Tag: Physics Analysis

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    See attached **files (2)** – ## DB2.docx – ## UnitII.pdf **## DB2.docx** DB 2: Your initial post should be at least 200 words in length. Discuss the main difference between science and pseudoscience. What are some examples of pseudoscience, and what are some of their claims? What makes pseudoscience…

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    **files (2)**

    – ## DB2.docx

    – ## UnitII.pdf

    **## DB2.docx**

    DB 2: Your initial post should be at least 200 words in length.

    Discuss the main difference between science and pseudoscience. What are some examples of pseudoscience, and what are some of their claims? What makes pseudoscience not real science? Is there your way to determine if it is a real science or not? Share your ideas with your classmates.

    Reply 1:

    Reply 2:
    **Monica Thompson**

    In Latin, science is scientia, meaning "knowledge". Science has become the most respectable kind of knowledge to date. It uses reliable data that has been proven, tested, and observed repeatedly, yielding the same results. Pseudoscience is a claim or system that appears scientific but is not based on testable, measurable, and repeatable evidence. It is literally armchair philosophy, so to speak; it is what one believes. It lacks solid evidence and can’t be tested or proven wrong. In other words, it is a belief or claim.

    An example of pseudoscience is eugenics. During a certain era, this movement claimed that by regulating who could have children would lead to a “better” human race. Can you guess when this occurred? How twisted and demented this was, and to think this practice may still be occurring today, without our knowledge. Obviously, in this example of pseudoscience, there was no solid evidence to support it as scientific. It was more racially biased and agenda-driven based on social and political beliefs, keyword beliefs. There was no objective evidence to support this claim, and in the long run, it only harmed the human race. My background is diverse, as is true for most people, and research shows that intelligence is not defined by race but by each individual.

    The use of observation and reasoning dates back to Aristotle, around 300 BCE, and the modern scientific method took hold in the early 1600’s, when Galileo and Francis Bacon applied it to their discoveries. So, the clear difference between science and pseudoscience is the scientific method. Science clearly uses the scientific method, and pseudoscience does not.

    Reply 2:
    **Victoria Papuchis**

    Top of Form

    There is a clear difference between science and pseudoscience, and the biggest distinction is that pseudoscience does not use the scientific method. The scientific method is a systematic, evidencebased process used to investigate questions, test ideas, and build reliable knowledge. Science depends on this structure because it reduces bias, follows logical steps, and allows researchers to test, analyze, and draw conclusions that can be repeated and verified by others.

    This is what makes scientific findings trustworthy and continually improving.

    From my understanding, pseudoscience is built more on personal beliefs, unsupported claims, and a refusal to accept criticism. These ideas often remain unchanged even when they have been proven wrong. Instead of being open to new evidence, supporters of pseudoscience tend to shift their explanations to avoid admitting that the claim doesn’t hold up. This resistance to scrutiny is exactly what makes pseudoscience invalid and frustrating. It often uses vague language, relies on testimonials instead of data, and cannot be independently verified. In real science, results must be consistent no matter who performs the experiment.

    Thinking back to my own science classes, from elementary school through my current college studies, the scientific method has always been required. Every experiment or project involved forming a hypothesis, collecting data, and drawing evidencebased conclusions. That consistency is what separates true scientific practice from pseudoscience.

    Bottom of Form

    Bottom of Form

    **## UnitII.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit II

    Upon completion of this unit, students should be able to:

    2. Illustrate the scientific method within everyday situations.
    2.1 Identify the appropriate formulas necessary to solve specific scenario questions.
    2.2 Calculate and analyze the acceleration and the force in various situations.

    3. Explain Newton's laws of motion at work in common phenomena.

    3.1 Solve problems using the law of motion.
    3.2 Explore the relationship between the first and second laws.
    3.3 Identify action-reaction pair in the third law.

    Required Unit Resources

    Chapter 3: Kinematics in Two Dimensions

    Chapter 4: Forces and Newton’s Laws of Motion

    Unit Lesson

    The Concept of Force and Mass

    A force is a push or pull, and it is a vector quantity. Contact forces arise from the physical contact between
    two objects. Noncontact forces are also called action-at-a-distance forces because they arise without physical
    contact between two objects. More sophisticatedly, a force is an external influence on an object that causes it
    to accelerate relative to an inertial reference frame where no acceleration occurs. The direction of the force is
    the same as that of the acceleration.

    Mass is an intrinsic property of an object that measures its resistance to acceleration. That is, mass is a
    measure of the object’s inertia. In dynamics, mass represents the inertia of a body or that body’s resistance to
    any change in its state of motion. Mass is a scalar quantity characterizing a body, and it does not depend on
    the body’s location or state of motion (Cutnell et al., 2022).

    Many people think that weight is the same term for mass; however, it is not! Mass is a measure of the amount
    of material contained in an object and does not depend on any gravitational force acting on it. Mass is
    constant. On the other hand, weight is the gravitational force acting on an object. It depends on the mass of
    the object and the gravitational acceleration of the object attracting it. That is, weight W is the product of the
    mass m and the acceleration g due to gravity.

    Your weight will vary if you measure it on other celestial bodies, but your mass does not change. For
    instance, if you measure your weight on Jupiter, it will be greater than that on Earth. On the moon, however,
    your weight will be less than that on Earth. The larger the gravitational pull, the greater your weight.

    UNIT II STUDY GUIDE
    Newton’s Laws

    W = mg

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Friction

    When an object is in contact with a surface, the surface exerts a force on the object. The perpendicular
    component of the force to the surface is called normal force (please look at Figure 4.13 in section 4.8 in
    Chapter 4 of your eTextbook) and the parallel component of the force to the surface is called frictional force,
    or simply friction. In order to initiate the motion of the object, the applied force must overcome the friction. The
    direction of the friction is opposite to that of the applied force. The frictional force is proportional to the normal
    force. Usually, the friction increases on an uneven surface.

    In general, there are two kinds of frictional forces: static and kinetic. The static friction prevents an object from
    moving. In order to move the object, the applied force must overcome the static friction. Once the object is in
    motion, the kinetic friction is now an obstacle for the moving object (Cutnell et al., 2022). Be sure to review the
    coefficients of friction for various surfaces in Table 4.2 located in section 4.9 of Chapter 4 in your eTextbook.

    Newton’s Three Laws

    Newton’s three laws are about forces that change the motion of objects and are based on total momentum
    conservation. In a simple expression, momentum is a quantity defined as mass multiplied by velocity. We will
    learn about momentum in detail in Unit IV.

    The first law is the law of inertia. This is the same law by Galileo’s inertia interpretation. Unless there is no
    external force, an object that is at rest stays at rest, and one that is in motion continues to move (Shipman,
    Wilson, & Todd, 2009). That is, there is a tendency for an object to keep its original property. The inertia of an
    object is measured by its mass. The more massive the object, the more inertia it has. For instance, when a
    bus driver stops suddenly, a man in the rear seat moves forward. This is because the man’s inertia that was
    previously in motion wants to stay in motion. In fact, this law of inertia is a special case of Newton’s second
    law.

    The second law is the law of motion or acceleration. Force (F) = mass (m) x acceleration (a) (Shipman et al.,
    2009). This is the most useful formula because it can describe the dynamics of motion in detail. We can
    picture how velocities change with respect to time when forces are applied. Simply, it can be expressed with
    the mathematical formula F = ma.

    The SI unit of force is the mass unit (kg) times the acceleration unit (m/s2), and thus [kg m/s2], which is
    defined as [N]. That is, 1 newton = 1 N = 1 kg m/s2

    Here, the force is summed over all applied forces exerted on an object. It accelerates the object relative to an
    inertial reference frame. Remember that the directions of force and acceleration are identical. Mass that
    opposes acceleration is an intrinsic property of an object and is a gauge of inertia. Acceleration is a change of
    velocity. That is, acceleration indicates how much the velocity has been increased or decreased during a
    certain amount of time. If the velocity is constant, which means the acceleration is zero, then there is no
    applied force. This is exactly the case for Newton’s first law. There is no distinction between an object that is
    in motion with a constant velocity and one that is at rest. One example of accelerated motion is circular
    motion. Even though the speed of the object is constant, the direction of the movement is changing, so the
    velocity is not constant.

    F = ma

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    The third law is the law of action-reaction. The forces exist in pairs. The magnitudes of forces are equal, but
    their directions are opposite (Shipman et al., 2009). That is, every action has an equal and opposite reaction.
    For example, when you push a wall on a slippery floor, you feel that your feet are moving backward. The push
    force exerted by you is the same as that by the wall. That is, the two forces are equal in magnitudes, but
    opposite in direction. Review Example 4 located in section 4.9 of Chapter 4 in your eTextbook (Cutnell et al.,
    2022).

    Sample Question 1: A lunar exploration vehicle was made by a research team, and it weighs about
    six times more than it will on the moon. In order to have the same acceleration on the moon as well
    as on the Earth, what will be the net force acting on the vehicle? Will it be greater than, less than, or
    the same as that required on Earth?

    Solution: According to Newton’s second law, the net force is F = ma. Here, m is the vehicle’s mass
    and a is the acceleration. The net force depends on the mass for a constant acceleration. Here, the
    mass of the vehicle is constant. Therefore, the same net force would be required.

    Sample Question 2: You and your friend are pushing a box in the same direction. Its mass is 200
    kg. You apply 20 N to the box, while your friend applies 30 N. A force created by friction is 40 N in the
    opposite direction. What is the acceleration of the box?

    Solution: There are three forces acting in this system, and the net force should be added up. Two of
    them are acting toward the same direction, and the remaining is acting in the opposite direction, as
    you can see below. Thus, the net force F = 30 N + 20 N – 40 N = 10 N. From Newton’s 2nd law, F =
    ma. The mass of box is given by 200 kg. Therefore, the acceleration a = F / m = 10 N / 200 kg = 0.05
    m/s2

    Sample Question 3: An astronaut pushes on the spacecraft with a force of 40 N near the
    International Space Station. His mass is 90 kg, and the spacecraft’s mass is 10,000 kg. What are
    the accelerations of the space craft and the astronaut?

    Solution: According to Newton’s third law, when the astronaut applies the force to the spacecraft,
    the spacecraft also applies the same amount of force to the astronaut. That is, the magnitudes of
    the two forces are identical, but the directions are different.

    From the second law of Newton, the acceleration of the spacecraft is as = 40 N / 10,000 kg = 0.004
    m/s2. The acceleration of the astronaut is aa = -40 N / 90 kg = -0.44 m/s2. Note that the magnitudes
    of the action and reaction forces are always equal, but they do not have the same acceleration
    unless they have the same mass.

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    Free Fall

    Every object on Earth falls downward because of Earth’s gravity. Earth’s gravity points to the center of Earth.
    If there is no air, all of the falling objects should experience the same acceleration. Recall from the Unit I
    Lesson that in order to prove this, an astronaut, David Scott of the Apollo 15 mission, performed the
    experiment on the moon.

    When we ignore the air resistance and assume the constant acceleration, we say that the falling object is in
    free fall. The acceleration of the freely falling body is due to gravitational acceleration g, and it is about 10
    m/s2 on the surface of the Earth. However, the object is at considerably high altitude, the acceleration due to
    gravity is not constant. The gravitational acceleration is inversely proportional to the square of the distance
    from the center of the Earth to the object.

    The Fundamental Forces

    All the different forces observed in nature can be explained in terms of four basic interactions that occur
    between elementary particles: gravitational force, electromagnetic force, strong nuclear force, and weak
    nuclear force. The associated particles for the strong force are mainly gluons and pi nucleons.

    The main role of the strong force is to hold the nuclei of atoms together. Principally, it is attractive; however, it
    can sometimes act in a repulsive way with proper conditions. The strong nuclear interaction has a very
    powerful strength of the force, but it is very short-ranged (about 10-13 cm). The attractive force only works well
    within the size of the nucleus. The role of a neutron is very important to maintain the structure of the nucleus.
    A repulsive force exists between two protons because they have the same charge. Remember that in the
    influence of electrostatic force two like charges repel, and two opposite charges attract. Thus, more neutrons
    than protons are needed to keep the stable nucleus as the atomic number increases. The maximum proton
    number is 83 with 126 neutrons for the stable structure in nature. All nuclei that have an atomic number
    greater than 83 are unstable because the binding force to hold the nuclei is weak and disintegrates or
    rearranges its structure. This is known as radioactivity. The radioactive decay or neutrino interactions can be
    explained by the weak force, and its strength is weak with a short range. The associated particles for the
    weak nuclear force are known as intermediated vector bosons such as W+, W-, and Z0. They are spin 1
    particles, and their masses are greater than 80 GeV.

    The electromagnetic force is responsible for electric and magnetic effects. The phenomena of electricity and
    magnetism look different; however, they are very closely related to each other. A changing electric field can
    create a magnetic field and vice versa. In order to combine electricity and magnetism, James Maxwell created
    the elegant mathematical formalism, Maxwell’s Equations, in the 19th century. Maxwell’s Equations contain
    integrated information regarding the fundamental properties of electricity and magnetism with their unique
    properties. It has both attractive and repulsive properties. For instance, there is an attractive force between
    opposite electric charges and a repulsive force between the same charges. It is a somewhat long-ranged
    force but much weaker than the strong force. The electromagnetic spectrum ranges from radio waves, which
    have long wavelengths and low frequencies, to gamma rays, which have short wavelengths and high
    frequencies. Thus, we can see that visible light, gamma rays, and microwaves are all the same; they are all
    types of electromagnetic radiations and differ only in their wavelengths. The associated particle for the
    electromagnetic force is the photon, and it is known as a massless spin 1 particle.

    The last fundamental force is gravity. We will study this force in detail in the next unit. The gravitational force
    works well on a larger scale and is long-ranged, and it has a distinctive property. It is always attractive, not
    repulsive. The gravitational force is well-described by Newton’s universal law of gravitation. It states that the
    attractive force between two objects in the universe is proportional to the product of the mass of one object
    and that of other objects and is inversely proportional to the square of the separation between two objects.
    The associated particle for the gravitational force is known as a massless spin 2 particle, a graviton; however,
    it has not been discovered yet. For a further detailed explanation, please visit the webpage “Fundamental
    Forces”.

    References

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html)

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html)

    PHY 1301, Physics I 5

    UNIT x STUDY GUIDE
    Title

    Hewitt, P. G. (2015). Conceptual physics (12th ed.). Pearson.

    Shipman, J. T., Wilson, J. D., & Todd, A. W. (2009). An introduction to physical science (12th ed.). Cengage

    Learning.

    Suggested Unit Resources

    In order to access the following resource, click the link below.

    You can explore the three laws by Newton with quantitative descriptions at the website below.

    Nave, R. (n.d.). Newton’s first law. HyperPhysics. http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html

    Learning Activities (Nongraded)

    Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
    them. If you have questions, contact your instructor for further guidance and information.

    1. Solve Questions 71–75 under Physics in Biology, Medicine, and Sports in Chapter 3 of your
    eTextbook.

    2. Solve Questions 91–103 under Physics in Biology, Medicine, and Sports in Chapter 4 of your
    eTextbook.

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html)

    – Course Learning Outcomes for Unit II
    – Required Unit Resources
    – Unit Lesson

    – The Concept of Force and Mass
    – Friction
    – Newton’s Three Laws
    – Free Fall
    – The Fundamental Forces
    – References

    – Suggested Unit Resources
    – Learning Activities (Nongraded)

    **## UnitII.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit II

    Upon completion of this unit, students should be able to:

    2. Illustrate the scientific method within everyday situations.
    2.1 Identify the appropriate formulas necessary to solve specific scenario questions.
    2.2 Calculate and analyze the acceleration and the force in various situations.

    3. Explain Newton's laws of motion at work in common phenomena.

    3.1 Solve problems using the law of motion.
    3.2 Explore the relationship between the first and second laws.
    3.3 Identify action-reaction pair in the third law.

    Required Unit Resources

    Chapter 3: Kinematics in Two Dimensions

    Chapter 4: Forces and Newton’s Laws of Motion

    Unit Lesson

    The Concept of Force and Mass

    A force is a push or pull, and it is a vector quantity. Contact forces arise from the physical contact between
    two objects. Noncontact forces are also called action-at-a-distance forces because they arise without physical
    contact between two objects. More sophisticatedly, a force is an external influence on an object that causes it
    to accelerate relative to an inertial reference frame where no acceleration occurs. The direction of the force is
    the same as that of the acceleration.

    Mass is an intrinsic property of an object that measures its resistance to acceleration. That is, mass is a
    measure of the object’s inertia. In dynamics, mass represents the inertia of a body or that body’s resistance to
    any change in its state of motion. Mass is a scalar quantity characterizing a body, and it does not depend on
    the body’s location or state of motion (Cutnell et al., 2022).

    Many people think that weight is the same term for mass; however, it is not! Mass is a measure of the amount
    of material contained in an object and does not depend on any gravitational force acting on it. Mass is
    constant. On the other hand, weight is the gravitational force acting on an object. It depends on the mass of
    the object and the gravitational acceleration of the object attracting it. That is, weight W is the product of the
    mass m and the acceleration g due to gravity.

    Your weight will vary if you measure it on other celestial bodies, but your mass does not change. For
    instance, if you measure your weight on Jupiter, it will be greater than that on Earth. On the moon, however,
    your weight will be less than that on Earth. The larger the gravitational pull, the greater your weight.

    UNIT II STUDY GUIDE
    Newton’s Laws

    W = mg

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Friction

    When an object is in contact with a surface, the surface exerts a force on the object. The perpendicular
    component of the force to the surface is called normal force (please look at Figure 4.13 in section 4.8 in
    Chapter 4 of your eTextbook) and the parallel component of the force to the surface is called frictional force,
    or simply friction. In order to initiate the motion of the object, the applied force must overcome the friction. The
    direction of the friction is opposite to that of the applied force. The frictional force is proportional to the normal
    force. Usually, the friction increases on an uneven surface.

    In general, there are two kinds of frictional forces: static and kinetic. The static friction prevents an object from
    moving. In order to move the object, the applied force must overcome the static friction. Once the object is in
    motion, the kinetic friction is now an obstacle for the moving object (Cutnell et al., 2022). Be sure to review the
    coefficients of friction for various surfaces in Table 4.2 located in section 4.9 of Chapter 4 in your eTextbook.

    Newton’s Three Laws

    Newton’s three laws are about forces that change the motion of objects and are based on total momentum
    conservation. In a simple expression, momentum is a quantity defined as mass multiplied by velocity. We will
    learn about momentum in detail in Unit IV.

    The first law is the law of inertia. This is the same law by Galileo’s inertia interpretation. Unless there is no
    external force, an object that is at rest stays at rest, and one that is in motion continues to move (Shipman,
    Wilson, & Todd, 2009). That is, there is a tendency for an object to keep its original property. The inertia of an
    object is measured by its mass. The more massive the object, the more inertia it has. For instance, when a
    bus driver stops suddenly, a man in the rear seat moves forward. This is because the man’s inertia that was
    previously in motion wants to stay in motion. In fact, this law of inertia is a special case of Newton’s second
    law.

    The second law is the law of motion or acceleration. Force (F) = mass (m) x acceleration (a) (Shipman et al.,
    2009). This is the most useful formula because it can describe the dynamics of motion in detail. We can
    picture how velocities change with respect to time when forces are applied. Simply, it can be expressed with
    the mathematical formula F = ma.

    The SI unit of force is the mass unit (kg) times the acceleration unit (m/s2), and thus [kg m/s2], which is
    defined as [N]. That is, 1 newton = 1 N = 1 kg m/s2

    Here, the force is summed over all applied forces exerted on an object. It accelerates the object relative to an
    inertial reference frame. Remember that the directions of force and acceleration are identical. Mass that
    opposes acceleration is an intrinsic property of an object and is a gauge of inertia. Acceleration is a change of
    velocity. That is, acceleration indicates how much the velocity has been increased or decreased during a
    certain amount of time. If the velocity is constant, which means the acceleration is zero, then there is no
    applied force. This is exactly the case for Newton’s first law. There is no distinction between an object that is
    in motion with a constant velocity and one that is at rest. One example of accelerated motion is circular
    motion. Even though the speed of the object is constant, the direction of the movement is changing, so the
    velocity is not constant.

    F = ma

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    The third law is the law of action-reaction. The forces exist in pairs. The magnitudes of forces are equal, but
    their directions are opposite (Shipman et al., 2009). That is, every action has an equal and opposite reaction.
    For example, when you push a wall on a slippery floor, you feel that your feet are moving backward. The push
    force exerted by you is the same as that by the wall. That is, the two forces are equal in magnitudes, but
    opposite in direction. Review Example 4 located in section 4.9 of Chapter 4 in your eTextbook (Cutnell et al.,
    2022).

    Sample Question 1: A lunar exploration vehicle was made by a research team, and it weighs about
    six times more than it will on the moon. In order to have the same acceleration on the moon as well
    as on the Earth, what will be the net force acting on the vehicle? Will it be greater than, less than, or
    the same as that required on Earth?

    Solution: According to Newton’s second law, the net force is F = ma. Here, m is the vehicle’s mass
    and a is the acceleration. The net force depends on the mass for a constant acceleration. Here, the
    mass of the vehicle is constant. Therefore, the same net force would be required.

    Sample Question 2: You and your friend are pushing a box in the same direction. Its mass is 200
    kg. You apply 20 N to the box, while your friend applies 30 N. A force created by friction is 40 N in the
    opposite direction. What is the acceleration of the box?

    Solution: There are three forces acting in this system, and the net force should be added up. Two of
    them are acting toward the same direction, and the remaining is acting in the opposite direction, as
    you can see below. Thus, the net force F = 30 N + 20 N – 40 N = 10 N. From Newton’s 2nd law, F =
    ma. The mass of box is given by 200 kg. Therefore, the acceleration a = F / m = 10 N / 200 kg = 0.05
    m/s2

    Sample Question 3: An astronaut pushes on the spacecraft with a force of 40 N near the
    International Space Station. His mass is 90 kg, and the spacecraft’s mass is 10,000 kg. What are
    the accelerations of the space craft and the astronaut?

    Solution: According to Newton’s third law, when the astronaut applies the force to the spacecraft,
    the spacecraft also applies the same amount of force to the astronaut. That is, the magnitudes of
    the two forces are identical, but the directions are different.

    From the second law of Newton, the acceleration of the spacecraft is as = 40 N / 10,000 kg = 0.004
    m/s2. The acceleration of the astronaut is aa = -40 N / 90 kg = -0.44 m/s2. Note that the magnitudes
    of the action and reaction forces are always equal, but they do not have the same acceleration
    unless they have the same mass.

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    Free Fall

    Every object on Earth falls downward because of Earth’s gravity. Earth’s gravity points to the center of Earth.
    If there is no air, all of the falling objects should experience the same acceleration. Recall from the Unit I
    Lesson that in order to prove this, an astronaut, David Scott of the Apollo 15 mission, performed the
    experiment on the moon.

    When we ignore the air resistance and assume the constant acceleration, we say that the falling object is in
    free fall. The acceleration of the freely falling body is due to gravitational acceleration g, and it is about 10
    m/s2 on the surface of the Earth. However, the object is at considerably high altitude, the acceleration due to
    gravity is not constant. The gravitational acceleration is inversely proportional to the square of the distance
    from the center of the Earth to the object.

    The Fundamental Forces

    All the different forces observed in nature can be explained in terms of four basic interactions that occur
    between elementary particles: gravitational force, electromagnetic force, strong nuclear force, and weak
    nuclear force. The associated particles for the strong force are mainly gluons and pi nucleons.

    The main role of the strong force is to hold the nuclei of atoms together. Principally, it is attractive; however, it
    can sometimes act in a repulsive way with proper conditions. The strong nuclear interaction has a very
    powerful strength of the force, but it is very short-ranged (about 10-13 cm). The attractive force only works well
    within the size of the nucleus. The role of a neutron is very important to maintain the structure of the nucleus.
    A repulsive force exists between two protons because they have the same charge. Remember that in the
    influence of electrostatic force two like charges repel, and two opposite charges attract. Thus, more neutrons
    than protons are needed to keep the stable nucleus as the atomic number increases. The maximum proton
    number is 83 with 126 neutrons for the stable structure in nature. All nuclei that have an atomic number
    greater than 83 are unstable because the binding force to hold the nuclei is weak and disintegrates or
    rearranges its structure. This is known as radioactivity. The radioactive decay or neutrino interactions can be
    explained by the weak force, and its strength is weak with a short range. The associated particles for the
    weak nuclear force are known as intermediated vector bosons such as W+, W-, and Z0. They are spin 1
    particles, and their masses are greater than 80 GeV.

    The electromagnetic force is responsible for electric and magnetic effects. The phenomena of electricity and
    magnetism look different; however, they are very closely related to each other. A changing electric field can
    create a magnetic field and vice versa. In order to combine electricity and magnetism, James Maxwell created
    the elegant mathematical formalism, Maxwell’s Equations, in the 19th century. Maxwell’s Equations contain
    integrated information regarding the fundamental properties of electricity and magnetism with their unique
    properties. It has both attractive and repulsive properties. For instance, there is an attractive force between
    opposite electric charges and a repulsive force between the same charges. It is a somewhat long-ranged
    force but much weaker than the strong force. The electromagnetic spectrum ranges from radio waves, which
    have long wavelengths and low frequencies, to gamma rays, which have short wavelengths and high
    frequencies. Thus, we can see that visible light, gamma rays, and microwaves are all the same; they are all
    types of electromagnetic radiations and differ only in their wavelengths. The associated particle for the
    electromagnetic force is the photon, and it is known as a massless spin 1 particle.

    The last fundamental force is gravity. We will study this force in detail in the next unit. The gravitational force
    works well on a larger scale and is long-ranged, and it has a distinctive property. It is always attractive, not
    repulsive. The gravitational force is well-described by Newton’s universal law of gravitation. It states that the
    attractive force between two objects in the universe is proportional to the product of the mass of one object
    and that of other objects and is inversely proportional to the square of the separation between two objects.
    The associated particle for the gravitational force is known as a massless spin 2 particle, a graviton; however,
    it has not been discovered yet. For a further detailed explanation, please visit the webpage “Fundamental
    Forces”.

    References

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html)

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html)

    PHY 1301, Physics I 5

    UNIT x STUDY GUIDE
    Title

    Hewitt, P. G. (2015). Conceptual physics (12th ed.). Pearson.

    Shipman, J. T., Wilson, J. D., & Todd, A. W. (2009). An introduction to physical science (12th ed.). Cengage

    Learning.

    Suggested Unit Resources

    In order to access the following resource, click the link below.

    You can explore the three laws by Newton with quantitative descriptions at the website below.

    Nave, R. (n.d.). Newton’s first law. HyperPhysics. http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html

    Learning Activities (Nongraded)

    Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
    them. If you have questions, contact your instructor for further guidance and information.

    1. Solve Questions 71–75 under Physics in Biology, Medicine, and Sports in Chapter 3 of your
    eTextbook.

    2. Solve Questions 91–103 under Physics in Biology, Medicine, and Sports in Chapter 4 of your
    eTextbook.

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html)

    – Course Learning Outcomes for Unit II
    – Required Unit Resources
    – Unit Lesson

    – The Concept of Force and Mass
    – Friction
    – Newton’s Three Laws
    – Free Fall
    – The Fundamental Forces
    – References

    – Suggested Unit Resources
    – Learning Activities (Nongraded)

    **## DB2.docx**

    DB 2: Your initial post should be at least 200 words in length.

    Discuss the main difference between science and pseudoscience. What are some examples of pseudoscience, and what are some of their claims? What makes pseudoscience not real science? Is there your way to determine if it is a real science or not? Share your ideas with your classmates.

    Reply 1:

    Reply 2:
    **Monica Thompson**

    In Latin, science is scientia, meaning "knowledge". Science has become the most respectable kind of knowledge to date. It uses reliable data that has been proven, tested, and observed repeatedly, yielding the same results. Pseudoscience is a claim or system that appears scientific but is not based on testable, measurable, and repeatable evidence. It is literally armchair philosophy, so to speak; it is what one believes. It lacks solid evidence and can’t be tested or proven wrong. In other words, it is a belief or claim.

    An example of pseudoscience is eugenics. During a certain era, this movement claimed that by regulating who could have children would lead to a “better” human race. Can you guess when this occurred? How twisted and demented this was, and to think this practice may still be occurring today, without our knowledge. Obviously, in this example of pseudoscience, there was no solid evidence to support it as scientific. It was more racially biased and agenda-driven based on social and political beliefs, keyword beliefs. There was no objective evidence to support this claim, and in the long run, it only harmed the human race. My background is diverse, as is true for most people, and research shows that intelligence is not defined by race but by each individual.

    The use of observation and reasoning dates back to Aristotle, around 300 BCE, and the modern scientific method took hold in the early 1600’s, when Galileo and Francis Bacon applied it to their discoveries. So, the clear difference between science and pseudoscience is the scientific method. Science clearly uses the scientific method, and pseudoscience does not.

    Reply 2:
    **Victoria Papuchis**

    Top of Form

    There is a clear difference between science and pseudoscience, and the biggest distinction is that pseudoscience does not use the scientific method. The scientific method is a systematic, evidencebased process used to investigate questions, test ideas, and build reliable knowledge. Science depends on this structure because it reduces bias, follows logical steps, and allows researchers to test, analyze, and draw conclusions that can be repeated and verified by others.

    This is what makes scientific findings trustworthy and continually improving.

    From my understanding, pseudoscience is built more on personal beliefs, unsupported claims, and a refusal to accept criticism. These ideas often remain unchanged even when they have been proven wrong. Instead of being open to new evidence, supporters of pseudoscience tend to shift their explanations to avoid admitting that the claim doesn’t hold up. This resistance to scrutiny is exactly what makes pseudoscience invalid and frustrating. It often uses vague language, relies on testimonials instead of data, and cannot be independently verified. In real science, results must be consistent no matter who performs the experiment.

    Thinking back to my own science classes, from elementary school through my current college studies, the scientific method has always been required. Every experiment or project involved forming a hypothesis, collecting data, and drawing evidencebased conclusions. That consistency is what separates true scientific practice from pseudoscience.

    Bottom of Form

    Bottom of Form

    **## UnitII.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit II

    Upon completion of this unit, students should be able to:

    2. Illustrate the scientific method within everyday situations.
    2.1 Identify the appropriate formulas necessary to solve specific scenario questions.
    2.2 Calculate and analyze the acceleration and the force in various situations.

    3. Explain Newton's laws of motion at work in common phenomena.

    3.1 Solve problems using the law of motion.
    3.2 Explore the relationship between the first and second laws.
    3.3 Identify action-reaction pair in the third law.

    Required Unit Resources

    Chapter 3: Kinematics in Two Dimensions

    Chapter 4: Forces and Newton’s Laws of Motion

    Unit Lesson

    The Concept of Force and Mass

    A force is a push or pull, and it is a vector quantity. Contact forces arise from the physical contact between
    two objects. Noncontact forces are also called action-at-a-distance forces because they arise without physical
    contact between two objects. More sophisticatedly, a force is an external influence on an object that causes it
    to accelerate relative to an inertial reference frame where no acceleration occurs. The direction of the force is
    the same as that of the acceleration.

    Mass is an intrinsic property of an object that measures its resistance to acceleration. That is, mass is a
    measure of the object’s inertia. In dynamics, mass represents the inertia of a body or that body’s resistance to
    any change in its state of motion. Mass is a scalar quantity characterizing a body, and it does not depend on
    the body’s location or state of motion (Cutnell et al., 2022).

    Many people think that weight is the same term for mass; however, it is not! Mass is a measure of the amount
    of material contained in an object and does not depend on any gravitational force acting on it. Mass is
    constant. On the other hand, weight is the gravitational force acting on an object. It depends on the mass of
    the object and the gravitational acceleration of the object attracting it. That is, weight W is the product of the
    mass m and the acceleration g due to gravity.

    Your weight will vary if you measure it on other celestial bodies, but your mass does not change. For
    instance, if you measure your weight on Jupiter, it will be greater than that on Earth. On the moon, however,
    your weight will be less than that on Earth. The larger the gravitational pull, the greater your weight.

    UNIT II STUDY GUIDE
    Newton’s Laws

    W = mg

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Friction

    When an object is in contact with a surface, the surface exerts a force on the object. The perpendicular
    component of the force to the surface is called normal force (please look at Figure 4.13 in section 4.8 in
    Chapter 4 of your eTextbook) and the parallel component of the force to the surface is called frictional force,
    or simply friction. In order to initiate the motion of the object, the applied force must overcome the friction. The
    direction of the friction is opposite to that of the applied force. The frictional force is proportional to the normal
    force. Usually, the friction increases on an uneven surface.

    In general, there are two kinds of frictional forces: static and kinetic. The static friction prevents an object from
    moving. In order to move the object, the applied force must overcome the static friction. Once the object is in
    motion, the kinetic friction is now an obstacle for the moving object (Cutnell et al., 2022). Be sure to review the
    coefficients of friction for various surfaces in Table 4.2 located in section 4.9 of Chapter 4 in your eTextbook.

    Newton’s Three Laws

    Newton’s three laws are about forces that change the motion of objects and are based on total momentum
    conservation. In a simple expression, momentum is a quantity defined as mass multiplied by velocity. We will
    learn about momentum in detail in Unit IV.

    The first law is the law of inertia. This is the same law by Galileo’s inertia interpretation. Unless there is no
    external force, an object that is at rest stays at rest, and one that is in motion continues to move (Shipman,
    Wilson, & Todd, 2009). That is, there is a tendency for an object to keep its original property. The inertia of an
    object is measured by its mass. The more massive the object, the more inertia it has. For instance, when a
    bus driver stops suddenly, a man in the rear seat moves forward. This is because the man’s inertia that was
    previously in motion wants to stay in motion. In fact, this law of inertia is a special case of Newton’s second
    law.

    The second law is the law of motion or acceleration. Force (F) = mass (m) x acceleration (a) (Shipman et al.,
    2009). This is the most useful formula because it can describe the dynamics of motion in detail. We can
    picture how velocities change with respect to time when forces are applied. Simply, it can be expressed with
    the mathematical formula F = ma.

    The SI unit of force is the mass unit (kg) times the acceleration unit (m/s2), and thus [kg m/s2], which is
    defined as [N]. That is, 1 newton = 1 N = 1 kg m/s2

    Here, the force is summed over all applied forces exerted on an object. It accelerates the object relative to an
    inertial reference frame. Remember that the directions of force and acceleration are identical. Mass that
    opposes acceleration is an intrinsic property of an object and is a gauge of inertia. Acceleration is a change of
    velocity. That is, acceleration indicates how much the velocity has been increased or decreased during a
    certain amount of time. If the velocity is constant, which means the acceleration is zero, then there is no
    applied force. This is exactly the case for Newton’s first law. There is no distinction between an object that is
    in motion with a constant velocity and one that is at rest. One example of accelerated motion is circular
    motion. Even though the speed of the object is constant, the direction of the movement is changing, so the
    velocity is not constant.

    F = ma

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    The third law is the law of action-reaction. The forces exist in pairs. The magnitudes of forces are equal, but
    their directions are opposite (Shipman et al., 2009). That is, every action has an equal and opposite reaction.
    For example, when you push a wall on a slippery floor, you feel that your feet are moving backward. The push
    force exerted by you is the same as that by the wall. That is, the two forces are equal in magnitudes, but
    opposite in direction. Review Example 4 located in section 4.9 of Chapter 4 in your eTextbook (Cutnell et al.,
    2022).

    Sample Question 1: A lunar exploration vehicle was made by a research team, and it weighs about
    six times more than it will on the moon. In order to have the same acceleration on the moon as well
    as on the Earth, what will be the net force acting on the vehicle? Will it be greater than, less than, or
    the same as that required on Earth?

    Solution: According to Newton’s second law, the net force is F = ma. Here, m is the vehicle’s mass
    and a is the acceleration. The net force depends on the mass for a constant acceleration. Here, the
    mass of the vehicle is constant. Therefore, the same net force would be required.

    Sample Question 2: You and your friend are pushing a box in the same direction. Its mass is 200
    kg. You apply 20 N to the box, while your friend applies 30 N. A force created by friction is 40 N in the
    opposite direction. What is the acceleration of the box?

    Solution: There are three forces acting in this system, and the net force should be added up. Two of
    them are acting toward the same direction, and the remaining is acting in the opposite direction, as
    you can see below. Thus, the net force F = 30 N + 20 N – 40 N = 10 N. From Newton’s 2nd law, F =
    ma. The mass of box is given by 200 kg. Therefore, the acceleration a = F / m = 10 N / 200 kg = 0.05
    m/s2

    Sample Question 3: An astronaut pushes on the spacecraft with a force of 40 N near the
    International Space Station. His mass is 90 kg, and the spacecraft’s mass is 10,000 kg. What are
    the accelerations of the space craft and the astronaut?

    Solution: According to Newton’s third law, when the astronaut applies the force to the spacecraft,
    the spacecraft also applies the same amount of force to the astronaut. That is, the magnitudes of
    the two forces are identical, but the directions are different.

    From the second law of Newton, the acceleration of the spacecraft is as = 40 N / 10,000 kg = 0.004
    m/s2. The acceleration of the astronaut is aa = -40 N / 90 kg = -0.44 m/s2. Note that the magnitudes
    of the action and reaction forces are always equal, but they do not have the same acceleration
    unless they have the same mass.

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    Free Fall

    Every object on Earth falls downward because of Earth’s gravity. Earth’s gravity points to the center of Earth.
    If there is no air, all of the falling objects should experience the same acceleration. Recall from the Unit I
    Lesson that in order to prove this, an astronaut, David Scott of the Apollo 15 mission, performed the
    experiment on the moon.

    When we ignore the air resistance and assume the constant acceleration, we say that the falling object is in
    free fall. The acceleration of the freely falling body is due to gravitational acceleration g, and it is about 10
    m/s2 on the surface of the Earth. However, the object is at considerably high altitude, the acceleration due to
    gravity is not constant. The gravitational acceleration is inversely proportional to the square of the distance
    from the center of the Earth to the object.

    The Fundamental Forces

    All the different forces observed in nature can be explained in terms of four basic interactions that occur
    between elementary particles: gravitational force, electromagnetic force, strong nuclear force, and weak
    nuclear force. The associated particles for the strong force are mainly gluons and pi nucleons.

    The main role of the strong force is to hold the nuclei of atoms together. Principally, it is attractive; however, it
    can sometimes act in a repulsive way with proper conditions. The strong nuclear interaction has a very
    powerful strength of the force, but it is very short-ranged (about 10-13 cm). The attractive force only works well
    within the size of the nucleus. The role of a neutron is very important to maintain the structure of the nucleus.
    A repulsive force exists between two protons because they have the same charge. Remember that in the
    influence of electrostatic force two like charges repel, and two opposite charges attract. Thus, more neutrons
    than protons are needed to keep the stable nucleus as the atomic number increases. The maximum proton
    number is 83 with 126 neutrons for the stable structure in nature. All nuclei that have an atomic number
    greater than 83 are unstable because the binding force to hold the nuclei is weak and disintegrates or
    rearranges its structure. This is known as radioactivity. The radioactive decay or neutrino interactions can be
    explained by the weak force, and its strength is weak with a short range. The associated particles for the
    weak nuclear force are known as intermediated vector bosons such as W+, W-, and Z0. They are spin 1
    particles, and their masses are greater than 80 GeV.

    The electromagnetic force is responsible for electric and magnetic effects. The phenomena of electricity and
    magnetism look different; however, they are very closely related to each other. A changing electric field can
    create a magnetic field and vice versa. In order to combine electricity and magnetism, James Maxwell created
    the elegant mathematical formalism, Maxwell’s Equations, in the 19th century. Maxwell’s Equations contain
    integrated information regarding the fundamental properties of electricity and magnetism with their unique
    properties. It has both attractive and repulsive properties. For instance, there is an attractive force between
    opposite electric charges and a repulsive force between the same charges. It is a somewhat long-ranged
    force but much weaker than the strong force. The electromagnetic spectrum ranges from radio waves, which
    have long wavelengths and low frequencies, to gamma rays, which have short wavelengths and high
    frequencies. Thus, we can see that visible light, gamma rays, and microwaves are all the same; they are all
    types of electromagnetic radiations and differ only in their wavelengths. The associated particle for the
    electromagnetic force is the photon, and it is known as a massless spin 1 particle.

    The last fundamental force is gravity. We will study this force in detail in the next unit. The gravitational force
    works well on a larger scale and is long-ranged, and it has a distinctive property. It is always attractive, not
    repulsive. The gravitational force is well-described by Newton’s universal law of gravitation. It states that the
    attractive force between two objects in the universe is proportional to the product of the mass of one object
    and that of other objects and is inversely proportional to the square of the separation between two objects.
    The associated particle for the gravitational force is known as a massless spin 2 particle, a graviton; however,
    it has not been discovered yet. For a further detailed explanation, please visit the webpage “Fundamental
    Forces”.

    References

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html)

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html)

    PHY 1301, Physics I 5

    UNIT x STUDY GUIDE
    Title

    Hewitt, P. G. (2015). Conceptual physics (12th ed.). Pearson.

    Shipman, J. T., Wilson, J. D., & Todd, A. W. (2009). An introduction to physical science (12th ed.). Cengage

    Learning.

    Suggested Unit Resources

    In order to access the following resource, click the link below.

    You can explore the three laws by Newton with quantitative descriptions at the website below.

    Nave, R. (n.d.). Newton’s first law. HyperPhysics. http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html

    Learning Activities (Nongraded)

    Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
    them. If you have questions, contact your instructor for further guidance and information.

    1. Solve Questions 71–75 under Physics in Biology, Medicine, and Sports in Chapter 3 of your
    eTextbook.

    2. Solve Questions 91–103 under Physics in Biology, Medicine, and Sports in Chapter 4 of your
    eTextbook.

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html)

    – Course Learning Outcomes for Unit II
    – Required Unit Resources
    – Unit Lesson

    – The Concept of Force and Mass
    – Friction
    – Newton’s Three Laws
    – Free Fall
    – The Fundamental Forces
    – References

    – Suggested Unit Resources
    – Learning Activities (Nongraded)

    **## DB2.docx**

    DB 2: Your initial post should be at least 200 words in length.

    Discuss the main difference between science and pseudoscience. What are some examples of pseudoscience, and what are some of their claims? What makes pseudoscience not real science? Is there your way to determine if it is a real science or not? Share your ideas with your classmates.

    Reply 1:

    Reply 2:
    **Monica Thompson**

    In Latin, science is scientia, meaning "knowledge". Science has become the most respectable kind of knowledge to date. It uses reliable data that has been proven, tested, and observed repeatedly, yielding the same results. Pseudoscience is a claim or system that appears scientific but is not based on testable, measurable, and repeatable evidence. It is literally armchair philosophy, so to speak; it is what one believes. It lacks solid evidence and can’t be tested or proven wrong. In other words, it is a belief or claim.

    An example of pseudoscience is eugenics. During a certain era, this movement claimed that by regulating who could have children would lead to a “better” human race. Can you guess when this occurred? How twisted and demented this was, and to think this practice may still be occurring today, without our knowledge. Obviously, in this example of pseudoscience, there was no solid evidence to support it as scientific. It was more racially biased and agenda-driven based on social and political beliefs, keyword beliefs. There was no objective evidence to support this claim, and in the long run, it only harmed the human race. My background is diverse, as is true for most people, and research shows that intelligence is not defined by race but by each individual.

    The use of observation and reasoning dates back to Aristotle, around 300 BCE, and the modern scientific method took hold in the early 1600’s, when Galileo and Francis Bacon applied it to their discoveries. So, the clear difference between science and pseudoscience is the scientific method. Science clearly uses the scientific method, and pseudoscience does not.

    Reply 2:
    **Victoria Papuchis**

    Top of Form

    There is a clear difference between science and pseudoscience, and the biggest distinction is that pseudoscience does not use the scientific method. The scientific method is a systematic, evidencebased process used to investigate questions, test ideas, and build reliable knowledge. Science depends on this structure because it reduces bias, follows logical steps, and allows researchers to test, analyze, and draw conclusions that can be repeated and verified by others.

    This is what makes scientific findings trustworthy and continually improving.

    From my understanding, pseudoscience is built more on personal beliefs, unsupported claims, and a refusal to accept criticism. These ideas often remain unchanged even when they have been proven wrong. Instead of being open to new evidence, supporters of pseudoscience tend to shift their explanations to avoid admitting that the claim doesn’t hold up. This resistance to scrutiny is exactly what makes pseudoscience invalid and frustrating. It often uses vague language, relies on testimonials instead of data, and cannot be independently verified. In real science, results must be consistent no matter who performs the experiment.

    Thinking back to my own science classes, from elementary school through my current college studies, the scientific method has always been required. Every experiment or project involved forming a hypothesis, collecting data, and drawing evidencebased conclusions. That consistency is what separates true scientific practice from pseudoscience.

    Bottom of Form

    Bottom of Form

    **## UnitII.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit II

    Upon completion of this unit, students should be able to:

    2. Illustrate the scientific method within everyday situations.
    2.1 Identify the appropriate formulas necessary to solve specific scenario questions.
    2.2 Calculate and analyze the acceleration and the force in various situations.

    3. Explain Newton's laws of motion at work in common phenomena.

    3.1 Solve problems using the law of motion.
    3.2 Explore the relationship between the first and second laws.
    3.3 Identify action-reaction pair in the third law.

    Required Unit Resources

    Chapter 3: Kinematics in Two Dimensions

    Chapter 4: Forces and Newton’s Laws of Motion

    Unit Lesson

    The Concept of Force and Mass

    A force is a push or pull, and it is a vector quantity. Contact forces arise from the physical contact between
    two objects. Noncontact forces are also called action-at-a-distance forces because they arise without physical
    contact between two objects. More sophisticatedly, a force is an external influence on an object that causes it
    to accelerate relative to an inertial reference frame where no acceleration occurs. The direction of the force is
    the same as that of the acceleration.

    Mass is an intrinsic property of an object that measures its resistance to acceleration. That is, mass is a
    measure of the object’s inertia. In dynamics, mass represents the inertia of a body or that body’s resistance to
    any change in its state of motion. Mass is a scalar quantity characterizing a body, and it does not depend on
    the body’s location or state of motion (Cutnell et al., 2022).

    Many people think that weight is the same term for mass; however, it is not! Mass is a measure of the amount
    of material contained in an object and does not depend on any gravitational force acting on it. Mass is
    constant. On the other hand, weight is the gravitational force acting on an object. It depends on the mass of
    the object and the gravitational acceleration of the object attracting it. That is, weight W is the product of the
    mass m and the acceleration g due to gravity.

    Your weight will vary if you measure it on other celestial bodies, but your mass does not change. For
    instance, if you measure your weight on Jupiter, it will be greater than that on Earth. On the moon, however,
    your weight will be less than that on Earth. The larger the gravitational pull, the greater your weight.

    UNIT II STUDY GUIDE
    Newton’s Laws

    W = mg

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Friction

    When an object is in contact with a surface, the surface exerts a force on the object. The perpendicular
    component of the force to the surface is called normal force (please look at Figure 4.13 in section 4.8 in
    Chapter 4 of your eTextbook) and the parallel component of the force to the surface is called frictional force,
    or simply friction. In order to initiate the motion of the object, the applied force must overcome the friction. The
    direction of the friction is opposite to that of the applied force. The frictional force is proportional to the normal
    force. Usually, the friction increases on an uneven surface.

    In general, there are two kinds of frictional forces: static and kinetic. The static friction prevents an object from
    moving. In order to move the object, the applied force must overcome the static friction. Once the object is in
    motion, the kinetic friction is now an obstacle for the moving object (Cutnell et al., 2022). Be sure to review the
    coefficients of friction for various surfaces in Table 4.2 located in section 4.9 of Chapter 4 in your eTextbook.

    Newton’s Three Laws

    Newton’s three laws are about forces that change the motion of objects and are based on total momentum
    conservation. In a simple expression, momentum is a quantity defined as mass multiplied by velocity. We will
    learn about momentum in detail in Unit IV.

    The first law is the law of inertia. This is the same law by Galileo’s inertia interpretation. Unless there is no
    external force, an object that is at rest stays at rest, and one that is in motion continues to move (Shipman,
    Wilson, & Todd, 2009). That is, there is a tendency for an object to keep its original property. The inertia of an
    object is measured by its mass. The more massive the object, the more inertia it has. For instance, when a
    bus driver stops suddenly, a man in the rear seat moves forward. This is because the man’s inertia that was
    previously in motion wants to stay in motion. In fact, this law of inertia is a special case of Newton’s second
    law.

    The second law is the law of motion or acceleration. Force (F) = mass (m) x acceleration (a) (Shipman et al.,
    2009). This is the most useful formula because it can describe the dynamics of motion in detail. We can
    picture how velocities change with respect to time when forces are applied. Simply, it can be expressed with
    the mathematical formula F = ma.

    The SI unit of force is the mass unit (kg) times the acceleration unit (m/s2), and thus [kg m/s2], which is
    defined as [N]. That is, 1 newton = 1 N = 1 kg m/s2

    Here, the force is summed over all applied forces exerted on an object. It accelerates the object relative to an
    inertial reference frame. Remember that the directions of force and acceleration are identical. Mass that
    opposes acceleration is an intrinsic property of an object and is a gauge of inertia. Acceleration is a change of
    velocity. That is, acceleration indicates how much the velocity has been increased or decreased during a
    certain amount of time. If the velocity is constant, which means the acceleration is zero, then there is no
    applied force. This is exactly the case for Newton’s first law. There is no distinction between an object that is
    in motion with a constant velocity and one that is at rest. One example of accelerated motion is circular
    motion. Even though the speed of the object is constant, the direction of the movement is changing, so the
    velocity is not constant.

    F = ma

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    The third law is the law of action-reaction. The forces exist in pairs. The magnitudes of forces are equal, but
    their directions are opposite (Shipman et al., 2009). That is, every action has an equal and opposite reaction.
    For example, when you push a wall on a slippery floor, you feel that your feet are moving backward. The push
    force exerted by you is the same as that by the wall. That is, the two forces are equal in magnitudes, but
    opposite in direction. Review Example 4 located in section 4.9 of Chapter 4 in your eTextbook (Cutnell et al.,
    2022).

    Sample Question 1: A lunar exploration vehicle was made by a research team, and it weighs about
    six times more than it will on the moon. In order to have the same acceleration on the moon as well
    as on the Earth, what will be the net force acting on the vehicle? Will it be greater than, less than, or
    the same as that required on Earth?

    Solution: According to Newton’s second law, the net force is F = ma. Here, m is the vehicle’s mass
    and a is the acceleration. The net force depends on the mass for a constant acceleration. Here, the
    mass of the vehicle is constant. Therefore, the same net force would be required.

    Sample Question 2: You and your friend are pushing a box in the same direction. Its mass is 200
    kg. You apply 20 N to the box, while your friend applies 30 N. A force created by friction is 40 N in the
    opposite direction. What is the acceleration of the box?

    Solution: There are three forces acting in this system, and the net force should be added up. Two of
    them are acting toward the same direction, and the remaining is acting in the opposite direction, as
    you can see below. Thus, the net force F = 30 N + 20 N – 40 N = 10 N. From Newton’s 2nd law, F =
    ma. The mass of box is given by 200 kg. Therefore, the acceleration a = F / m = 10 N / 200 kg = 0.05
    m/s2

    Sample Question 3: An astronaut pushes on the spacecraft with a force of 40 N near the
    International Space Station. His mass is 90 kg, and the spacecraft’s mass is 10,000 kg. What are
    the accelerations of the space craft and the astronaut?

    Solution: According to Newton’s third law, when the astronaut applies the force to the spacecraft,
    the spacecraft also applies the same amount of force to the astronaut. That is, the magnitudes of
    the two forces are identical, but the directions are different.

    From the second law of Newton, the acceleration of the spacecraft is as = 40 N / 10,000 kg = 0.004
    m/s2. The acceleration of the astronaut is aa = -40 N / 90 kg = -0.44 m/s2. Note that the magnitudes
    of the action and reaction forces are always equal, but they do not have the same acceleration
    unless they have the same mass.

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    Free Fall

    Every object on Earth falls downward because of Earth’s gravity. Earth’s gravity points to the center of Earth.
    If there is no air, all of the falling objects should experience the same acceleration. Recall from the Unit I
    Lesson that in order to prove this, an astronaut, David Scott of the Apollo 15 mission, performed the
    experiment on the moon.

    When we ignore the air resistance and assume the constant acceleration, we say that the falling object is in
    free fall. The acceleration of the freely falling body is due to gravitational acceleration g, and it is about 10
    m/s2 on the surface of the Earth. However, the object is at considerably high altitude, the acceleration due to
    gravity is not constant. The gravitational acceleration is inversely proportional to the square of the distance
    from the center of the Earth to the object.

    The Fundamental Forces

    All the different forces observed in nature can be explained in terms of four basic interactions that occur
    between elementary particles: gravitational force, electromagnetic force, strong nuclear force, and weak
    nuclear force. The associated particles for the strong force are mainly gluons and pi nucleons.

    The main role of the strong force is to hold the nuclei of atoms together. Principally, it is attractive; however, it
    can sometimes act in a repulsive way with proper conditions. The strong nuclear interaction has a very
    powerful strength of the force, but it is very short-ranged (about 10-13 cm). The attractive force only works well
    within the size of the nucleus. The role of a neutron is very important to maintain the structure of the nucleus.
    A repulsive force exists between two protons because they have the same charge. Remember that in the
    influence of electrostatic force two like charges repel, and two opposite charges attract. Thus, more neutrons
    than protons are needed to keep the stable nucleus as the atomic number increases. The maximum proton
    number is 83 with 126 neutrons for the stable structure in nature. All nuclei that have an atomic number
    greater than 83 are unstable because the binding force to hold the nuclei is weak and disintegrates or
    rearranges its structure. This is known as radioactivity. The radioactive decay or neutrino interactions can be
    explained by the weak force, and its strength is weak with a short range. The associated particles for the
    weak nuclear force are known as intermediated vector bosons such as W+, W-, and Z0. They are spin 1
    particles, and their masses are greater than 80 GeV.

    The electromagnetic force is responsible for electric and magnetic effects. The phenomena of electricity and
    magnetism look different; however, they are very closely related to each other. A changing electric field can
    create a magnetic field and vice versa. In order to combine electricity and magnetism, James Maxwell created
    the elegant mathematical formalism, Maxwell’s Equations, in the 19th century. Maxwell’s Equations contain
    integrated information regarding the fundamental properties of electricity and magnetism with their unique
    properties. It has both attractive and repulsive properties. For instance, there is an attractive force between
    opposite electric charges and a repulsive force between the same charges. It is a somewhat long-ranged
    force but much weaker than the strong force. The electromagnetic spectrum ranges from radio waves, which
    have long wavelengths and low frequencies, to gamma rays, which have short wavelengths and high
    frequencies. Thus, we can see that visible light, gamma rays, and microwaves are all the same; they are all
    types of electromagnetic radiations and differ only in their wavelengths. The associated particle for the
    electromagnetic force is the photon, and it is known as a massless spin 1 particle.

    The last fundamental force is gravity. We will study this force in detail in the next unit. The gravitational force
    works well on a larger scale and is long-ranged, and it has a distinctive property. It is always attractive, not
    repulsive. The gravitational force is well-described by Newton’s universal law of gravitation. It states that the
    attractive force between two objects in the universe is proportional to the product of the mass of one object
    and that of other objects and is inversely proportional to the square of the separation between two objects.
    The associated particle for the gravitational force is known as a massless spin 2 particle, a graviton; however,
    it has not been discovered yet. For a further detailed explanation, please visit the webpage “Fundamental
    Forces”.

    References

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html)

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html)

    PHY 1301, Physics I 5

    UNIT x STUDY GUIDE
    Title

    Hewitt, P. G. (2015). Conceptual physics (12th ed.). Pearson.

    Shipman, J. T., Wilson, J. D., & Todd, A. W. (2009). An introduction to physical science (12th ed.). Cengage

    Learning.

    Suggested Unit Resources

    In order to access the following resource, click the link below.

    You can explore the three laws by Newton with quantitative descriptions at the website below.

    Nave, R. (n.d.). Newton’s first law. HyperPhysics. http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html

    Learning Activities (Nongraded)

    Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
    them. If you have questions, contact your instructor for further guidance and information.

    1. Solve Questions 71–75 under Physics in Biology, Medicine, and Sports in Chapter 3 of your
    eTextbook.

    2. Solve Questions 91–103 under Physics in Biology, Medicine, and Sports in Chapter 4 of your
    eTextbook.

    [http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html](http://hyperphysics.phy-astr.gsu.edu/hbase/Newt.html)

    – Course Learning Outcomes for Unit II
    – Required Unit Resources
    – Unit Lesson

    – The Concept of Force and Mass
    – Friction
    – Newton’s Three Laws
    – Free Fall
    – The Fundamental Forces
    – References

    – Suggested Unit Resources
    – Learning Activities (Nongraded)

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  • [Solved] Physics DB V

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    See attached **files (3)** – ## DBV.docx – Screenshot2026-04-22194417.png – Screenshot2026-04-22194252.png **## DBV.docx** DB V • Your initial post should be at least 200 words in length. As we have learned in this unit, Hooke’s law cannot perfectly predict the relationship between stress and strain…

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    **files (3)**

    – ## DBV.docx

    – Screenshot2026-04-22194417.png
    – Screenshot2026-04-22194252.png

    **## DBV.docx**

    DB V

    • Your initial post should be at least 200 words in length.

    As we have learned in this unit, Hooke’s law cannot perfectly predict the relationship between stress and strain of materials. See Figure 10.33 in Section 10.8 of Chapter 10 in the eTextbook. Can you provide your opinion why the materials behave like this on the elastic limit? Do you think we need new physics to explain that? Share your idea with other students.

    Response 1:
    Allison Burch

    Top of Form

    Hooke’s Law states that stress and strain are proportional within the elastic limit, meaning a material should return to its original shape when the force is removed. However, in real life, materials don’t always follow this perfectly, especially as they approach the elastic limit. After looking at this concept, I think the main reason materials behave this way is because real materials are not perfectly uniform at the microscopic level. At small levels of stress, atoms in a material stretch like tiny springs, which is why Hooke’s Law works well at first. But as the force increases, the internal structure of the material starts to change. Bonds between atoms begin to stretch unevenly, and some internal defects (like dislocations or imperfections) start to move. This causes the material to gradually stop behaving in a perfectly linear way before it actually reaches its breaking point. That’s why the stress-strain curve starts to curve instead of staying a straight line. I don’t think we need completely new physics to explain this behavior. Instead, it seems more like an extension of what we already know. Classical mechanics and materials science already account for things like plastic deformation, molecular bonding, and material imperfections. So rather than new physics, it feels more like more detailed modeling of what’s already happening at the microscopic level. Overall, Hooke’s Law is still very useful, but it works best as an approximation for small deformations, not for real-world materials under higher stress conditions.

    Bottom of Form

    Response 2: Bailey Phillips

    Hooke’s law is kind of like a “good approximation” for how materials behave, but only at the beginning. When you first stretch or compress something, the atoms inside are just slightly moving from their normal positions. I would compare this to like tiny springs. In that range, everything is neat and predictable. The stress and strain have a straight-line relationship. But once you push the material too fast, past the elastic limit, things get messy. This kind of makes me think of what happens between students and teachers. Teachers are always nice and easy going, until the moment that they are pushed too far. The atoms aren’t just stretching anymore; they start slipping past each other or breaking their original arrangement. I have spent a lot think about this and I have been picturing it like pulling on a paperclip. At first, it springs back, but if you bend it too much, it stays bent. That’s because its internal structure has permanently changed. It’s going through complex changes inside, like bonds rearranging.

    I don’t think we need new physics to explain this. The physics we already know, like atomic forces and material structure, can explain it. Hooke’s law is just a simplified model, and real materials are more complicated than that. It gives us a nice example and jumping off point, but we have the physics to get it done.

    **Screenshot2026-04-22194417.png**

    This file is too large to display. [View in new window](https://www.sweetstudy.com/files/screenshot2026-04-22194417-png-10538499)

    **Screenshot2026-04-22194252.png**

    This file is too large to display. [View in new window](https://www.sweetstudy.com/files/screenshot2026-04-22194252-png)

    **Screenshot2026-04-22194252.png**

    This file is too large to display. [View in new window](https://www.sweetstudy.com/files/screenshot2026-04-22194252-png)

    **## DBV.docx**

    DB V

    • Your initial post should be at least 200 words in length.

    As we have learned in this unit, Hooke’s law cannot perfectly predict the relationship between stress and strain of materials. See Figure 10.33 in Section 10.8 of Chapter 10 in the eTextbook. Can you provide your opinion why the materials behave like this on the elastic limit? Do you think we need new physics to explain that? Share your idea with other students.

    Response 1:
    Allison Burch

    Top of Form

    Hooke’s Law states that stress and strain are proportional within the elastic limit, meaning a material should return to its original shape when the force is removed. However, in real life, materials don’t always follow this perfectly, especially as they approach the elastic limit. After looking at this concept, I think the main reason materials behave this way is because real materials are not perfectly uniform at the microscopic level. At small levels of stress, atoms in a material stretch like tiny springs, which is why Hooke’s Law works well at first. But as the force increases, the internal structure of the material starts to change. Bonds between atoms begin to stretch unevenly, and some internal defects (like dislocations or imperfections) start to move. This causes the material to gradually stop behaving in a perfectly linear way before it actually reaches its breaking point. That’s why the stress-strain curve starts to curve instead of staying a straight line. I don’t think we need completely new physics to explain this behavior. Instead, it seems more like an extension of what we already know. Classical mechanics and materials science already account for things like plastic deformation, molecular bonding, and material imperfections. So rather than new physics, it feels more like more detailed modeling of what’s already happening at the microscopic level. Overall, Hooke’s Law is still very useful, but it works best as an approximation for small deformations, not for real-world materials under higher stress conditions.

    Bottom of Form

    Response 2: Bailey Phillips

    Hooke’s law is kind of like a “good approximation” for how materials behave, but only at the beginning. When you first stretch or compress something, the atoms inside are just slightly moving from their normal positions. I would compare this to like tiny springs. In that range, everything is neat and predictable. The stress and strain have a straight-line relationship. But once you push the material too fast, past the elastic limit, things get messy. This kind of makes me think of what happens between students and teachers. Teachers are always nice and easy going, until the moment that they are pushed too far. The atoms aren’t just stretching anymore; they start slipping past each other or breaking their original arrangement. I have spent a lot think about this and I have been picturing it like pulling on a paperclip. At first, it springs back, but if you bend it too much, it stays bent. That’s because its internal structure has permanently changed. It’s going through complex changes inside, like bonds rearranging.

    I don’t think we need new physics to explain this. The physics we already know, like atomic forces and material structure, can explain it. Hooke’s law is just a simplified model, and real materials are more complicated than that. It gives us a nice example and jumping off point, but we have the physics to get it done.

    **Screenshot2026-04-22194417.png**

    This file is too large to display. [View in new window](https://www.sweetstudy.com/files/screenshot2026-04-22194417-png-10538499)

    **Screenshot2026-04-22194252.png**

    This file is too large to display. [View in new window](https://www.sweetstudy.com/files/screenshot2026-04-22194252-png)

    **## DBV.docx**

    DB V

    • Your initial post should be at least 200 words in length.

    As we have learned in this unit, Hooke’s law cannot perfectly predict the relationship between stress and strain of materials. See Figure 10.33 in Section 10.8 of Chapter 10 in the eTextbook. Can you provide your opinion why the materials behave like this on the elastic limit? Do you think we need new physics to explain that? Share your idea with other students.

    Response 1:
    Allison Burch

    Top of Form

    Hooke’s Law states that stress and strain are proportional within the elastic limit, meaning a material should return to its original shape when the force is removed. However, in real life, materials don’t always follow this perfectly, especially as they approach the elastic limit. After looking at this concept, I think the main reason materials behave this way is because real materials are not perfectly uniform at the microscopic level. At small levels of stress, atoms in a material stretch like tiny springs, which is why Hooke’s Law works well at first. But as the force increases, the internal structure of the material starts to change. Bonds between atoms begin to stretch unevenly, and some internal defects (like dislocations or imperfections) start to move. This causes the material to gradually stop behaving in a perfectly linear way before it actually reaches its breaking point. That’s why the stress-strain curve starts to curve instead of staying a straight line. I don’t think we need completely new physics to explain this behavior. Instead, it seems more like an extension of what we already know. Classical mechanics and materials science already account for things like plastic deformation, molecular bonding, and material imperfections. So rather than new physics, it feels more like more detailed modeling of what’s already happening at the microscopic level. Overall, Hooke’s Law is still very useful, but it works best as an approximation for small deformations, not for real-world materials under higher stress conditions.

    Bottom of Form

    Response 2: Bailey Phillips

    Hooke’s law is kind of like a “good approximation” for how materials behave, but only at the beginning. When you first stretch or compress something, the atoms inside are just slightly moving from their normal positions. I would compare this to like tiny springs. In that range, everything is neat and predictable. The stress and strain have a straight-line relationship. But once you push the material too fast, past the elastic limit, things get messy. This kind of makes me think of what happens between students and teachers. Teachers are always nice and easy going, until the moment that they are pushed too far. The atoms aren’t just stretching anymore; they start slipping past each other or breaking their original arrangement. I have spent a lot think about this and I have been picturing it like pulling on a paperclip. At first, it springs back, but if you bend it too much, it stays bent. That’s because its internal structure has permanently changed. It’s going through complex changes inside, like bonds rearranging.

    I don’t think we need new physics to explain this. The physics we already know, like atomic forces and material structure, can explain it. Hooke’s law is just a simplified model, and real materials are more complicated than that. It gives us a nice example and jumping off point, but we have the physics to get it done.

    **Screenshot2026-04-22194417.png**

    This file is too large to display. [View in new window](https://www.sweetstudy.com/files/screenshot2026-04-22194417-png-10538499)

    **Screenshot2026-04-22194252.png**

    This file is too large to display. [View in new window](https://www.sweetstudy.com/files/screenshot2026-04-22194252-png)

  • [Solved] JL IV

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    See attached **files (2)** – ## PhyicsUnitIVJournal.docx – ## UnitIV.pdf **## PhyicsUnitIVJournal.docx** Unit IV Journal Have you experienced or observed momentum conservation in your daily life? Describe three examples with a detailed explanation. Your journal entry must be at least 200 words in le…

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    **files (2)**

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    **## PhyicsUnitIVJournal.docx**

    Unit IV Journal

    Have you experienced or observed momentum conservation in your daily life? Describe three examples with a detailed explanation.

    Your journal entry must be at least 200 words in length. No references or citations are necessary.

    **## UnitIV.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit IV

    Upon completion of this unit, students should be able to:

    4. Apply the concept of momentum conservations to daily life.
    4.1 Relate impulse-momentum theorem to Newton’s second law.
    4.2 Show the relationship between linear momentum conservation and Newton’s third law.
    4.3 Apply momentum conservation to rotational kinematics.

    5. Identify the total mechanical energy conservation.

    5.1 Interpret total kinetic energy conservation in elastic collision.

    Required Unit Resources

    Chapter 7: Impulse and Momentum

    Chapter 8: Rotational Kinematics

    Unit Lesson

    Impulse-Momentum Relation and Newton’s Second Law

    Momentum, or the linear momentum (p) of an object, is defined by its mass (m) times velocity (v): p = mv.
    Mass is an intrinsic property of an object, and it is the quantity of matter in the object. Mass is a scalar
    quantity and velocity is a vector. Thus, the direction of momentum is the same as the velocity points.

    Momentum can be related to force. The average force <F> acting on a body can be obtained by the
    momentum change ∆p per a time interval ∆t. That is, <F> = m <a> = m ∆v / ∆t = ∆p / ∆t. The product of <F>
    and ∆t is the impulse J: J = <F> ∆t = ∆p. The impulse is the same as the change in momentum, which is the
    final momentum minus initial momentum; this is the impulse-momentum theorem. The impulse-momentum
    theorem is another expression of Newton’s second law. Note that both are measured in kg m/s, which is
    equal to N s.

    Suppose a stuntman is trying to jump from a tall building, and he has a choice to land on sand or water. If he
    lands on water, his stopping time is much greater than if he lands on the sand. The water exerts a smaller
    average force according to the impulse-momentum theorem because the sand and the water exert the same
    impulse on him to halt the motion. Therefore, he should land on water to diminish damage.

    UNIT IV STUDY GUIDE
    Kinematics of Collisions

    Sample Question 1: What impulse occurs when an average force of 20 N is exerted on a boxcar for
    3 seconds?

    Solution: The impulse J = <F> ∆t. The average force <F> is given by 20 N, and the time interval ∆t
    is 3 s. So, the impulse J = 20 x 3 = 60 N s.

    Impulse = Average Force x Time Interval = Change in Momentum

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Linear Momentum Conservation and Newton’s Third Law

    The total linear momentum of a system without external forces is conserved, that is, constant. This is one of
    the most important and fundamental principles to support the action-reaction law of Newton. Suppose that
    there are two colliding balls, A and B, whose masses are mA and mB. There are two forces: the force exerted
    on A by B, FAB and the force exerted on B by A, FBA.

    According to Newton’s second law, the total force of this system is F = FAB + FBA. Notice that F is 0 due to
    Newton’s third law; FAB = -FBA. That is, F = ma = m ∆v / ∆t = ∆p / ∆t = 0; momentum is constant (Cutnell et al.,
    2022). The total momentum before the collision is the same as the total momentum after collision; the total
    linear momentum is conserved. If it is not conserved, Newton’s law of action-reaction will no longer be valid.

    Elastic and Inelastic Collision

    In general, there are two kinds of collision: elastic and inelastic. If the total kinetic energy of the system is
    conserved, it is called an elastic collision. If it is not conserved, it is called inelastic collision (Cutnell et al.,
    2022). In any case, the total linear moment is conserved. When two atoms collide, the total kinetic energy
    remains constant before and after the collision in most cases. This is an example of an elastic collision. If the
    first atom gains kinetic energy, the second atom lost the energy. On the other hand, when two cars collide,
    the kinetic energy is lost mainly because of friction. The total kinetic energy is not constant in this inelastic
    collision.

    Sample Question 3: Boxcar A whose mass mA = 5,000 kg and velocity vA = 1 m/s will link to
    boxcar B whose mass mB = 15,000 kg and velocity vB = 4 m/s. Find the final velocity V after they
    are connected. Assume that there is no friction in this system.

    Solution: According to the momentum conservation, the total momentum before linking, mAvA +
    mBvB should be equal to the total momentum after linking, (mA+ mB)V. That is, V = (mAvA + mBvB) /
    (mA + mB) = (5000 x 1 + 15000 x 4) / 20000 = 13 / 4 = 3.25 m/s.

    Sample Question 2: A 0.2 kg model rocket is constructed with a motor that can provide a total impulse
    of 30 N s. What is the speed that this rocket achieves when launched from rest? Ignore the effects of
    gravity and air resistance.

    Solution: According to the impulse-momentum theorem, the rocket’s final momentum mV differs from
    its initial momentum mv by an amount equal to the impulse J of the net force exerted on it. That is J
    = mV – mv. Here, m is 0.2 kg, and v is 0 m/s because the rocket starts from rest. We are ignoring
    gravitational and frictional forces, so this impulse is due entirely to the force generated by the motor.
    The magnitude of the motor’s impulse is given as J = 30 N s. Therefore, the rocket’s final speed V is
    the magnitude of the motor impulse divided by the rocket’s mass:
    V = J / m = 30 / 0.2 = 150 m/s.

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    Angular Velocity, Angular Acceleration, and Angular Displacement

    When describing a rotational motion such as spinning top, rotating fan blades, and rolling wheels, we use
    angular velocity, angular acceleration, and angular displacement. Remember that we have studied the
    relationship between velocity, acceleration, and displacement in the first unit to describe the linear motion of
    an object. The angular displacement has an analogous idea of linear displacement. The angular displacement
    is defined as the angular velocity multiplied by the elapsed time of the motion.

    Angular Displacement = Angular Velocity x Elapsed Time

    Sample Question 4: Consider an elastic head-on collision between a green ball and an orange ball.
    The mass m of the green ball is 1 kg and the mass M of the orange ball is 3 kg. Before the collision,
    the orange ball was at rest, and the initial velocity of the green ball was 5 m/s. After the collision, the
    final speed V of the orange ball is identical to that of the green ball. However, they have a different
    velocity; the direction of the green ball is the opposite to that of the orange ball. What are the
    velocities after the collision?

    Solution: The linear momentum is conserved during the collisional process. The linear momentum
    before the collision is the same as that after the collision: mv = MV + (-mV)

    So, the final speed V = mv / (M – m) = (1 x 5) / (3 – 1) = 5 / 2 = 2.5 m/s. The final velocity of the
    orange ball is +2.5 m/s and that of the green ball is -2.5 m/s.

    In an elastic collision, the total kinetic energy is conserved. That is, total kinetic energy before the
    collision is equal to that after the collision.

    Total kinetic energy before the collision is ½ mv2 = 0.5 x 1 x 25 = 12.5 J.

    Total kinetic energy after the collision is ½ mV2 + ½ MV2 = (0.5 x 1 x 6.25) + (0.5 x 3 x 6.25) = 12.5
    J.

    That is, total kinetic energy of the system is conserved.

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    The SI unit of angular displacement is radian or rad. Please note, 1 radian is equivalent to 360 degrees
    divided by 2π radians. So, 1 radian is equal to 57.3 degrees. The SI unit of angular velocity is radian per
    second or rad/s. The angular velocity is related to the angular acceleration as follows. The change in angular
    velocity is the product of average angular acceleration and the elapsed time of the motion. The SI unit of
    angular acceleration is radian per second squared or rad/s2.

    Small Angle Formula

    What is parallax? In order to see what parallax is, hold your index finger in front of your nose, a few
    centimeters away, and look at it first with the left eye closed, and then with the right eye closed. If you
    compare its position with some background objects, you will notice that, apparently, the finger moves when
    you switch from one eye to the other! This explanation is obvious: your eyes are not in the same place, and
    hence they have a different line of sight to the finger and have to look at a different angle in order to see the
    finger. The angles at which your eyes have to look obviously depend on the distance of the finger to your
    nose. What is more important is that the difference in the angles depends on the distance; the angle gets
    smaller as the distance gets greater. Therefore, the difference in angles, the parallax, can be used to
    measure the distance. This method is known as trigonometric or heliocentric parallax and can be used to
    measure up to 100 parsecs (pc) or 0.01 arcseconds.

    The distance between your eyes is called the baseline; the bigger the baseline (here, D), the bigger the
    parallax, for the same distance, d, of the observed object to the baseline.

    We define the trigonometric parallax of the star as the angle θ subtended as seen from the star by the orbital
    radius of the Earth, 1 AU: in other words D = 1 AU. One parsec (pc), parallax second, is the distance d when
    the star has the parallax angle θ = 1 arcsecond = 1 / 206,264.81 radian.
    That is, d = D / θ = 206, 265 AU = 3.086 × 1016 m = 3.26 ly = 1 pc

    Sample Question 5: The nearest star, Proxima Centauri, has a parallax of 0.76". Calculate the

    distance of the star from us.

    Solution: θ = 0.76” = 0.76/206264.81 = 3.684 × 10-6 radian

    d = 206264.81 / 0.76 = 271,401 AU = 1.32 pc = 4.3 ly

    PHY 1301, Physics I 5

    UNIT x STUDY GUIDE
    Title

    Centripetal Acceleration and Tangential Acceleration

    When an object moves in a curved path, its total acceleration is the vector sum of the centripetal acceleration
    and the tangential acceleration. The magnitude of the total acceleration squared is equal to the centripetal
    acceleration squared plus the tangential acceleration squared. As we have learned about centripetal
    acceleration in Unit III, it maintains the circular motion. The direction of the centripetal acceleration points to
    the center of the circle. The direction of the tangential acceleration is perpendicular to that of the centripetal
    acceleration. The tangential acceleration exists because the tangential velocity of the object is increasing as it
    moves. The tangential acceleration is the product of the orbital radius and the angular acceleration in rad/s2.
    The tangential velocity is the product of the orbital radius and the angular velocity in rad/s.

    Motions of Celestial Bodies

    We can determine some obvious results from short observations of the sky. All the celestial objects move
    from the direction where the sun rises to the direction where the sun sets every day; this is known as diurnal
    motion. Also, the entire sky turns around any fixed points once a day. When one observes the night sky at the
    northern hemisphere, one would notice that the stars move counterclockwise around a fixed star in the north:
    the North Star. The North Star does not move because its direction aligns with the rotational axis of the Earth.
    The North Star is now the Polaris, but in the future, it will not be so, due to the Earth’s precession.

    The diurnal motion of stars is one of the consequences of the Earth’s rotation. The direction of the rotation of
    the Earth is the opposite of the diurnal motion’s direction. The common center of the circle of stars is the
    North Star, and stars are moving 15 degrees an hour in a counterclockwise direction. Diurnal motions of stars
    differ in different places. If one lives in the northern hemisphere at the mid-latitude area, then one would
    observe rising stars in the eastern sky, setting stars in the western sky, moving stars from east to west in the
    southern sky , and counterclockwise circling around Polaris in the northern sky at night. At the equator, the
    stars move from the east to west, perpendicular to the horizon. At the poles, the stars move parallel to the
    horizon.

    The sidereal day is defined as the time for the vernal equinox to return to your celestial meridian. It is the time,
    in fact, that it takes the Earth to complete one rotation with respect to the stars. It is the true rotation period of
    the Earth. The solar day is the time for the sun to return to your celestial meridian. Locally, it is the time from
    high noon to the next one. The Earth is also revolving around the sun, so the solar day is about 4 minutes
    longer than the sidereal day. The orbital period of the Earth around the sun is about 365 days, and the angle
    needed to complete one circle is 360 degrees. Therefore, the angle per day for the Earth to see the sun
    overhead position is 360 / 365 = 0.986 degrees. That is, the sun appears to move eastward by about 1
    degree per day with respect to the background stars. There are 24 hours in a full circle (= 360o). Also, 1 hour
    is equal to 60 minutes. That is, 24 x 60 / 360 = 4 minutes. One degree is 4 minutes. After the Earth rotates
    from east to west for about 4 minutes (= 1 degree), the sun will rise over your head, your celestial meridian.
    However, if the Earth’s rotation direction suddenly changed from west to east, you will see the high sun 4
    minutes before the complete 1 sidereal day. So, 4 + 4 = 8. The solar day will be about 8 minutes shorter than
    the original value.

    In the case of the moon, the sidereal period is 27.3 days. The moon must move eastward among the
    background stars to go completely around the sky (360 degrees) in the sidereal period. The moon moves
    about 13 degrees (= 360 / 27.3) a day. Since the celestial sphere appears to turn 1 degree about every 4
    minutes, the moon crosses our celestial meridian about 13.2 x 4 = 52.8 minutes later each day. That is, the
    sun rises about one degree (about 4 minutes) and the moon rises about 13 degrees (about 53 minutes) later
    each day, and, in the long run, they are moving from the west to east along the ecliptic.

    The rotational axis of the Earth has an angle of 23.5 degrees from the celestial north. It is not fixed like a
    spinning top and undergoes a very slow turn, or precession, over time. Its precession period is about 26,000
    years. The gravitational attraction of the sun and the moon on the Earth’s equatorial bulge is responsible for
    this effect. As a consequence, the vernal equinox, the intersection of the equator and ecliptic, regresses by
    about 50” per year along the ecliptic. In the present, the North Star is the Polaris, but it will change to Vega in
    13,000 years. In another 13,000 years, Polaris will be the North Star again. The position of the equinoxes
    changes, as well, with respect to the background stars, so the coordinates of celestial objects change as well,
    over a period of 26,000 years. This might be a negligible amount, but it is significant for an exact calculation.
    Because the moon and the sun move above and below the Earth’s equatorial plane, periodic variations occur

    PHY 1301, Physics I 6

    UNIT x STUDY GUIDE
    Title

    in the torques acting on the Earth’s equatorial bulge. These variations lead to a nutation or wobbling, of the
    Earth’s rotation axis. The nutation is a periodic oscillation of the rotational axis of the Earth around its “mean”
    position. Small contributions to the nutation have monthly and yearly periods, but the principal contribution
    (discovered by James Bradley) of amplitude 9” and period 18.6 years is from the regression of the nodes of
    the moon’s orbit.

    Reference

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    Learning Activities (Nongraded)

    Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
    them. If you have questions, contact your instructor for further guidance and information.

    1. Solve Questions 53–60 under Physics in Biology, Medicine, and Sports in Chapter 7 of your
    eTextbook.

    2. Solve Questions 62–69 under Physics in Biology, Medicine, and Sports in Chapter 8 of your
    eTextbook.

    – Course Learning Outcomes for Unit IV
    – Required Unit Resources
    – Unit Lesson

    – Impulse-Momentum Relation and Newton’s Second Law
    – Linear Momentum Conservation and Newton’s Third Law
    – Elastic and Inelastic Collision
    – Angular Velocity, Angular Acceleration, and Angular Displacement
    – Small Angle Formula
    – Centripetal Acceleration and Tangential Acceleration
    – Motions of Celestial Bodies
    – Reference

    – Learning Activities (Nongraded)

    **## UnitIV.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit IV

    Upon completion of this unit, students should be able to:

    4. Apply the concept of momentum conservations to daily life.
    4.1 Relate impulse-momentum theorem to Newton’s second law.
    4.2 Show the relationship between linear momentum conservation and Newton’s third law.
    4.3 Apply momentum conservation to rotational kinematics.

    5. Identify the total mechanical energy conservation.

    5.1 Interpret total kinetic energy conservation in elastic collision.

    Required Unit Resources

    Chapter 7: Impulse and Momentum

    Chapter 8: Rotational Kinematics

    Unit Lesson

    Impulse-Momentum Relation and Newton’s Second Law

    Momentum, or the linear momentum (p) of an object, is defined by its mass (m) times velocity (v): p = mv.
    Mass is an intrinsic property of an object, and it is the quantity of matter in the object. Mass is a scalar
    quantity and velocity is a vector. Thus, the direction of momentum is the same as the velocity points.

    Momentum can be related to force. The average force <F> acting on a body can be obtained by the
    momentum change ∆p per a time interval ∆t. That is, <F> = m <a> = m ∆v / ∆t = ∆p / ∆t. The product of <F>
    and ∆t is the impulse J: J = <F> ∆t = ∆p. The impulse is the same as the change in momentum, which is the
    final momentum minus initial momentum; this is the impulse-momentum theorem. The impulse-momentum
    theorem is another expression of Newton’s second law. Note that both are measured in kg m/s, which is
    equal to N s.

    Suppose a stuntman is trying to jump from a tall building, and he has a choice to land on sand or water. If he
    lands on water, his stopping time is much greater than if he lands on the sand. The water exerts a smaller
    average force according to the impulse-momentum theorem because the sand and the water exert the same
    impulse on him to halt the motion. Therefore, he should land on water to diminish damage.

    UNIT IV STUDY GUIDE
    Kinematics of Collisions

    Sample Question 1: What impulse occurs when an average force of 20 N is exerted on a boxcar for
    3 seconds?

    Solution: The impulse J = <F> ∆t. The average force <F> is given by 20 N, and the time interval ∆t
    is 3 s. So, the impulse J = 20 x 3 = 60 N s.

    Impulse = Average Force x Time Interval = Change in Momentum

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Linear Momentum Conservation and Newton’s Third Law

    The total linear momentum of a system without external forces is conserved, that is, constant. This is one of
    the most important and fundamental principles to support the action-reaction law of Newton. Suppose that
    there are two colliding balls, A and B, whose masses are mA and mB. There are two forces: the force exerted
    on A by B, FAB and the force exerted on B by A, FBA.

    According to Newton’s second law, the total force of this system is F = FAB + FBA. Notice that F is 0 due to
    Newton’s third law; FAB = -FBA. That is, F = ma = m ∆v / ∆t = ∆p / ∆t = 0; momentum is constant (Cutnell et al.,
    2022). The total momentum before the collision is the same as the total momentum after collision; the total
    linear momentum is conserved. If it is not conserved, Newton’s law of action-reaction will no longer be valid.

    Elastic and Inelastic Collision

    In general, there are two kinds of collision: elastic and inelastic. If the total kinetic energy of the system is
    conserved, it is called an elastic collision. If it is not conserved, it is called inelastic collision (Cutnell et al.,
    2022). In any case, the total linear moment is conserved. When two atoms collide, the total kinetic energy
    remains constant before and after the collision in most cases. This is an example of an elastic collision. If the
    first atom gains kinetic energy, the second atom lost the energy. On the other hand, when two cars collide,
    the kinetic energy is lost mainly because of friction. The total kinetic energy is not constant in this inelastic
    collision.

    Sample Question 3: Boxcar A whose mass mA = 5,000 kg and velocity vA = 1 m/s will link to
    boxcar B whose mass mB = 15,000 kg and velocity vB = 4 m/s. Find the final velocity V after they
    are connected. Assume that there is no friction in this system.

    Solution: According to the momentum conservation, the total momentum before linking, mAvA +
    mBvB should be equal to the total momentum after linking, (mA+ mB)V. That is, V = (mAvA + mBvB) /
    (mA + mB) = (5000 x 1 + 15000 x 4) / 20000 = 13 / 4 = 3.25 m/s.

    Sample Question 2: A 0.2 kg model rocket is constructed with a motor that can provide a total impulse
    of 30 N s. What is the speed that this rocket achieves when launched from rest? Ignore the effects of
    gravity and air resistance.

    Solution: According to the impulse-momentum theorem, the rocket’s final momentum mV differs from
    its initial momentum mv by an amount equal to the impulse J of the net force exerted on it. That is J
    = mV – mv. Here, m is 0.2 kg, and v is 0 m/s because the rocket starts from rest. We are ignoring
    gravitational and frictional forces, so this impulse is due entirely to the force generated by the motor.
    The magnitude of the motor’s impulse is given as J = 30 N s. Therefore, the rocket’s final speed V is
    the magnitude of the motor impulse divided by the rocket’s mass:
    V = J / m = 30 / 0.2 = 150 m/s.

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    Angular Velocity, Angular Acceleration, and Angular Displacement

    When describing a rotational motion such as spinning top, rotating fan blades, and rolling wheels, we use
    angular velocity, angular acceleration, and angular displacement. Remember that we have studied the
    relationship between velocity, acceleration, and displacement in the first unit to describe the linear motion of
    an object. The angular displacement has an analogous idea of linear displacement. The angular displacement
    is defined as the angular velocity multiplied by the elapsed time of the motion.

    Angular Displacement = Angular Velocity x Elapsed Time

    Sample Question 4: Consider an elastic head-on collision between a green ball and an orange ball.
    The mass m of the green ball is 1 kg and the mass M of the orange ball is 3 kg. Before the collision,
    the orange ball was at rest, and the initial velocity of the green ball was 5 m/s. After the collision, the
    final speed V of the orange ball is identical to that of the green ball. However, they have a different
    velocity; the direction of the green ball is the opposite to that of the orange ball. What are the
    velocities after the collision?

    Solution: The linear momentum is conserved during the collisional process. The linear momentum
    before the collision is the same as that after the collision: mv = MV + (-mV)

    So, the final speed V = mv / (M – m) = (1 x 5) / (3 – 1) = 5 / 2 = 2.5 m/s. The final velocity of the
    orange ball is +2.5 m/s and that of the green ball is -2.5 m/s.

    In an elastic collision, the total kinetic energy is conserved. That is, total kinetic energy before the
    collision is equal to that after the collision.

    Total kinetic energy before the collision is ½ mv2 = 0.5 x 1 x 25 = 12.5 J.

    Total kinetic energy after the collision is ½ mV2 + ½ MV2 = (0.5 x 1 x 6.25) + (0.5 x 3 x 6.25) = 12.5
    J.

    That is, total kinetic energy of the system is conserved.

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    The SI unit of angular displacement is radian or rad. Please note, 1 radian is equivalent to 360 degrees
    divided by 2π radians. So, 1 radian is equal to 57.3 degrees. The SI unit of angular velocity is radian per
    second or rad/s. The angular velocity is related to the angular acceleration as follows. The change in angular
    velocity is the product of average angular acceleration and the elapsed time of the motion. The SI unit of
    angular acceleration is radian per second squared or rad/s2.

    Small Angle Formula

    What is parallax? In order to see what parallax is, hold your index finger in front of your nose, a few
    centimeters away, and look at it first with the left eye closed, and then with the right eye closed. If you
    compare its position with some background objects, you will notice that, apparently, the finger moves when
    you switch from one eye to the other! This explanation is obvious: your eyes are not in the same place, and
    hence they have a different line of sight to the finger and have to look at a different angle in order to see the
    finger. The angles at which your eyes have to look obviously depend on the distance of the finger to your
    nose. What is more important is that the difference in the angles depends on the distance; the angle gets
    smaller as the distance gets greater. Therefore, the difference in angles, the parallax, can be used to
    measure the distance. This method is known as trigonometric or heliocentric parallax and can be used to
    measure up to 100 parsecs (pc) or 0.01 arcseconds.

    The distance between your eyes is called the baseline; the bigger the baseline (here, D), the bigger the
    parallax, for the same distance, d, of the observed object to the baseline.

    We define the trigonometric parallax of the star as the angle θ subtended as seen from the star by the orbital
    radius of the Earth, 1 AU: in other words D = 1 AU. One parsec (pc), parallax second, is the distance d when
    the star has the parallax angle θ = 1 arcsecond = 1 / 206,264.81 radian.
    That is, d = D / θ = 206, 265 AU = 3.086 × 1016 m = 3.26 ly = 1 pc

    Sample Question 5: The nearest star, Proxima Centauri, has a parallax of 0.76". Calculate the

    distance of the star from us.

    Solution: θ = 0.76” = 0.76/206264.81 = 3.684 × 10-6 radian

    d = 206264.81 / 0.76 = 271,401 AU = 1.32 pc = 4.3 ly

    PHY 1301, Physics I 5

    UNIT x STUDY GUIDE
    Title

    Centripetal Acceleration and Tangential Acceleration

    When an object moves in a curved path, its total acceleration is the vector sum of the centripetal acceleration
    and the tangential acceleration. The magnitude of the total acceleration squared is equal to the centripetal
    acceleration squared plus the tangential acceleration squared. As we have learned about centripetal
    acceleration in Unit III, it maintains the circular motion. The direction of the centripetal acceleration points to
    the center of the circle. The direction of the tangential acceleration is perpendicular to that of the centripetal
    acceleration. The tangential acceleration exists because the tangential velocity of the object is increasing as it
    moves. The tangential acceleration is the product of the orbital radius and the angular acceleration in rad/s2.
    The tangential velocity is the product of the orbital radius and the angular velocity in rad/s.

    Motions of Celestial Bodies

    We can determine some obvious results from short observations of the sky. All the celestial objects move
    from the direction where the sun rises to the direction where the sun sets every day; this is known as diurnal
    motion. Also, the entire sky turns around any fixed points once a day. When one observes the night sky at the
    northern hemisphere, one would notice that the stars move counterclockwise around a fixed star in the north:
    the North Star. The North Star does not move because its direction aligns with the rotational axis of the Earth.
    The North Star is now the Polaris, but in the future, it will not be so, due to the Earth’s precession.

    The diurnal motion of stars is one of the consequences of the Earth’s rotation. The direction of the rotation of
    the Earth is the opposite of the diurnal motion’s direction. The common center of the circle of stars is the
    North Star, and stars are moving 15 degrees an hour in a counterclockwise direction. Diurnal motions of stars
    differ in different places. If one lives in the northern hemisphere at the mid-latitude area, then one would
    observe rising stars in the eastern sky, setting stars in the western sky, moving stars from east to west in the
    southern sky , and counterclockwise circling around Polaris in the northern sky at night. At the equator, the
    stars move from the east to west, perpendicular to the horizon. At the poles, the stars move parallel to the
    horizon.

    The sidereal day is defined as the time for the vernal equinox to return to your celestial meridian. It is the time,
    in fact, that it takes the Earth to complete one rotation with respect to the stars. It is the true rotation period of
    the Earth. The solar day is the time for the sun to return to your celestial meridian. Locally, it is the time from
    high noon to the next one. The Earth is also revolving around the sun, so the solar day is about 4 minutes
    longer than the sidereal day. The orbital period of the Earth around the sun is about 365 days, and the angle
    needed to complete one circle is 360 degrees. Therefore, the angle per day for the Earth to see the sun
    overhead position is 360 / 365 = 0.986 degrees. That is, the sun appears to move eastward by about 1
    degree per day with respect to the background stars. There are 24 hours in a full circle (= 360o). Also, 1 hour
    is equal to 60 minutes. That is, 24 x 60 / 360 = 4 minutes. One degree is 4 minutes. After the Earth rotates
    from east to west for about 4 minutes (= 1 degree), the sun will rise over your head, your celestial meridian.
    However, if the Earth’s rotation direction suddenly changed from west to east, you will see the high sun 4
    minutes before the complete 1 sidereal day. So, 4 + 4 = 8. The solar day will be about 8 minutes shorter than
    the original value.

    In the case of the moon, the sidereal period is 27.3 days. The moon must move eastward among the
    background stars to go completely around the sky (360 degrees) in the sidereal period. The moon moves
    about 13 degrees (= 360 / 27.3) a day. Since the celestial sphere appears to turn 1 degree about every 4
    minutes, the moon crosses our celestial meridian about 13.2 x 4 = 52.8 minutes later each day. That is, the
    sun rises about one degree (about 4 minutes) and the moon rises about 13 degrees (about 53 minutes) later
    each day, and, in the long run, they are moving from the west to east along the ecliptic.

    The rotational axis of the Earth has an angle of 23.5 degrees from the celestial north. It is not fixed like a
    spinning top and undergoes a very slow turn, or precession, over time. Its precession period is about 26,000
    years. The gravitational attraction of the sun and the moon on the Earth’s equatorial bulge is responsible for
    this effect. As a consequence, the vernal equinox, the intersection of the equator and ecliptic, regresses by
    about 50” per year along the ecliptic. In the present, the North Star is the Polaris, but it will change to Vega in
    13,000 years. In another 13,000 years, Polaris will be the North Star again. The position of the equinoxes
    changes, as well, with respect to the background stars, so the coordinates of celestial objects change as well,
    over a period of 26,000 years. This might be a negligible amount, but it is significant for an exact calculation.
    Because the moon and the sun move above and below the Earth’s equatorial plane, periodic variations occur

    PHY 1301, Physics I 6

    UNIT x STUDY GUIDE
    Title

    in the torques acting on the Earth’s equatorial bulge. These variations lead to a nutation or wobbling, of the
    Earth’s rotation axis. The nutation is a periodic oscillation of the rotational axis of the Earth around its “mean”
    position. Small contributions to the nutation have monthly and yearly periods, but the principal contribution
    (discovered by James Bradley) of amplitude 9” and period 18.6 years is from the regression of the nodes of
    the moon’s orbit.

    Reference

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    Learning Activities (Nongraded)

    Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
    them. If you have questions, contact your instructor for further guidance and information.

    1. Solve Questions 53–60 under Physics in Biology, Medicine, and Sports in Chapter 7 of your
    eTextbook.

    2. Solve Questions 62–69 under Physics in Biology, Medicine, and Sports in Chapter 8 of your
    eTextbook.

    – Course Learning Outcomes for Unit IV
    – Required Unit Resources
    – Unit Lesson

    – Impulse-Momentum Relation and Newton’s Second Law
    – Linear Momentum Conservation and Newton’s Third Law
    – Elastic and Inelastic Collision
    – Angular Velocity, Angular Acceleration, and Angular Displacement
    – Small Angle Formula
    – Centripetal Acceleration and Tangential Acceleration
    – Motions of Celestial Bodies
    – Reference

    – Learning Activities (Nongraded)

    **## PhyicsUnitIVJournal.docx**

    Unit IV Journal

    Have you experienced or observed momentum conservation in your daily life? Describe three examples with a detailed explanation.

    Your journal entry must be at least 200 words in length. No references or citations are necessary.

    **## UnitIV.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit IV

    Upon completion of this unit, students should be able to:

    4. Apply the concept of momentum conservations to daily life.
    4.1 Relate impulse-momentum theorem to Newton’s second law.
    4.2 Show the relationship between linear momentum conservation and Newton’s third law.
    4.3 Apply momentum conservation to rotational kinematics.

    5. Identify the total mechanical energy conservation.

    5.1 Interpret total kinetic energy conservation in elastic collision.

    Required Unit Resources

    Chapter 7: Impulse and Momentum

    Chapter 8: Rotational Kinematics

    Unit Lesson

    Impulse-Momentum Relation and Newton’s Second Law

    Momentum, or the linear momentum (p) of an object, is defined by its mass (m) times velocity (v): p = mv.
    Mass is an intrinsic property of an object, and it is the quantity of matter in the object. Mass is a scalar
    quantity and velocity is a vector. Thus, the direction of momentum is the same as the velocity points.

    Momentum can be related to force. The average force <F> acting on a body can be obtained by the
    momentum change ∆p per a time interval ∆t. That is, <F> = m <a> = m ∆v / ∆t = ∆p / ∆t. The product of <F>
    and ∆t is the impulse J: J = <F> ∆t = ∆p. The impulse is the same as the change in momentum, which is the
    final momentum minus initial momentum; this is the impulse-momentum theorem. The impulse-momentum
    theorem is another expression of Newton’s second law. Note that both are measured in kg m/s, which is
    equal to N s.

    Suppose a stuntman is trying to jump from a tall building, and he has a choice to land on sand or water. If he
    lands on water, his stopping time is much greater than if he lands on the sand. The water exerts a smaller
    average force according to the impulse-momentum theorem because the sand and the water exert the same
    impulse on him to halt the motion. Therefore, he should land on water to diminish damage.

    UNIT IV STUDY GUIDE
    Kinematics of Collisions

    Sample Question 1: What impulse occurs when an average force of 20 N is exerted on a boxcar for
    3 seconds?

    Solution: The impulse J = <F> ∆t. The average force <F> is given by 20 N, and the time interval ∆t
    is 3 s. So, the impulse J = 20 x 3 = 60 N s.

    Impulse = Average Force x Time Interval = Change in Momentum

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Linear Momentum Conservation and Newton’s Third Law

    The total linear momentum of a system without external forces is conserved, that is, constant. This is one of
    the most important and fundamental principles to support the action-reaction law of Newton. Suppose that
    there are two colliding balls, A and B, whose masses are mA and mB. There are two forces: the force exerted
    on A by B, FAB and the force exerted on B by A, FBA.

    According to Newton’s second law, the total force of this system is F = FAB + FBA. Notice that F is 0 due to
    Newton’s third law; FAB = -FBA. That is, F = ma = m ∆v / ∆t = ∆p / ∆t = 0; momentum is constant (Cutnell et al.,
    2022). The total momentum before the collision is the same as the total momentum after collision; the total
    linear momentum is conserved. If it is not conserved, Newton’s law of action-reaction will no longer be valid.

    Elastic and Inelastic Collision

    In general, there are two kinds of collision: elastic and inelastic. If the total kinetic energy of the system is
    conserved, it is called an elastic collision. If it is not conserved, it is called inelastic collision (Cutnell et al.,
    2022). In any case, the total linear moment is conserved. When two atoms collide, the total kinetic energy
    remains constant before and after the collision in most cases. This is an example of an elastic collision. If the
    first atom gains kinetic energy, the second atom lost the energy. On the other hand, when two cars collide,
    the kinetic energy is lost mainly because of friction. The total kinetic energy is not constant in this inelastic
    collision.

    Sample Question 3: Boxcar A whose mass mA = 5,000 kg and velocity vA = 1 m/s will link to
    boxcar B whose mass mB = 15,000 kg and velocity vB = 4 m/s. Find the final velocity V after they
    are connected. Assume that there is no friction in this system.

    Solution: According to the momentum conservation, the total momentum before linking, mAvA +
    mBvB should be equal to the total momentum after linking, (mA+ mB)V. That is, V = (mAvA + mBvB) /
    (mA + mB) = (5000 x 1 + 15000 x 4) / 20000 = 13 / 4 = 3.25 m/s.

    Sample Question 2: A 0.2 kg model rocket is constructed with a motor that can provide a total impulse
    of 30 N s. What is the speed that this rocket achieves when launched from rest? Ignore the effects of
    gravity and air resistance.

    Solution: According to the impulse-momentum theorem, the rocket’s final momentum mV differs from
    its initial momentum mv by an amount equal to the impulse J of the net force exerted on it. That is J
    = mV – mv. Here, m is 0.2 kg, and v is 0 m/s because the rocket starts from rest. We are ignoring
    gravitational and frictional forces, so this impulse is due entirely to the force generated by the motor.
    The magnitude of the motor’s impulse is given as J = 30 N s. Therefore, the rocket’s final speed V is
    the magnitude of the motor impulse divided by the rocket’s mass:
    V = J / m = 30 / 0.2 = 150 m/s.

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    Angular Velocity, Angular Acceleration, and Angular Displacement

    When describing a rotational motion such as spinning top, rotating fan blades, and rolling wheels, we use
    angular velocity, angular acceleration, and angular displacement. Remember that we have studied the
    relationship between velocity, acceleration, and displacement in the first unit to describe the linear motion of
    an object. The angular displacement has an analogous idea of linear displacement. The angular displacement
    is defined as the angular velocity multiplied by the elapsed time of the motion.

    Angular Displacement = Angular Velocity x Elapsed Time

    Sample Question 4: Consider an elastic head-on collision between a green ball and an orange ball.
    The mass m of the green ball is 1 kg and the mass M of the orange ball is 3 kg. Before the collision,
    the orange ball was at rest, and the initial velocity of the green ball was 5 m/s. After the collision, the
    final speed V of the orange ball is identical to that of the green ball. However, they have a different
    velocity; the direction of the green ball is the opposite to that of the orange ball. What are the
    velocities after the collision?

    Solution: The linear momentum is conserved during the collisional process. The linear momentum
    before the collision is the same as that after the collision: mv = MV + (-mV)

    So, the final speed V = mv / (M – m) = (1 x 5) / (3 – 1) = 5 / 2 = 2.5 m/s. The final velocity of the
    orange ball is +2.5 m/s and that of the green ball is -2.5 m/s.

    In an elastic collision, the total kinetic energy is conserved. That is, total kinetic energy before the
    collision is equal to that after the collision.

    Total kinetic energy before the collision is ½ mv2 = 0.5 x 1 x 25 = 12.5 J.

    Total kinetic energy after the collision is ½ mV2 + ½ MV2 = (0.5 x 1 x 6.25) + (0.5 x 3 x 6.25) = 12.5
    J.

    That is, total kinetic energy of the system is conserved.

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    The SI unit of angular displacement is radian or rad. Please note, 1 radian is equivalent to 360 degrees
    divided by 2π radians. So, 1 radian is equal to 57.3 degrees. The SI unit of angular velocity is radian per
    second or rad/s. The angular velocity is related to the angular acceleration as follows. The change in angular
    velocity is the product of average angular acceleration and the elapsed time of the motion. The SI unit of
    angular acceleration is radian per second squared or rad/s2.

    Small Angle Formula

    What is parallax? In order to see what parallax is, hold your index finger in front of your nose, a few
    centimeters away, and look at it first with the left eye closed, and then with the right eye closed. If you
    compare its position with some background objects, you will notice that, apparently, the finger moves when
    you switch from one eye to the other! This explanation is obvious: your eyes are not in the same place, and
    hence they have a different line of sight to the finger and have to look at a different angle in order to see the
    finger. The angles at which your eyes have to look obviously depend on the distance of the finger to your
    nose. What is more important is that the difference in the angles depends on the distance; the angle gets
    smaller as the distance gets greater. Therefore, the difference in angles, the parallax, can be used to
    measure the distance. This method is known as trigonometric or heliocentric parallax and can be used to
    measure up to 100 parsecs (pc) or 0.01 arcseconds.

    The distance between your eyes is called the baseline; the bigger the baseline (here, D), the bigger the
    parallax, for the same distance, d, of the observed object to the baseline.

    We define the trigonometric parallax of the star as the angle θ subtended as seen from the star by the orbital
    radius of the Earth, 1 AU: in other words D = 1 AU. One parsec (pc), parallax second, is the distance d when
    the star has the parallax angle θ = 1 arcsecond = 1 / 206,264.81 radian.
    That is, d = D / θ = 206, 265 AU = 3.086 × 1016 m = 3.26 ly = 1 pc

    Sample Question 5: The nearest star, Proxima Centauri, has a parallax of 0.76". Calculate the

    distance of the star from us.

    Solution: θ = 0.76” = 0.76/206264.81 = 3.684 × 10-6 radian

    d = 206264.81 / 0.76 = 271,401 AU = 1.32 pc = 4.3 ly

    PHY 1301, Physics I 5

    UNIT x STUDY GUIDE
    Title

    Centripetal Acceleration and Tangential Acceleration

    When an object moves in a curved path, its total acceleration is the vector sum of the centripetal acceleration
    and the tangential acceleration. The magnitude of the total acceleration squared is equal to the centripetal
    acceleration squared plus the tangential acceleration squared. As we have learned about centripetal
    acceleration in Unit III, it maintains the circular motion. The direction of the centripetal acceleration points to
    the center of the circle. The direction of the tangential acceleration is perpendicular to that of the centripetal
    acceleration. The tangential acceleration exists because the tangential velocity of the object is increasing as it
    moves. The tangential acceleration is the product of the orbital radius and the angular acceleration in rad/s2.
    The tangential velocity is the product of the orbital radius and the angular velocity in rad/s.

    Motions of Celestial Bodies

    We can determine some obvious results from short observations of the sky. All the celestial objects move
    from the direction where the sun rises to the direction where the sun sets every day; this is known as diurnal
    motion. Also, the entire sky turns around any fixed points once a day. When one observes the night sky at the
    northern hemisphere, one would notice that the stars move counterclockwise around a fixed star in the north:
    the North Star. The North Star does not move because its direction aligns with the rotational axis of the Earth.
    The North Star is now the Polaris, but in the future, it will not be so, due to the Earth’s precession.

    The diurnal motion of stars is one of the consequences of the Earth’s rotation. The direction of the rotation of
    the Earth is the opposite of the diurnal motion’s direction. The common center of the circle of stars is the
    North Star, and stars are moving 15 degrees an hour in a counterclockwise direction. Diurnal motions of stars
    differ in different places. If one lives in the northern hemisphere at the mid-latitude area, then one would
    observe rising stars in the eastern sky, setting stars in the western sky, moving stars from east to west in the
    southern sky , and counterclockwise circling around Polaris in the northern sky at night. At the equator, the
    stars move from the east to west, perpendicular to the horizon. At the poles, the stars move parallel to the
    horizon.

    The sidereal day is defined as the time for the vernal equinox to return to your celestial meridian. It is the time,
    in fact, that it takes the Earth to complete one rotation with respect to the stars. It is the true rotation period of
    the Earth. The solar day is the time for the sun to return to your celestial meridian. Locally, it is the time from
    high noon to the next one. The Earth is also revolving around the sun, so the solar day is about 4 minutes
    longer than the sidereal day. The orbital period of the Earth around the sun is about 365 days, and the angle
    needed to complete one circle is 360 degrees. Therefore, the angle per day for the Earth to see the sun
    overhead position is 360 / 365 = 0.986 degrees. That is, the sun appears to move eastward by about 1
    degree per day with respect to the background stars. There are 24 hours in a full circle (= 360o). Also, 1 hour
    is equal to 60 minutes. That is, 24 x 60 / 360 = 4 minutes. One degree is 4 minutes. After the Earth rotates
    from east to west for about 4 minutes (= 1 degree), the sun will rise over your head, your celestial meridian.
    However, if the Earth’s rotation direction suddenly changed from west to east, you will see the high sun 4
    minutes before the complete 1 sidereal day. So, 4 + 4 = 8. The solar day will be about 8 minutes shorter than
    the original value.

    In the case of the moon, the sidereal period is 27.3 days. The moon must move eastward among the
    background stars to go completely around the sky (360 degrees) in the sidereal period. The moon moves
    about 13 degrees (= 360 / 27.3) a day. Since the celestial sphere appears to turn 1 degree about every 4
    minutes, the moon crosses our celestial meridian about 13.2 x 4 = 52.8 minutes later each day. That is, the
    sun rises about one degree (about 4 minutes) and the moon rises about 13 degrees (about 53 minutes) later
    each day, and, in the long run, they are moving from the west to east along the ecliptic.

    The rotational axis of the Earth has an angle of 23.5 degrees from the celestial north. It is not fixed like a
    spinning top and undergoes a very slow turn, or precession, over time. Its precession period is about 26,000
    years. The gravitational attraction of the sun and the moon on the Earth’s equatorial bulge is responsible for
    this effect. As a consequence, the vernal equinox, the intersection of the equator and ecliptic, regresses by
    about 50” per year along the ecliptic. In the present, the North Star is the Polaris, but it will change to Vega in
    13,000 years. In another 13,000 years, Polaris will be the North Star again. The position of the equinoxes
    changes, as well, with respect to the background stars, so the coordinates of celestial objects change as well,
    over a period of 26,000 years. This might be a negligible amount, but it is significant for an exact calculation.
    Because the moon and the sun move above and below the Earth’s equatorial plane, periodic variations occur

    PHY 1301, Physics I 6

    UNIT x STUDY GUIDE
    Title

    in the torques acting on the Earth’s equatorial bulge. These variations lead to a nutation or wobbling, of the
    Earth’s rotation axis. The nutation is a periodic oscillation of the rotational axis of the Earth around its “mean”
    position. Small contributions to the nutation have monthly and yearly periods, but the principal contribution
    (discovered by James Bradley) of amplitude 9” and period 18.6 years is from the regression of the nodes of
    the moon’s orbit.

    Reference

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    Learning Activities (Nongraded)

    Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
    them. If you have questions, contact your instructor for further guidance and information.

    1. Solve Questions 53–60 under Physics in Biology, Medicine, and Sports in Chapter 7 of your
    eTextbook.

    2. Solve Questions 62–69 under Physics in Biology, Medicine, and Sports in Chapter 8 of your
    eTextbook.

    – Course Learning Outcomes for Unit IV
    – Required Unit Resources
    – Unit Lesson

    – Impulse-Momentum Relation and Newton’s Second Law
    – Linear Momentum Conservation and Newton’s Third Law
    – Elastic and Inelastic Collision
    – Angular Velocity, Angular Acceleration, and Angular Displacement
    – Small Angle Formula
    – Centripetal Acceleration and Tangential Acceleration
    – Motions of Celestial Bodies
    – Reference

    – Learning Activities (Nongraded)

    **## PhyicsUnitIVJournal.docx**

    Unit IV Journal

    Have you experienced or observed momentum conservation in your daily life? Describe three examples with a detailed explanation.

    Your journal entry must be at least 200 words in length. No references or citations are necessary.

    **## UnitIV.pdf**

    PHY 1301, Physics I 1

    Course Learning Outcomes for Unit IV

    Upon completion of this unit, students should be able to:

    4. Apply the concept of momentum conservations to daily life.
    4.1 Relate impulse-momentum theorem to Newton’s second law.
    4.2 Show the relationship between linear momentum conservation and Newton’s third law.
    4.3 Apply momentum conservation to rotational kinematics.

    5. Identify the total mechanical energy conservation.

    5.1 Interpret total kinetic energy conservation in elastic collision.

    Required Unit Resources

    Chapter 7: Impulse and Momentum

    Chapter 8: Rotational Kinematics

    Unit Lesson

    Impulse-Momentum Relation and Newton’s Second Law

    Momentum, or the linear momentum (p) of an object, is defined by its mass (m) times velocity (v): p = mv.
    Mass is an intrinsic property of an object, and it is the quantity of matter in the object. Mass is a scalar
    quantity and velocity is a vector. Thus, the direction of momentum is the same as the velocity points.

    Momentum can be related to force. The average force <F> acting on a body can be obtained by the
    momentum change ∆p per a time interval ∆t. That is, <F> = m <a> = m ∆v / ∆t = ∆p / ∆t. The product of <F>
    and ∆t is the impulse J: J = <F> ∆t = ∆p. The impulse is the same as the change in momentum, which is the
    final momentum minus initial momentum; this is the impulse-momentum theorem. The impulse-momentum
    theorem is another expression of Newton’s second law. Note that both are measured in kg m/s, which is
    equal to N s.

    Suppose a stuntman is trying to jump from a tall building, and he has a choice to land on sand or water. If he
    lands on water, his stopping time is much greater than if he lands on the sand. The water exerts a smaller
    average force according to the impulse-momentum theorem because the sand and the water exert the same
    impulse on him to halt the motion. Therefore, he should land on water to diminish damage.

    UNIT IV STUDY GUIDE
    Kinematics of Collisions

    Sample Question 1: What impulse occurs when an average force of 20 N is exerted on a boxcar for
    3 seconds?

    Solution: The impulse J = <F> ∆t. The average force <F> is given by 20 N, and the time interval ∆t
    is 3 s. So, the impulse J = 20 x 3 = 60 N s.

    Impulse = Average Force x Time Interval = Change in Momentum

    PHY 1301, Physics I 2

    UNIT x STUDY GUIDE
    Title

    Linear Momentum Conservation and Newton’s Third Law

    The total linear momentum of a system without external forces is conserved, that is, constant. This is one of
    the most important and fundamental principles to support the action-reaction law of Newton. Suppose that
    there are two colliding balls, A and B, whose masses are mA and mB. There are two forces: the force exerted
    on A by B, FAB and the force exerted on B by A, FBA.

    According to Newton’s second law, the total force of this system is F = FAB + FBA. Notice that F is 0 due to
    Newton’s third law; FAB = -FBA. That is, F = ma = m ∆v / ∆t = ∆p / ∆t = 0; momentum is constant (Cutnell et al.,
    2022). The total momentum before the collision is the same as the total momentum after collision; the total
    linear momentum is conserved. If it is not conserved, Newton’s law of action-reaction will no longer be valid.

    Elastic and Inelastic Collision

    In general, there are two kinds of collision: elastic and inelastic. If the total kinetic energy of the system is
    conserved, it is called an elastic collision. If it is not conserved, it is called inelastic collision (Cutnell et al.,
    2022). In any case, the total linear moment is conserved. When two atoms collide, the total kinetic energy
    remains constant before and after the collision in most cases. This is an example of an elastic collision. If the
    first atom gains kinetic energy, the second atom lost the energy. On the other hand, when two cars collide,
    the kinetic energy is lost mainly because of friction. The total kinetic energy is not constant in this inelastic
    collision.

    Sample Question 3: Boxcar A whose mass mA = 5,000 kg and velocity vA = 1 m/s will link to
    boxcar B whose mass mB = 15,000 kg and velocity vB = 4 m/s. Find the final velocity V after they
    are connected. Assume that there is no friction in this system.

    Solution: According to the momentum conservation, the total momentum before linking, mAvA +
    mBvB should be equal to the total momentum after linking, (mA+ mB)V. That is, V = (mAvA + mBvB) /
    (mA + mB) = (5000 x 1 + 15000 x 4) / 20000 = 13 / 4 = 3.25 m/s.

    Sample Question 2: A 0.2 kg model rocket is constructed with a motor that can provide a total impulse
    of 30 N s. What is the speed that this rocket achieves when launched from rest? Ignore the effects of
    gravity and air resistance.

    Solution: According to the impulse-momentum theorem, the rocket’s final momentum mV differs from
    its initial momentum mv by an amount equal to the impulse J of the net force exerted on it. That is J
    = mV – mv. Here, m is 0.2 kg, and v is 0 m/s because the rocket starts from rest. We are ignoring
    gravitational and frictional forces, so this impulse is due entirely to the force generated by the motor.
    The magnitude of the motor’s impulse is given as J = 30 N s. Therefore, the rocket’s final speed V is
    the magnitude of the motor impulse divided by the rocket’s mass:
    V = J / m = 30 / 0.2 = 150 m/s.

    PHY 1301, Physics I 3

    UNIT x STUDY GUIDE
    Title

    Angular Velocity, Angular Acceleration, and Angular Displacement

    When describing a rotational motion such as spinning top, rotating fan blades, and rolling wheels, we use
    angular velocity, angular acceleration, and angular displacement. Remember that we have studied the
    relationship between velocity, acceleration, and displacement in the first unit to describe the linear motion of
    an object. The angular displacement has an analogous idea of linear displacement. The angular displacement
    is defined as the angular velocity multiplied by the elapsed time of the motion.

    Angular Displacement = Angular Velocity x Elapsed Time

    Sample Question 4: Consider an elastic head-on collision between a green ball and an orange ball.
    The mass m of the green ball is 1 kg and the mass M of the orange ball is 3 kg. Before the collision,
    the orange ball was at rest, and the initial velocity of the green ball was 5 m/s. After the collision, the
    final speed V of the orange ball is identical to that of the green ball. However, they have a different
    velocity; the direction of the green ball is the opposite to that of the orange ball. What are the
    velocities after the collision?

    Solution: The linear momentum is conserved during the collisional process. The linear momentum
    before the collision is the same as that after the collision: mv = MV + (-mV)

    So, the final speed V = mv / (M – m) = (1 x 5) / (3 – 1) = 5 / 2 = 2.5 m/s. The final velocity of the
    orange ball is +2.5 m/s and that of the green ball is -2.5 m/s.

    In an elastic collision, the total kinetic energy is conserved. That is, total kinetic energy before the
    collision is equal to that after the collision.

    Total kinetic energy before the collision is ½ mv2 = 0.5 x 1 x 25 = 12.5 J.

    Total kinetic energy after the collision is ½ mV2 + ½ MV2 = (0.5 x 1 x 6.25) + (0.5 x 3 x 6.25) = 12.5
    J.

    That is, total kinetic energy of the system is conserved.

    PHY 1301, Physics I 4

    UNIT x STUDY GUIDE
    Title

    The SI unit of angular displacement is radian or rad. Please note, 1 radian is equivalent to 360 degrees
    divided by 2π radians. So, 1 radian is equal to 57.3 degrees. The SI unit of angular velocity is radian per
    second or rad/s. The angular velocity is related to the angular acceleration as follows. The change in angular
    velocity is the product of average angular acceleration and the elapsed time of the motion. The SI unit of
    angular acceleration is radian per second squared or rad/s2.

    Small Angle Formula

    What is parallax? In order to see what parallax is, hold your index finger in front of your nose, a few
    centimeters away, and look at it first with the left eye closed, and then with the right eye closed. If you
    compare its position with some background objects, you will notice that, apparently, the finger moves when
    you switch from one eye to the other! This explanation is obvious: your eyes are not in the same place, and
    hence they have a different line of sight to the finger and have to look at a different angle in order to see the
    finger. The angles at which your eyes have to look obviously depend on the distance of the finger to your
    nose. What is more important is that the difference in the angles depends on the distance; the angle gets
    smaller as the distance gets greater. Therefore, the difference in angles, the parallax, can be used to
    measure the distance. This method is known as trigonometric or heliocentric parallax and can be used to
    measure up to 100 parsecs (pc) or 0.01 arcseconds.

    The distance between your eyes is called the baseline; the bigger the baseline (here, D), the bigger the
    parallax, for the same distance, d, of the observed object to the baseline.

    We define the trigonometric parallax of the star as the angle θ subtended as seen from the star by the orbital
    radius of the Earth, 1 AU: in other words D = 1 AU. One parsec (pc), parallax second, is the distance d when
    the star has the parallax angle θ = 1 arcsecond = 1 / 206,264.81 radian.
    That is, d = D / θ = 206, 265 AU = 3.086 × 1016 m = 3.26 ly = 1 pc

    Sample Question 5: The nearest star, Proxima Centauri, has a parallax of 0.76". Calculate the

    distance of the star from us.

    Solution: θ = 0.76” = 0.76/206264.81 = 3.684 × 10-6 radian

    d = 206264.81 / 0.76 = 271,401 AU = 1.32 pc = 4.3 ly

    PHY 1301, Physics I 5

    UNIT x STUDY GUIDE
    Title

    Centripetal Acceleration and Tangential Acceleration

    When an object moves in a curved path, its total acceleration is the vector sum of the centripetal acceleration
    and the tangential acceleration. The magnitude of the total acceleration squared is equal to the centripetal
    acceleration squared plus the tangential acceleration squared. As we have learned about centripetal
    acceleration in Unit III, it maintains the circular motion. The direction of the centripetal acceleration points to
    the center of the circle. The direction of the tangential acceleration is perpendicular to that of the centripetal
    acceleration. The tangential acceleration exists because the tangential velocity of the object is increasing as it
    moves. The tangential acceleration is the product of the orbital radius and the angular acceleration in rad/s2.
    The tangential velocity is the product of the orbital radius and the angular velocity in rad/s.

    Motions of Celestial Bodies

    We can determine some obvious results from short observations of the sky. All the celestial objects move
    from the direction where the sun rises to the direction where the sun sets every day; this is known as diurnal
    motion. Also, the entire sky turns around any fixed points once a day. When one observes the night sky at the
    northern hemisphere, one would notice that the stars move counterclockwise around a fixed star in the north:
    the North Star. The North Star does not move because its direction aligns with the rotational axis of the Earth.
    The North Star is now the Polaris, but in the future, it will not be so, due to the Earth’s precession.

    The diurnal motion of stars is one of the consequences of the Earth’s rotation. The direction of the rotation of
    the Earth is the opposite of the diurnal motion’s direction. The common center of the circle of stars is the
    North Star, and stars are moving 15 degrees an hour in a counterclockwise direction. Diurnal motions of stars
    differ in different places. If one lives in the northern hemisphere at the mid-latitude area, then one would
    observe rising stars in the eastern sky, setting stars in the western sky, moving stars from east to west in the
    southern sky , and counterclockwise circling around Polaris in the northern sky at night. At the equator, the
    stars move from the east to west, perpendicular to the horizon. At the poles, the stars move parallel to the
    horizon.

    The sidereal day is defined as the time for the vernal equinox to return to your celestial meridian. It is the time,
    in fact, that it takes the Earth to complete one rotation with respect to the stars. It is the true rotation period of
    the Earth. The solar day is the time for the sun to return to your celestial meridian. Locally, it is the time from
    high noon to the next one. The Earth is also revolving around the sun, so the solar day is about 4 minutes
    longer than the sidereal day. The orbital period of the Earth around the sun is about 365 days, and the angle
    needed to complete one circle is 360 degrees. Therefore, the angle per day for the Earth to see the sun
    overhead position is 360 / 365 = 0.986 degrees. That is, the sun appears to move eastward by about 1
    degree per day with respect to the background stars. There are 24 hours in a full circle (= 360o). Also, 1 hour
    is equal to 60 minutes. That is, 24 x 60 / 360 = 4 minutes. One degree is 4 minutes. After the Earth rotates
    from east to west for about 4 minutes (= 1 degree), the sun will rise over your head, your celestial meridian.
    However, if the Earth’s rotation direction suddenly changed from west to east, you will see the high sun 4
    minutes before the complete 1 sidereal day. So, 4 + 4 = 8. The solar day will be about 8 minutes shorter than
    the original value.

    In the case of the moon, the sidereal period is 27.3 days. The moon must move eastward among the
    background stars to go completely around the sky (360 degrees) in the sidereal period. The moon moves
    about 13 degrees (= 360 / 27.3) a day. Since the celestial sphere appears to turn 1 degree about every 4
    minutes, the moon crosses our celestial meridian about 13.2 x 4 = 52.8 minutes later each day. That is, the
    sun rises about one degree (about 4 minutes) and the moon rises about 13 degrees (about 53 minutes) later
    each day, and, in the long run, they are moving from the west to east along the ecliptic.

    The rotational axis of the Earth has an angle of 23.5 degrees from the celestial north. It is not fixed like a
    spinning top and undergoes a very slow turn, or precession, over time. Its precession period is about 26,000
    years. The gravitational attraction of the sun and the moon on the Earth’s equatorial bulge is responsible for
    this effect. As a consequence, the vernal equinox, the intersection of the equator and ecliptic, regresses by
    about 50” per year along the ecliptic. In the present, the North Star is the Polaris, but it will change to Vega in
    13,000 years. In another 13,000 years, Polaris will be the North Star again. The position of the equinoxes
    changes, as well, with respect to the background stars, so the coordinates of celestial objects change as well,
    over a period of 26,000 years. This might be a negligible amount, but it is significant for an exact calculation.
    Because the moon and the sun move above and below the Earth’s equatorial plane, periodic variations occur

    PHY 1301, Physics I 6

    UNIT x STUDY GUIDE
    Title

    in the torques acting on the Earth’s equatorial bulge. These variations lead to a nutation or wobbling, of the
    Earth’s rotation axis. The nutation is a periodic oscillation of the rotational axis of the Earth around its “mean”
    position. Small contributions to the nutation have monthly and yearly periods, but the principal contribution
    (discovered by James Bradley) of amplitude 9” and period 18.6 years is from the regression of the nodes of
    the moon’s orbit.

    Reference

    Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley.

    Learning Activities (Nongraded)

    Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
    them. If you have questions, contact your instructor for further guidance and information.

    1. Solve Questions 53–60 under Physics in Biology, Medicine, and Sports in Chapter 7 of your
    eTextbook.

    2. Solve Questions 62–69 under Physics in Biology, Medicine, and Sports in Chapter 8 of your
    eTextbook.

    – Course Learning Outcomes for Unit IV
    – Required Unit Resources
    – Unit Lesson

    – Impulse-Momentum Relation and Newton’s Second Law
    – Linear Momentum Conservation and Newton’s Third Law
    – Elastic and Inelastic Collision
    – Angular Velocity, Angular Acceleration, and Angular Displacement
    – Small Angle Formula
    – Centripetal Acceleration and Tangential Acceleration
    – Motions of Celestial Bodies
    – Reference

    – Learning Activities (Nongraded)