Momentum, Impulse, and Conservation Laws in Linear and Angular Systems

Upcoming Exam and Logistics

• Administrative Note: A practice test for the exam coming up on Monday will be conducted.
• Exam Coverage: The exam will cover all material since the last test, including:     * Rotation.     * Static equilibrium.     * Momentum concepts (covered in this session).

Introduction to Momentum and Impulse

• Definition of Momentum ($p$): Momentum is defined as the product of an object's mass and its velocity.     * Formula: $\mathbf{p} = m \times \mathbf{v}$.     * Units: $\text{kilograms} \times \text{meters per second}$ ($kg\,m/s$).     * Vector Nature: Momentum is a vector quantity, meaning the direction is significant in all calculations.
• Definition of Impulse ($J$): An impulse is an interaction between a system and its environment that results in a change in the system's momentum. Anything with mass that is moving possesses momentum; impulse is what changes that momentum.

The Impulse-Momentum Theorem

• First Way to Express Impulse ($J$): Impulse is simply the change in momentum ($\Delta p$).     * Formula: $J = \Delta p = p_{final} - p_{initial}$.     * Note: Because these are vectors, one must be careful with signs to ensure directions are calculated correctly.
• Second Way to Express Impulse ($J$): Impulse can be detailed as an average force ($F_{avg}$) acting over a specific time interval ($\Delta t$).     * Formula: $J = F_{avg} \times \Delta t$.
• Impulse-Momentum Theorem: Combining these definitions, if more than one force acts on a system, the sum of those forces over a time interval equals the change in momentum.     * Formula: $\sum F_{avg} \times \Delta t = \Delta p$.

Graphical Interpretation of Impulse

• Relationship to Calculus: While this is not a calculus-based course, the impulse exerted by a force is technically the area under the curve of a Force vs. Time graph.
• Integration: The impulse is the integral of the force function from $t_{initial}$ to $t_{final}$.
• Simplified Averaging: To avoid complex integration, the concept of average force is used. The average force over a time interval ($\Delta t$) creates a rectangle whose area ($F_{avg} \times \Delta t$) is exactly equal to the area under the complex force curve. This allows the calculation of impulse without knowing the specific function of the force over time.

Example Problem: Baseball and Bat

• Scenario: A baseball is hit by a bat. In this system, the ball is the system of interest.
• Given Data:     * Initial Velocity ($v_i$): $40\,m/s$ (coming from the left, defined as $-40\,m/s$ in the negative x-direction).     * Final Velocity ($v_f$): $50\,m/s$ (moving to the right, defined as $+50\,m/s$ in the positive x-direction).     * Time Interval ($\Delta t$): $0.01\,s$.     * Mass of Baseball ($m$): $0.15\,kg$.
• Task 1: Find the Impulse ($J$):     * Using $J = \Delta p = m(v_f - v_i)$.     * Calculation: $0.15\,kg \times (50\,m/s - (-40\,m/s)) = 0.15 \times (50 + 40) = 0.15 \times 90$.     * Result: $13.5\,kg\,m/s$ in the positive x-direction ($+13.5\,\mathbf{\hat{x}}$).     * Units Note: Units for impulse are the same as momentum: $kg\,m/s$.
• Task 2: Find the Average Force ($F_{avg}$):     * Using $F_{avg} = \frac{J}{\Delta t}$.     * Calculation: $F_{avg} = \frac{13.5}{0.01}$.     * Result: $1350\,N$.     * Verification of Units: $\frac{kg\,m/s}{s} = kg\,m/s^2$, which is equivalent to Newtons ($N$).

Conservation of Momentum

• Condition for Conservation: The momentum of a system remains constant (it does not change) if there is no net external force acting on it.
• Isolated System: This principle applies to an isolated system where the sum of external forces is zero ($\sum F_{ext} = 0$).
• Conservation Statement: The total initial momentum equals the total final momentum ($\sum p_{initial} = \sum p_{final}$).

Example Problem: Archer on Ice

• Scenario: An archer stands on a frictionless, ice-covered pond and fires an arrow horizontally.
• Given Data:     * Archer Mass ($m_1$): $60\,kg$.     * Arrow Mass ($m_2$): $0.5\,kg$.     * Arrow Final Speed ($v_{2f}$): $50\,m/s$.
• Analysis:     * System: Archer and the arrow together.     * Horizontal Dimension: There are no horizontal external forces (friction is zero). Gravity and normal force act vertically but cancel each other out ($F_{net,y} = 0$).     * Initial State: Archer and arrow are stationary; $\sum p_{initial} = 0$.     * Final State: Final momentum must also be zero ($m_1 v_{1f} + m_2 v_{2f} = 0$).
• Recoil Velocity Calculation:     * Setting up the equation: $0 = m_1 v_{1f} + m_2 v_{2f}$.     * $(60\,kg) v_{1f} + (0.5\,kg)(50\,m/s) = 0$.     * $60 v_{1f} = -25$.     * Result: Archer's final speed is approximately $0.42\,m/s$ (moving in the opposite direction of the arrow, or the $-x$ direction).
• System Definition Note: If the system were defined only as the arrow, momentum would not be conserved because the bow exerts an external force on the arrow. Additionally, over time, gravity acts as an external force changing the arrow's vertical momentum.

Collisions and Kinetic Energy

• Elastic Collisions:     * Definition: Kinetic energy ($KE$) of the system is conserved.     * Note: These will be discussed after the exam once work and energy concepts are introduced. Elastic collisions are rare (e.g., a ball dropped will usually not bounce back to its exact initial height).
• Inelastic Collisions:     * Definition: Total kinetic energy of the system changes (it is not conserved).     * Mechanism: Energy is transformed into other types, such as thermal energy (heating up), sound energy, or energy used to deform the objects.     * Momentum Note: Conservation of momentum still typically applies to these systems.
• Perfectly Inelastic Collisions:     * Definition: The colliding objects stick together and move as a single combined mass after the collision.     * Energy: These collisions involve the maximum loss of kinetic energy.

Example Problem: Perfectly Inelastic Car Collision

• Scenario: A parked car at a traffic light is hit by another car, and they stick together.
• Given Data:     * Parked Car Mass ($m_1$): $1800\,kg$ ($v_{1i} = 0$).     * Moving Car Mass ($m_2$): $900\,kg$.     * Incoming Speed ($v_{2i}$): $20\,m/s$.
• Assumptions: We approximate horizontal friction as zero during the small time interval of the collision, as it is insignificant compared to the impact forces.
• Calculation:     * $m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f$.     * $(1800)(0) + (900)(20) = (1800 + 900) v_f$.     * $18000 = 2700 v_f$.     * Result: $v_f = 6.67\,m/s$.     * Direction: If the initial car was moving in the positive x-direction, the final velocity is also in the positive x-direction ($+6.67\,\mathbf{\hat{x}}$).

Two-Dimensional Collisions

• Approach: For 2D collisions, the conservation of momentum is applied to each dimension (x and y) independently.
• Steps:     1. Analyze the x-dimension momentum conservation.     2. Analyze the y-dimension momentum conservation.     3. Combine the resulting components to find the final velocity vector's magnitude and direction.

Example Problem: 2D Car Collision at an Intersection

• Scenario: Two cars collide at an intersection and stick together (Perfectly Inelastic).
• Aside/Dialogue Interruption:     * In the middle of discussing the angle in the diagram, an interruption occurs: "Mom, mom looking at some angle. Can I please have a slice of toast? Can I please have a slice of toast with butter?"
• Given Data:     * Car 1 Mass ($m_1$): $1500\,kg$, traveling at $25\,m/s$ to the right ($+x$).     * Car 2 Mass ($m_2$): $2500\,kg$, traveling at $20\,m/s$ upward ($+y$).
• X-Dimension Conservation:     * $m_1 v_{1ix} + m_2 v_{2ix} = (m_1 + m_2) v_{fx}$.     * $(1500)(25) + (2500)(0) = (1500 + 2500) v_{fx}$.     * $37500 = 4000 v_{fx}$.     * $v_{fx} = 9.375\,m/s$.
• Y-Dimension Conservation:     * $m_1 v_{1iy} + m_2 v_{2iy} = (m_1 + m_2) v_{fy}$.     * $(1500)(0) + (2500)(20) = (4000) v_{fy}$.     * $50000 = 4000 v_{fy}$.     * $v_{fy} = 12.5\,m/s$.
• Combining Components:     * Magnitude: $v_f = \sqrt{v_{fx}^2 + v_{fy}^2} = \sqrt{9.375^2 + 12.5^2} = 15.6\,m/s$.     * Angle: $\tan(\theta) = \frac{v_{fy}}{v_{fx}}$ results in $\theta = 53.1^\circ$ counterclockwise from the positive x-axis.

Angular Momentum

• Definitions of Angular Momentum ($L$):     * Cross Product Definition: $L = r \times p$.     * Magnitude: $L = |r| \times |p| \times \sin(\text{angle between them})$.     * Rotation Definition: $L = I \times \omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity.
• Direction: Determined by the right-hand rule. Curl fingers in the direction of rotation, and the thumb points in the direction of both $\omega$ and $L$.

Conservation of Angular Momentum

• Condition: The net external torque acting on the system must be zero ($\sum \tau_{ext} = 0$).
• Isolated System Statement: The total initial angular momentum equals the total final angular momentum ($\sum L_{initial} = \sum L_{final}$, or $I_i \omega_i = I_f \omega_f$).

Example Problem: Rotating Cylinders

• Scenario: A rotating cylinder (Moment of Inertia $I_1$, speed $\omega_{1i}$) has a stationary cylinder ($I_2$, speed $0$) dropped onto it along the same axis.
• Outcome: The two eventually reach a common final angular speed ($\omega_{f}$).
• Calculation:     * $I_1 \omega_{1i} = (I_1 + I_2) \omega_f$.     * $\omega_f = \omega_{1i} \times \left( \frac{I_1}{I_1 + I_2} \right)$.     * Result: Since the factor in parentheses is less than 1, the final angular speed is less than the initial speed of the bottom cylinder.

Example Problem: Figure Skater with Weights

• Scenario: A figure skater rotates on frictionless ice while holding two weights. They bring weights in from their arms out position to closer to their body.
• Initial Data:     * Skater Moment of Inertia ($I_{skater}$): $3\,kg\,m^2$.     * Initial Weight Radius ($r_i$): $1\,m$.     * Weight Mass ($m_{weight}$): $3\,kg$ each (two weights total).     * Initial Angular Speed ($\omega_i$): $0.75\,rad/s$.
• Final Data:     * Final Weight Radius ($r_f$): $0.3\,m$.
• Moment of Inertia of Weights (Point Particles):     * Initial ($I_{wi}$): $2 \times m r_i^2 = 2 \times (3) \times (1)^2 = 6\,kg\,m^2$.     * Final ($I_{wf}$): $2 \times m r_f^2 = 2 \times (3) \times (0.3)^2 = 0.54\,kg\,m^2$.
• Calculation:     * $L_i = L_f$.     * $(I_{skater} + I_{wi})\omega_i = (I_{skater} + I_{wf})\omega_f$.     * $(3 + 6)(0.75) = (3 + 0.54)\omega_f$.     * $9(0.75) = 3.54 \omega_f$.     * $6.75 = 3.54 \omega_f$.     * Result: $\omega_f = 1.91\,rad/s$.
• Conclusion: Bringing arms in reduces the moment of inertia, which causes an increase in angular speed to conserve angular momentum.

Questions & Discussion

• Q: But don't any questions so far?     * A: (None recorded in transcript; moves to calculus interpretation).
• Q: Any questions on that graphical interpretation of impulse?     * A: (None recorded; moves to baseball example).
• Q: Any questions on how this works here on this example problem?     * A: (None recorded; moves to conservation of momentum).
• Q: Any questions on this example? (Referring to Archer).     * A: (None recorded; emphasizes system definition).
• Q: Any questions so far [regarding collisions]?     * A: (Speaker makes aside about rarity of elastic collisions).
• Q: Are there external forces acting on the two cars? During the collision. What would you say? Are there external?     * A: Gravity and normal force act vertically but cancel out. Friction exists horizontally but is approximated as zero for the duration of the short collision.
• Q: Any questions on this collision example?     * A: (None recorded; moves to 2D collisions).
• Q: Any questions about this example? (Intersection collision).     * A: No. Everybody's good.
• Q: Any questions on this concept? (Rotating cylinders).     * A: (None recorded; moves to figure skater example).
• Q: Any questions on that example? (Figure skater).     * A: (Speaker proceeds to open learning analytics).