:

$\mathrm{PE_{el}}=\frac{1}{2}kx^2$

where $\mathrm{PE_{el}}$ is the elastic potential energy of the spring.

Review an example of calculating the potential energy of a spring in Example 16.2: Calculating Stored Energy: A Tranquilizer Gun Spring.

4i. Identify the fundamental physical properties of a simple pendulum, and describe the relationships among them

• Describe and sketch a simple pendulum.
• What forces are involved in the swinging of a simple pendulum?
• How do you describe the period of a simple pendulum?

A simple pendulum consists of a small mass on the end of a string. The string can swing in an arc in the x-y axis. Figure 16.14 shows a figure of a simple pendulum swinging. With a simple pendulum, there are two main opposing forces. First, the weight of the small mass on the string acts as a force in the down direction. There is the restoring force from the string opposing the weight of the mass. As the pendulum swings, the restoring force relates to the displacement and the angle of the pendulum with respect to the x-y coordinate system:

$F=-mg\theta$, where $m$ is mass, $g$ is gravity, and $\theta$ is angle

$\theta = \frac{s}{L}$

$F=\frac{-mg}{L}s$,

where $L$ is the length of the string on the pendulum.

Because $m$, $g$, and $L$ are constants, this takes the form of Hooke's Law: $F=-kx$.

The period ($T$) of a simple pendulum is the time it takes for the pendulum to swing. It is given by the following equation:

$T = 2\pi\sqrt{\frac{L}{g}}$

Review a worked example of using the pendulum equations to calculate the acceleration of a given pendulum's motion in Example 16.5: Measuring Acceleration Due to Gravity: The Period of a Pendulum.

Unit 4 Vocabulary

• Acceleration
• External force
• Force constant
• Free body diagram
• Friction
• Hooke's Law
• Inertia
• Kinetic friction
• Law of Action and Reaction
• Law of Inertia
• Mass
• Net force
• Newton (N)
• Newton's First Law of Motion
• Newton's Second Law of Motion
• Newton's Third Law of Motion
• Normal force
• Period
• Potential energy
• Restoring force
• Simple pendulum
• Static friction
• System
• Tension
• Unit for force
• Weight

5a. Compare and contrast the physical properties associated with linear motion and rotational motion

• Define the rotational angle, arc length, and radius of curvature.
• What is the meaning of the unit radians (rad)?

In rotational motion we deal with two dimensional motion. Unlike with linear motion, we need to define angles and distances associated with circular motion. To understand circular or rotational motion, picture a spinning disk, such as the picture of a CD in Figure 6.2. This figure shows a CD with a line drawn from the center to the edge. All the points along this line travel the same travel the same angle in the same amount of time as the CD spins. This is called the rotational angle, which is defined as:

$\Delta \theta=\frac{\Delta s}{r}$

where $\Delta s$ is the circumference traveled and $r$ is the radius.

We call the curvature traveled ($\Delta s$) the arc length, and we call the radius ($r$) the radius of curvature. 

When describing angles, we often use the unit radian, abbreviated as rad or r. We define radians as:

1 revolution = $2\pi$ rad

Radians can be converted to the more familiar degrees. Review Table 6.1: Comparison of Angular Units for conversions between radians and degrees. 

5b. Explain why an object moving at a constant speed in a circle is accelerating

• Define angular velocity. How is angular velocity related to linear velocity?
• Define centripetal acceleration. In what direction is centripetal acceleration?

We define angular velocity, $\omega$ (the Greek letter omega), as the rate at which the angle changes while an object is rotating. We can write it as:

$\omega=\frac{\Delta \theta}{\Delta t}$

where $\Delta \theta$ is the change in angle and $\Delta t$ is the change in time.

We can relate angular velocity to linear velocity using the following relation:

$v = r \omega$

with $r$ being the radius of curvature.

Review the derivation of how angular velocity relates to linear velocity in equations 6.6, 6.7, 6.8, and 6.9 in Rotational Angle and Angular Velocity.

When an object moves in a circular motion, the net change in velocity points toward the center of the object, based on the vector addition of the velocity vectors as the object rotates. Whenever there is a change in velocity, there is acceleration in the direction of the change of the change in velocity. This means that for a rotating object, there is an acceleration toward the center of the object. We call this centripetal acceleration, $a_c$.

The equation for centripetal acceleration is:

$a_c=\frac{v^2}{r}$

where $v$ is speed and $r$ is radius.

Review an example of this in Figure 6.8. In this example, a disk is rotating at a constant speed. As the disk rotates, the velocity vector at any given point on the disk changes because the direction changes. As shown in the free body diagram at the top of the figure, the velocity vectors add to make a net velocity vector toward the center of the disk. This leads to centripetal acceleration because there is a net change in acceleration toward the disk.

5c. Apply Newton's Law of Gravity

• State Newton's Universal Law of Gravity.
• How does gravitational force relate to object mass? Distance between objects?
• Calculate the acceleration of an object due to gravity.

Newton's Universal Law of Gravity states that all objects in the universe attract each other in straight force lines between them. Review an example of how two objects exert gravitational forces on each other in a straight line in Figure 6.21. 

The force between two objects is directly related to the product of the masses. It is inversely proportional to the distance between the objects squared.

For two objects with masses $M$ and $m$ and radius $r$, this can be written as:

$F = G\frac{Mm}{r^2}$

where $G$ is the gravitational constant, $6.674\times 10^{-11} \mathrm{\frac{Nm^2}{kg^2}}$

Review a worked example of using Newton's Universal Law of Gravity to calculate the acceleration of an object due to gravity in Example 6.6: Earth's Gravitational Force is the Centripetal Force Making the Moon Move in a Curved Path.

5d. Solve problems involving planets and satellites

• State Kepler's Three Laws.
• Use Kepler's Laws to determine how long it takes for a satellite to make its orbit.

Kepler's Laws of Planetary Motion describe the motion of planets around the sun. They can also be extended to describe the motion of satellites around planets.

Kepler's First Law of Planetary Motion states that planets move around the sun in an ellipse shaped orbit with the sun at the center of the ellipse. Review Figure 6.29.

Kepler's Second Law of Planetary Motion states that planets move such that a point on the planet sweeps an equal area in equal times. Review Figure 6.30.

Kepler's Third Law of Planetary Motion relates the time it takes for two planets to revolve around the sun to their distances from the sun in the following equation:

$\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3}$

where $T_1$ and $T_2$ are periods and $r_1$ and $r_2$ are radii for planets 1 and 2.

We can use Kepler's Third Law to solve problems to determine the period for planetary or satellite orbits.

Review a worked example of using the equation from Kepler's Third Law to determine the period of a satellite in Example 6.7: Find the Time for One Orbit of an Earth Satellite.

5e. Explain what it means when an astronaut in earth's orbit is described as being "weightless"

• Define microgravity.
• Is an astronaut in space really experiencing zero gravity? Why or why not?

Microgravity occurs when the net acceleration on an object is significantly less than the net acceleration the object would experience on the earth's surface. It occurs for astronauts in orbit. The astronauts are in free fall toward the earth. This means they are experiencing the earth's gravitational force and are therefore not "weightless" or at "zero gravity". 

Unit 5 Vocabulary

• Acceleration
• Angular velocity
• Arc length
• Centripetal acceleration
• Circular motion
• Gravitational force
• Kepler's First Law of Planetary Motion
• Kepler's Second Law of Planetary Motion
• Kepler's Third Law of Planetary Motion
• Linear motion
• Microgravity
• Newton's Universal Law of Gravity
• Object mass
• Radian
• Radius of curvature
• Rotational angle
• Zero gravity

6a. Calculate the work done on an object by a force

• Define work.
• What is the unit used for work done on an object by a force?
• Use the equation describing work to calculate the work done on an object by a force.

Work is done on a system when a constant applied force causes the system to be displaced or moved in the direction of the applied force. We can describe work using the following equation:

$w = Fd\cos\theta$

where $F$ is force, $d$ is displacement, and $\theta$ (the Greek letter theta) is the angle between $F$ and $d$.

From the equation for work, we can see that the unit for work must be Newton meter, as the unit for force is Newton and the unit for displacement (distance) is meter. We define the Newton meter as the unit joule. We use joules as the unit for work and energy.

Review a worked example of calculating work for a given force and displacement in Example 7.1: Calculating the Work You Do to Push a Lawn Mower across a Large Lawn. This example also asks you to convert from joules to the commonly used energy unit kilocalories (kcal).

6b. Use the relationship between work done and the change in kinetic energy to make calculations

• Define net work and kinetic energy.
• Calculate the kinetic energy of an object given mass and velocity.
• Calculate work done on an accelerating object.

We define kinetic energy as the energy associated with motion. Kinetic energy is calculated as 

$\mathrm{KE} = \frac{1}{2}mv^2$.

When work is done on a system, energy is transferred to the system.

We define net work as the total of all work done on a system by all external forces. We can think of the of the sum of all the external forces acting on a system as a net force, or $F_\mathrm{net}$.

We can write the equation for net work in a similar way to how we wrote the equation for work in learning outcome 6a:

$w_\mathrm{net} = F_\mathrm{net}d\cos\theta$

where $w_\mathrm{net}$ is net work, $F_\mathrm{net}$ is net force, $d$ is displacement, and $\theta$ is the angle between force and displacement.

Review an example of the forces on a box going across a conveyor belt in Figure 7.4. In this figure, we see different forces acting on the box. First, gravitational force is always present, which affects the weight ($w$) of the box. The normal force ($N$) balances the weight of the box. There is the applied force of the moving conveyor belt going to the right. Lastly, there is a horizontal frictional force from the rollers on the conveyor belt going back to the left. The weight and normal force cancel out. Therefore, the net force is the applied force minus the frictional force.

Review a worked example of calculating the kinetic energy for this box on a conveyor belt in Example 7.3: Determining the Work to Accelerate a Package. 

Review a worked example in which the net force is calculated and used to determine the net work for the same system of the box on the conveyor belt in Example 7.4: Determining Speed from Work and Energy.

6c. State the work-energy theorem

•  What is the work-energy theorem?

The work-energy theorem states that the net work on a system is the change of $\frac{1}{2}mv^2$. That is: 

work $= \frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i$

where $m$ is mass, $v_f$ is final velocity, and $v_i$ is initial velocity. 

Review the section Net Work and the Work-Energy Theorem.

6d. Describe the concept of potential energy and how it relates to work

• Define conservative force, potential energy, and mechanical energy.
• Calculate the potential energy of a spring.
• What is meant by conservation of mechanical energy?
• Use conservation of mechanical energy to calculate the velocity (or speed) of an object.

A conservative force is a force that only depends on the beginning point and the end point of the system. That is, it does not depend on the path the system takes to get from beginning to end.

We define potential energy as stored energy due to a system's position. An example of an object with high potential energy is a compressed or stretched spring. When you let go of the compressed or stretched spring, the spring will release its potential energy as kinetic energy and go back to its usual shape.

Mechanical energy is the sum of potential energy and kinetic energy of a system.

To calculate the potential energy of a spring, $\mathrm{PE_s}$, we use the following equation:

$\mathrm{PE_s} = \frac{1}{2}kx^2$

where $k$ is the spring constant and $x$ is displacement.

Review an example of a spring being stretched in Figure 7.10. The figure shows the work and potential energy associated with this.

Conservation of Mechanical Energy states that the sum of potential energy ($\mathrm{PE}$) and kinetic energy ($\mathrm{KE}$) is constant for a given system if only conservative forces act upon the system. We can write this in two different forms:

$\mathrm{KE + PE = Constant}$

$\mathrm{KE_i + PE_i = KE_f + PE_f}$

The second version of the equation can be more useful in describing changes from initial conditions ($\mathrm{KE_i}$ and $\mathrm{PE_i}$) to final conditions ($\mathrm{KE_f}$ and $\mathrm{PE_f}$). 

Review the derivation of the conservation of mechanical energy from the work-energy theorem in equations 7.43, 7.44, 7.45, 7.46, 7.47, and 7.48 in Conservative Forces and Potential Energy.

Review a worked example of using conservation of mechanical energy to determine an object's speed in Example 7.8: Using Conservation of Mechanical Energy to Calculate the Speed of a Toy Car. In this example, we use the conservation of mechanical energy and the definitions of potential and kinetic energy to determine velocity. In these types of problems, it can be helpful to make a list of the information given in the problem to help determine what variable you can solve for.

6e. Compare and contrast conservative and non-conservative forces

• What are non-conservative forces? How does this differ from conservavive forces? 
• Give examples of conservative forces and non-conservative forces.

A non-conservative force is a force that does depend on the path taken of the object. That is, a non-conservative force depends on how an object got from its initial state to its final state. Non-conservative forces change the amount of mechanical energy in a system. This differs from conservative forces, which do not depend on the path taken from initial to final state and do not change the amount of mechanical energy in a system. 

An important example of a non-conservative force is friction. Friction is the force between two surfaces. We see friction when rolling a ball on a carpet versus a hardwood floor. The ball rolls farther on the hardwood floor than it does on a carpet. This is because the fuzzy carpet has more friction than the smooth hardwood. Friction converts some of the kinetic energy of the ball to thermal energy, or heat. As kinetic energy is converted to thermal energy, the balls slows to a stop. 

Conservative forces exist in ideal systems with no friction. An idealized spring that does not experience friction would be an example of conservative forces.

Review Figure 7.15 for a comparison of conservative and non-conservative forces. In Figure 7.15 (a) a rock is being "bounced" on an ideal spring with no friction. The mechanical energy does not change and the rock will continue bouncing indefinitely. In Figure 7.15 (b) the rock is thrown and lands on the ground. When it hits the ground, its kinetic energy is converted to thermal energy and sound. The rock can not "bounce" back up because its mechanical energy is not conserved.

6f. Solve dynamics problems using conservation of energy

• State the Law of Conservation of Energy.
• What are some examples of other energy?
• List the problem solving steps for solving dynamics problems.

The Law of Conservation of Energy states that the total energy in any process is constant. Energy can be transformed between different forms, and energy can be transferred between objects. However, energy cannot be created or destroyed. This is a broader law than the conservation of mechanical energy because this applies to all energy, not just energy when only conservative forces are applied.

We can write the Law of Conservation of Energy in the following ways:

$\mathrm{KE + PE + OE = Constant}$

$\mathrm{KE_i + PE_i + OE_i = KE_f + PE_f + OE_f}$

In the second equation, the $\mathrm{KE_i}$, $\mathrm{PE_i}$, and $\mathrm{OE_i}$ are initial conditions and $\mathrm{KE_f}$, $\mathrm{PE_f}$, and $\mathrm{OE_f}$ are final conditions. The new term, $\mathrm{OE}$, is other energy. This is a collected term for all forms of energy that are not kinetic energy or potential energy.

Other energy includes forms such as thermal energy (heat), nuclear energy (used in nuclear power plants), electrical energy (used to power electronics), radiant energy (light), and chemical energy (energy from chemical reactions). 

When solving Conservation of Energy problems, it is important to identify the system of interest, and all forms of energy that can occur in the system. To do this, we need to first identify all forces acting on the system. Then, we can plug equations for different types of energy into the Law of Conservation of Energy equation to solve for the unknown in the problem.

Review Problem Solving Strategies for Energy for a step-by-step guide for solving these types of problems.

6g. Illustrate what is meant by power

• Define power. What is the unit for power?
• Calculate power for a given system given energy and time.

We define power as the rate at which work is done. We can write this as:

$P = \frac{w}{t}$

where $w$ is work and $t$ is time.

The unit for power is the watt, W, with one W = one joule/second.

Higher power means more work is done in a shorter time. This also means that more energy is given off in a shorter time. For example, a 60 W light bulb uses 60 J of work in a second, and also gives off 60 J of radiant and heat energy every second. 

See a worked example of calculating the power of an object in motion in Example 7.11: Calculating the Power to Climb Stairs. In this problem, the power of a person going up a flight of stairs is calculated. First, the work of going up the stairs is calculated using the equation: work $\mathrm{= KE + PE}$. Then, power is calculated given the time it took the person to go up the stairs.

Review this material in Power and Work, Energy, and Power in Humans.

Unit 6 Vocabulary

• Chemical energy
• Conservation of mechanical energy
• Conservative force
• Electrical energy
• Fossil fuels
• Friction
• Joule
• Kinetic energy
• Law of Conservation of Energy
• Mechanical energy
• Net force
• Net work
• Newton meter
• Nonconservative force
• Nonrenewable energy
• Nuclear energy
• Other energy
• Potential energy
• Power
• Radiant energy
• Renewable energy
• Spring
• Thermal energy 
• Watt
• Work
• Work-energy theorem

7a. Define linear momentum

• Define linear momentum and the equation we use to calculate it.
• Why is linear momentum a vector?
• What are the units for linear momentum?

We define linear momentum as the product of an object's mass and velocity. It can be written as $p = mv$, where $p$ is linear momentum, $m$ is mass, and $v$ is velocity.

Linear momentum is a vector quantity because velocity is a vector quantity, and the linear momentum will have the same direction as the velocity. The units for linear momentum are kgm/s.

7b. Use Newton's Second Law in terms of momentum

• Describe how we can reframe Newton's Second Law of Motion in terms of momentum.
• Use the equations from Newton's Second Law of Motion to calculate force.

We reviewed in Unit 4 learning outcome 4b that we can write Newton's Second Law of Motion as $F= ma$. While this is the most common way to write and use this law, it was not how Newton originally wrote it. Newton wrote this law in terms of momentum rather than force and acceleration:

$F_\mathrm{net} = \frac{\Delta p}{\Delta t}$

This shows that the net force equals the change in momentum divided by the change in time. This equation certainly appears different from the familiar $F=ma$ we used in Unit 4.

Review the derivation of how $F=ma$ can be obtained from the Second Law in terms of momentum in equations 8.9, 8.10, 8.11, and 8.12 in Linear Momentum and Force.

Review a worked example of using Newton's Second Law in terms of momentum in Example 8.2: Calculating Force: Venus Williams' Racquet. This problem calculates the force applied to a tennis ball: there is a change of velocity of the ball but no change in mass, so pay special attention to how change in momentum is calculated in equation 8.14.

7c. Describe the relationship between impulse and momentum

• Define impulse.
• What is the assumption we make when calculating impulse?

We define impulse as change in momentum. Using Newton's Second Law of Motion, we can write this as:

$\Delta p=F_\mathrm{net} \Delta t$

When we calculate impulse, we assume the net force is constant during the time we are interested in. In reality, force is rarely constant. For example, in Example 8.2, we assumed the force on the tennis ball was constant over time. In reality, the force on the tennis ball probably changed from the beginning to the end of the swing of the tennis racquet. However, the change in force was probably not significant, and we assume it is constant to make our calculations easier. 

7d. Define elastic, inelastic, and totally inelastic collisions

•  Define elastic, inelastic, and totally inelastic collisions. 
•  Give examples of each type of collision.

When two or more objects physically interact, we say the objects collide or have a collision. There are three types of collisions we can consider when solving physics problems, which are all based on energy transfer in the collisions. 

By definition, an elastic collision is a collision in which the internal kinetic energy is conserved in the interaction. Internal kinetic energy is the sum of the kinetic energy of all of the objects colliding. So, in an elastic collision, all the kinetic energy remains kinetic energy. That is, no kinetic energy is converted to heat, friction, or other types of energy. In reality, no collisions are perfectly elastic because some kinetic energy is always "lost" by being converted to other forms of energy. A close example of an elastic collision is if two balls collide on a smooth icy surface. Because the ice has almost no friction, little kinetic energy would be lost to friction.

Review a diagram of two metal boxes interacting in an elastic collision on an ice surface in Figure 8.6.

Unlike an elastic collision, an inelastic collision is a collision in which the internal kinetic energy is not conserved. In inelastic collisions, some kinetic energy of the colliding objects is lost to friction, heat, or even work. Inelastic collisions are what we mostly observe in the real world.

Review a good example of an inelastic collision in Figure 8.9. In this example, a hockey goalie stops a puck in the net. Although the ice surface is essentially frictionless, some kinetic energy of the puck is converted to heat and sound as the goalie stops it. A totally inelastic collision (also called perfectly inelastic collision) is an inelastic collision in which the objects "stick together" upon colliding.

Review an example of two blocks experiencing a totally inelastic collision in Figure 8.8. 

7e. Use conservation of momentum to solve collision problems

• Calculate velocities of objects following an elastic collision.
• Calculate velocities of objects and energy lost following an inelastic collision.

When solving problems for elastic collisions, it is important to remember that the internal kinetic energy is conserved. Therefore, the total kinetic energy at the start of the collision must equal the total kinetic energy at the end of the collision.

We can write this as: $\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}m_1 v_1^{'2} + \frac{1}{2}m_2 v_2^{'2}$

Moreover, we know that momentum must be conserved in the collision. Therefore, the total momentum at the start of the collision must equal the total momentum at the end of the collision. That is, for two objects (object one and two) colliding, we can write: $\frac{1}{2}m_1 v_1 + \frac{1}{2}m_2 v_2 = \frac{1}{2}m_1 v_1^{'} + \frac{1}{2}m_2 v_2^{'}$. Using conservation of momentum, we can usually set up these problems so we only have to solve for one unknown.

Review a worked example of this type of problem in Example 8.4: Calculating Velocities Following an Elastic Collision. In this example, one of the objects is initially at rest (its velocity equals zero), so it does not have an initial momentum. This lets us simplify the conservation of energy momentum equations. Then, by using the equations for conservation of energy and momentum, we can solve for final velocity after collision.

We can also solve for final velocities after inelastic collisions. In these problems, it is important to remember that kinetic energy is not conserved but momentum is conserved. Example 8.5 (a): Calculating Velocity and Change in Kinetic Energy: Inelastic Collision of a Puck and a Goalie shows a worked example of this type of problem. Here, the conservation of momentum equation is used to determine the final velocity of the object (the hockey goalie) in an inelastic collision.

In inelastic collisions, some kinetic energy is converted to other forms of energy. The energy difference before and after collision can be calculated to determine how much kinetic energy was lost. Example 8.5 (b): Calculating Velocity and Change in Kinetic Energy: Inelastic Collision of a Puck and a Goalie shows a worked example of calculating the energy lost in this inelastic collision. Here, the total kinetic energy in the system is calculated before and after collision based on the mass and velocities of the objects. The difference in kinetic energy shows how much kinetic energy was converted to other forms of energy during the collision.

Example 8.6: Calculating Final Velocity and Energy Release: Two Carts Collide is a similar worked example.

7f. Demonstrate the physics behind rocket propulsion

• Which law of motion governs rocket propulsion? Why?
• Give some examples of rocket propulsion in nature and technology.
• What makes a rocket accelerate?

Rocket propulsion is directly tied to Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction (review learning outcome 4d). In rocket propulsion, matter is removed from a system with a strong force. This causes an equal force in the opposite direction on the system. An example is the recoil of a gun. When a gun shoots, the bullet is removed from the system (the gun) with a strong force. Because of Newton's Third Law, the gun itself experiences recoil and pushes back away from the bullet's path. Another example is the movement of squid. To move, squid push water out of their bodies with a strong force. This causes the squid to move in the opposite direction of the water being pushed out of the squid's body.

Figure 8.13 shows a diagram of a rocket going straight up. When the rocket launches, it expels a large mass of hot gas (from the fuel) down toward the earth. As the gas is expelled, the rocket's velocity increases because its overall mass decreases. The acceleration of the rocket is proportional to its change in mass as it burns its fuel. The faster the fuel is burned, the greater the acceleration of the rocket.

Unit 7 Vocabulary

• Elastic collision
• Impulse 
• Inelastic collision
• Internal kinetic energy
• Linear momentum
• Momentum
• Newton's Second Law of Motion (in terms of momentum)
• Rocket propulsion
• Totally inelastic collision

8a. Define the conditions necessary for a rigid body to be in equilibrium

• Define equilibrium.
• What is the study of statics?
• Compare and contrast static and dynamic equilibrium.
• What are the two major conditions for a body to be in equilibrium?

When an object is in equilibrium, the forces acting upon the object are balanced. That is, the net force on the object is zero. For this to occur, the object must either not be moving, or it must be moving at a constant velocity.

There are two types of equilibrium: static equilibrium and dynamic equilibrium.

Review an illustration of static equilibrium in Figure 9.3 and an illustration of dynamic equilibrium in Figure 9.4. An object in static equilibrium is completely motionless. An object in dynamic equilibrium is moving at constant velocity. The study of statics is the study of objects that are in equilibrium.

Two important conditions must be met for an object to be in equilibrium. First, the net force on the object must be zero. Secondly, a rotating object does not experience rotational acceleration. That is, a rotating object can be in equilibrium if its rotational velocity does not change.

8b. Define torque, remembering that it is a vector physical quantity

• Define torque. What is the unit for torque? 
• How is torque related to the second condition necessary for an object to be in equilibrium? 
• How does the direction of applied force impact torque?

We define torque as the effectiveness of a force to turn or twist an object, thus changing its rotational velocity. We can write the definition of torque as:

$\tau = rF\sin\theta$

where $\tau$ (the Greek letter tau) is torque, $r$ is radius of force, $F$ is magnitude, and $\theta$ is the angle between the force and the pivot point.

The unit for torque is the Newton meter (Nm).

We can restate the second condition for an object to be in equilibrium in terms of torque. In learning outcome 8a above, we said that an object in equilibrium must have no rotational acceleration. We can restate this by saying that an object in equilibrium must have a torque of zero.

Review a diagram showing the torque on a rotating plank of wood, secured at a pivot point at one end in Figure 9.6. This diagram shows how the direction of the force impacts the rotation of the plank of wood. When the force is perpendicular to the length of the plank of wood, the plank experiences torque and it rotates. When the force is parallel to the length of the plank of wood, it does not experience a net force and therefore does not experience torque or rotate. When the force is at an angle other than 90° from the length of the plank, the plank experiences less torque than if the force was at 90° from the plank's length.

8c. Solve statics problems

• Briefly outline the steps needed to solve statics problems.
• Solve statics problems to determine the force needed to support a weight.
• Solve statics problems to determine the torque needed to produce rotational motion for a given system.

When performing statics calculations, the first step is to determine if the system is in fact in equilibrium. Recall from learning outcome 8a that for a system to be in equilibrium, two conditions must be met: the system must not be accelerating and the torque must be zero. The second step is to draw a free body diagram of the system. It is important to determine all the forces acting upon the system. The third step is to solve the problem by applying the relevant conditions of equilibrium: force is zero, and torque is zero.

Review more details in Problem Solving Strategy: Static Equilibrium Situations.

Example 9.1: She Saw Torques on a Seesaw shows a worked example of a statics problem. Here, children are balanced on a seesaw. We are given information about the masses of both children, and how far from the pivot point one child is sitting. We are asked to determine where the second child is sitting to balance.

In Figure 9.8 we see that the children are balanced and therefore are at equilibrium. The free body diagram shows that there is no net force, and no net rotational acceleration. To determine the distance of the second child from the pivot point, we use the torque equation, and set torque equal to zero. To determine the upward balancing force from the pivot point, we use the fact that net force equals zero to solve for the individual force at the pivot point.

Example 9.2: What Force is Needed to Support a Weight Held Near its CG? shows a similar worked example of a statics problem. Here, a pole vaulter holds a pole at one end and we are asked to calculate the forces from each of the pole vaulter's hands. We take the same approach as in Example 9.1: She Saw Torques on a Seesaw.

Unit 8 Vocabulary

• Dynamic equilibrium
• Equilibrium
• Newton meter (Nm)
• Static equilibrium
• Statics
• Torque

9a. Solve kinematics problems involving rotational motion

• Define angular velocity and the relationship between angular velocity and linear velocity.
• Define angular acceleration.
• Define kinematics.
• Calculate angular acceleration or deceleration.
• Given angular acceleration, determine the time needed for a system to accelerate or decelerate.
• Calculate the distance traveled by a rotating object

We define angular velocity (or rotational velocity) as:

$\omega = \frac{\Delta \theta}{\Delta t}$

where $\omega$ represents angular velocity.

We can relate angular velocity to linear velocity with the equation:

$v = r \omega$

where $r$ is the radius of curvature.

See a diagram of a rotating object showing the relationship between linear and angular velocity in Figure 10.3.

Angular acceleration is defined as the change in angular velocity with respect to time:

$\alpha = \frac{\Delta \omega}{\Delta t}$

where $\alpha$ represents angular acceleration.

Kinematics is the study of motion. When solving kinematics problems of rotational motion, we look at the relationships between angular velocity, linear velocity, and angular acceleration. See equations 10.17, 10.18, and 10.19 in Kinematics of Rotational Motion for kinematics equations discussed in previous sections and equations modified for rotational motion.

Example 10.1: Calculating the Angular Acceleration and Deceleration of a Bike Wheel shows a worked example of how to calculate the angular acceleration of a bike wheel. In the first part of the problem, we calculate the angular acceleration of the wheel given the change in angular velocity and time. In the second part of the problem, we calculate the time needed to stop an already spinning wheel given angular deceleration an initial velocity, using the same angular acceleration equation.

Example 10.3: Calculating the Acceleration of a Fishing Reel shows a worked example of how to calculate the kinematics of an accelerating fishing reel. Here, equation 10.19 is used to determine how the angular velocity changes with time. This result is used to calculate linear speed.

Example 10.4: Calculating the Duration When the Fishing Reel Slows Down and Stops is a worked example in which the fishing reel decelerates. Using equation 10.19, we solve for time rather than angular velocity.

Review for additional worked examples of these types of problems in Example 10.5: Calculating the Slow Acceleration of Trains and their Wheels and Example 10.6: Calculating the Distance Traveled by a Fly on the Edge of a Microwave Oven Plate.

9b. Solve dynamics problems involving rotational motion

• Define dynamics.
• List the steps for solving dynamics problems for rotational motion.

The study of dynamics involves calculations of how force and mass impact an object's motion. When solving dynamics problems, first we need to identify the system and draw a free body diagram of all the forces acting upon the system. Once the forces acting upon the system are defined, we can use the following torque equation and angular acceleration equations to solve the problem:

$\mathrm{net\ }\tau=I\alpha$, and $\alpha=\frac{\mathrm{net\ }\tau}{I}$

where $I$ is the moment of inertia.

Example 10.7: Calculating the Effect of Mass Distribution on a Merry-Go-Round shows a worked example of using the above equations to determine the angular acceleration of a person pushing a merry-go-round. Here, the first step is to calculate torque. The next step is to calculate the moment of inertia. Lastly, torque and moment of inertia are used to calculate the angular acceleration on the merry-go-round.

9c. Define rotational inertia

• Define rotational inertia.
• Define moment of inertia.

We define rotational inertia as $mr^2$, where $m$ is the mass of the object being rotated and $r$ is the radius from the pivot point to the end of the mass.

Review a diagram of an object rotating on a frictionless table in Figure 10.11. We can see the radius from the center of the table (the pivot point) and the mass at the end of the radius. 

The moment of inertia, $I$, is the sum of all the rotational inertia acting upon an object. We can write the following summation:

$I=\sum mr^2$

9d. Compare and contrast the dynamics of linear and rotational motion

•  How are the definitions of linear and rotational motion related to each other? 

Rotational velocity is related to linear velocity in the following equation:

$v=r\omega$

where $r$ is the radius of curvature.

9e. Apply energy concepts to rotational motion

• Define work for rotational motion.
• Define rotational kinetic energy.

For rotational motion to occur, work must be done. However, we cannot use the simple definitions of work that we used earlier in this course, because we now must account for rotational motion.

Review a diagram of a spinning disk in Figure 10.15.

For the disk to spin, work must be done on the disk. The force acting upon the disk must be perpendicular to the radius of the disk, which we know is torque. We also know torque is related to moment of inertia. We can relate the work done on the disk to moment of inertia using the following equations:

$\mathrm{net\ }\omega=\left ( \mathrm{net\ }\tau \right )\theta=I\alpha\theta$

We can write an equation for the rotational kinetic energy (the energy of rotational motion) as:

$\mathrm{KE_{rot}}=\frac{1}{2}I\omega^2$

Example 10.8: Calculating the Work and Energy for Spinning a Grindstone is a worked example for calculating the net work for a rotating disk using the above work equation. In the second part of the example, the rotational velocity is determined using the equation for rotational acceleration and moment of inertia. Lastly, the rotational kinetic energy is calculated using the above equation.

9f. Give several everyday examples of conservation of angular momentum

• Define angular momentum.
• Describe everyday examples of conservation of angular momentum.

We define angular momentum as $L=I\omega$.

It is similar to the momentum defined for linear motion. As such, angular momentum in a system is conserved in the same way that linear momentum is conserved. Therefore, we can say that $L=L'$, where $L$ is the initial angular momentum in a system and $L'$ is the final angular momentum in the system. We can also write this as:

$I\omega=I'\omega'$

We see conservation of angular momentum in many everyday examples. One example is the figure skater spinning in Figure 10.23. In the first picture, the figure skater is spinning with her arms out on a frictionless ice surface. In the second picture, she pulls her arms in, and her rotational velocity increases. When the figure skater pulls in her arms, she lowers her moment of inertia. Because angular momentum is conserved, because her moment of inertia decreases, her angular velocity must therefore increase.

Unit 9 Vocabulary

• Angular acceleration
• Angular momentum
• Angular velocity
• Dynamics
• Kinematics
• Linear velocity
• Moment of inertia
• Rotational inertia
• Rotational kinetic energy