Momentum, Conservation Laws, and Center of Mass Study

Administrative Overview and Test Two Review

• Performance Review for Test Two:
  • Results for the second test varied; some students performed exceptionally well while others struggled.
  • The instructor reminded students that test scores are only one component of the overall grade.
  • Students were encouraged to keep trying and to put forth their best effort despite the difficulty of the material.
• Summertime Curricular Compression:
  • The instructor noted that the summertime schedule is heavily compressed, making the tests and the pace of the class particularly challenging.
• Intervention and Improvement:
  • Students with grades that are not meeting their expectations were advised to talk to the instructor to review their approach to the class.
  • Tutoring and other academic resources were recommended.
• Office Hours and Test Review:
  • Tests will not be reviewed during class time to preserve instructional periods.
  • Students are welcome to visit the instructor during office hours or anytime they are in the office to review their tests together and address specific questions.

Introduction to the Concept of Momentum

• Frameworks for Describing Motion:
  • The instructor identified three main physical approaches to describing motion:
    1. Force-Acceleration Approach: This involves looking at how forces cause acceleration ($F_{net} = m \times a$).
    2. Work-Energy Approach: This examines how work transfers energy between systems and forms.
    3. Momentum Approach: The third way to describe motion, which is the focus of Chapter 7.
• Verbatim Definition of Momentum:
  • Momentum is the property of an object describing its motion as a vector that is conserved.
• Nature of Momentum:
  • Momentum is a physical phenomenon through space, not an emotional or metaphorical concept.
  • It is a property of an object, not the object itself. Just as energy is a property, momentum depends on an object being in motion.
  • If an object is stationary (not moving), it possesses zero momentum. A small amount of motion corresponds to little momentum, while high motion corresponds to significant momentum.

Comparative Analysis: Momentum vs. Kinetic Energy

• The Overlap of Descriptions:
  • Both kinetic energy ($K$) and momentum ($P$) describe the state of an object in motion. However, they are fundamentally and independently meaningful in the way the universe operates.
• Mathematical Review of Kinetic Energy:
  • The kinetic energy formula is defined as:         $K = \frac{1}{2} \times m \times v^2$
  • In this equation, kinetic energy equals one-half times the mass ($m$) of the object times the magnitude of the velocity ($v$) squared (the speed squared).
  • If velocity is zero, kinetic energy is zero.
• Key Differences Between Momentum and Kinetic Energy:
  • Vector vs. Scalar: Momentum is a vector. It has both magnitude and direction. Kinetic energy is a scalar; it has no directionality.
  • Negativity and Cancellation: Kinetic energy can never be negative because mass is never negative and velocity is squared. Consequently, different parts of a system cannot have kinetic energies that cancel each other out. The total kinetic energy of a system is always greater than or equal to the kinetic energy of its constituent parts.
  • Vector Cancellation in Momentum: Because momentum is a vector, it can cancel out ($30\,\text{units East} + 30\,\text{units West} = 0$). Much like force vectors, two momentum vectors can partially or fully cancel each other out, leading to a net system momentum of zero even if individual parts are moving.
• Conceptual Interpretation of Momentum:
  • Magnitude: Roughly thought of as "how hard it is to stop the moving object."
  • Direction: The direction of the momentum vector is always the direction of motion (the same direction as the velocity vector).
  • Note on Directionality: While acceleration vectors can point forward, backward, or sideways relative to motion, the momentum vector by definition always points "forward" in the direction of the object's movement.

The Mechanics and Equations of Momentum

• The Momentum Equation:
  • The symbol for momentum is $p$ (derived from historical contexts as $m$ was already taken for mass).
  • The definitional equation is:         $p = m \times v$
  • Where $p$ is the momentum vector, $m$ is the mass of the object, and $v$ is the velocity vector.
• Units of Momentum:
  • Mass is measured in kilograms ($kg$).
  • Velocity is measured in meters per second ($m/s$).
  • Therefore, the units for momentum are kilogram meters per second ($kg \, m/s$).
  • Unlike the Newton (force) or Joule (energy), there is no official single-word nickname for $kg \, m/s$, though students are free to invent one for personal use.
• Relationships in the Equation:
  • A faster object has more momentum than a slower object of the same mass.
  • A more massive object has more momentum than a lighter object at the same velocity (e.g., a runaway semi-truck at $5\,\text{miles per hour}$ is much harder to stop than a bird at the same speed).
  • Interesting Possibility: A car with half the mass but twice the speed of another car will have the identical momentum magnitude ($\frac{1}{2}m \times 2v = mv$).
• Momentum as the Preferred Tool for Collisions:
  • While collisions can be analyzed using force-acceleration or work-energy, the momentum approach is the most efficient and natural method for these scenarios.

Newton's Second Law in Momentum Form

• Derivation from the Force-Acceleration Form:
  • Starting with $F_{net} = m \times a$.
  • Substituting acceleration: $a = \frac{\Delta v}{\Delta t}$, where $\Delta t$ is small (instantaneous).
  • Since $\Delta v = v_f - v_i$, the equation becomes:         $F_{net} = m \times \frac{(v_f - v_i)}{\Delta t}$
  • Distributing the mass:         $F_{net} = \frac{(m \times v_f - m \times v_i)}{\Delta t}$
  • Recognizing that $m \times v_f = p_f$ and $m \times v_i = p_i$:         $F_{net} = \frac{\Delta p}{\Delta t}$
• The Fundamental Law:
  • The momentum form of Newton's second law is actually more fundamental than the acceleration form.
  • It defines a non-zero net force as the cause of a change in an object's momentum vector.
• Implications of the Net Force:
  • A net force can increase magnitude (speeding up), decrease magnitude (slowing down), or change the direction of the momentum vector (turning).
  • The strength (magnitude) of the net force equals the rate at which the momentum vector changes.
  • Strong Force: Causes a rapid change in momentum.
  • Weak Force: Causes a gradual change in momentum.

Newton's First Law and the Conservation of Momentum

• First Law as a Special Case:
  • Newton's first law states that if $F_{net} = 0$, then $\frac{\Delta p}{\Delta t} = 0$.
  • This implies that momentum is constant ($p_f = p_i = \text{constant}$).
• Application to Systems:
  • A system can have many parts (even parts not physically connected, like two billiard balls).
  • For a closed system ($F_{net} = 0$), the total momentum of the system is constant:         $p_{\text{system, final}} = p_{\text{system, initial}}$
• Law of Conservation of Momentum:
  • In a closed system, momentum cannot be created or destroyed, but it can be transferred between parts.
  • This is distinct from kinetic energy, which is not conserved (it can be created or destroyed through energy transformations).

Symmetries and the Foundations of Physics

• Conservation Laws and Symmetries:
  • Conservation laws are the most fundamental laws of the universe and result from inherent symmetries in space and time.
• Time Symmetry:
  • The laws of physics do not change from one day to the next. This time symmetry leads to the Law of Conservation of Total Energy.
• Spatial Translation Symmetry:
  • The laws of physics are the same at every point in space. This symmetry leads to the Law of Conservation of Momentum.

Momentum Creation and Recoil (Guns and Bullets)

• Net Momentum vs. Individual Momentum:
  • While individual objects can gain or lose momentum, the total momentum of a closed system cannot change.
  • If you start with zero momentum (gun and bullet at rest), firing the bullet creates momentum for the bullet. To conserve total momentum, an equal and opposite momentum must be created for the gun.
• Verbatim Rule of Recoil:
  • Momentum must be created in equal and opposite amounts so that no overall (net) momentum is created.
• The Example of Gun Recoil:
  • If a bullet gains $20\,\text{units}$ of momentum East, the gun MUST gain $20\,\text{units}$ of momentum West.
  • Total momentum: $20\,\text{East} + 20\,\text{West} = 0$.

Mathematical Application to Collision Problems

• Problem-Solving Framework:
  1. Define "Before" (right before the collision).
  2. Define "After" (right after the collision).
  3. Set $p_{\text{system, initial}} = p_{\text{system, final}}$.
• Expansion of Terms:
  • $p_{1i} + p_{2i} = p_{1f} + p_{2f}$
  • $m_1 \times v_{1i} + m_2 \times v_{2i} = m_1 \times v_{1f} + m_2 \times v_{2f}$
  • Note: Subscripts (1, 2, i, f) are mandatory to keep track of the specific momentum values.
• Recoil Velocity Derivation:
  • If starting from rest ($v_{1i} = 0, v_{2i} = 0$):         $0 = m_1 \times v_{1f} + m_2 \times v_{2f}$$m_1 \times v_{1f} = -m_2 \times v_{2f}$$v_{1f} = -(\frac{m_2}{m_1}) \times v_{2f}$
  • Recoil Velocity ($v_{1f}$): The final velocity of the object (gun or thrower) that moves in the opposite direction of the projectile.
  • The negative sign confirms that the objects move in opposite directions.

Strategies for Minimizing Recoil Velocity

• To make the recoil velocity ($v_{1f}$) as small as possible, one can manipulate three variables in the equation $v_{1f} = (\frac{m_2}{m_1}) \times v_{2f}$:
  1. Decrease $m_2$ (Bullet Mass): Use low-mass bullets (e.g., training a child with a BB gun).
  2. Decrease $v_{2f}$ (Bullet Velocity): Use "slow" bullets with less muzzle velocity; however, this reduces penetrative power.
  3. Increase $m_1$ (Gun Mass):
     • Use heavier/beefier guns.
     • Locking the Arm/Body: If you lock your arm rigidly, the gun effectively becomes part of your body mass. Instead of just the gun's mass ($5\,kg$) recoiling, it's the gun plus the user's mass ($100\,kg$), significantly lowering the recoil velocity.
     • Propping against the Earth: By laying down or propping against a wall, you rigidly connect yourself to the Earth. In this case, $m_1$ is effectively the mass of the gun + the mass of the human + the mass of the Earth. The recoil velocity becomes so small it is almost unnoticeable, as the momentum is absorbed by the planet.
• Battleship Application: Large guns are bolted to the deck of a battleship. This transfers recoil momentum to the entire ship. Springs on these guns do not reduce the total recoil momentum; they simply transfer it to the ship more gradually to prevent mechanical wear and tear.

Theory of Center of Mass (CM)

• Definition: The center of mass is a special point associated with an object that represents the motion of the object as a whole.
• Simplification of Systems: Instead of solving trillion-trillion projectile motion equations for every atom in a baseball, we solve one equation for the center of mass.
• Conceptual Meaning:
  • It is the balance point of an object.
  • Every object has only one correct center of mass.
• Motion of Irregular Objects:
  • A thrown tennis racket might tumble and spin through the air, appearing complicated. However, its center of mass travels on a smooth, predictable parabolic path.
  • Tumbling motion = Projectile motion of the CM + Rotation of the object about its CM.
• Free Rotation: Objects forced to spin but not bolted to an axle will naturally rotate around their center of mass.

Mathematical Identification of Center of Mass

• Coordinate Context: Center of mass is a point in space described by $x$, $y$, and $z$ coordinates relative to a fixed origin.
• Weighted Average Concept: The center of mass is a weighted average of the positions of all parts of the system based on their respective masses.
• General Equations:
  • $x_{cm} = \frac{(m_1 \times x_1 + m_2 \times x_2 + m_3 \times x_3 + ...)}{(m_1 + m_2 + m_3 + ...)}$
  • $y_{cm} = \frac{(m_1 \times y_1 + m_2 \times y_2 + m_3 \times y_3 + ...)}{(m_1 + m_2 + m_3 + ...)}$
  • $z_{cm} = \frac{(m_1 \times z_1 + m_2 \times z_2 + m_3 \times z_3 + ...)}{(m_1 + m_2 + m_3 + ...)}$
• Logic for the weighted average:
  • If one part has more mass, it contributes more to the weighted average, pulling the center of mass closer to that location (e.g., the scoop of a spoon).
  • Units: The kilograms on top ($m \times x$) cancel with the kilograms on the bottom ($\sum m$), leaving the result in meters.

Practical Methodologies and Tips for Finding CM

• Measurement Relative to Origin: All values for $x$, $y$, and $z$ must be measured from a fixed origin, NOT relative to the end of the object.
• Symmetric Objects: For objects with uniform density and shape (e.g., a solid steel sphere), the center of mass is the geometric middle.
• Subdividing Complex Objects: Break an irregular object into sub-parts where you can identify the CM of each part, then use the weighted average formula on those parts.
• Visual Check: After calculating, visualize whether the calculated point makes sense as a balance point. It should be closer to the denser/heavier side of the object.
• Location Outside the Object: The center of mass does not have to be physically touching the object (e.g., the center of mass of a horseshoe is in the empty space between the arms).

Questions & Discussion

• Question: "Are you talking about the firing pin hitting the bullet is the collision or the bullet hitting the object is the collision?"
• Response: The instructor clarified that for the recoil analysis, they are referring to the collision between the bullet and the gun during firing. The firing pin triggers an explosion which then fires the bullet; that explosive interaction is the collision of interest.
• Question: "So would you essentially be adding in your body into the system?"
• Response: Yes. If you lock your arm, you and the gun act as one single rigid object in the physics calculation.
• Question: "Would it be like $p_1 + p_2 + p_3$ because now you have added your body as part of the system?"
• Response: Effectively no, because you and the gun move as a single unit. Therefore, you are treated as one single massive object ($m_1$) versus the bullet ($m_2$). If your arms were not rigid ("spaghetti arms"), you wouldn't count as part of the gun, and the gun would recoil independently.