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IAL M2-Kinematics & Dynamics
šÆ Projectile Motion - Complete Student Notes
Key Idea: When you throw something, its motion can be split into two separate parts: sideways and up-and-down. We treat these separately to make the math easy!
š§ The Two Components of Motion
1. Horizontal Motion (Sideways)
No acceleration (if we ignore air resistance). Velocity is constant.
General Case (Launched at Angle Īø)
| Property | Equation |
|---|---|
| Velocity | \( v_x = u \cos \theta \) |
| Displacement | \( x = (u \cos \theta) \times t \) |
| Acceleration | \( a_x = 0 \) |
Horizontal Launch (Īø = 0Ā°)
| Property | Equation |
|---|---|
| Velocity | \( v_x = u \) |
| Displacement | \( x = u \times t \) |
| Acceleration | \( a_x = 0 \) |
2. Vertical Motion (Up-and-Down)
Acceleration is due to gravity (`g`). `g` is approximately 9.8 m/sĀ² downward. In equations, we use `-g` because we take "up" as positive.
General Case (Launched at Angle Īø)
| Property | Equation |
|---|---|
| Velocity | \( v_y = u \sin \theta - gt \) |
| Displacement | \( y = (u \sin \theta)t - \frac{1}{2}gt^2 \) |
| Acceleration | \( a_y = -g \) |
Horizontal Launch (Īø = 0Ā°)
| Property | Equation |
|---|---|
| Velocity | \( v_y = -gt \) |
| Displacement | \( y = -\frac{1}{2}gt^2 \) |
| Acceleration | \( a_y = -g \) |
š” Remember: `u cos Īø` is the initial horizontal velocity. `u sin Īø` is the initial vertical velocity.
š Key Results & Formulas
Time of Flight (T)
What it is: Total time the object is in the air.
\( T = \frac{2u \sin \theta}{g} \)
Maximum Height (H)
What it is: The highest point it reaches.
\( H = \frac{u^2 \sin^2 \theta}{2g} \)
Range (R)
What it is: Total horizontal distance traveled.
\( R = \frac{u^2 \sin 2\theta}{g} \)
Fun Fact: The maximum range is at Īø = 45Ā°
š Equation of Trajectory (The Path)
What we want: An equation that directly relates the vertical position (`y`) to the horizontal position (`x`) without time (`t`). This will show us the shape of the path.
1
Start with the two displacement equations
Horizontal: \( \quad x = (u \cos \theta) t \) ...(1)
Vertical: \( \quad y = (u \sin \theta) t - \frac{1}{2} g t^2 \) ...(2)
2
Make `t` the subject from equation (1)
From (1): \( \quad t = \frac{x}{u \cos \theta} \)
3
Substitute this `t` into equation (2)
\[ y = (u \sin \theta) \left( \frac{x}{u \cos \theta} \right) - \frac{1}{2} g \left( \frac{x}{u \cos \theta} \right)^2 \]
4
Simplify the equation
The `u` cancels in the first term:
\[ y = (\tan \theta) x - \frac{1}{2} g \left( \frac{x^2}{u^2 \cos^2 \theta} \right) \]
5
Final form
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
ā
This is the Equation of Trajectory! It has the form \( y = ax - bx^2 \), which is the equation of a parabola.
šÆ Problem-Solving Steps
Follow these steps to solve any projectile motion problem!
1
Resolve the Velocity
Split the initial velocity (`u`) into its components:
\( u_x = u \cos \theta \)
\( u_y = u \sin \theta \)
2
List Your Knowns
Write down what you know for the horizontal and vertical directions separately.
3
Find the Link
Time (`t`) is the same for both directions. Use it to connect the horizontal and vertical motions.
4
Pick Your Equation
Choose the right kinematic equation based on what you need to find.
š§ Pro-Tip: The 4 Universal Kinematic Equations
These work for any motion with constant acceleration (like the vertical part of projectile motion!).
\( v = u + at \)
\( s = ut + \frac{1}{2}at^2 \)
\( s = \frac{1}{2}(u+v)t \)
\( v^2 = u^2 + 2as \)
(s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time)
š The Big Takeaways
Horizontal motion is constant speed.
Vertical motion is constant acceleration (just like an object dropped or thrown straight up).
The path of a projectile is a parabola.
The motion is symmetric if it lands at the same height it was launched from.
š Happy Studying! You've got this! šŖAbel.Masitsa.com Works!
ā” Dynamics Fundamentals
The study of forces and motion - Why objects move the way they do
šÆ Newton's Laws of Motion
1st
Law of Inertia
"An object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force."
No net force ā constant velocity
At rest ā stays at rest
Inertia: resistance to change in motion
2nd
F = ma
"The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass."
\[ \vec{F}_{net} = m\vec{a} \]
Force causes acceleration
More mass ā less acceleration
Direction matters (vectors!)
3rd
Action-Reaction
"For every action, there is an equal and opposite reaction."
\[ \vec{F}_{AāB} = -\vec{F}_{BāA} \]
Forces always come in pairs
Equal magnitude, opposite direction
Act on different objects
š Common Forces in Mechanics
Weight (Gravity)
\[ W = mg \]
Direction: Always vertically downward
Value: \( g = 9.8 \, m/s^2 \)
Acts on: All objects with mass
Note: Mass ā Weight! Mass is scalar, weight is force
Normal Force
\[ N = \text{Perpendicular to surface} \]
Direction: Perpendicular to contact surface
Value: Whatever prevents sinking
Acts on: Objects touching surfaces
Note: Not always equal to weight!
Friction
Static Friction
\[ f_s \leq \mu_s N \]
Prevents starting motion
Kinetic Friction
\[ f_k = \mu_k N \]
Opposes sliding motion
Direction: Parallel to surface, opposes motion
Depends on: Surface materials (Ī¼) and normal force
Note: \( \mu_s > \mu_k \) (harder to start than keep moving)
Tension
\[ T = \text{Pull through rope/string} \]
Direction: Along the rope, away from object
Value: Same throughout massless rope
Acts on: Objects connected by ropes
Note: Ropes can only pull, not push
š Free-Body Diagrams: Your #1 Tool
1
Isolate the Object
Draw the object as a dot or simple shape. Forget everything else!
2
Identify All Forces
List every force acting ON the object (not by the object!)
3
Draw Force Vectors
Draw arrows showing direction and relative magnitude
ā
Weight (always down)
ā
Normal (ā„ to surface)
ā
Friction (opposes motion)
4
Choose Coordinate System
Align axes with motion when possible
Example: Box on Horizontal Surface
W
N
F
f
Vertical: \( N - W = 0 \) ā \( N = mg \)
Horizontal: \( F - f = ma \)
š Applying F=ma: Step-by-Step
Draw Free-Body Diagram
Identify all forces with proper directions
Choose Coordinate System
Typically: x-horizontal, y-vertical (align with acceleration if known)
Resolve Forces into Components
Break forces into x and y components
\[ F_x = F\cos\theta \quad F_y = F\sin\theta \]
Write F=ma for Each Direction
\( \sum F_x = ma_x \)
\( \sum F_y = ma_y \)
Solve the Equations
Find unknown forces, accelerations, or other quantities
šÆ Special Cases & Common Scenarios
Object on Horizontal Surface
\[ \begin{cases} \sum F_y = N - mg = 0 \\ \sum F_x = F_{applied} - f = ma \end{cases} \]
Key: Normal force equals weight only if no vertical acceleration
Object on Incline
\[ \begin{cases} \sum F_\parallel = mg\sin\theta - f = ma \\ \sum F_\perp = N - mg\cos\theta = 0 \end{cases} \]
Key: Use tilted coordinates! Weight components: \( mg\sin\theta \) (down incline), \( mg\cos\theta \) (into incline)
Vertical Motion
\[ \sum F_y = T - mg = ma \]
Key: Acceleration up: T > mg, Acceleration down: T < mg, Constant velocity: T = mg
š” Dynamics Problem-Solving Tips
Forces are Vectors
Always consider direction. Use + and - signs consistently in your coordinate system.
Net Force = ma
Not just F = ma! It's the vector sum of ALL forces that equals mass times acceleration.
Action-Reaction Pairs
Remember: equal and opposite forces act on DIFFERENT objects. Don't add them in FBD!
Check Your Units
Force in Newtons (N), mass in kilograms (kg), acceleration in m/sĀ². 1 N = 1 kgĀ·m/sĀ²
š Dynamics & Kinematics Integration Guide
How to connect forces to motion in Mechanics problems
šÆ The Core Connection: F=ma Links Dynamics and Kinematics
1
Dynamics (Forces)
Analyze all forces acting on object
ā
2
F=ma
Calculate acceleration
ā
3
Kinematics Equations
Describe the resulting motion
š Essential Dynamics Formulas for Kinematics Problems
Newton's Second Law
\[ \vec{F}_{net} = m\vec{a} \]
ā¢ This is your gateway from forces to kinematics
ā¢ Component Form: \( F_{net,x} = ma_x \) and \( F_{net,y} = ma_y \)
Common Force Models You MUST Know
Gravity (Weight)
\[ W = mg \]
ā¢ Always acts downward
ā¢ \( g = 9.8 \, \text{m/s}^2 \)
Normal Force
\[ N = \text{Perpendicular to surface} \]
ā¢ Balances other perpendicular forces
ā¢ Magnitude depends on situation
Friction
\[ f_k = \mu_k N \]
ā¢ Kinetic: \( \mu_k N \) (sliding)
ā¢ Static: \( \leq \mu_s N \) (not moving)
ā¢ Always opposes motion
š Problem-Solving Strategy
Step 1: Analyze the Problem
- Identify all phases of motion
- Draw diagrams for each phase
- List knowns and unknowns for each phase
Step 2: Apply Dynamics (F=ma)
For each motion phase:
1. Draw free-body diagram
2. Write F_net = ma for each direction
3. Solve for acceleration
Step 3: Apply Kinematics
Use the acceleration from Step 2 in:
ā¢ v = u + at
ā¢ s = ut + Ā½atĀ²
ā¢ vĀ² = uĀ² + 2as
Step 4: Connect the Phases
- Final conditions of one phase become initial conditions of next
- Time is often the connecting variable
- Velocity at end of one phase = velocity at start of next
š§© Multi-Stage Problem Example
Phase 1: On Rough Surface
Dynamics: \( f_k = \mu_k N \), \( F_{net} = -\mu_k mg = ma \)
Acceleration: \( a = -\mu_k g \)
Kinematics: Use \( v^2 = u^2 + 2as \) to find exit velocity
Phase 2: Projectile Motion
Initial velocity: Same as final velocity from Phase 1
Horizontal: \( a_x = 0 \), \( v_x = \text{constant} \)
Vertical: \( a_y = -g \), use \( s_y = u_yt + \frac{1}{2}a_yt^2 \)
š Quick Reference Table
| Situation | Forces Involved | Acceleration | Key Equations |
|---|---|---|---|
| Sliding with Friction | Weight, Normal, Friction | \( a = -\mu g \) | \( v^2 = u^2 - 2\mu g s \) |
| Free Fall | Weight only | \( a_y = -g \) | \( s_y = ut - \frac{1}{2}gt^2 \) |
| Projectile from Height | Weight only | \( a_x = 0, a_y = -g \) | \( x = u_xt, y = u_yt - \frac{1}{2}gt^2 \) |
š” Pro Tips
Start with the phase you know most about - often the projectile motion part has more known values
Time is your best connector - total time = sum of individual phase times
Work backwards if needed - sometimes solving the last phase first gives you information for earlier phases
š Calculus in Kinematics & Dynamics
Connecting displacement, velocity, acceleration, and forces through calculus
š The Calculus Relationships
Displacement ā Velocity
Differentiate
\[ v = \frac{ds}{dt} \]
Velocity is the rate of change of displacement with respect to time
ā
Velocity ā Acceleration
Differentiate
\[ a = \frac{dv}{dt} = \frac{d^2s}{dt^2} \]
Acceleration is the rate of change of velocity with respect to time
ā
Acceleration ā Force
F = ma
\[ F = m \cdot \frac{d^2s}{dt^2} \]
Force is mass times the second derivative of displacement
Acceleration ā Velocity
Integrate
\[ v = \int a \, dt \]
Velocity is the integral of acceleration with respect to time
ā
Velocity ā Displacement
Integrate
\[ s = \int v \, dt \]
Displacement is the integral of velocity with respect to time
š§® Solving Kinematics Equations
Type 1: Given displacement s(t), find velocity and acceleration
Example:
Given \( s = 2t^3 - 4t^2 + 3t - 1 \)
Velocity: \( v = \frac{ds}{dt} = 6t^2 - 8t + 3 \)
Acceleration: \( a = \frac{dv}{dt} = 12t - 8 \)
Type 2: Given acceleration a(t), find velocity and displacement
Example:
Given \( a = 4t + 2 \), with initial conditions \( v(0) = 1 \), \( s(0) = 0 \)
Velocity: \( v = \int a \, dt = \int (4t + 2) \, dt = 2t^2 + 2t + C \)
Using \( v(0) = 1 \): \( 1 = 0 + 0 + C \) ā \( C = 1 \)
ā“ \( v = 2t^2 + 2t + 1 \)
Displacement: \( s = \int v \, dt = \int (2t^2 + 2t + 1) \, dt = \frac{2}{3}t^3 + t^2 + t + D \)
Using \( s(0) = 0 \): \( 0 = 0 + 0 + 0 + D \) ā \( D = 0 \)
ā“ \( s = \frac{2}{3}t^3 + t^2 + t \)
šÆ Vector Calculus in Kinematics
Position Vector
\[ \vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k} \]
Describes position as a function of time in 2D or 3D space
Velocity Vector
\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} + \frac{dz}{dt}\hat{k} \]
Differentiate each component separately
Acceleration Vector
\[ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d^2x}{dt^2}\hat{i} + \frac{d^2y}{dt^2}\hat{j} + \frac{d^2z}{dt^2}\hat{k} \]
Second derivative of position vector
Worked Example
Given: \( \vec{r} = t^2\hat{i} + t^{\frac{3}{2}}\hat{j} \)
Velocity:
\( \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(t^2)\hat{i} + \frac{d}{dt}(t^{\frac{3}{2}})\hat{j} = 2t\hat{i} + \frac{3}{2}t^{\frac{1}{2}}\hat{j} \)
Acceleration:
\( \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(2t)\hat{i} + \frac{d}{dt}(\frac{3}{2}t^{\frac{1}{2}})\hat{j} = 2\hat{i} + \frac{3}{4}t^{-\frac{1}{2}}\hat{j} \)
At t = 4 seconds:
\( \vec{v} = 2(4)\hat{i} + \frac{3}{2}(4^{\frac{1}{2}})\hat{j} = 8\hat{i} + 3\hat{j} \)
\( \vec{a} = 2\hat{i} + \frac{3}{4}(4^{-\frac{1}{2}})\hat{j} = 2\hat{i} + \frac{3}{8}\hat{j} \)
ā” Dynamics with Calculus
Force as a Function of Time
\[ \vec{F}(t) = m\vec{a}(t) = m\frac{d^2\vec{r}}{dt^2} \]
If you know the position function, you can find the force required
Example 1: Variable Force
Given position: \( \vec{r} = (t^3 + 2t)\hat{i} + (4t^2)\hat{j} \)
Mass: m = 2 kg
\( \vec{v} = \frac{d\vec{r}}{dt} = (3t^2 + 2)\hat{i} + (8t)\hat{j} \)
\( \vec{a} = \frac{d\vec{v}}{dt} = (6t)\hat{i} + (8)\hat{j} \)
\( \vec{F} = m\vec{a} = 2[(6t)\hat{i} + (8)\hat{j}] = (12t)\hat{i} + (16)\hat{j} \) N
Example 2: Finding Motion from Force
Given: \( \vec{F} = (6t)\hat{i} + (4)\hat{j} \) N, m = 2 kg, \( \vec{r}(0) = \vec{0} \), \( \vec{v}(0) = \vec{0} \)
\( \vec{a} = \frac{\vec{F}}{m} = (3t)\hat{i} + (2)\hat{j} \) m/sĀ²
\( \vec{v} = \int \vec{a} \, dt = \int (3t)\,dt\,\hat{i} + \int (2)\,dt\,\hat{j} = (\frac{3}{2}t^2 + C_1)\hat{i} + (2t + C_2)\hat{j} \)
Using \( \vec{v}(0) = \vec{0} \): \( C_1 = 0, C_2 = 0 \)
ā“ \( \vec{v} = (\frac{3}{2}t^2)\hat{i} + (2t)\hat{j} \)
\( \vec{r} = \int \vec{v} \, dt = (\frac{1}{2}t^3 + D_1)\hat{i} + (t^2 + D_2)\hat{j} \)
Using \( \vec{r}(0) = \vec{0} \): \( D_1 = 0, D_2 = 0 \)
ā“ \( \vec{r} = (\frac{1}{2}t^3)\hat{i} + (t^2)\hat{j} \)
š Problem-Solving Framework
1
Identify What's Given
- Position function r(t)?
- Velocity function v(t)?
- Acceleration function a(t)?
- Force function F(t)?
- Initial conditions?
2
Choose the Right Operation
Given s(t)
ā Differentiate ā v(t)
ā Differentiate ā a(t)
Given a(t)
ā Integrate ā v(t)
ā Integrate ā s(t)
Given F(t)
ā Divide by m ā a(t)
3
Apply Initial/Boundary Conditions
Use given values to find constants of integration:
- Initial position: s(0) = sā
- Initial velocity: v(0) = vā
- Known values at specific times
4
Solve for Required Quantities
Substitute specific time values or solve equations as needed
š” Key Insights
Constants of Integration
Every integration introduces a constant. You need initial/boundary conditions to determine these constants.
Vector Independence
i, j, k components are independent. You can differentiate/integrate each component separately.
Physical Meaning
Derivatives = rates of change, Integrals = accumulation. Connect the math to physical motion.