University Physics: Comprehensive Notes on Momentum, Collisions, and Impulse

Course Logistics and Upcoming Schedule

• Homework Assignments
      - The current homework is due on the 3rd (this Sunday) at 5:00 PM.
      - If a student feels comfortable with the dynamics questions on the homework, they should feel prepared for the upcoming quiz.
      - Submission process: Students should submit the homework by Sunday. The instructor will then send back the solutions, which can be used to study for the quiz on Monday.
      - Homework for the new topic, Momentum, will be posted this evening.
• Quiz Details
      - Quiz 4 is scheduled for the 4th (next Monday).
      - The format is very similar to the kinematics quiz: it will consist of four multiple-choice questions with multiple parts.
      - It will be graded and returned shortly after.
• Midterm Examination
      - The midterm is scheduled for the 11th (the Monday after next).
      - Unlike the quizzes, the midterm will be free response rather than multiple choice.
      - It covers two main topics: Kinematics and Dynamics.
      - There will be four questions total, with two questions dedicated to each topic.
      - Difficulty: It will be slightly harder and longer than the quizzes, but students will have the entire class time to complete it.
• Practice Midterm
      - A practice midterm session will be held on the 6th.
      - This session replaces the scheduled lab time.
      - Attendance is optional/not mandatory.
      - The practice exam will be sent out on the 4th for review, and the solutions will be discussed in person on the 6th. Eventually, these solutions will be posted online.

Conservation of Momentum in One Dimension

• Conceptual Overview
      - Momentum is conserved because, according to Newton's Third Law, if mass 1 ($m_1$) pushes on mass 2 ($m_2$) with a force ($F_{12}$), mass 2 must push back on mass 1 with an equal and opposite force ($F_{21}$).
      - These internal forces cancel each other out, resulting in a total net force of zero for the system.
      - When the net force ($F_{\text{net}}$) is zero, total momentum is conserved: $P_{\text{initial, total}} = P_{\text{final, total}}$.
• The Astronaut Scenario
      - Two astronauts are initially at rest in space ($v_1 = 0$, $v_2 = 0$).
      - Given data:
          - Mass 1 ($m_1$): $55\,kg$
          - Mass 2 ($m_2$): $85\,kg$
          - Final velocity of mass 1 ($v_3$): $-5\,m/s$ (the negative sign indicates movement to the left).
      - Conservation Equation:
          - $m_1 v_1 + m_2 v_2 = m_1 v_3 + m_2 v_4$
          - Since initial velocities are zero: $0 = m_1 v_3 + m_2 v_4$
      - Solving for the final velocity of mass 2 ($v_4$):
          - $v_4 = \frac{-(m_1 v_3)}{m_2}$
          - $v_4 = \frac{-(55\,kg \times -5\,m/s)}{85\,kg}$
          - $v_4 \approx 3.234\,m/s$
• Sanity Check
      - It makes sense that $m_2$ has a smaller velocity than $m_1$ because $m_2$ is more massive. Applying the same force to a larger mass results in smaller acceleration/velocity ($F = ma$).
      - In space, when two objects push off each other, the smaller object moves faster and the heavier object moves slower.
• Distance Calculation
      - To find the total distance between them after $10\,s$:
          - Displacement of Mass 1 ($\Delta d_A$): $-5\,m/s \times 10\,s = -50\,m$
          - Displacement of Mass 2 ($\Delta d_B$): $3.24\,m/s \times 10\,s = 32.4\,m$
          - Total distance is the sum of the absolute values: $|-50| + |32.4| = 82.4\,m$.

Types of Collisions

• Definition of a Collision
      - A collision occurs when two or more objects apply force to each other for a very short period of time (an instant).
• Elastic Collisions
      - Objects collide and bounce off each other, remaining separated afterward.
      - Momentum is conserved: $m_1 v_1 + m_2 v_2 = m_1 v_3 + m_2 v_4$.
      - Example: Bullet passing through a block
          - Bullet mass ($m_1$): $22\,g = 22 \times 10^{-3}\,kg$
          - Bullet initial velocity ($v_1$): $240\,m/s$
          - Block mass ($m_2$): $212\,kg$
          - Block initial velocity ($v_2$): $0\,m/s$
          - Bullet final velocity ($v_3$): $150\,m/s$
          - Solve for final block velocity ($v_4$): $v_4 = \frac{m_1 v_1 - m_1 v_3}{m_2}$.
• Inelastic Collisions
      - Objects collide and stick together, becoming one single mass.
      - Momentum is conserved: $m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_3$.
      - Example: Bullet embedded in a block
          - Bullet mass ($m_1$): $25\,g = 25 \times 10^{-3}\,kg$
          - Bullet initial velocity ($v_1$): $230\,m/s$
          - Block ($m_2$) is stationary ($v_2 = 0$).
          - Calculation of system velocity ($v_3$): $v_3 = \frac{m_1 v_1}{m_1 + m_2} \approx 4.04\,m/s$.
      - Height Calculation (Post-collision Kinematics)
          - Once the bullet-block system has its shared velocity ($v_3$), it may rise to a certain height ($\Delta d_y$).
          - This becomes a kinematics problem using the equation: $(v_{fy})^2 = (v_{iy})^2 + 2 a_y \Delta d_y$.
          - At the maximum height, final velocity in the y-direction is zero ($v_{fy} = 0$).

Conservation of Momentum in Two Dimensions

• Principle of Independence
      - If the net force in the x and y directions is zero, conservation of momentum occurs in both axes independently.
      - During a 2D collision between mass A and mass B, the forces exerted ($F_{AB}$ and $F_{BA}$) are equal and opposite in magnitude and direction.
      - This implies the components are also equal and opposite: $F_{AB,x} = -F_{BA,x}$ and $F_{AB,y} = -F_{BA,y}$.
• Equations for 2D Problem Solving
      - X-axis: $m_1 v_{1x} + m_2 v_{2x} = m_1 v_{3x} + m_2 v_{4x}$
      - Y-axis: $m_1 v_{1y} + m_2 v_{2y} = m_1 v_{3y} + m_2 v_{4y}$
• Vector Decomposition Example (Pool Balls)
      - Ball A ($3\,kg$): $5\,m/s$ at $30^\circ$ north of east.
          - $v_{1x} = 5 \cos(30^\circ) \approx 4.33\,m/s$
          - $v_{1y} = 5 \sin(30^\circ) = 2.5\,m/s$
      - Ball B ($3\,kg$): $4\,m/s$ at $140^\circ$ north of east.
          - $v_{2x} = 4 \cos(140^\circ) \approx -3.06\,m/s$
          - $v_{2y} = 4 \sin(140^\circ) \approx 2.57\,m/s$
      - Ball B Final ($v_4$): $3\,m/s$ at $80^\circ$.
          - $v_{4x} = 3 \cos(80^\circ) \approx 0.52\,m/s$
          - $v_{4y} = 3 \sin(80^\circ) \approx 2.95\,m/s$
      - Solving for Ball A's final velocity ($v_3$):
          - Solve the x-momentum equation to find $v_{3x}$.
          - Solve the y-momentum equation to find $v_{3y}$.
          - Combine components using the Pythagorean theorem: $v_3 = \sqrt{(v_{3x})^2 + (v_{3y})^2}$.
          - Find direction: $\theta = \tan^{-1}(\frac{v_{3y}}{v_{3x}})$.
• 2D Inelastic Collisions
      - Objects stick and move with a shared velocity $v_3$ in a specific 2D direction.
      - X-momentum: $m_1 v_{1x} + m_2 v_{2x} = (m_1 + m_2) v_{3x}$
      - Y-momentum: $m_1 v_{1y} + m_2 v_{2y} = (m_1 + m_2) v_{3y}$

Impulse and Change in Momentum

• Definition of Impulse ($I$)
      - Impulse is the change in momentum ($\Delta P$) experienced by a single object.
• Formulas for Impulse
      - Momentum method: $I = \Delta P = m v_f - m v_i$
      - Force method: Since $F_{\text{net}} = \frac{\Delta P}{\Delta t}$, then $I = F_{\text{net}} \times \Delta t$.
      - Here, $\Delta t$ is the "contact time" describing how long the net force acts on the object.
• One-Dimensional Impulse Examples
      - Horizontal: A $5\,kg$ ball traveling east at $5\,m/s$ hits a wall and bounces back at $-2\,m/s$.
          - $I = 5(-2) - 5(5) = -35\,kg \cdot m/s$.
      - Vertical (Falling Vase): A $4\,kg$ vase is dropped from $15\,m$ and stops when hitting the ground.
          - Scene 1 (The Fall): Find velocity just before impact ($v_{f1}$).
              - $(v_{f1})^2 = 0 + 2(-9.8)(-15)$
              - $v_{f1} \approx -17.15\,m/s$
          - Scene 2 (The Impact): Initial velocity ($v_{i2}$) = $-17.15\,m/s$. Final velocity ($v_{f2}$) = $0$.
              - $I = m(v_{f2} - v_{i2}) = 4(0 - (-17.15)) = 68.6\,kg \cdot m/s$.
• Force Analysis During Impact
      - Net Force Calculation: Using contact time $\Delta t = 10\,ms = 10 \times 10^{-3}\,s$.
          - $F_{\text{net}} = \frac{I}{\Delta t} = \frac{68.6}{0.01} = 6,860\,N$.
      - Normal Force Calculation: $F_{\text{net}}$ is the vector sum of the normal force ($F_n$) and gravity ($F_g$).
          - $F_{\text{net}} = F_n - F_g \Rightarrow F_n = F_{\text{net}} + mg$
          - $F_n = 6860 + (4 \times 9.8) = 6,899.2\,N$.
      - Vase Integrity: If the max force the vase can take is $100\,N$, it will certainly break under $6,899.2\,N$.
• Effect of a Cushion
      - A cushion increases the contact time ($\Delta t$).
      - If $\Delta t$ increases, $F_{\text{net}}$ decreases even if the total impulse ($I$) remains the same.
      - Example: With a cushion, $F_{\text{net}} \approx 27.44\,N$ and $F_n \approx 67\,N$. In this case, the vase survives because 67\,N < 100\,N.

Impulse in Two Dimensions

• Conceptual Model
      - Finding impulse in 2D involves calculating the change in momentum for both the x and y components.
      - X-impulse: $I_x = m v_{fx} - m v_{ix} = F_{\text{net},x} \Delta t$
      - Y-impulse: $I_y = m v_{fy} - m v_{iy} = F_{\text{net},y} \Delta t$
• Example: Diagonal Bounce
      - A $2\,kg$ ball strikes a surface at $3\,m/s$ at a $45^\circ$ angle and bounces back at the same speed and angle toward the left.
      - Components:
          - $v_{ix} = 3 \cos(45^\circ) \approx 2.12\,m/s$
          - $v_{iy} = 3 \sin(45^\circ) \approx 2.12\,m/s$
          - $v_{fx} = -3 \cos(45^\circ) \approx -2.12\,m/s$
          - $v_{fy} = 3 \sin(45^\circ) \approx 2.12\,m/s$
      - Calculated Impulses:
          - $I_x = 2(-2.12 - 2.12) = -8.48\,kg \cdot m/s$
          - $I_y = 2(2.12 - 2.12) = 0\,kg \cdot m/s$ (the vertical component of velocity remained unchanged).

Questions & Discussion

• Student Question: "Where'd you get the four from for the equation?"
• Instructor Response: That's the mass ($m$). The mass of the vase is $4\,kg$.
• Student Question: "Exactly?"
• Instructor Response: "Yes."