Engineering Physics 1A: Momentum and Impulse Study Guide

Calculate the momentum vector of a moving particle.
Apply conservation of momentum to solve problems.
Apply the relationship between a particle’s momentum and the net force acting on it.
Apply the relationship between impulse, average force, and the time interval of an interaction.
Given net force as a function of time, calculate the impulse of a particle by integration, using both algebraic and graphical methods.

Test Grading: Tests will be graded within 10 business days. Full worked solutions will be posted on Friday.
Supplemental Test: An optional supplemental test will be held in Week 7 after the teaching break for students who achieved a score of less than $50\%$ .
Quiz 5: Due at $2359$ next Monday. It includes an unmarked reflection question concerning the test and student revision.
Laboratory: The first lab rotation continues this week for all students.

Interaction of Particles: Consider two particles with masses $m_1$ and $m_2$ colliding with each other. Their velocities are represented as $\mathbf{v}_1$ and $\mathbf{v}_2$ .
Newton’s Third Law: The contact forces between the particles behave as a Newton’s Third Law pair. If no other forces act on the system, then:
- $\mathbf{F}_{12} = -\mathbf{F}_{21}$
- $\mathbf{F}_{12} + \mathbf{F}_{21} = 0$
Newton’s Second Law Application: Substituting force with mass times acceleration ( $\mathbf{F} = m\mathbf{a}$ ):
- $m_1 \mathbf{a}_1 + m_2 \mathbf{a}_2 = 0$
- This represents the same type of force acting on different objects at different locations.
Transition to Velocity: Rewriting acceleration as the derivative of velocity ( $\mathbf{a} = \frac{d\mathbf{v}}{dt}$ ):
- $m_1 \frac{d\mathbf{v}_1}{dt} + m_2 \frac{d\mathbf{v}_2}{dt} = 0$
Constant Mass Assumption: Assuming the masses are constant, the equation can be written as:
- $\frac{d}{dt}(m_1 \mathbf{v}_1) + \frac{d}{dt}(m_2 \mathbf{v}_2) = 0$
General Principle: This implies that for a system of particles, the sum of their (mass $\times$ velocity) is conserved.

Momentum (Particle): A vector quantity defined as the product of mass and velocity.
- $\mathbf{p} = m\mathbf{v}$
Units: The units for momentum are $\text{kg\,m/s}$ , which is equivalent to $\text{N\,s}$ (since $1\,\text{N} = 1\,\text{kg\,m/s}^2$ ).

Commonalities: Both momentum and kinetic energy involve the mass and velocity of a particle.
Key Differences:
- Vector vs. Scalar: Momentum is a vector quantity, whereas kinetic energy is a scalar.
- Conservation: Kinetic energy is not always conserved in an interaction (e.g., in an inelastic collision). However, momentum is always conserved in an isolated system with no net external forces acting.
- Scope: These comparisons apply to non-relativistic particles only.
Formulas:
- Magnitude of momentum: $|\mathbf{p}| = p = mv$
- Kinetic energy in terms of momentum: $K = \frac{1}{2}mv^2 = \frac{p^2}{2m}$
Problem-Solving Application: Momentum provides an independent tool for analysis alongside energy, dynamics (forces), and kinematics.

Principle: Whenever two or more particles in an isolated system interact (apply force on each other), the total momentum of the system remains constant.
Vector Nature: Total momentum is conserved in both magnitude and direction.
- $\sum \mathbf{p} = \mathbf{p}_{tot} = \text{constant vector}$
N-Particle Interaction (where N=2):
- $\mathbf{p}_{1,i} + \mathbf{p}_{2,i} = \mathbf{p}_{1,f} + \mathbf{p}_{2,f}$
Component Conservation (2D): Momentum is conserved independently along the x and y axes.
- x-axis: $p_{1,i,x} + p_{2,i,x} = p_{1,f,x} + p_{2,f,x}$
- y-axis: $p_{1,i,y} + p_{2,i,y} = p_{1,f,y} + p_{2,f,y}$

Problem Scenario: A cannon on wheels (on frictionless tracks) has a mass of $1000\,\text{kg}$ . It fires a $20\,\text{kg}$ cannonball at an angle of $45^{\circ}$ with a speed of $400\,\text{m/s}$ .
Recoil Analysis: Putting a cannon on wheels is highlighted as a "bad idea" due to recoil velocity.
Vertical Momentum Question: The initial vertical momentum of the system is zero. After firing, the cannonball has an upwards momentum component.
Explanation of Vertical Stability: The cannon does not move downwards to conserve momentum because it is not an isolated system in the vertical direction. The ground (via normal force $F_N$ ) provides an external upward force.

Momentum Formulation: The net force acting on a particle is equal to the rate of change of its momentum over time.
- $\sum \mathbf{F} = m\mathbf{a} = m \frac{d\mathbf{v}}{dt} = \frac{d(m\mathbf{v})}{dt} = \frac{d\mathbf{p}}{dt}$
Variable Mass Systems: This general form allows for systems where mass changes with time (e.g., a rocket or a conveyor belt).
Differentiation Product Rule:
- $\sum \mathbf{F} = \frac{d(m\mathbf{v})}{dt} = m \frac{d\mathbf{v}}{dt} + \mathbf{v} \frac{dm}{dt}$

Definition: The integral of the net force over a time interval gives the change in momentum, defined as the impulse ( $\mathbf{J}$ ).
Derivation:
- $\mathbf{F}_{net} = \frac{d\mathbf{p}}{dt}$
- $d\mathbf{p} = \mathbf{F}_{net}(t) dt$
- $\int_{\mathbf{p}_i}^{\mathbf{p}_f} d\mathbf{p} = \Delta \mathbf{p} = \int_{t_i}^{t_f} \mathbf{F}_{net}(t) dt$
Impulse Equation:
- $\mathbf{J} = \Delta \mathbf{p} = \int_{t_i}^{t_f} \mathbf{F}_{net}(t) dt$
Distinction: Impulse is force integrated over time, whereas Work is force integrated over displacement.
Graphical Interpretation: Impulse magnitude is equal to the area under a net force-time graph. The force does not need to be constant.

Time-Average Force: The impulse can be expressed using a constant average force ( $\mathbf{F}_{avg}$ ) that would produce the same impulse over the same time interval ( $\Delta t$ ).
- $\mathbf{J} = \int_{t_i}^{t_f} \mathbf{F}_{net}(t) dt = \mathbf{F}_{avg} \int_{t_i}^{t_f} dt = \mathbf{F}_{avg} \Delta t$
Application (Airbags and Crumple Zones): These safety features exist to slow down the change in momentum (increase $\Delta t$ ). By increasing the time over which the impact occurs, the average net force acting on the occupant is reduced.

Vertical Momentum of the Cannon: The behavior in the vertical direction is explained by the impulse-momentum theorem. The increased normal force during firing provides an upwards impulse responsible for the cannonball’s vertical momentum.
- Equation: $p_{\text{cannon+ball},i} + (F_N - mg) \Delta t = p_{\text{cannon},f} + p_{\text{ball},f}$
Car Crash Analysis:
- Data: Mass $m = 1500\,\text{kg}$ . Initial velocity $v_i = -15.0\,\text{m/s}$ . Final velocity $v_f = +2.60\,\text{m/s}$ (away from wall). Collision duration $\Delta t = 150\,\text{ms} = 0.150\,\text{s}$ .
- Objective: Find average net force exerted by the wall.
Lucky Paratrooper Case Study (February 1955):
- Scenario: A paratrooper fell $370\,\text{m}$ without a functional parachute and landed in snow.
- Data: Impact speed (terminal speed) $v = 58\,\text{m/s}$ . Mass $m = 85\,\text{kg}$ . Maximum survivable force from snow $F = 1.2 \times 10^5\,\text{N}$ .
- Objective: Calculate the minimum depth of snow required to stop him safely and determine the duration of the impact.