Physics 10-08 Friday Linear Momentum, Impulse, and Conservation Laws

Course Announcements and Pedagogical Framework

• Teaching Analysis Poll and Course Adjustments:

  • Aggregate student feedback provided through coach Kell was utilized to adjust course delivery.

  • Roster photos have been printed and organized to assist in memorizing all student names, despite minor changes in student appearances since photos were taken.

  • Lecture transitions are explicitly signposted to prevent confusion, summarizing completed concepts before outlining subsequent topics.

  • A take-home practice examination will be distributed next Wednesday to prepare students for the final examination.

• Foundational Conceptual Context:

  • Conservation laws—specifically conservation of mass, energy, and momentum—serve as a foundational "zeroth law" underpinning Newton's three laws of motion.

  • By analogy to the Ten Commandments and the zeroth law of Moses, conservation principles represent the fundamental physical realities from which dynamical laws emerge.

  • Introductory music played prior to class was John Michael Talbot from Isaiah.

• Linear Momentum Lecture Outline:

  • Review of inertia and the inertia concept map.

  • Definition of linear momentum.

  • Integration of the mass-velocity product into Newton's second law.

  • Derivation of impulse as a force applied across time ($\Delta p = F_{\text{net}} \Delta t$).

  • Derivation of conservation of linear momentum by combining Newton's first and third laws.

  • Practical scenarios: recoil from rest and perfectly inelastic collisions.

  • Joint analysis of conservation of momentum and conservation of energy is reserved for Monday's session.

The Inertia Map: Kinematic and Dynamical Interconnections

• Central Role of Velocity:

  • Velocity ($v$) serves as the core physical quantity from which dynamical variables branch.

  • The time rate of change of velocity is acceleration ($a = \frac{\Delta v}{\Delta t}$).

• Acceleration Across Dimensions:

  • Across Time: Acceleration acting over a duration produces a change in velocity ($\Delta v = a \Delta t$).

  • Across Space: Acceleration acting across displacement produces a change in half-squared velocity ($\Delta (\frac{1}{2} v^2) = a \Delta x$).

• Mass Coupling (Dynamics):

  • Multiplying velocity change by mass yields a change in linear momentum ($\Delta p = m \Delta v$).

  • Dividing change in momentum by elapsed time yields the net force ($F_{\text{net}} = \frac{\Delta p}{\Delta t} = m a$).

  • Multiplying half-squared velocity by mass yields kinetic energy ($K = \frac{1}{2} m v^2$).

  • Net force integrated across displacement produces mechanical work, which equals the change in kinetic energy ($W_{\text{net}} = F_{\text{net}} \Delta x = \Delta K$).

  • Dividing dynamic quantities (momentum, force, energy) by mass returns the purely kinematic quantities (velocity, acceleration).

Definition and Directionality of Linear Momentum

• Mathematical Definition:

  • Linear momentum ($p$) is the product of inertial mass ($m$) and velocity ($v$):     $p = m v$

  • The variable symbol $p$ originates from classical linguistic conventions (Latin petere / impetus, alongside Greek roots) avoiding the already occupied variable $m$ for mass.

  • In relativistic regimes where velocity approaches the speed of light ($c$), momentum incorporates the Lorentz factor ($\gamma$); however, classical mechanics treats momentum strictly as $p = m v$.

  • Momentum is an intrinsically vector quantity, possessing identical direction and orientation to the velocity vector.

  • The standard SI unit of linear momentum is kilograms times meters per second ($\text{kg}\,\text{m/s}$).

• Two-Dimensional Vector Example:

  • Consider an object with mass $m = 10\,\text{kg}$ and velocity vector components $v_x = 5\,\text{m/s}$ and $v_y = -7\,\text{m/s}$.

  • The components of the linear momentum vector are calculated via scalar multiplication:     $p_x = m v_x = (10\,\text{kg})(5\,\text{m/s}) = 50\,\text{kg}\,\text{m/s}$     $p_y = m v_y = (10\,\text{kg})(-7\,\text{m/s}) = -70\,\text{kg}\,\text{m/s}$

  • Represented as a vector coordinate pair:     $p = (50, -70)\,\text{kg}\,\text{m/s}$

  • The scalar magnitude of momentum can be retrieved via the Pythagorean theorem:     $\|p\| = \sqrt{p_x^2 + p_y^2} = \sqrt{(50\,\text{kg}\,\text{m/s})^2 + (-70\,\text{kg}\,\text{m/s})^2} = \sqrt{2500 + 4900} = \sqrt{7400} \approx 86.02\,\text{kg}\,\text{m/s}$

• Velocity Comparison: Tiger Versus Gazelle:

  • Scenario: A $400\,\text{kg}$ tiger and a $200\,\text{kg}$ gazelle both run with a linear momentum of $20{,}000\,\text{kg}\,\text{m/s}$.

  • Rearranging the momentum definition for velocity:     $v = \frac{p}{m}$

  • Velocity calculation for the tiger:     $v_{\text{tiger}} = \frac{20{,}000\,\text{kg}\,\text{m/s}}{400\,\text{kg}} = 50\,\text{m/s}$

  • Velocity calculation for the gazelle:     $v_{\text{gazelle}} = \frac{20{,}000\,\text{kg}\,\text{m/s}}{200\,\text{kg}} = 100\,\text{m/s}$

  • Comparative analysis: With half the mass, the gazelle maintains exactly twice the velocity of the tiger ($v_{\text{gazelle}} = 2 v_{\text{tiger}}$). Under realistic biological limits, gazelles sprint up to roughly $100\,\text{km/h}$ rather than $100\,\text{m/s}$. Given predator-prey dynamics, both vectors are directed along the same collinear line of action.

Impulse and Newton's Second Law

• Derivation from Newton's Second Law:

  • Newton's second law defines acceleration as net force divided by mass:     $a = \frac{\Delta v}{\Delta t} = \frac{F_{\text{net}}}{m}$

  • Multiplying both sides by mass:     $m \frac{\Delta v}{\Delta t} = F_{\text{net}}$

  • Because mass is constant in classical formulations, $m \Delta v = \Delta (m v) = \Delta p$:     $\frac{\Delta p}{\Delta t} = F_{\text{net}}$

  • Multiplying across the interaction time interval ($\Delta t$) yields the impulse-momentum relationship:     $\Delta p = F_{\text{net}} \Delta t$

• Definition of Impulse:

  • Impulse is defined as the change in momentum ($\Delta p$) produced by an applied net force acting over a discrete time interval ($\Delta t$).

  • The units of impulse are equivalent to momentum: $\text{kg}\,\text{m/s}$ or Newton-seconds ($\text{N}\,\text{s}$).

• Single-Axis Impulse Application (Golf Ball off a Tee):

  • Given parameters: A golf ball of mass $m = 0.045\,\text{kg}$ sits at rest ($v_i = 0\,\text{m/s}$) on a tee and is launched at a final velocity $v_f = 38\,\text{m/s}$. The club-ball contact duration is $\Delta t = 3.5\,\text{ms} = 3.5 \times 10^{-3}\,\text{s}$.

  • Change in momentum (impulse):     $\Delta p = m (v_f - v_i) = (0.045\,\text{kg})(38\,\text{m/s} - 0\,\text{m/s}) = 1.71\,\text{kg}\,\text{m/s}$

  • Average net force exerted by the club on the ball:     $F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{1.71\,\text{kg}\,\text{m/s}}{3.5 \times 10^{-3}\,\text{s}} \approx 488.57\,\text{N} \approx 489\,\text{N}$

  • The direction of the force is identical to the forward trajectory vector of the launched ball.

Two-Dimensional Vector Impulse and Elastic Rebound

• Problem Setup:

  • A ball with mass $m = 0.06\,\text{kg}$ approaches a reflective barrier parallel to the y-axis at an initial speed of $28\,\text{m/s}$ directed at an absolute angle of $45^\circ$.

  • The ball rebounds at the same speed ($28\,\text{m/s}$) along a trajectory rotated by an additional $90^\circ$, making its absolute angle $135^\circ$.

• Momentum Magnitudes and Components:

  • Magnitude of momentum before and after impact:     $\|p_1\| = \|p_2\| = m v = (0.06\,\text{kg})(28\,\text{m/s}) = 1.68\,\text{kg}\,\text{m/s}$

  • Initial momentum vector components ($\theta_1 = 45^\circ$):     $p_{1x} = \|p_1\| \cos(45^\circ) = 1.68 \left(\frac{\sqrt{2}}{2}\right) = \frac{1.68}{\sqrt{2}}\,\text{kg}\,\text{m/s}$     $p_{1y} = \|p_1\| \sin(45^\circ) = 1.68 \left(\frac{\sqrt{2}}{2}\right) = \frac{1.68}{\sqrt{2}}\,\text{kg}\,\text{m/s}$

  • Final momentum vector components ($\theta_2 = 135^\circ$):     $p_{2x} = \|p_2\| \cos(135^\circ) = 1.68 \left(-\frac{\sqrt{2}}{2}\right) = -\frac{1.68}{\sqrt{2}}\,\text{kg}\,\text{m/s}$     $p_{2y} = \|p_2\| \sin(135^\circ) = 1.68 \left(\frac{\sqrt{2}}{2}\right) = \frac{1.68}{\sqrt{2}}\,\text{kg}\,\text{m/s}$

• Impulse Vector Determination:

  • Impulse is defined as final state minus initial state ($\Delta p = p_2 - p_1$).

  • Component in the x-direction:     $\Delta p_x = p_{2x} - p_{1x} = -\frac{1.68}{\sqrt{2}} - \frac{1.68}{\sqrt{2}} = -2 \left(\frac{1.68}{\sqrt{2}}\right) = -1.68 \sqrt{2} \approx -2.38\,\text{kg}\,\text{m/s}$

  • Component in the y-direction:     $\Delta p_y = p_{2y} - p_{1y} = \frac{1.68}{\sqrt{2}} - \frac{1.68}{\sqrt{2}} = 0\,\text{kg}\,\text{m/s}$

  • Physical validation check: The initial momentum was oriented toward positive x and the final momentum toward negative x; therefore, the impulse vector must be negative along x. The reflection boundary is parallel to the y-axis, meaning no vertical forces act on the ball during contact, ensuring zero change in vertical momentum ($\Delta p_y = 0$).

Conservation of Momentum Derived from Newton's Laws

• Synthesis of First and Third Laws:

  • Newton's First Law: An isolated entity maintains constant velocity unless subjected to a net external force. In the absence of external forces, $\Delta p = 0$.

  • Newton's Third Law: Forces always occur in matched action-reaction pairs between interacting bodies:     $F_{12} = -F_{21}$

  • If two interacting bodies exert forces on one another over a mutual contact duration $\Delta t$:     $F_{12} \Delta t = -F_{21} \Delta t$

  • Consequently, the impulse delivered to body 1 is equal in magnitude and opposite in sign to the impulse delivered to body 2:     $\Delta p_1 = -\Delta p_2$

  • Grouping the bodies into a single closed thermodynamic/mechanical system:     $\Delta p_{\text{total}} = \Delta p_1 + \Delta p_2 = 0$

  • Internal forces cancel out within system boundaries. Total momentum is strictly conserved across time unless net external forces breach the system boundary.

• Macroscopic Example: The Physics of Jumping:

  • A person jumping pushes downward against the Earth with a normal force exceeding their gravitational weight.

  • By Newton's third law, the Earth exerts an identical upward force on the jumper over the same contact duration.

  • The upward momentum gained by the jumper is exactly balanced by downward momentum delivered to the Earth ($\Delta p_{\text{jumper}} = -\Delta p_{\text{Earth}}$).

  • Because velocity change is governed by $v = \frac{p}{m}$, and the mass of the Earth is extraordinarily large ($m_{\text{Earth}} \approx 5.97 \times 10^{24}\,\text{kg}$), the resultant downward velocity change imparted to the planet is imperceptibly close to zero.

Recoil Dynamics: Package Thrown from a Boat

• Problem Formulation:

  • A boy in a rowboat at rest throws a package horizontally.

  • Boy mass: $m_{\text{boy}} = 24\,\text{kg}$

  • Boat mass: $m_{\text{boat}} = 35\,\text{kg}$

  • Package mass: $m_b = 5.3\,\text{kg}$

  • Velocity of package relative to ground: $v_b = 10\,\text{m/s}$ in the positive x-direction.

  • Combined mass of remainder (system a): $m_a = m_{\text{boy}} + m_{\text{boat}} = 24\,\text{kg} + 35\,\text{kg} = 59\,\text{kg}$

• Conservation Equation:

  • System initial state is at rest:     $p_{\text{initial}} = p_a + p_b = 0$

  • System final state post-throw:     $p_{\text{final}} = m_a v_a + m_b v_b = 0$

  • Solving for the recoil velocity of the boat and boy ($v_a$):     $v_a = -\frac{m_b v_b}{m_a}$     $v_a = -\frac{(5.3\,\text{kg})(10\,\text{m/s})}{59\,\text{kg}} = -\frac{53\,\text{kg}\,\text{m/s}}{59\,\text{kg}} \approx -0.898\,\text{m/s}$

  • The negative sign indicates the boat and boy recoil backward in the negative x-direction, counterbalancing the forward package momentum.

Ballistic Pendulum: Inelastic Collision and Energy Constraints

• Two-Phase Separation:

  • A ballistic pendulum consists of a bullet of mass $m$ fired at high speed $v$ into a stationary wooden block of mass $M$ suspended by a rod or string of length $L$.

  • The physical process divides into two distinct operational phases: the collision phase (governed strictly by conservation of momentum) followed by the swing phase (governed by conservation of mechanical energy).

• Phase 1: Perfectly Inelastic Collision:

  • Definition: A collision is perfectly inelastic when colliding objects stick together post-impact, moving with a single shared common velocity.

  • Kinetic energy is non-conserved during collision due to thermal dissipation and deformation; momentum is strictly conserved.

  • Initial momentum:     $p_i = m v + M(0) = m v$

  • Final momentum immediately post-embedding:     $p_f = (m + M) v_1$

  • Equating initial and final momentum to find post-collision speed ($v_1$):     $v_1 = \frac{m v}{m + M}$

• Phase 2: Post-Collision Swing Phase (Height $h = 2L$):

  • Following impact, kinetic energy transforms into gravitational potential energy as the system ascends through height $h = 2L$.

  • Case 1: Rigid Rod Suspension:

  • A rigid rod supports compression and will not buckle when velocity drops to zero at the top.

  • The minimum condition to complete the swing to the apex ($2L$) is that the system velocity reaches zero at the top ($v_{\text{top}} = 0$).

  • All initial kinetic energy is converted fully to potential energy:       $\frac{1}{2}(m + M) v_1^2 = (m + M) g (2L)$       $v_1 = \sqrt{4 g L} = 2\sqrt{g L}$

  • Case 2: Flexible String Suspension:

  • A string cannot support compressive loads and will go slack if tension falls to zero before reaching the top.

  • To maintain a continuous circular arc, tension must satisfy $T \ge 0$ at the apex, dictating that centripetal acceleration must at minimum match gravity:       $\frac{v_{\text{top}}^2}{L} \ge g \implies v_{\text{top}}^2 \ge g L$

  • Non-zero kinetic energy is required at the apex; hence, the initial kinetic energy post-collision must be substantially greater for a string than for a rigid rod.

Questions and Classroom Dialogue

• Question on Momentum Vector Versus Scalar Magnitude:

  • Student Inquiry: How do students know whether an assessment question requires general momentum or the magnitude of momentum?

  • Instructor Response: If the question requires spatial orientation for further analysis, directional components must be supplied. Science emphasizes retaining maximal scenario information rather than reporting minimum scalars. If asked exclusively for the x-component of momentum, one reports only $p_x = m v_x$ (e.g., $50\,\text{kg}\,\text{m/s}$) and disregards the orthogonal axes.

• Question on Algebraic Elimination in Velocity Ratio:

  • Student Inquiry: When manipulating $p = m v$ to $\frac{p}{m} = v$, is the mass factor eliminated or canceled out?

  • Instructor Response: Dividing both sides of the momentum definition by the mass isolates velocity on one side, algebraically clearing mass from the velocity expression.

• Question on Units of Momentum and Impulse:

  • Student Inquiry (Cash): During calculation of the golf ball momentum change ($1.71$), the units were suggested to be $\text{m}^2/\text{s}$.

  • Instructor Response: Multiplying mass ($\text{kg}$) by velocity ($\text{m/s}$) yields units of kilograms times meters per second ($\text{kg}\,\text{m/s}$), not $\text{m}^2/\text{s}$.

• Question on Mathematical Order and Sign Error (Melanie):

  • Student Inquiry (Melanie): In the 2D rebound calculation, an error appeared where subtracting the values resulted in zero or an inverted sign.

  • Instructor Response: The initial calculation mistakenly subtracted the later event from the earlier event rather than subtracting initial from final ($\Delta p = p_{\text{final}} - p_{\text{initial}}$). Reversing the subtraction corrected the sign error, yielding the requisite negative impulse along the x-axis ($-1.68\sqrt{2}\,\text{kg}\,\text{m/s}$).

• Question on Boat Recoil Evaluation (Sophia):

  • Student Inquiry (Sophia): Clarification requested on the arithmetic division for the boat recoil equation, as entering the values yielded unexpected numbers.

  • Instructor Response: The calculator entry accidentally inverted the numerator and denominator ($59 / 53$ instead of $53 / 59$). The explicit correct expression is $-(5.3\,\text{kg} \times 10\,\text{m/s}) / (24\,\text{kg} + 35\,\text{kg}) = -53 / 59\,\text{m/s}$, with final units of meters per second ($\text{m/s}$).