Meditaliano IMAT Prep
Lesson 2: Physics: Dynamics, Energy, Gravitation & Oscillations
Lesson 2: Comprehensive Dynamics
Welcome back! In this lesson, we will expand our study of dynamics to include periodic motion and the universal force of gravity. We'll explore Simple Harmonic Motion, the underlying physics of oscillations, and Newton's Law of Universal Gravitation. We will then revisit the powerful concepts of Energy and Momentum, which provide alternative and often simpler ways to solve complex dynamics problems.
Learning Objective (LO P2.1):
By the end of this lesson, you will be able to apply principles of gravitation, simple harmonic motion, energy, and momentum conservation to solve problems, including complex collision scenarios.
Part 1: Advanced Dynamics
1.1 Newton's Third Law: Action-Reaction
This law describes a fundamental symmetry of forces in nature:
For every action, there is an equal and opposite reaction.
This means that forces always occur in pairs. If Object A exerts a force on Object B, then Object B simultaneously exerts a force on Object A that is equal in magnitude and opposite in direction.
Crucial Point: The action and reaction forces act on different objects. They never cancel each other out because they don't act on the same object. When analyzing the motion of an object, you only consider the forces acting *on* it, not the forces *it exerts*. For example, when a rocket expels gas downwards (action), the gas pushes the rocket upwards (reaction). We analyze the upward force on the rocket to understand its motion.
1.2 Centripetal Force & Centripetal Acceleration
When an object moves in a circle at a constant speed, its direction is constantly changing. A change in direction means a change in velocity (since velocity is a vector), which means it's accelerating. This acceleration, directed towards the center of the circle, is called centripetal acceleration ($a_c$).
According to Newton's Second Law, if there's an acceleration, there must be a net force causing it. The centripetal force ($F_c$) is the net force that causes centripetal acceleration, always pointing towards the center of the circular path. This net force can be provided by tension, gravity, friction, or any other real force.
Centripetal Force & Acceleration Formulas
$$ a_c = \frac{v^2}{r} \quad \text{and} \quad F_c = ma_c = \frac{mv^2}{r} $$
Where: $v$ = tangential speed, $r$ = radius of the circle, $m$ = mass.
Derivation of $a_c = \frac{v^2}{r}$ (Geometric Similar Triangles)
How do we prove that centripetal acceleration is equal to $v^2/r$? We can derive it geometrically using the concept of **similar triangles** formed by the position vectors and velocity vectors.
Step 1: Analyzing the Position Triangle
Consider a particle moving along a circular path of radius $r$ at a constant speed $v$. Let its position vector be $\vec{r}_1$ at time $t_1$, and $\vec{r}_2$ at a slightly later time $t_2$.
The change in position is represented by the vector: $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1$.
Since the radius is constant, the magnitudes of the position vectors are equal: $|\vec{r}_1| = |\vec{r}_2| = r$. The triangle formed by $\vec{r}_1$, $\vec{r}_2$, and $\Delta \vec{r}$ is an isosceles triangle with vertex angle $\Delta \theta$. For a very small time interval $\Delta t$, the chord length $|\Delta \vec{r}|$ is approximately equal to the arc length traveled: **$|\Delta \vec{r}| \approx v \Delta t$**.
Step 2: Analyzing the Velocity Triangle
At any point, the velocity vector is tangent to the circle (perpendicular to the position vector). Therefore, the angle between $\vec{v}_1$ and $\vec{v}_2$ is also exactly **$\Delta \theta$**.
Since the speed is constant, the magnitudes of the velocity vectors are equal: $|\vec{v}_1| = |\vec{v}_2| = v$. If we translate the velocity vectors so their tails meet at a common origin, the change in velocity is: $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$. This forms another isosceles triangle with sides of length $v$ and base $|\Delta \vec{v}|$, and a vertex angle of $\Delta \theta$.
Step 3: Similar Triangles and Ratios
Because both triangles are isosceles and share the same vertex angle $\Delta \theta$, they are **similar triangles** ($\Delta OAB \sim \Delta PQR$ in vector space). Therefore, the ratio of their bases to their sides must be equal:
$$ \frac{|\Delta \vec{v}|}{v} = \frac{|\Delta \vec{r}|}{r} $$
Substituting $|\Delta \vec{r}| \approx v \Delta t$ into the equation:
$$ \frac{|\Delta \vec{v}|}{v} = \frac{v \Delta t}{r} \quad \Rightarrow \quad \frac{|\Delta \vec{v}|}{\Delta t} = \frac{v^2}{r} $$
Taking the limit as $\Delta t \to 0$ gives the instantaneous centripetal acceleration:
$$ a_c = \lim_{\Delta t \to 0} \frac{|\Delta \vec{v}|}{\Delta t} = \frac{v^2}{r} $$
Geometric representation of similar triangles showing that $\Delta v / v = \Delta r / r$.
1.3 Angular Velocity & Detailed Derivations
When describing circular motion, it is often much simpler to express the position in terms of the angle rather than linear distance. We introduce Angular Velocity ($\omega$), which measures how fast an object rotates through an angle.
Angular Velocity Definition
$$ \omega = \frac{\Delta \theta}{\Delta t} $$
Where $\Delta \theta$ is the change in angle in **radians**, and $\Delta t$ is the time interval. The standard unit is **radians per second (rad/s)**.
Detailed Derivations of Angular Formulas
1. Derivation of $v = r\omega$ (Relationship to Linear Speed):
Consider an object moving along a circular arc. The arc length $\Delta s$ traveled by the object is related to the radius $r$ and the subtended angle $\Delta \theta$ (in radians) by the definition of radian measure:
$$ \Delta s = r \Delta \theta $$
Dividing both sides by the time interval $\Delta t$:
$$ \frac{\Delta s}{\Delta t} = r \frac{\Delta \theta}{\Delta t} $$
In the limit as $\Delta t \to 0$, $\frac{\Delta s}{\Delta t}$ is the tangential linear speed $v$, and $\frac{\Delta \theta}{\Delta t}$ is the angular velocity $\omega$. Therefore:
$$ v = r\omega $$
2. Derivation of $\omega = \frac{2\pi}{T} = 2\pi f$ (Relationship to Period & Frequency):
The time required for an object to complete one full revolution ($360^\circ$ or $2\pi$ radians) is called the **Period ($T$)**.
Since one full revolution corresponds to an angular displacement of $\Delta \theta = 2\pi$ radians in a time interval of $\Delta t = T$, we have:
$$ \omega = \frac{2\pi}{T} $$
The frequency **$f$** is the number of revolutions per second, defined as $f = \frac{1}{T}$. Substituting this relation in, we get:
$$ \omega = 2\pi f $$
3. Centripetal Acceleration in terms of $\omega$:
Using the relation $v = r\omega$, we can substitute this into the centripetal acceleration formula:
$$ a_c = \frac{v^2}{r} = \frac{(r\omega)^2}{r} = \frac{r^2\omega^2}{r} = \omega^2 r $$
Angular velocity $\omega$ indicates the rate of rotation.
Challenge an IMAT Question!
Official Paper: 2021 - Q57
row 1: 25 m/s, 250 m/s²
row 2: 25 m/s, 40 m/s²
row 3: 15 m/s, 50 m/s²
row 4: 10 m/s, 40 m/s²
row 5: 10 m/s, 0 m/s²
worked solution & explanation
Concept Circular Motion Kinematics. Convert angular parameters into linear parameters using standard formulas.
Step 1 Calculate linear tangential speed ($v$). $v = \omega \cdot r = 10\text{ rad/s} \times 2.5\text{ m} = 25\text{ m/s}$.
Step 2 Calculate centripetal acceleration ($a_c$). Formula: $a_c = \omega^2 \cdot r$.
Step 3 Substitute values: $a_c = (10)^2 \times 2.5 = 100 \times 2.5 = 250\text{ m/s}^2$.
Challenge an IMAT Question!
Official Paper: 2015 - Q56
Row 1: 0kN, R
Row 2: 18kN, Q
Row 3: 18kN, R
Row 4: 30kN, Q
Row 5: 30kN, P
worked solution & explanation
Concept Centripetal Force Dynamics. Even though speed is constant, the changing direction implies acceleration, demanding a net inward force.
Step 1 Calculate magnitude using $F_c = \frac{m \cdot v^2}{r}$.
Step 2 Substitute values: $F_c = \frac{1000 \cdot (30)^2}{50} = \frac{1000 \cdot 900}{50} = 18,000\text{ N} = 18\text{ kN}$.
Step 3 Direction: Centripetal literally means 'center-seeking'. The net force strictly points towards the geometric center of the curve (direction R).
Part 2: Gravitation and Oscillations
2.1 Newton's Law of Universal Gravitation
Newton proposed that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Universal Gravitation
$$ F_g = G \frac{m_1 m_2}{r^2} $$
Where $F_g$ is the gravitational force, $m_1$ and $m_2$ are the masses, $r$ is the distance between their centers, and $G$ is the gravitational constant ($G \approx 6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$).
Gravitational Potential Energy and Conservation
On a planetary scale, the simple $PE_g=mgh$ is insufficient. The gravitational potential energy ($U_g$) of a system of two masses is defined to be zero when they are infinitely far apart.
Gravitational Potential Energy
$$ U_g = -G \frac{m_1 m_2}{r} $$
The negative sign is crucial: it indicates that gravity is an attractive force and that work must be done on the system to separate the masses to infinity (i.e., to bring the energy to zero).
For a satellite orbiting a planet, the total mechanical energy is the sum of its kinetic and potential energies. Since gravity is a conservative force, this total energy is conserved.
Total Orbital Energy: $E_{total} = KE + U_g = \frac{1}{2}mv^2 - G\frac{Mm}{r}$
Challenge an IMAT Question!
Official Paper: 2011 - Q75
$G = 7\times10^{-11}\text{ Nm}^2\text{kg}^{-2}$
$r = 4\times10^8\text{ m}$
$m_1 = 6\times10^{24}\text{ kg}$
$m_2 = 7\times10^{22}\text{ kg}$.
Calculate the force.
worked solution & explanation
Concept Newton's Law of Universal Gravitation ($F = \frac{Gm_1m_2}{r^2}$). Purely a test of executing scientific notation math without a calculator.
Step 1 Setup: $F = \frac{(7\times10^{-11}) \cdot (6\times10^{24}) \cdot (7\times10^{22})}{(4\times10^8)^2}$.
Step 2 Multiply coefficients in numerator: $7 \times 6 \times 7 = 42 \times 7 = 294$.
Step 3 Sum exponents in numerator: $10^{-11} \cdot 10^{24} \cdot 10^{22} = 10^{-11+24+22} = 10^{35}$.
Step 4 Denominator: $(4\times10^8)^2 = 16 \times 10^{16}$.
Step 5 Divide: Coefficient = $294 / 16 = 18.375$. Exponent = $10^{35} / 10^{16} = 10^{19}$.
Step 6 Result: $18.375 \times 10^{19}$, which adjusts to $1.8375 \times 10^{20}\text{ N}$.
2.2 Simple Harmonic Motion (SHM)
Simple Harmonic Motion (SHM) is oscillatory motion where the restoring force is proportional to the displacement.
Condition for SHM
$$ F_{restoring} = -kx $$
Where $k$ is a constant of proportionality (the spring constant in the case of a mass-spring system) and $x$ is the displacement from equilibrium.
In SHM, there is a continuous conversion between potential energy (maximum at the endpoints, or maximum displacement) and kinetic energy (maximum at the equilibrium position).
The time for one complete oscillation is the period (T). For a mass-spring system, the period is given by:
$$ T = 2\pi\sqrt{\frac{m}{k}} $$
2.2.1 The Mathematics of SHM (Calculus Approach)
To fully understand SHM, we combine Newton's Second Law with the restoring force to derive the differential equation.
1. Differential Equation:
$$ F = ma \quad \Rightarrow \quad -kx = m \frac{d^2x}{dt^2} $$
$$ \frac{d^2x}{dt^2} + \frac{k}{m}x = 0 $$
We define the angular frequency ($\omega$) as $\omega = \sqrt{\frac{k}{m}}$. Thus, the equation becomes: $\frac{d^2x}{dt^2} = -\omega^2 x$.
2. General Solution (Displacement):
The solution to this differential equation is a sinusoidal function describing the position $x(t)$:
$$ x(t) = A \cos(\omega t + \phi) $$
Where $A$ is the amplitude and $\phi$ is the phase constant (determined by initial conditions).
3. Velocity and Acceleration (Derivatives):
- Velocity ($v$): differentiate position with respect to time.
$$ v(t) = \frac{dx}{dt} = \frac{d}{dt}[A \cos(\omega t + \phi)] = -\omega A \sin(\omega t + \phi) $$ - Acceleration ($a$): differentiate velocity with respect to time.
$$ a(t) = \frac{dv}{dt} = \frac{d}{dt}[-\omega A \sin(\omega t + \phi)] = -\omega^2 A \cos(\omega t + \phi) $$
Notice that $a(t) = -\omega^2 [A \cos(\omega t + \phi)] = -\omega^2 x(t)$, which confirms our original differential equation.
Dynamics of a Mass-Spring System.
Challenge an IMAT Question!
Official Paper: 2013 - Q53
1. The kinetic energy of the mass is at a maximum half way up.
2. The potential energy of the system is at a maximum at the top of the mass's motion.
3. The potential energy of the system is at a maximum at the bottom of the mass's motion.
worked solution & explanation
Concept Energy transformations in Vertical Simple Harmonic Motion. Total mechanical energy ($KE + PE_{\text{gravity}} + PE_{\text{elastic}}$) remains perfectly constant.
Step 1 Statement 1: The 'halfway up' point is the equilibrium position where velocity is maximum. Since $KE = \frac{1}{2}mv^2$, KE is maximized here. (True)
Step 2 Statements 2 & 3: At the extreme top and bottom, the mass momentarily stops ($v=0, KE=0$). Since Total Energy is constant, Potential Energy must be at its absolute maximum at both extremes. (True)
Challenge an IMAT Question!
Official Paper: 2011 - Q70
worked solution & explanation
Concept Definition of SHM. Simple Harmonic Motion STRICTLY requires a restoring force that is directly proportional to the displacement from equilibrium ($F = -kx$).
Step 1 A bouncing ball falls under constant gravitational acceleration ($F = mg$), which is NOT proportional to displacement.
Step 2 Upon hitting the floor, it experiences a sudden impulse force. This non-proportional, discontinuous force profile disqualifies it from SHM.
2.3 The Simple Pendulum
A simple pendulum (a mass suspended from a light string) is a classic example of oscillatory motion. For small angles of displacement (typically < 15°), its motion is an excellent approximation of Simple Harmonic Motion.
Period of a Simple Pendulum
$$ T = 2\pi\sqrt{\frac{L}{g}} $$
Where $T$ is the period (time for one full swing), $L$ is the length of the pendulum, and $g$ is the acceleration due to gravity. Notably, for small angles, the period does not depend on the mass of the bob or the amplitude of the swing.
Challenge an IMAT Question!
Official Paper: 2024 - Q60
worked solution & explanation
Concept Conservation of Mechanical Energy. A pendulum converts between kinetic energy and gravitational potential energy.
Step 1 Mechanical energy ($KE + PE$) is perfectly conserved unless acted upon by a non-conservative force like friction or air resistance.
Step 2 If friction is strictly defined as 'absent', there is zero mechanism for energy dissipation. The pendulum will oscillate endlessly with constant amplitude.
Challenge an IMAT Question!
Official Paper: 2022 - Q60
worked solution & explanation
Concept Period of a Simple Pendulum. The formula is $T = 2\pi \sqrt{\frac{L}{g}}$. Carefully define $T$ from the text.
Step 1 'Passes through the vertical position at 1.0s intervals'. Passing the equilibrium happens twice per full cycle. Thus, time between passes is half a period: $T/2 = 1.0\text{s} \implies T = 2.0\text{s}$.
Step 2 Substitute into formula: $2.0 = 2\pi \sqrt{\frac{L}{10}} \implies 1.0 = \pi \sqrt{\frac{L}{10}}$.
Step 3 Square both sides: $1.0 = \pi^2 \cdot \frac{L}{10}$. Since $\pi^2 = 10$, this becomes $1.0 = 10 \cdot \frac{L}{10} \implies L = 1.0\text{ m}$.
Part 3: Work, Energy, and Power
3.1 Work
In physics, Work (a scalar quantity) is done on an object when a force causes it to move a certain distance. Importantly, only the component of the force that is parallel to the direction of motion does work. The unit of work is the Joule (J), where $1 J = 1 N \cdot m$.
Work Done by a Constant Force
$$ W = Fd\cos(\theta) $$
Where: $W$ = work, $F$ = magnitude of the force, $d$ = magnitude of the displacement, $\theta$ = angle between the force and displacement vectors.
3.2 Energy
Energy is the capacity to do work. In mechanics, we focus on:
- Kinetic Energy ($KE$): The energy of motion.
- Potential Energy ($PE$): Stored energy, such as Gravitational Potential Energy ($PE_g=mgh$) or Elastic Potential Energy in a spring ($PE_s = \frac{1}{2}kx^2$).
Kinetic Energy
$$ KE = \frac{1}{2}mv^2 $$
Gravitational Potential Energy (Local)
$$ PE_g = mgh $$
Elastic Potential Energy
$$ PE_s = \frac{1}{2}kx^2 $$
3.3 Work-Energy Theorem & Conservation of Energy
The Work-Energy Theorem states that the net work done on an object equals its change in kinetic energy ($W_{net} = \Delta KE$).
The Principle of Conservation of Mechanical Energy applies when only conservative forces (like gravity or spring forces) do work. In this case, the total mechanical energy ($E = KE + PE$) of a system remains constant.
Conservation of Mechanical Energy (No Friction)
$$ E_{initial} = E_{final} \quad \implies \quad KE_i + PE_i = KE_f + PE_f $$
On a frictionless roller coaster, total energy is conserved. PE (blue) converts to KE (orange) and back again.
3.4 Power
Power is the rate at which work is done or energy is transferred. The unit of power is the Watt (W), where $1 \text{ Watt} = 1 \text{ Joule/second}$.
Power
$$ P_{avg} = \frac{W}{t} = \frac{\Delta E}{t} \quad \text{and (Instantaneous)} \quad P = Fv\cos(\theta) $$
(Where $\theta$ is the angle between Force and Velocity. Often $P=Fv$ if they are parallel).
Challenge an IMAT Question!
Official Paper: 2024 - Q55
worked solution & explanation
Concept Work calculation and Italian translation nuances. The IMAT often suffers from slight translation inaccuracies from the original Italian test.
Step 1 The question asks for 'power' but provides force and distance, and options implicitly use the scalar value of Work. In Italian, 'lavoro' (Work) is sometimes loosely confused with power.
Step 2 Calculate Work: $W = F \cdot d \cdot \cos(\theta)$. Since braking is anti-parallel, magnitude is $W = 210 \cdot 5 = 1050\text{ Joules}$.
Step 3 The test makers equated the numerical magnitude of Work ($1050$) to the answer choice $1050\text{ W}$. You must adapt to these systemic quirks.
Challenge an IMAT Question!
Official Paper: 2022 - Q52
worked solution & explanation
Concept Work-Energy Theorem. The net work done on an object equals its change in kinetic energy: $W_{\text{net}} = \Delta KE$.
Step 1 Work done by friction is $F \cdot d = 4.0\text{ N} \times 2.0\text{ m} = 8.0\text{ Joules}$ (energy removed from the system).
Step 2 This work removes all initial kinetic energy. Thus, Initial $KE = 8.0\text{ J}$.
Step 3 Set up KE equation: $8.0 = \frac{1}{2} m v^2 \implies 8.0 = \frac{1}{2} m (2.0)^2 \implies 8.0 = 2m \implies m = 4.0\text{ kg}$.
Part 4: Momentum and Collisions
4.1 Momentum and Impulse
Linear momentum ($\vec{p}$) is a measure of an object's "quantity of motion." It's a vector quantity ($p = mv$). Impulse ($\vec{J}$) is the change in momentum of an object ($J = \Delta p = F_{avg}\Delta t$).
Momentum
$$ \vec{p} = m\vec{v} $$
Impulse
$$ \vec{J} = \Delta \vec{p} = \vec{F}_{avg} \Delta t $$
4.2 Conservation of Momentum
For an isolated system (one with no external net forces), the total momentum before an event (like a collision) is equal to the total momentum after the event.
Conservation of Momentum
$$ \vec{p}_{\text{total, initial}} = \vec{p}_{\text{total, final}} $$
(For two objects colliding):
$$ m_1\vec{u}_1 + m_2\vec{u}_2 = m_1\vec{v}_1 + m_2\vec{v}_2 $$
4.3 Types of Collisions
We classify collisions based on whether kinetic energy is conserved. In all collisions (assuming no external forces), momentum is conserved.
Elastic Collision
- Momentum: Conserved
- Kinetic Energy: Conserved
- Example: Billiard balls, gas molecules.
- Objects bounce off perfectly.
Inelastic Collision
- Momentum: Conserved
- Kinetic Energy: NOT Conserved (Lost to heat, sound)
- Example: Car crash, ball dropping on floor.
Perfectly Inelastic
- Momentum: Conserved
- Kinetic Energy: Maximum Loss
- Objects stick together and move with a common velocity ($v_f$).
- $m_1u_1 + m_2u_2 = (m_1+m_2)v_f$
Challenge an IMAT Question!
Official Paper: 2023 - Q56
worked solution & explanation
Concept 1D Elastic Collisions. Both momentum and kinetic energy are conserved. Standard derived formulas for target at rest: $v_1' = \frac{m_1 - m_2}{m_1 + m_2} v_1$ and $v_2' = \frac{2m_1}{m_1 + m_2} v_1$.
Step 1 Calculate $v_A$ (incoming mass $m$): $v_A = \frac{m - 2m}{m + 2m} v = \frac{-m}{3m} v = -\frac{1}{3} v$. Negative sign indicates it rebounds backward.
Step 2 Calculate $v_B$ (target mass $2m$): $v_B = \frac{2(m)}{m + 2m} v = \frac{2m}{3m} v = \frac{2}{3} v$.
4.4 Collisions in 2 Dimensions
When objects collide at an angle (glancing collisions), momentum is conserved separately in the X direction and the Y direction.
How to solve 2D Collision Problems:
- Decompose Velocities: Break all initial velocity vectors into x and y components using $\cos\theta$ and $\sin\theta$.
- X-Axis Conservation: Write the equation $\Sigma p_{ix} = \Sigma p_{fx}$.
e.g., $m_1 u_{1x} + m_2 u_{2x} = m_1 v_{1x} + m_2 v_{2x}$ - Y-Axis Conservation: Write the equation $\Sigma p_{iy} = \Sigma p_{fy}$.
e.g., $m_1 u_{1y} + m_2 u_{2y} = m_1 v_{1y} + m_2 v_{2y}$ - Solve: You now have a system of two equations to solve for the unknowns.
- Recombine (if needed): Use Pythagorean theorem ($v = \sqrt{v_x^2 + v_y^2}$) and inverse tangent ($\theta = \tan^{-1}(v_y/v_x)$) to find final magnitude and direction.
Decomposition of velocity vectors after collision.
4.5 Coefficient of Restitution ($e$)
Coefficient of Restitution
$$ e = \frac{|\text{Relative Velocity of Separation}|}{|\text{Relative Velocity of Approach}|} = \frac{|v_2 - v_1|}{|u_1 - u_2|} $$
- $e = 1$: Perfectly Elastic Collision (No KE lost).
- $0 < e < 1$: Inelastic Collision (Real world collisions).
- $e = 0$: Perfectly Inelastic Collision (Objects stick together).
Part 5: Practice Quiz
Apply your knowledge of dynamics, energy, and momentum. Answer all questions and then click "Submit Answers" to see your score.
Part 6: Key Formula Summary
Dynamics
- Newton's 3rd Law: $\vec{F}_{AB} = -\vec{F}_{BA}$
- Centripetal Force: $F_c = \frac{mv^2}{r}$
- Angular Velocity: $\omega = 2\pi f$
- Centripetal Accel: $a_c = \omega^2 r$
Gravitation & Oscillations
- Universal Gravitation: $F_g = G \frac{m_1 m_2}{r^2}$
- Gravitational PE: $U_g = -G \frac{m_1 m_2}{r}$
- Pendulum Period: $T = 2\pi\sqrt{\frac{L}{g}}$
- Spring Period: $T = 2\pi\sqrt{\frac{m}{k}}$
- SHM: $x(t) = A\cos(\omega t + \phi)$
Work & Power
- Work: $W = Fd\cos(\theta)$
- Power: $P = \frac{W}{t} = Fv\cos(\theta)$
Energy & Momentum
- Kinetic Energy: $KE = \frac{1}{2}mv^2$
- Potential Energy (Grav): $PE_g = mgh$
- Potential Energy (Spring): $PE_s = \frac{1}{2}kx^2$
- Momentum: $\vec{p} = m\vec{v}$
- Impulse: $\vec{J} = \Delta\vec{p} = F_{avg}\Delta t$
- Restitution: $e = \frac{|v_2 - v_1|}{|u_1 - u_2|}$
Conservation Laws
- Conservation of Energy (Mech): $KE_i + PE_i = KE_f + PE_f$
- Work-Energy Theorem: $W_{net} = \Delta KE$
- Conservation of Momentum (1D): $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
- Conservation of Momentum (2D): Solve X and Y components independently.