Impulse and Momentum Physics Study Guide

Definition of Linear Impulse: Linear impulse is a vector quantity that serves to measure the overall effect of a force during the specific interval of time that the force is acting upon an object.
Directionality and Scalarity:
- Time is categorized as a positive scalar quantity.
- Consequently, the impulse vector acts in the exact same direction as the force vector.
Units and Magnitude: The magnitude of impulse is characterized by the units of force multiplied by time, typically expressed as Newton-seconds ( $\text{N} \cdot \text{s}$ ) or $\text{kg} \cdot \text{m/s}$ .
Mathematical Determination via Integration:
- When force is provided as a function of time ( $F(t)$ ), the impulse is determined by evaluating the integral of the force over the time interval.
- In the specific scenario where the force is constant in both its magnitude and its direction, the resulting impulse simplifies to the product of the force and the duration of time ( $I = F \Delta t$ ).

Conceptual Definition: The scientific definition of linear momentum aligns with intuitive physical understanding: a large, fast-moving object possesses significantly greater momentum than a smaller, slower-moving object.
Formal Definition: Linear momentum is defined as the product of a system's total mass ( $m$ ) and its instantaneous velocity ( $\mathbf{v}$ ).
Mathematical Symbolism: Momentum is generally expressed using the symbol $L$ (or $\mathbf{p}$ ), calculated as:
- $L = m \times \mathbf{v}$

Sample Problem 1:
- Scenario: A $3\,\text{kg}$ ball is rolling with a velocity of $2\,\text{m/s}$ .
- Objective: Determine the linear momentum.
- Calculation: $L = 3\,\text{kg} \times 2\,\text{m/s}$ .
- Result: $L = 6\,\text{kg} \cdot \text{m/s}$ .
Sample Problem 2:
- Scenario: A $2.5\,\text{kg}$ ball is moving at $6\,\text{m/s}$ . The ball’s velocity subsequently changes to $2\,\text{m/s}$ .
- Objective: Calculate the change in linear momentum ( $\Delta L$ ).
- Process:
  - Initial Momentum ( $L_i$ ): $2.5\,\text{kg} \times 6\,\text{m/s} = 15\,\text{kg} \cdot \text{m/s}$ .
  - Final Momentum ( $L_f$ ): $2.5\,\text{kg} \times 2\,\text{m/s} = 5\,\text{kg} \cdot \text{m/s}$ .
  - Change ( $\Delta L$ ): $L_f - L_i = 5 - 15 = -10\,\text{kg} \cdot \text{m/s}$ .

Principles of Particle Systems: When applying the principle of impulse and momentum to a system containing multiple particles, collisions between those particles generate internal impulses.
Internal vs. External Impulses:
- Internal Impulses: These are equal, opposite, and collinear. Due to Newton's Third Law, they cancel each other out within the system equation.
- External Impulses: If an external impulse is considered small—defined by a small force acting over a very short duration of time—it is classified as "nonimpulsive."
Negligibility and Conservation: Nonimpulsive forces can be neglected in calculations. Consequently, the total momentum for the system of particles is conserved.
Application: The conservation-of-momentum equation is primarily used to find the final velocity of a particle after internal impulses (collisions) occur, provided the initial velocities are known.
Isolated Analysis: If the specific value of an internal impulse needs to be determined, one must isolate a single particle from the system and apply the principle of impulse and momentum to that specific particle individually.

Sample Problem 3 (System Conservation):
- Instruction: "A $15\,\text{kg}$ ball moving at $12\,\text{m/s}$ collides with a stationary $8\,\text{kg}$ ball, and after the collision, the $10\,\text{kg}$ ball continues in the same direction at $4\,\text{m/s}$ . Find the velocity of the $5\,\text{kg}$ ball post-collision."
- Note: This transcript entry includes conflicting mass values ( $15 \to 10$ and $8 \to 5$ ) for the interaction.
Practice Problem (Cart Collision):
- Scenario: A $5\,\text{kg}$ cart moving at $10\,\text{m/s}$ to the right collides with a $3\,\text{kg}$ cart moving at $4\,\text{m/s}$ to the left.
- Post-Collision: After the impact, the $5\,\text{kg}$ cart reverses direction and moves at $2\,\text{m/s}$ to the left.
- Objective: Find the final velocity of the $3\,\text{kg}$ cart.
- Setup Solution:
  - $m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$
  - $(5)(10) + (3)(-4) = (5)(-2) + (3)(v_2')$
  - $50 - 12 = -10 + 3v_2'$
  - $38 + 10 = 3v_2'$
  - $v_2' = 16\,\text{m/s}$ (to the right).

Definition of Elastic Collisions: Collisions where both momentum and internal kinetic energy are conserved.
Governing Equations:
- Conservation of Momentum: $m_A \mathbf{v}_A + m_B \mathbf{v}_B = m_A \mathbf{v}_A' + m_B \mathbf{v}_B'$
- Conservation of Internal Kinetic Energy: $\frac{1}{2} m_A (\mathbf{v}_A)^2 + \frac{1}{2} m_B (\mathbf{v}_B)^2 = \frac{1}{2} m_A (\mathbf{v}_A')^2 + \frac{1}{2} m_B (\mathbf{v}_B')^2$
Sample Problem 4:
- Scenario: A soccer ball with a mass of $10\,\text{kg}$ moving at $24\,\text{m/s}$ collides with a stationary $12\,\text{kg}$ soccer ball and comes to a complete rest.
- Objective: Calculate the final velocity of the $12\,\text{kg}$ soccer ball after this elastic collision.
- Solution: $10(24) + 12(0) = 10(0) + 12(v_f) \rightarrow 240 = 12v_f \rightarrow v_f = 20\,\text{m/s}$ .

Characteristics of Inelastic Collisions:
- Momentum is conserved: $m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2 = (m_1 + m_2) \mathbf{v}_f$ .
- Internal kinetic energy is NOT conserved ( $KE_{\text{final}} < KE_{\text{initial}}$ ). Energy is typically lost to heat, deformation, or sound.
Sample Problem 5 (Perfectly Inelastic):
- Scenario: A $32\text{-g}$ ( $0.032\,\text{kg}$ ) bullet traveling horizontally at $65\,\text{m/s}$ hits a $12\,\text{kg}$ block at rest on a table.
- Interaction: The bullet becomes embedded in the block after the collision.
- Objective: Determine the speed of the block (and embedded bullet) after the collision.
- Solution Formula: $m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f$
- Calculation: $(0.032)(65) + (12)(0) = (0.032 + 12) (v_f)$
- Finalizing: $2.08 = (12.032) v_f \rightarrow v_f \approx 0.1728\,\text{m/s}$ .